PRINCIPLES OF TURBOMACHINERY SOLUTIONS MANUAL by Seppo A. Korpela Department of Mechanical and Aerospace Engineering January 2012 c Copyright ⃝2011-2012, Seppo A. Korpela Chapter 2 Exercise 2.1 Steam flows through a bank of nozzles shown in the figure below, with wall thickness t2 = 2 mm, spacing s = 4 cm, blade height b = 2.5 cm, and exit angle α2 = 68◦ . The exit velocity V2 = 400 m/s, pressure is p2 = 1.5 bar, and temperature is T2 = 200 C. Find the mass flow rate. Given: b = 2.5 C s = 4 cm T2 = 200 C = 473.15 K t2 = 0.2 cm α2 = 68◦ V2 = 400 m/s v2 = 1.4443 m3 /kg p1 = 1.50 bar = 150 kPa Find: Mass flow rate. Solution: A2 = b(2 cos α2 − t2 ) = 2.5(4 cos(68◦ ) − 0.2) = 3.25 cm A2 V2 3.25 · 400 = 4 = 0.09 kg/s ⇐ v2 10 · 1.4443 Note that the specific volume could also be approximated as ṁ = v2 = R̄T2 8.314 · 473.15 = = 1.457 m3 /kg Mp2 18 · 150 V1 t1 s 1 t2 2 α2 V2 2 Exercise 2.2 Air enters a compressor from atmosphere at pressure 102 kPa and temperature 42 C. Assuming that its density remains constant determine the specific compression work required to raise its pressure to 140 kPa in a reversible adiabatic process, if the exit velocity is 50 m/s. Given: Since the air is stagnant in the atmosphere, its conditions are the stagnation conditions. The flow is isentropic and T2 = 42 C = 315.15 K p1 = 102 kPa p2 = 140 kPa V2 = 50 m/s Find: Specific work done. Solution: For isentropic flow T ds = dh − vdp leads to dh = dp/ρ. In addition h0 = h + V 2 /2 is constant. The flow is assumed incompressible because the pressure changes only slightly and the exit velocity is small. p 2 − p1 1 2 w= + V2 ρ 2 The density is p1 102 ρ= = = 1.128 kg/m2 RT1 0.287 · 315.156 so that 502 140 − 102 + = 43.95 kJ/kg ⇐ w= 1.128 2 · 1000 Exercise 2.3 Steam flows through a turbine at the rate of ṁ = 9000 kg/h. The rate at which power is delivered by the turbine is Ẇ = 440 hp. The inlet total pressure is p01 = 70 bar and total temperature is T01 = 420 C. For a reversible and adiabatic process find the total pressure and temperature. leaving the turbine. Given: The flow is isentropic and T01 = 420 C p01 = 70 bar ṁ = 9000 kg/h Ẇ = 440 hp Find: T02 and p02 . Solution: Ẇ 440 · 0.7457 · 3600 = = 131.2 kJ/kg ṁ 9000 From steam tables h)1 = 3209.8 kJ/kg, s1 = 6.5270 kJ/kg K. At the exit w= h02 = h01 − w = 3209.8 − 131.2 = 3078.6 kJ/kg s2 = 6.5270 kJ/kg K From the superheated steam tables, or using EES, p02 = 43.58 bar T02 = 348.5 C 3 ⇐ Exercise 2.4 Water enters a pump as saturated liquid at total pressure of p01 = 0.08 bar and leaves it at p02 = 30 bar. If the mass flow rate is ṁ = 10, 000 kg/h and the process can be assumed to take place reversibly and adiabatically, determine the power required. Given: The flow is isentropic and p01 = 0.08 bar p02 = 30 bar ṁ = 10.000 kg/h Find: Ẇ . Solution: For isentropic flow w= p02 − p01 30 − 0.08) · 105 = = 2.998 kJ/kg ρ 998 Therefore Ẇ = ṁw = 10, 000 · 2.998 = 8.33 kW 3600 ⇐ Exercise 2.5 Liquid water at 700 kPa and temperature 20 C flows at velocity 15 m/s. Find the stagnation temperature and stagnation pressure. Given: The flow is isentropic and p = 700 kPa T = 20 K V = 15 m/s Find: T0 and p0 . Solution: Since water is incompressible T0 = T + V2 152 = 20 + = 20.027 = 20.027 C 2cp 2 · 4187 ⇐ 998 · 152 1 p0 = p + ρV 2 = 700 + = 700 + 122.75 = 812.3 kPa 2 2 · 1000 ⇐ Exercise 2.6 Water at temperature T1 = 20 C flows through a turbine with inlet velocity V1 = 3 m/s, static pressure p1 = 780 kPa and elevation z1 = 2 m. At the exit the conditions are V2 = 6 m/s, p2 = 100 kPa and z2 = 1.2 m. Find the specific work delivered by the turbine. Given: Assuming that the process is isentropic and given that T1 = 20 C, and p1 = 780 kPa z1 = 2 m 4 V1 = 3 m/s p2 = 100 kPa z1 = 1.2 m V1 = 6 m/s Find: w Solution: Since water is incompressible w = h01 − h02s w= p01 − p02 V12 − V22 = + + g(z1 − z2 ) ρ 2 780 − 100) 103 32 − 62 + + 9.81 (2 − 2.12) = 681.36 − 13.50 + 7.85 998 2 w = 675.71 J/kg ⇐ Note the small contributions from potential and kinetic energy. Exercise 2.7 Air at static pressure of 2 bar and static temperature of 300 K flows with velocity 60 m/s. Find total temperature and pressure. Given: p1 = 2 bar T1 = 300 K V1 = 60 m/s Find: T01 and p01 Solution: The total temperature is T01 = T1 + V12 602 = 300 + = 301.8 K 2 cp 2 · 1004.5 ⇐ and since the velocity is quite small, air may be taken to be incompressible with density p1 2 · 105 ρ= = = 2.323 kg/m3 RT1 287 · 300 and the stagnation pressure is 1 2.323 · 602 p01 = p1 + ρV12 = 2 · 105 + = 204.2 kPa 2 2 ⇐ Exercise 2.8 Air at static temperature of 300 K and static pressure of 140 kPa flow with velocity 60 m/s. Evaluate the total temperature and total pressure of air. Repeat the calculation if the air speed is 300 m/s. Given: p1 = 140 kPa T1 = 300 K V1 = 60 m/s 5 Find: T01 and p01 Solution: T01 = T1 + V12 602 = 301.8 K = 300 + 2 cp 2 · 1004.5 ⇐ and since the velocity is quite small, air may be taken to be incompressible with density p1 2 · 105 ρ= = = 2.323 kg/m3 RT1 287 · 300 and the stagnation pressure is 1 2.323 · 602 p01 = p1 + ρV12 = 140 · 103 + = 142.93 kPa 2 2 ⇐ and if V1 = 300 m/s then T01 = T1 + and ( p01 = p1 T01 T1 V12 3002 = 300 + = 344.8 K 2 cp 2 · 1004.5 )γ/(γ−1) ( = 140 344.8 300 ⇐ )3/5 = 227.9 kPa ⇐ Exercise 2.9 Air undergoes an increase of 1.75 kPa in total pressure through a blower. The inlet total pressure is one atmosphere and the inlet total temperature is 21 C. Evaluate the exit total temperature if the process is reversible adiabatic. Evaluate the energy added to the air per unit mass flow. Given: p01 = 101.325 kPa T01 = 294.15 K ∆p0 = 1.74 kPa Find: T02 Solution: The stagnation density is ρ01 = p01 101.325 · 103 = = 1.200 kg/m3 RT01 287 · 294.15 Since the process is reversible and adiabatic and the pressure increase is quite small, the work done by the blower may be calculated from w= 1.75 p02 − p01 = = 1.458 kJ/kg ρ01 1.200 6 ⇐ and the stagnation temperature leaving the blower is T02 = T01 + w 1458 = 294.15 + = 295.6 K = 22.4 C cp 1004.5 ⇐ Exercise 2.10 Air enters a blower from the atmosphere where pressure is 101.3 kPa and temperature is 27 C. Its velocity at the inlet is and with velocity 46 m/s. At the exit the total temperature is 28 C and the velocity is 123 m/s. Assuming that the flow is reversible and adiabatic, determine (a) the change in total pressure in millimeters of water and (b) the change in static pressure, also in millimeters of water. Given: The inlet stagnation pressure is p01 = 101.3 kPa and stagnation temperature and velocity are T01 = 300.15 K V1 = 46 m/s T02 = 301.15 K V2 = 123 m/s At the exit Find: ∆p0 in millimeters of water. Solution: Work done by the blower is w = cp (T02 − T01 ) = 1004.5(28 − 27) = 1004.5 kJ/kg The stagnation density at the inlet is ρ01 = p01 101.3 · 103 = = 1.176 kg/m3 RT01 287 · 300.15 The flow is isentropic and the pressure rise is small. Therefore ws = ∆p0 ρ01 ∆p0 = ρ01 w = 1.176 · 1.0045 = 1.181 kPa In millimeters of water this is ∆H = ∆p0 1182 = = 0.121 mm ρw g 998 · 9.81 ⇐ (a) To calculate the difference in the static pressure, first calculate 1 462 p1 = p01 − ρ01 V12 = 300.15 − = 100.06 kPa 2 2 · 1004.5 7 The static temperatures are T1 = T01 − V12 462 = 300.15 − = 299.10 K 2cp 2 · 1004.5 T2 = T02 − V22 1232 = 301.15 − = 293.62 K 2cp 2 · 1004.5 The exit stagnation pressure is p02 = 101.300 + 1.182 = 102.482 kPa. and the static pressure is ( p2 = p02 T2 T02 )γ/(γ−1) ( = 102.48 293.62 301.15 )3.5 = 93.789 kPa and thus ∆Hs = p2 − p1 93.789 − 100.06 = = −641 mm ρw g 998 · 9.81 ⇐ (b) Exercise 2.11 The total pressure, static pressure and the total temperature of air at a certain point in a flow are 700 kPa, 350 kPa, and 450 K, respectively. Find the velocity at that point. Given: The properties are T0 = 450 K p = 350 kPa p0 = 700 kPa Find: The velocity Solution: ( T = T0 p p0 )(γ−1)/γ ( = 450 350 700 )1/3.5 = 369.15 K and √ √ V = 2cp (T0 − T ) = 2 · 1004.5(450 − 369.15) = 403.0 m/s Exercise 2.12 Air has a static pressure of 2 bar and static temperature 300 K while it flows at speed 1000 m/s. (a) Assuming air obeys the ideal gas law with constant specific heats, determine its stagnation temperature and stagnation pressure. (b) Repeat part (a) using the air tables. 8 Given: The properties are T1 = 300 K p1 = 2 bar V1 = 1000 m/s Find: p01 and T01 Solution: 1 10002 2c = 300 + = 797.8 K ⇐ (a) p V12 2 · 1004.5 ( )γ/(γ−1) ( )3/5 T01 797.8 = p1 = 200 = 61.33 bar ⇐ (a) T1 300 T01 = T1 + p01 Using air tables: At the inlet pr1 = 1.386 h01 V12 10002 =h+ = 300.19 + = 800.19 kJ/kg 2 2cdot1000 ⇐ (b) Thus with the same entropy and this stagnation enthalpy pr0 = 43.38, and T01 = 780.15 K. Therefore p01 = p1 43.38 pr0 =2 = 6.26 bar pr 1.386 ⇐ (b) Exercise 2.13 At a certain location the velocity of air flowing in a duct is 321.5 m/s. At that location the stagnation pressure is 700 kPa and stagnation temperature is 450 K. What is the static density at this location. Given: The properties are T01 = 450 K Find: ρ1 . Solution: p1 = 700 kPa V1 = 321.5 m/s V12 321.52 T1 = T01 − = 450 − = 398.55 kPa 2cp 2 · 1004.5 ( ( )γ/(γ−1) )3.5 389.55 T1 = 700 p1 = p01 = 457.66 kPa T01 430 The density is then ρ1 = 398550 p1 = = 4.00 kg/m3 RT1 287 · 457.66 9 ⇐ Exercise 2.14 Air flows in a circular duct of diameter 4 cm at the rate of 0.5 kg/s. The flow is adiabatic with stagnation temperature 288 K. At certain location the static pressure is 110 kPa. Find the velocity at this location. Given: The properties are T0 = 288 K p = 110 kPa ṁ = 0.5 kg/s D = 0.04 m Find: The velocity Solution: Solving ṁ = ρAV = pAV /RT for temperature gives T = pAV ṁR Since T = T0 − V 2 /2cp a quadratic equation for velocity if found V2+ 2cp pA V − 2cp T0 = 0 ṁR the solution of which is cp pA V =− ṁR √ ( 1− 2T0 ṁ2 R2 1+ cp p2 A2 ) The terms are 2cp ṁ2 R2 2 · 288 · 0.52 · 2872 = = 0.618 cp p2 A2 1004.5 · 1100002 (π · 0.022 )2 cp RA 10004.5 · 110000 · π · 0.02 = = 967.61 ṁR 0.5 · 287 Therefore V = −967.61(1 − √ 1 + 0.6180) = 263.2 m/s ⇐ Exercise 2.15 Saturated steam enters a nozzle at static pressure 14 bar at velocity 52 m/s. It expands isentropically to pressure 8.2 bar. Mass flow rate is ṁ = 0.7 kg/s. Find the exit area if, (a) steam is assumed to behave as an ideal gas with γ = 1.135, and cp = 2731 J/kg K; (b) the end state is calculated with properties obtained from the steam tables. 10 Given: The properties are T1 = 468.2 K p1 = 14 bar V1 = 52 m/s ṁ = 0.7 kg/s γ = 1.135 Find: A2 Solution: From steam tables the value of enthalpy is h1 = 2789.4 kJ/kg and the specific volume is V1 = 0.1408 m3 /kg. Then assuming ideal gas behavior with constant specific heats, gives T01 = T1 + V12 522 = 468.2 + = 468.7 K 2cp 2 · 2731 and the stagnation pressure is 1 522 p01 = p1 + ρ1 V12 = 14 · 105 + = 14.096 bar 2 2 · 0.1408 State 2 has p2 = 820 bar, so that ( )0.135 ( )(γ−1)/γ 820 p2 T2 = T1 = 468.2 1.135 = 439.3 K p1 1400 Since no work is done T02 = T01 and √ √ V2 = 2cp (T02 − T2 ) = 2 · 2731(468.7 − 439.3) = 400.5 m/s The specific volume is ( v2 = v1 p2 p1 )1/γ ( = 0.1408 1400 820 ) 1/1.315 = 0.2256 m3 /kg Hence A = ṁv2 /V2 = 0.7 · 0.2256/400.5 = 3.944 cm2 ⇐ Using steam tables x2 = 6.4684 − 2.0559 s2 − sf = = 0.9596 sg − Sf 6.6546 − 2.0559 h2 = hf + x2 hf g = 725.6 + 0.9596(2770.0 − 725.36) = 2687.0 kJ/kg v2 = vf + x2 vf g = 1.116 · 10−3 + 0.9596(0.2354 − 1.116 · 10−3) = 0.2259 m3 /kg 11 Therefore V2 √ √ 2(h02 − h2 ) = 2(2791.0 − 2687.0 = 454.9 m2 /s so that A = ṁv2 /V2 = 0.7 · 0.2259/54509 = 0.3469 cm2 ⇐ Exercise 2.16 A fluid enters a turbine with total temperature of 330 K and total pressure of 700 kPa. The outlet total pressure is 100 kPa. If the expansion process through the turbine is isentropic, evaluate (a) the work per unit mass flow if the fluid is incompressible and having a density equal to 1000 kg/m3 , (b) the work per unit mass flow if the fluid is air Given: The properties are T01 = 330 K p01 = 700 kPa p02 = 100 kPa Find: w for water and air. Solution: Since the process is isentropic, for water ws = p01 − p02 700 − 100 = = 0.6 kJ/kg ρ 998 For air ( T02 = T01 p02 p01 )(γ−1)/γ ( = 330 100 700 ⇐ )1/3/5 = 189.3 K so that ws = h01 − h02 = cp (T01 − T02 ) = 1.0045(330 − 189.3) = 141.4 kJ/kg ⇐ Note that the air temperature leaving the turbine is very low, and operation such as this would be unusual. It might happen in an expander in cryogenic applications. Exercise 2.17 Air flows through a turbine which has a total pressure ratio 5 to 1. The total-to-total efficiency is 80% and the flow rate is 1.5 kg/s. The desired output power is to be 250 hp(186.4 kW). Determine: (a) The inlet total temperature; (b) The outlet total temperature; (c) the outlet static temperature if the exit velocity is 90 m/s; (d) draw the process on a T S-diagram and determine the total-to-static efficiency of the turbine. 12 Given: The properties of air are p01 /p02 = 5 ηtt = 0.8 ṁ = 1.5 kg/s V2 = 90 m/s Ẇ = 186.4 kW Find: T01 , T02 , T2 , the TS-diagram and ηts . Solution: w= Ẇ 186.4 = = 124.27 kJ/kg ṁ 1.5 ws = 124.27 w = = 155.33 kJ/kg ηtt 0.8 ws = cp (T01 − T02s ) T01 = ws 155.33 = = 419.51 K (γ−1)/γ cp [1 − (p02 /p01 ) ] 1.0045(1 − 5−1/3.5 ) T02 = T01 − T2 = T02 − w 124.27 = 419.51 − = 295.80 K cp 1.0045 ⇐ V22 902 = 295.80 − = 291.76 K 2cp 2 · 1.0045 T02s = T01 − ⇐ ⇐ w 155.33 = 419.51 − = 264.9 K 2cp 2 · 1.0045 Assuming that V2s = V2 T2s = T02s − Thus ηts = V2s2 902 = 264.9 − = 260.8 K 2cp 2 · 1.0045 T01 − T02 419.51 − 295.80 = = 0.78 T01 − T2s 419.51 − 260.8 ⇐ Exercise 2.18 A blower has a change in total enthalpy of 6000 J/kg, an inlet total temperature 288 K, and inlet total pressure 101.3 kPa. Find: (a) the exit total temperature if the working fluid is air; (b) the total pressure ratio across the machine if the total-to-total efficiency is 75%. Given: The properties of air are h02 − h01 = 6000 J/kg T01 = 288 K Find: T02 , p02 /p01 . 13 p01 = 101.3 kPa ηtt = 0.75 Solution: T02 = T01 + h02 − h01 6000 = 288 + = 294.0 K cp 1004.5 Next ηtt = ⇐ T02s − T01 T02 − T01 T02s = T01 + ηtt (T02 − T01 ) = 288 + 0.75 · 6 = 292.5 K so that p02 = p01 ( T02s T01 )γ/(γ−1) ( = 292.5 288 )3/5 = 1.0555 ⇐ Exercise 2.19 A multi-stage turbine has a total pressure ratio 2.5 across each of four stages. The inlet total temperature is T01 = 1200 K and the total-to-total efficiency of each stage is 0.87. Evaluate the overall total-to-total efficiency of the turbine if steam is flowing through it. Steam can be assumed to behave as a perfect gas with γ = 1.3. Why is the overall efficiency higher than the stage efficiency? Given: The properties of air are p02 = 2.5 p01 T01 = 1200 K ηtt = 0.87 Find: the total-to-total efficiency for the multistage turbine. Solution: p01 p01 p02 p03 p04 = = 2.54 = 39 p05 p02 p03 p04 p05 Also [ ( )γ/(γ−1) ] p02 T02 = T01 1 − ηtt p01 and since the efficiency of each stage is the same [ ( )γ/(γ−1) ] p02 T03 = T02 1 − ηtt p01 [ ( T03 = T01 1 − ηtt 14 p02 p01 )γ/(γ−1) ]2 and so forth to [ so that T05 ( )γ/(γ−1) ]4 T05 = T01 1 − ηtt p02 p01 [ ( = 1200 1 − 0.87 1 − 1 2.50.3/1.3 )]4 = 581.1 K The overall efficiency is therefore η0 = T01 − T05 1 − T05 /T01 = T01 − T05s 1 − (p05 /p01 )γ/(γ−1) η0 = 1 − 581.1/1200 = 0.904 1 − (1/2.54 )0.3/1.3 The overall efficiency is larger than the stage efficiency because the internal energy at the exit of each stage is higher because of internal heating and this becomes available during the next expansion process. Exercise 2.20 Gases from a combustion chamber enter a gas turbine at total pressure of 700 kPa and total temperature of 1100 K. The total pressure and total temperature at the exit of the turbine are 140 kPa and 780 K. If γ = 4/3 is used for the mixture of combustion gases, which has a molecular mass of 28.97, find the total-to-total efficiency and the total-to-static efficiency of the turbine, if the exit velocity is 210 m/s. Given: The inlet and exit conditions are and V2 = 210 m/s. p01 = 700 kPa p02 = 140 kPa T01 = 1100 K T02 = 780 K Combustion gases with γ = 4/3 and R = 287 J/kg. Find: the total-to-total efficiency. Solution: γR = 4 · 287 = 1148 J/kg cp = γ−1 ( )(γ−1)/γ ( )1/4 p02 140 T02s = T01 = 1100 = 735.6 K p01 700 ηtt = T01 − T02 1100 − 780 = = 0.878 T01 − T02s 1100 − 735.6 15 ⇐ T2s = T02s − ηts = V2s2 2102 = 735.6 − = 716.4 K 2cp 2 · 1148 T01 − T02 1100 − 780 = = 0.834 T01 − T2s 1100 − 716.4 ⇐ Exercise 2.21 Air enters a compressor from atmosphere at 101.3 kPa, 288 K. It is compressed to static pressure of 420 kPa and at the exit its velocity is 300 m/s. The compressor total-to-total efficiency is 0.82. (a) Find the exit static temperature, without making the assumption that V2s = V2 . (b) Find the exit static temperature by assuming that V2s ̸= V2 . Given: The exit velocity of air is V2 = 300 m and the properties of air are p01 = 101.3 kPa T01 = 288 K p2 = 420 kPa ηtt = 0.82 Find: T2 is two different ways. Solution: Assume first that V2s = V2 , then ( T2s = T01 T02s p2 p01 )(γ−1)/γ ( = 288 420 1013.3 )1/3.5 = 432.4 K V22 3002 = T2s + = 432.4 + = 477.1 K 2cp 2 · 1004.5 As a result T02 = T01 + 1 1 (T02s − T01 ) = 432.4 + (477.1 − 288) = 518.7 K ηtt 0.82 and T2 = T02 − V22 3002 = 518.7 − = 473.9 K 2cp 2 · 1004.5 If V2 ̸= V2s , then T02s = T2s + ⇐ (a) V2s2 V 2 T2s = T2s + 2 2cp 2cp T2 and 1 1 V22 T2s V22 T2 = T01 + (T02s − T01 ) = T01 + (T2s − T01 + )− ηtt ηtt 2cp T2 2cp 16 from which T22 1 V22 V22 T2s − [T01 + (T2s − T01 ) − ] T2 − =0 ηtt 2cp 2ηtt cp which when solved for T2 gives √ 419.22 419.2 T2 = − + + 23620 = 469.5 K 2 4 ⇐ (b) Exercise 2.22 Liquid water issues at velocity V1 = 20 m/s from a bank of five oblique nozzles shown in Figure 1. The nozzles with wall thickness t = 0.2 cm are spaced s = 4 cm apart. The nozzle angle is α1 = 70 deg. Using the mass and momentum balance: (a) Find the downstream velocity V2 . (b) Find the pressure increase in the flow. (c) Show how to deduce this result from the final equation in the example on mixing from the text. (d). If the thickness of the wall is vanishingly small, what is the change in pressure. Given: t = 0.2 cm s = 4 cm V1 = 20 m/s α1 = 70◦ Find: the pressure increase if δp = p2 − p1 , and also the pressure increase is t = 0. Solution: From the mass balance V2 V1 (cos α1 − t/s) From the momentum balance ρV2 (V2 − V1 cos α1 ) = p1 − p2 and after eliminating V2 between this and the mass balance lead to ( ) ( ) t 0.2 2 0.2 ◦ 2t cos α1 − = 998·20 cos(70 ) − = 5.84 kPa p2 −p1 = ρV1 s s 4 4 ⇐ Since Va = V1 cos α1 ) from p2 −p1 = ρV (Va −V ) V = t t Va A a = V1 cos α1 (1− ) = V1 (cos α1 − ) Aa + Ab s cos α1 s 17 V1 V2 α1 Figure 1: Nozzles with an oblique discharge. and Va − V = V1 cos α1 − V1 cos α1 + V1 so that p2 − p1 = t ρV12 s ( t cos α1 − s t s ) ⇐ Exercise 2.23 Consider the flow shown in Figure 1. Prove that the kinetic energy lost in the flow as it moves to the downstream section is equal to that associated with the transverse component of the velocity. Given: The channel width s and angle α1 , and V . Find: the expression for the loss in terms of the transverse component. Solution: From continuity, after canceling s V2 = V1 cos α1 From x-component of momentum equation, after canceling s V2 (V2 − V1 cos α1 ) = p1 − p2 Substituting from continuity gives p1 = p2 . From energy equation p1 V12 p2 V22 V2 + = + +ζ 1 ρ 2 ρ 2 2 Canceling the pressure terms and from continuity V2 = V1 cos α1 yields 1 − cos2 α1 = ζ V12 = V22 + ζV12 so that ζ V 2 sin2 α1 V2 V12 = 1 = t 2 2 2 18 ⇐ Chapter 3 Exercise 3.1 Conditions in an air reservoir are 680 kPa and 560 K. From there the air flows isentropically though a convergent nozzle to a back pressure of 101.3 kPa. Find the velocity at the exit plane of the nozzle. Given: The properties are T01 = 560 K p01 = 630 kPa p2 = 101.3 kPa Find: The velocity V2 . Solution: Since 101.3 p2 = = 0.149 p01 630 the flow is choked and the velocity at the exit is the sonic velocity. To find it first calculate 2 · 560 2T01 = = 466.7 K T∗ = γ+1 2.4 and then √ √ V ∗ = γRT ∗ = 1.4 · 287 · 466.7 = 433 m/s ⇐ Exercise 3.2 Air flows in a converging duct. At a certain location, where the area is A1 = 6.5 cm2 , pressure is p1 = 140 kPa and Mach number is M1 = 0.6. The mass flow rate is ṁ = 0.25 kg/s. (a) Find the stagnation temperature. (b) If the flow is choked find the size of the throat area. (c) Give the percent reduction in area from station 1 to the throat. (d) Find the pressure at the throat. Given: At the location A1 = 6.5 cm2 and the properties are M1 = 0.6 p1 = 140 kPa ṁ = 0.25 kg/m Find: The velocity T01 , A∗ , percent reduction in area, and p∗ . Solution: Solve the equation ṁ = ρ1 A1 V1 = p 1 A1 √ M1 γRT1 RT1 for T1 . It yields ( ( )2 )2 p1 A1 M1 140000 · 0.00065 · 0.6 γ 1.4 T1 = = = 232.7 K ṁ R 0.25 287 19 and then ( T01 = T1 ) γ−1 2 1+ M1 = 232.7(1 + 0.2 · 0.62 ) = 249.4 K 2 ( )(γ+1)/2(γ+1) 1 2 γ−1 2 A∗ = + M A1 M1 γ + 1 γ + 1 1 so that ⇐ )−(γ+1)/2(γ+1) γ−1 2 2 A = A1 M 1 + M γ+1 γ+1 1 2 0.4 = 6.5 · 0.6( + · 0.62 )−3 = 5.47 cm2 ⇐ 2.4 2.4 Percent reduction is 6.5 − 5.47 A1 − A∗ = = 0.1584 15.84 % ⇐ A1 6.5 ( ∗ To calculate the exit pressure, first )γ/(γ−1) ( γ−1 2 p01 = p1 1 + M1 = 140(10.2 · 0.62 )3.5 = 178.6 kPa 2 and then ( ∗ p = p01 2 γ−1 )γ/(γ−1) ( = 178.6 2 2.4 )3/5 = 94.3 kPa ⇐ Exercise 3.3 Air flows in a convergent nozzle. At a certain location, where the area is A1 = 5 cm2 , pressure is p1 = 240 kPa and temperature is T1 = 360 K. Mach number at this location is M1 = 0.4. Find the mass flow rate. Given: At the location A1 = 5 cm2 and the properties are T1 = 360 K p1 = 140 kPa M1 = 0.4 Find: Find the mass flow rate. Solution: √ √ V1 = M1 γRT1 = 0.4 1.4 · 287 · 360 = 152.1 m/s ρ1 = p1 240000 = = 2.323 kg/m3 RT1 287 · 360 so that ṁ = ρ1 A1 V1 = 2.323 · 5 · 10−4 · 152.1 = 0.177 kg/s 20 ⇐ Exercise 3.4 The area of a throat in a circular nozzle is At = 1 cm2 . For a choked flow find the diameter where M1 = 0.5. Determine the value of Mach number at a location where the diameter is D2 = 1.941 cm. Assume the flow to be isentropic and γ = 1.4. Given: At the throat A∗ = 1.0 cm2 and D2 = 1.941 cm. Find: The D1 and M2 Solution: From ( )(γ+1)/2(γ+1) ( )3 1 1 A∗ 2 γ−1 2 2 0.4 2 = + M = + 0.5 ) = 1.34 A1 M1 γ + 1 γ + 1 1 0.5 2.4 2.4 so that √ A1 = 1.34A∗ = 1.34 · 1 = 1.34 cm2 D1 = 4A1 = 1.31 cm π ⇐ For D2 = 1.941 cm, the area is A2 = πD22 /4 = 2.96 cm2 , then solve ( )(γ+1)/2(γ+1) 2 2.96 1 γ−1 2 A2 = = 2.96 = + M A∗ 1 M2 γ + 1 γ + 1 2 for M2 . Carrying out the solution by iteration gives M2 = 0.2 ⇐ Exercise 3.5 In a location in a circular nozzle where the area is A1 = 4 , Mach number has the value M1 = 0.2. Find the diameter at a location where M = 0.6. Given: At the location A1 = 4 cm2 and M1 = 0.2 Find: The area at which M2 = 0.6. Solution: From ( )(γ+1)/2(γ+1) ( )3 A1 2 2 1 γ−1 2 1 0.4 2 = + M = + 0.2 = 2.964 A∗ M1 γ + 1 γ + 1 1 0.2 2.4 2.4 therefore A∗ = 4/2.964 = 1.350 cm2 and then from ( )(γ+1)/2(γ+1) )3 ( 1 1 2 γ−1 2 0.4 2 A2 2 = = 1.188 = + M + 0.6 A∗ M2 γ + 1 γ + 1 2 0.6 2.4 2.4 so that A2 = 1.188 · 1.350 = 1.604 cm2 and √ √ D2 = 4A2 /π = 4 · 1.604/π = 1.43 cm 21 ⇐ Exercise 3.6 Air flows through a circular duct 15 cm in diameter with a flow rate 2.25 kg/s. The total temperature and static pressure at a certain location in the duct are 30 C and 106 kPa, respectively. Evaluate (a) the flow velocity, (b) the static temperature, (c) the total pressure, and (d) the density at this location. Given: At the location where D1 = 15.0 cm2 and the flow rate is ṁ = 2.25 kg/s, the properties are T01 = 303.15 K p1 = 106 kPa Find: The velocity, the static temperature, the total pressure, the static density. Solution: πD12 A1 = = 0.01767 m2 4 From the mass balance ( )1/2 √ γ p1 A 1 M 1 √ T01 ṁ = ρ1 A1 V1 = γRT1 = p1 A1 M1 RT1 RT01 T1 or M14 2 2 + M12 − γ−1 γ−1 ( ṁ p1 A1 )2 RT01 =0 γ Substituting the numerical values into this gives √ M14 + 5M12 − 0.4483 = 0 M12 = −2.5 + 2.52 + 0.4483 M1 = 0.297 Therefore ( T1 = T01 and V1 = M 1 γ−1 2 M1 1+ 2 √ )−1 = 303.15 = 297.9 K 1 + 0.2 · 0.2972 √ γRT1 = 0.297 1.4 · 287 · 297.9 = 102.8 m/s The stagnation pressure is ( )γ/(γ−1) )3.5 ( 303.15 T01 = 106 = 112.7kPa p01 = p1 T1 297.9 and the static density is ρ1 = 106 p1 = = 1.24 kg/m3 RT1 0.287 · 297.9 22 ⇐ ⇐ ⇐ Exercise 3.7 Conditions in an air reservoir are 380 kPa and 460 K. From there the air flows though a convergent nozzle to a back pressure of 101.3 kPa. The polytropic efficiency of the nozzle is ηp = 0.98. Find: (a) Exit plane pressure, (b) Exit plane temperature, (c) Velocity at the exit plane of the nozzle. Given: The properties of air are. T01 = 460 K p01 = 380 kPa pb = 101.3 kPa ηp = 0.98 Find: pe , Te , Ve . Solution: Since pb /p01 = 101.3/380 = 0.266, the flow is choked. The polytropic exponent is n= Then pe = p01 ( 2 γ+1 γ 1/4 = = 1.389 ηp + γ(1 − ηp ) 0.98 + 1.4 · 0.02 )n/(n−1) ( = 380 2 2.4 ) 1/389/0.389 = 198.1 kPa ⇐ and 2 2 T01 = 460 = 383.3 K ⇐ γ+1 2.4 √ √ and the Mach number is Me = (n − 1)/(γ − 1) = 0.389/0.4 = 0.986 so that √ √ Ve = Me γRTe = 0.986 1.4 · 287 · 383.3 = 387.0 m/s ⇐ Te = Exercise 3.8 Air issues from a reservoir at conditions 260 kPa and 540 K into a converging nozzle. The nozzle efficiency is estimated to be ηN = 0.986. The back pressure is pb = 101.3kPa. Find: (a) The exit Mach number, (b) Exit plane temperature, (c) Exit plane pressure, (d) Exit velocity. Given: The properties of air are. T01 = 540 K p01 = 260 kPa pb = 101.3 kPa ηN = 0.986 Find: Me , pe , Te , Ve . Solution: Since pb /p01 = 101.3/260 = 0.389, the flow is choked. The Mach number can be calculated from ]1/2 [ √ (γ − 1)(3ηN − 1) − 2γ + [(γ − 1)(3ηN − 1) − 2γ]2 + 8ηN (1 − ηN )(γ − 1)2 Me = 2(γ − 1)2 (1 − ηN ) 23 [ Me = 0.4 · 1.958 − 2.8 + √ (0.4 · 1.958 − 2.8)2 + 8 · 0.989 · 0.014 · 0.42 2 · 0.16 · 0.014 ]1/2 = 0.9883 The exit temperature is Te = 2T01 2 · 540 = = 450 K γ+1 2.4 ⇐ The polytropic exponent is n = 1 + (γ − 1)Me2 = 1.3907 and the exit pressure is ( pe = p01 2 γ+1 )n/(n−1) ( = 260 2 2.4 )1.3907 0.3907 = 135.9 kPa ⇐ and the velocity is Ve = Me √ √ γRTe = 0.9882 1.4 · 287 · 450 = 420.2 m/s ⇐ Exercise 3.9 At the inlet to a nozzle the conditions are M1 = 0.3, p01 = 320 kPa, and T01 = 430 K. The flow is irreversible with polytropic exponent n = 1.396. Show that ( )γ/(γ−1) ( )(n−1)/n p01 p1 T02 = T2 p1 p2 Find the Mach number at a location where p2 = 210 kPa. Given: T01 = 430 K, p01 = 320 kPa, M1 = 0.3, and n = 1.396. Find: Show that the expression in the statement of the exercise is correct. Also find the Mach number at the location where p2 = 210 kPa. Solution: From ( )(n−1)/n p1 T1 T1 T01 = = p2 T2 T01 T2 Therefore T01 T01 = T2 T1 ( p1 p2 )(n−1)/n and since T01 − T02 T02 = T2 ( p01 p1 )(γ−1)/γ ( 24 p1 p2 )(n−1)/n ⇐ From p01 = p1 ( γ−1 2 1+ M1 2 so that p1 = and )γ/(γ−1) = (1 + 0.2 · 0.32 )3/5 = 1.0644 320 = 300.63 kPa 1.0644 ( ) ( )n/(n−1) γ−1 2 p1 1− M1 2 p2 ( )0.396 T02 300.62 = (1 + 0.2 · 0.32 ) 1.396 = 1.127 T2 210 T01 T1 T02 = = T2 T1 T2 and also T02 γ−1 2 =1+ M2 T2 2 so that √ M2 = 2 γ−1 ( ) √ T02 2 −1 = (1.127 − 1) = 0.797 T2 0.4 ⇐ Exercise 3.10 Flow from a reservoir with p01 = 260 kPa and T01 = 530 K flows through a nozzle. It is estimated that the static enthalpy loss coefficient is ζ = 0.020. The exit pressure is p2 = 180 kPa. (a) Find the exit Mach number. (b) Find the polytropic efficiency of the nozzle. Given: The conditions are T01 = 530 K p01 = 260 kPa p2 = 180 kPa ζ = 0.020 Find: ηp . Solution: The static temperature for an isentropic process is ( ( )(γ−1)/γ )1/3.5 180 p2 = 530 T2s = T01 = 4771 K p01 260 The static temperature is obtained from ζ= T2 − T2s T02 − T2 T2 = ζT02 + T 2s 0.02 · 530 + 477.1 = = 478.2 K 1+ζ 1.002 25 From T02 γ−1 2 = 1+ M2 M2 = T2 2 √ 2 T02 ( − 1) = γ − 1 T2 √ 2 530 ( − 1) = 0.736 ⇐ 0.4 478.2 The polytropic exponent is obtained from T2 = T01 ( p2 p01 )(n−1)/n therefore n= ( ηN = n−1 n )( 1 1− γ γ−1 ln(T2 /T01 ) ln(p2 /p01 ) ) = = 1 1− ln(478.2/530) ln(180/260) 0.3885 1.4 = 0.9793 1.3885 0.4 = 1.3885 ⇐ Exercise 3.11 A two dimensional nozzle has a shape 4 x(x − 1) + 5 2(x + 2) √ The √ nozzle stretches from 0 < x/a < 2( 2 − 1). The throat is at xt /a = 2( 2 − 1). The scale factor a is chosen such that the half-width of the nozzle at x = 0 is 4a/5. Assume that 4f¯ = 0.02 and the inlet Mach number is M1 = 0.5. Calculate and plot p/p1 as a function of x. Calculate the Mach number along the nozzle and graph it on the same plot. Given: The shape of the flow channel and the friction factor. Find: Plot p/p1 and the Mach number. Solution: y= %Nozzle Calculation %Nozzle Calculation clear all; clf k=1.4; f=0.02/4; w=1; x=[0:0.005:2*(sqrt(2)-1)]; y=4/5+x.*(x-2)./(4+2*x); n=length(x); M(1)=0.5 mx(1)=M(1)^2; d(1)=4*y(1)*w/(w+y(1)); A(1)=2*y(1)*w; 26 p0(1)=1; p(1)=1; for i=2:n d(i)=4*y(i)*w/(w+y(i)); A(i)=2*y(i)*w; mx(i)=mx(i-1)-2*mx(i-1)*(1+0.5*(k-1)* ... mx(i-1))*(A(i)/A(i-1)-1)/(1-mx(i-1))+ ... mx(i-1)*k*mx(i-1)*(1+0.5*(k-1)*mx(i-1))* ... 4*f*(x(i)-x(i-1))/(d(i)*(1-mx(i-1))); M(i)=sqrt(mx(i)); p0(i)=p0(i-1)-p0(i-1)*k*mx(i-1)*2*f*(x(i)-x(i-1))/d(i); r(i)=((2+(k-1)*mx(i))/(2+(k-1)*mx(1)))^(k/(k-1)); p(i)=p0(i)*p(1)/(p0(1)*r(i)); end plot(x,p,x,M);grid; Exercise 3.12 Steam enters a nozzle from a steam chest at saturated vapor state at pressure p0 = 0.8 bar. It expands isentropically through a steam nozzle. Find the degree of supersaturation when it crosses the Wilson line at x = 0.96. Given: Saturated steam at p01 = 80 kPa. Find: The degree of saturation if it isentropically expands to x2 = 0.96. Solution: The temperature of saturated steam at this pressure is T01 = 366.6 K. Assuming that the adiabatic index according to Zeuner’s formula is γ = 1.135, then for an isentropic process ( T2 = T01 p2 p01 )(γ−1)/γ ( = 366.6 36.6 80 )0.135/1.135 = 333.7 K The pressure of saturated steam at this temperature is pss = 20.43 kPa. Hence the degree of supersaturation is S= p2 36.6 = = 1.774 pss 20.43 ⇐ Exercise 3.13 Consider a supersonic flow over a convex corner with angle θ2 = 5◦ , when the inflow moves in the direction of θ1 = 0◦ . The upstream Mach number is M1 = 1.1, pressure is p1 = 130 kPa and T1 = 310 K. Find: (a) Mach number, (b) temperature, and (c) pressure after the expansion is complete. 27 Given: The conditions at the inlet are M1 = 1.1 p1 = 130 kPa T1 = 310 K and the wall makes an angle θ2 = 15◦ . Find: M2 , T2 and p2 . Solution: The Prandtl-Meyer function a the inlet is ( ) ) ( 1 1 −1 −1 √ √ µ1 = tan = tan = 65.38◦ 1.12 − 1 M12 − 1 then √ ) (√ √ γ+1 γ−1 −1 −1 2 ν1 = tan (M − 1) − tan M12 − 1 γ−1 γ+1 1 (√ ) √ √ 2.4 0.4 tan−1 (1.12 − 1) − tan−1 1.12 − 1 = 1.336◦ ν1 = 0.4 2.4 so that ν2 = ν1 + θ2 = 1.336◦ + 5◦ = 6.336◦ Since √ ν2 = γ+1 tan−1 γ−1 (√ ) √ γ−1 −1 2 (M − 1) − tan M22 − 1 γ+1 2 With γ = 1.4 and ν1 = 6.336◦ , this equation can be solved by iterations for M2 . It gives M2 = 1.306 ⇐ (a) Next T0 = T1 (1 + and T2 = T0 (1 + and ( p2 = p 1 T2 T1 γ−1 2 M1 ) = 310(1 + 0.2 · 1.12 ) = 385.0 K 2 γ − 1 2 −1 M1 ) = 385(1 + 0.2 · 1.12 )−1 = 287.1 K 2 ( )γ/(γ−1) = 130 287.1 310 28 )3/5 = 99.36 kPa ⇐ Exercise 3.14 Consider the steam flow from a low pressure nozzle at an angle α = 65◦ . At the inlet of the nozzle steam is saturated vapor at pressure p0 = 18 kPa. Steam exhausts into the inter-blade space, where pressure is 7 kPa. Find the angle θ by which the flow turns on leaving the nozzle, the far downstream velocity, and its direction. Given: Steam, with γ = 1.135 and R = 8314/18 = 461.9 J/kg K at the inlet as saturated steam at p0 = 18 kPa and leaves at the nozzle angle α1 = 65◦ . Its far downstream pressure is p2 = 7 kPa. Find: The turning angle θ, V2 and α2 . Solution: The nozzle is choked because p2 /p0 = 0.389. The saturation temperature is T0 = 330.9 K. Then ( T2 = T0 From p2 p0 )(γ−1)/γ ( = 330.9 7 18 )0.135/1.135 = 295.7 K T0 γ−1 2 =1+ M2 T2 2 √ ( ) √ ( ) 2 T0 2 330.9 M2 = −1 = − 1 = 1.327 γ − 1 T2 0.135 295.37 so that V2 = M2 √ √ γRT2 = 1.327 1.135 · 461.9 · 295/7 = 522.6 m/s √ √ γ+1 γ−1 ϕ= tan−1 (M 2 − 1) γ−1 γ+1 2 √ √ 3.135 0.135 −1 ϕ= tan (1.3282 − 1) = 49.2◦ 0.135 2.135 Next In addition ( −1 µ2 = tan √ therefore θ2 = ϕ − 1 M22 − 1 ) ( −1 = tan 1 √ 1.3282 − 1 ) π + µ2 = 49.36 − 90 + 4839 = 8.1◦ 2 29 ⇐ = 48.86◦ ⇐ ⇐ Exercise 3.15 Consider the steam flow from a low pressure nozzle at an angle α = 65◦ . At the inlet of the nozzle steam is saturated vapor at pressure p0 = 18 kPa. Steam exhausts into the inter-blade space, where pressure is 7 kPa. Using the continuity equation, find the angle θ by which the flow turns on leaving the nozzle, the far downstream velocity, and its direction. Given: Steam, with γ = 1.135 and R = 8314/18 = 461.9 J/kg K at the inlet as saturated steam at p0 = 18 kPa and leaves at the nozzle angle α1 = 65◦ . Its far downstream pressure is p2 = 7 kPa Find: The turning angle θ, V2 and α2 . Solution: The flow is choked because p2 /p0 = 0.389. The specific volume of saturated steam at p0 = 18 kPa is v0 = 8.46 m3 /kg and the temperature is T0 = 330.2 K. The specific volume at the throat is ( )1/0.135 γ + 1 1/(γ−1) 2.135 v1 = v0 = 8.46 = 13.7 m3 /kg 2 2 and the temperature is T1 = 2T 0 2 · 330.2 = = 309.3 K γ+1 1.135 and the velocity is V1 = √ √ γRT1 = 1.135 · 461.9 · 309.3 = 403.1 m/s The temperature far downstream is ( )0.135/1.135 ( )(γ−1)/γ 7 p2 T2 = T0 = 330.9 = 295.7 K p0 18 and the Mach number can be calculated from T0 γ−1 2 =1+ M2 T2 2 √ ) √ ) ( ( 2 2 T0 330.9 −1 = − 1 = 1.327 M2 = γ − 1 T2 0.135 295.7 so that V2 = M 2 √ √ γRT2 = 1.327 1.135 · 461.9 · 295.7 = 522.6 m/s 30 ⇐ The specific volume is v2 = RT2 461.9 · 295.7 = = 19.5 m3 /kg p2 7000 and by continuity the angle is ) V1 v2 cos α1 α2 = cos V2 v1 ( ) 403.1 · 19.5 −1 ◦ α2 = cos cos(65 ) = 62.5◦ 522.6 · 13.77 ( −1 θ = α1 − α2 = 65◦ − 62.73◦ = 2.5◦ 31 ⇐ ⇐ Chapter 4 Exercise 4.1 Steam enters a rotor of an axial turbine with an absolute velocity V2 = 320 m/s at the angle α2 = 73◦ . The axial velocity remains constant. The blade speed is U = 165 m/s. The rotor blades are equiangular so that β3 = −β2 and the magnitude of the relative velocity remains constant across the rotor. Draw the velocity triangles. Find, (a) the relative flow angle β2 , (b) the magnitude of the velocity V3 after the flow leaves the rotor, and (c) the flow angle α3 which V3 makes with the axial direction. Given: The rotor blades are equiangular with β3 = −β2 and W3 = W2 . At the inlet of the rotor V2 320 m/s α2 = 73◦ U = 165 m/s Find: β2 , V3 , α3 . Solution: The axial and tangential components of the velocity are Vx2 = V2 cos α2 = 320 cos(73◦ ) = 93.56 m/s Vu2 = V2 sin α2 = 320 sin(73◦ ) = 306.0 m/s The the tangential component of the relative velocity is Wu2 = Vu2 − U = 306.0 − 165.0 = 141.0 m/s √ so that W2 = 2 2 Wx2 + Wu2 = √ 93.562 + 141.02 = 169.2 m/s and β2 = tan−1 (Wu2 /Wx2 ) = tan−1 (141.0/93.56) = 56.44◦ Since W3 = W2 β3 = −56.44◦ ⇐ (a) Wu3 = −W u2 = −141.0m/s and √ Vu3 = Wu3 +U = 141.0+165.0 = 24.0 m/s V3 = 2 2 Vx3 + Vu3 = √ 93.562 + 24.02 = 96.6 m/s ⇐ so that α3 = tan−1 (Vu3 /Vx3 ) = tan−1 (24.0/93.56) = 14.4◦ 32 ⇐ (c) Exercise 4.2 Water with density 998 kg/m3 flows in centrifugal pump at the rate of 22 liters per second. The impeller radius is r2 = 7.7 cm and the blade width at the impeller exit is b2 = 0.8 cm. If the flow angles at the impeller exit are α2 = 67◦ and β2 = −40◦ , find the rotational speed of the shaft in rpm. Given: The flow rate, the impeller radius and the height of its blade. The flow angles are also known at the exit. Find: The rotational speed of the shaft. Solution: From Q = 2πr2 b2 Vr2 Then Vr2 = Q 0.022 · 104 = = 5/68 m/s 2πr2 b2 2π · 7.7 · 0.8 Vu2 = Vr2 tan α2 = 5.68 tan(67◦ ) = 13.39 m/s Wu2 = Wr2 tan β2 = 5.68 tan(−40◦ ) = −4.77 m/s and U = Vu2 − Wu2 = 13.39 + 4.77 = 18.16 m/s So that Ω= U 18.16 · 100 · 30 = = 2252 rpm r2 7.7 · π ⇐ Exercise 4.3 In a velocity diagram at the inlet of a turbine the angle of the absolute velocity is 60◦ and the flow angle of the relative velocity is −51.7◦ . Draw the velocity diagram and find the value of U/V and Vx /U . Given: The angles α = 60◦ and β = −51.7◦ . Find: U/V and Vx /U . Solution: Let V = 1, then Vx = V cos α = cos(60◦ ) = 0.5 and Vu = V sin α = sin(60◦ ) = 0.866 Wu = Wx tan(β) = 0.5 tan(−51.7◦ ) = −0.633 so that U = Vu − Wu = 0.866 + 0.633 = 1.5 and U/V = 1/5 ⇐ The ratio Vx /U = 0.866/1.5 = 1/3 ⇐ 33 Exercise 4.4 A small axial-flow turbine has an output power of 37 kW when handling one kg of air per second with an inlet total temperature of 335 K. The totalto-total efficiency of the turbine is 80%. The rotor operates at 50, 000 rpm and the mean blade diameter is 10 cm. Evaluate (a) the average driving force on the turbine blades, (b) the change in the tangential component of the absolute velocity across the rotor, and (c) the required total pressure ratio across the turbine. Given: Ẇ = 37 kW, ṁ = 1 kg/s, T01 = 335 K, ηtt = 0.8, and r2 = 0.05 m Ω = 55, 000 rpm. Find: The force Fu , Vu2 − Vu3 and p03 /p01 . Solution: The blade speed is U = r2 Ω = 0.05 · 50, 000 · π = 261.8 m/s 30 so that from Ẇ = Fu U , the force is Fu = Ẇ /U = 37, 000/261.8 = 141.3 N ⇐ The specific work is w = Ẇ /ṁ = 37 kJ/kg and the change in tangential velocity is from w = u(VU 2 − Vu3 ) Vu2 − Vu3 = w 37, 000 = = 141.3 m/s U 261.8 ⇐ The isentropic work is ws = w 37 ws 46.25 = = 46.25 kJ/kg rmso that T02s = T01 − = 335− = 289 K ηtt 0.8 cp 1.0045 and p01 = p03 ( T01 T02s )γ/(γ−1) ( = 335 289 )3.5 = 1.678 ⇐ Exercise 4.5 The exit flow angle of stator in an axial steam turbine is 68◦ . The flow angle of the relative velocity leaving the rotor is −67◦ . Steam leaves the stator at V2 = 120 m/s and the axial velocity is Vx2 = 0.41U . At the exit of the rotor blades the axial steam velocity is Vx3 = 0.42U . The mass flow rate is ṁ = 2.2 kg/s. Find, (a) the flow angle entering the stator assuming it to be the same as the absolute flow angle leaving the rotor, (b) the flow angle of the relative velocity entering the rotor, (c) the reaction, and (d) the power delivered by the stage. Given: α2 = 68◦ , β3 = −67◦ , V2 = 120 m/s, Vx2 = 0.41U , Vx3 = 0.42U , and ṁ = 2.2 kg/s. 34 Find: β2 , α3 , and R. Solution: From Vx2 = V2 cos α2 = 120 cos(68◦ ) = 44.95 m/s Vu2 = V1 sin α2 = 120 sin(68◦ ) = 111.3 m/s and U = Vx2 /0.41 = 44.95/0.41 = 109.64 m/s. Thus Wu2 = Vu2 − U = 111.26 − 109.64 = 1.62 m/s ( and −1 β2 = tan Wu2 Wx2 ) ( −1 = tan 1.62 44.95 ) = 2.1◦ ⇐ (b) Next Vx3 = 0.42U = 0.42 · 109.64 = 46.05 m/s Wu3 = Wx3 tan β3 = 46.05 tan(−67◦ ) = −108.5 m/s so that Vu3 = Wu3 + U = −108.5 + 109.64 = 1.16 m/s ( and −1 α3 = tan Vu3 Vu2 ) ( −1 = tan 1.16 46.05 ) = 1.44◦ ⇐ (a) Next w = U (Vu2 − Vu3 ) = 109.64(112.26 − 1.16) = 12072 J/kg and Ẇ = ṁw = 202 · 12072 = 26.6 kW The reaction is R= ⇐ (d). 108.52 + 46.042 − 1.622 − 44.952 W32 − W22 = = 0.492 2e 2 · 12072 ⇐ (c) Exercise 4.6 The axial component of air flow leaving a stator in an axial flow turbine is Vx2 = 175 m/s and its flow angle is 64◦ . The axial velocity is constant, the reaction of the stage is R = 0.5, and the blade speed is U = 140 m/s. Since the reaction is fifty percent the relationships between the flow angles are β2 = −α3 and α2 = −β3 . Find the flow angle of the velocity entering the stator. Given: Vx2 , U , α2 , U and R. Find: α2 . Solution: The tangential velocity is Vu2 = Vx2 = tan α2 = 175 tan(64◦ ) = 358.8 m/s 35 and Wu2 = Vu2 − U = 358.8 − 140 = 218.8 m/s Therefore ( −1 β2 = tan Wu2 Wx2 ( ) −1 = tan 218.8 175 ) = 51.35◦ α3 = −54.3◦ ⇐ Exercise 4.7 The air flow leaving the rotor of an axial flow turbine is Vx3 = 140 m/s and its flow angle is 0◦ . The axial velocity is constant and equal to the blade speed. The inlet flow angle to the rotor is α2 = 60◦ . Find the reaction. Given: Vx , U = Vx , α1 = 0◦ , and α2 = 60◦ . Find: R Solution: Since the flow at the inlet and exit of the stage is axial √ √ √ W3 = Vx2 + U 2 = U 2 = 140 2 = 198 m/s and Vu2 = Vx2 tan α2 = 140 tan(60◦ ) = 242.5 m/s so that Wu2 = Vu2 − U = 242.5 − 140 = 102.58 m/s and √ √ 2 W2 = Wx2 + Wu2 = 1402 + 102.52 = 173.5 m/s The specific work is w = U (Vu2 − Vu3 ) = U Vu2 = 140 · 242.5 = 33, 950 J/kg so that R+ W32 − W22 1982 − 173.52 = 0.134 2w 2 · 33, 950 ⇐ Exercise 4.8 A large centrifugal pump operates at 6000 rpm and produces a head of 800 m while the flow rate is 30, 000 liters per minute. (a) Find the value of the specific speed. (b) Estimate the efficiency of the pump. Given: H, Ω, Q. Find: The specific speed and η. Solution: The flow rate is standard units is Q= 30, 000 = 0.5 m3 /s 1000 · 60 36 The specific speed is √ √ Ω Q 6000π 0.5 Ωs = = = 0.533 (gH)3/4 30 · (9.81 · 800)3/4 From the figure for Q = 500 liters/s, the efficiency is 0.84 ⇐ ⇐ (a) (b). Exercise 4.9 A fan handles air at the rate of 500 liters per second when operating at 1800 rpm. (a) What is the flow rate if the same fan is operated at 3600 rpm? (b) What is the percentage increase in total pressure rise of the air assuming incompressible flow? (c) What is the power input required at 3600 rpm relative to that at 1800 rpm. Assume that the operating point of the fan in terms of the dimensionless parameters is the same in both cases. Given: Q1 , Ω1 , Ω2 , and D1 = D2 . Find: Q2 , increase in total pressure, Ẇ2 . Solution: From Q1 Q2 Ω2 3600 = Q2 = Q1 = 500 = 1000 l/s 3 3 Ω1 D1 Ω2 D2 Ω1 1800 (a) Next Ẇ2 Ẇ1 = 3 5 ρΩ1 D1 ρΩ32 D25 so that Ẇ2 = Ẇ1 Then Ẇ = Ω32 = 8Ẇ1 Ω31 ⇐ (c) ṁ (p02 − p01 )ηtt = Q∆p0 ηtt ρ and Ẇ1 Q2 ηtt (∆p0 )1 2 1 = = = (∆p0 )2 Q1 ηtt Ẇ2 8 4 so that the percent increase is 300 % ⇐ (b). Exercise 4.10 An axial flow pump having a rotor diameter of 20 cm handles water at the rate of 60 liters per second when operating at 3550 rpm. The corresponding increase in total enthalpy of the water is 120 J/kg and the total-to-total efficiency is 75%. Suppose a second pump in the same series is to be designed to handle 37 water having a rotor diameter of 30 cm and operating at 1750 rpm. For this second pump what will be the predicted values for (a) the flow rate, (b) the change in the total pressure of water, (c) the input power. Given: Q1 , D1 , Ω1 , ηt t, (∆h0 )1 , D2 and Ω2 . Find: Q2 , (∆p0 )2 , and Ẇ2 . Solution: From ) ( )3 ( ) ( )3 ( 30 Q1 Q2 Ω2 D2 1750 = 99.8 l/s ⇐ = Q2 = Q1 = 60 3 3 Ω1 D1 Ω2 D2 Ω1 D1 3550 20 Next Ẇ1 = ṁ1 (∆h0 )1 = ρQ1 (∆h0 )1 = 998 · 60 · 120 = 7.18 kW 1000 then from Ẇ2 Ẇ1 = 3 5 ρΩ1 D1 ρΩ32 D25 ( )3 ( )5 ( )3 ( )5 D2 1750 30 Ω1 Ẇ2 = Ẇ1 = 7.18 = 6.54 kW Lef tarrow (b) Ω2 D1 3550 20 Then Ẇ2 6540 (∆h0 )2 = = = 65.6 J/kg ρQ2 99.8 and (∆p0 )2 = ρ (∆h0 )2 = 998 · 65.6 = 49.1 kPa ⇐ (c) Exercise 4.11 A small centrifugal pump handles water at the rate of 6 liters per second with input power of 5 hp and total-to-total efficiency of 70%. Suppose the fluid being handled is changed to gasoline having specific gravity 0.70. What are the predicted values for (a) flow rate, (b) input power, (c) total pressure rise of the gasoline? Given: Q1 , Ẇ1 , ηt t, ρ1 , Q2 = Q1 and the specific gravity of gasoline. Find: Ẇ2 , (∆p0 )2 . Solution: Since the diameter and the shaft speed are the same, then from Q2 Q1 = 3 Ω 1 D1 Ω2 D23 Q2 = Q1 ⇐ (a) and from Ẇ1 Ẇ2 = 3 5 ρ 1 Ω 1 D1 ρ2 Ω32 D25 38 (a) Ẇ2 = Ẇ1 ρ2 = 5 · 0.7 = 3.5 hp ρ1 Next from Ẇ2 = ṁ2 (∆h0 )2 = ⇐ (b) Q2 (∆p0 )2 ηtt so that (∆p0 ) = ηtt Ẇ2 3/5 · 0.7457 · 1000 = = 304.5 kPa Q2 6 ⇐ (c) Exercise 4.12 A blower handling air at the rate of 240 liters per second at the inlet conditions of 103.1 kPa for total pressure and 288 K for total temperature. It produces a pressure rise of air equal to 250 mm of water. If the blower is operated at the same rotational speed, but with an inlet total pressure and total temperature of 20 kPa and 253 K, find; (a) the predicted value for the mass flow rate, (b) the total pressure rise. Given: Ω2 = Ω1 , Q1 , p01 , T01 , (∆p0 )1 , p02 , and T02 . Find: ṁ2 , (∆p0 )2 . Solution: From Q1 Q2 = Q2 = Q1 3 Ω1 D1 Ω2 D23 The stagnation densities are ρ01 = p01 101, 300 = = 1.226 kg/m3 RT01 287 · 288 p02 20, 000 = = 0.275 kg/m3 RT02 287 · 253 Density across the blower is assumed small so that ρ02 = ṁ2 = ρ02 Q2 = 0.275 · 0.24 = 0.066 kg/s Next ⇐ (a) (∆p0 )2 (∆p0 )1 = 2 2 ρ01 Ω1 D1 ρ02 Ω22 D22 so that (∆p0 )2 = (∆p0 )1 0.275 ρ02 = 250 = 56.2 mm H2 O ρ01 1.226 39 ⇐ (b) Exercise 4.13 Consider a fan with a flow rate of 1500 cfm, (cubic feet per minute) and has a shaft speed 3600 rpm. If a similar fan one half its size is to have the same tip speed, what will the flow rate be at a dynamically similar operating condition? What is the ratio of power consumption of the second fan compared to the first one? Given: D2 = D1 /2, U2 = U1 , Ω1 , Q1 . Find: Q2 , Ẇ2 /Ẇ1 . Solution: Since U2 = U1 , then Ω2 D2 = Ω1 D1 and Ω2 = 2Ω1 . Next from Q1 Q2 1500 · 2 Ω2 D23 = Q = Q = = 375 cfm ⇐ (a) 2 1 3 3 3 Ω1 D1 Ω2 D2 Ω 1 D1 8 and since Ω1 D1 = Ω2 D2 , from dotW1 dotW2 D22 1 = Ẇ = Ẇ = Ẇ1 2 1 3 5 3 5 2 ρΩ1 D1 ρΩ2 D2 D1 4 ⇐ (b) Exercise 4.14 A fan operating at 1750 rpm at a volumetric flow of 4.25 m3 /s develops a head of 153 mm of water. It is required to build a larger, geometrically similar fan which will deliver the same head at the same efficiency as the existing fan, but at the rotational speed of 1440 rpm. (a) Determine the volume flow rate of the larger fan. (b) If the diameter of the original fan is 40 cm, what is the diameter of the larger fan. (c) What are the specific speed of these fans. Given: Ω1 , Ω2 , Q1 , D1 , and H2 = H1 . Find: Q2 , D2 , and Ωs . Solution: From H1 H2 = 2 2 2 2 Ω 1 D1 Ω 2 D2 since H2 = H1 , then D2 = D1 Ω1 /Ω2 . Next from Q1 Q2 = 3 Ω 1 D1 Ω2 D23 Q2 = Q1 Ω2 D23 Ω21 17502 = Q = 4.25 = 6.28 m3 /s (a) 1 Ω1 D13 Ω22 14402 Next convert the head to standard units. (ws )1 == 998 · 9.81 · 0.153 ρH2 O gH1 = = 1246 J/kg ρ 1.20 40 and the density is the standard density at T = 293 K and p = 101.3215 kPa. Then √ √ Ω Q 1750 · π 4.25 Ωs = 3/4 = = 1.80 ⇐ (b) 30 12463/4 ws Exercise 4.15 The impeller of a centrifugal pump, with an outlet radius r2 = 8.75 cm and a blade width b2 = 0.7 cm, operates at 3550 rpm and produces a pressure rise of 522 kPa at the flow rate of 1.5 liters per minute. Assume that the inlet flow is axial and that the pump efficiency is 0.63. Find: (a) The specific 2 speed. (b). Show that the expression (??) for work reduces to w = (Vu2 + U22 − 2 Wu2 )/2 and calculate the work two ways and confirm that they are equal. Given: r2 , b2 , Ω, ∆p, Q, α1 and η. Find: Ωs , and the work in terms of kinetic energies. Solution: The radial velocity component at the exit is Q 1/45 · 104 = = = 6.28 m/s 2πr2 b2 60 · 2π · 0.7 Vr2 The head is H= ∆p0 522, 000 = = 53.37 m ρg 998 · 9.81 and the specific speed is √ √ 3550 · π 1.45/60 Ω Q Ωs = = = 0.528 (gH)3/4 30(9.81 · 53.37)3/4 ⇐ (a) Next the blade speed is U2 = r2 Ω = 8.75 · 3550π = 32.53 m/s 100 · 30 and the isentropic and actual work are ws = gH = 981 · 53.37 = 523 J/kg w= w2 523 = = 830 J/kg η 0.63 Next the kinetic energy terms are Vu2 = w 830 = = 25.52 m/s U2 32.53 41 and Wu2 = Vu2 − U2 = 25.52 − 32.53 = −7.0 m/s and √ √ 2 2 W2 = Wr2 + Wu2 = 6.282 + 7.02 = 9.41 m/s √ √ 2 = 6.282 + 25.522 = 26.28 m/s V2 = Vr22 + Vu2 thus 1 1 w = (V22 + U22 − W22 ) = (26.283 + 32.532 − 9.412 ) = 830 J/kg 2 2 42 ⇐ (b) Chapter 5 Exercise 5.1 Steam leaves the nozzles of a de Laval turbine with the velocity V2 = 1000 m/s. The flow angle from the nozzle is α2 = 70◦ . The blade velocity is U = 360 m/s and the mass flow rate is 800 kg/h. Take the rotor velocity coefficient to be cR = 0.8. The rotor blade is equiangular. Draw the velocity diagrams and determine: (a) The flow angles of the relative velocity at the inlet to the rotor, (b) the flow angle leaving the rotor, (c) the tangential force on the blades, (d) the axial thrust on the blades, (e) the power developed, and (f) the rotor efficiency. Given: V2 , α2 , U , cR , and ṁ. Find: β2 , α3 , Fu , Fx , and ηR . Solution: The components of the velocities are Vu2 = V2 sin α2 = 1000 sin(70◦ ) = 939.7 m/s Wu2 = Vu2 − U = 939.7 − 360 = 579.7 m/s Vx2 = V2 cos α2 = 1000 cos(70◦ ) = 342, m/s √ and W2 = 2 2 Wx2 + Wu2 = √ Wx2 = Vx2 3422 + 579.72 = 673.1 m/s The relative flow angle is ( ) ( ) Wu2 579.7 −1 −1 β2 = tan = tan = 59.46◦ Wx2 342.0 ⇐ (a) and W2 = cR W2 = 0.8 · 673.1 = 538.5 m/s so that Wu3 = W3 sin β3 = 583.5 sin(−59.46◦ ) = −463.8 m/s Vu3 = U + Wu3 = 360 − 463.8 = −103.8 m/s The flow angle at the exit of the rotor is ) ( ) ( −103.8 Vu3 −1 −1 = tan α3 = tan = −20.77◦ Vx3 273.6 ⇐ (b) Next the force on the blades is obtained. First w = U (Vu2 − V u3) = 360(939.7 + 103.8) = 375.64 kJ/kg 43 so that Ẇ = ṁw = 800 · 375.64 = 83.476 kW 3600 ⇐ (e) then Fu = 83, 476 Ẇ = = 231.9 N U 360 ⇐ (c) The axial thrust is Fx = ṁ(Vx2 − Vx3 ) = 800 (342 − 273.6) = 15.2 N 3600 ⇐ (d) The rotor efficiency is ηR = w 2 · 375, 640 = = 0.751 V2 /2 10002 ⇐ (f ) Exercise 5.2 The diameter of a wheel of a single stage impulse turbine is 1060 mm and its shaft speed is 3000 rpm. The nozzle angle is 72◦ and the ratio of the blade speed to the speed at which steam issues from the nozzles is 0.42. The ratio of the relative velocity leaving the blades is 0.84 of that entering the blades. The outlet flow angle of the relative velocity is 3◦ more than the inlet flow angle. The mass flow rate of steam is 7.23 kg/s. Draw the velocity diagram for the blades and determine: (a) The axial thrust on the blades. (b) The tangential force on the blades. (c) Power developed by the blade row. (d) Rotor efficiency. Given: D, α2 , Ω, U/V2 , W3 /W2 , the difference between β2 and β3 , and ṁ. Find: Fx , Fu , Ẇ , and ηR . Solution: The blade speed is U = r2 Ω = 0.53 · 3000 · π = 166.5 m/s 30 and the nozzle velocity is V2 = U/0.42 = 166.5/0.42 = 396.4 m/s. Also W3 = 0.84W2 . Next the components of the velocities are Vu2 = V2 sin α2 = 396.4 sin(72◦ ) = 377.0 m/s Wu2 = Vu2 − U = 377.0 − 166.5 = 210.5 m/s Vx2 = V2 cos α2 = 396.4 cos(72◦ ) = 122.5, m/s √ and W2 = 2 2 Wx2 + Wu2 = √ Wx2 = Vx2 122.52 + 210.52 = 243.6 m/s 44 The relative flow angle is ( ( ) ) 210.5 Wu2 −1 −1 β2 = tan = tan = 59.81◦ β3 = −59.81◦ −3◦ = −62.81◦ Wx2 122.5 and W3 = cR W2 = 0.8 · 243.6 = 204.6 m/s so that Wu3 = W3 sin β3 = 204.6 sin(−62.81◦ ) = −182.0 m/s Wx3 = W3 cos β3 = 204.6 cos(−62.81◦ ) = 93.5 m/s and Vu3 = Wu3 + U = −182 + 166.5 = −15.5 m/s Vx3 = Wx3 = 93.5 m/s The flow angle at the exit of the rotor is ( ) ( ) Vu3 −15.5 −1 −1 α3 = tan = tan = −9.41◦ Vx3 93.5 Next the force on the blades is obtained. Fx = ṁ(Vx2 − Vx3 ) = 7.23(93.5 − 122.5) = −209.7 N ⇐ (a) Fu = ṁ(Vu2 − Vu3 ) = 7.23(−15.5 − 377.0) = 2838 N ⇐ (b) The specific work done is w = U (Vu2 − V u3) = 166.5(377.0 + 15.5) = 65.35 kJ/kg so that Ẇ = ṁw = 7.23 · 65.35 = 472.5 kW then the rotor efficiency is ηR = w 2 · 65, 350 = = 0.832 V2 /2 396.42 ⇐ (c) Exercise 5.3 The wheel diameter of a single stage impulse steam turbine is 40 cm and the shaft speed is 3000 rpm. The steam issues from nozzles at velocity 275 m/s at the nozzle angle of 70◦ . The rotor blades are equiangular and friction reduces the relative velocity as the steam flows through the blades to 0.86 45 times the entering velocity. Find the power developed by the wheel when the axial thrust is Fx = 120 N. Given: D, α2 , Ω, V2 , cR , and Fx . Find: Ẇ . Solution: The blade speed is U = r2 Ω = 0.2 · 3000 · π = 62.83 m/s 30 Next the components of the velocities are Vu2 = V2 sin α2 = 275 sin(70◦ ) = 258.4 m/s Wu2 = Vu2 − U = 258.4 − 62.83 = 195.6 m/s Vx2 = V2 cos α2 = 275 cos(70◦ ) = 94.06, m/s Wx2 = Vx2 and Wu3 = −cR Wu2 = −0.86 · 195.6 = −168.2 m/s and Vu3 = Wu3 + U = −168.2 + 62.83 = −105.4 m/s so that w = U (Vu2 − Vu3 ) = 62.83(258.54 + 105.4) = 22.86 kJ/kg In addition V x3 = cN Vx2 = 0.86 · 94.06 = 80.89 m/s. From the expression for the axial component of the blade force Fx = ṁ(Vx2 − Vx3 ) ṁ = Fx 12) = = 9.11 kg/s Vx2 − Vx3 94.06 − 80.89 so that Ẇ = ṁw = 9.11 · 22.86 = 208.3 kW Exercise 5.4 Steam issues from the nozzles of a single stage impulse turbine with velocity 400 m/s. The nozzle angle is at 74◦ . The absolute velocity at the exit is 94 m/s and its direction is −8.2◦ . If the blades are equiangular: (a) Find the power developed by the blade row when the steam flow rate is 7.3 kg/s. (b) The the rate of irreversible energy conversion per kg of steam flowing through the rotor. Given: V2 , V3 , α2 , α3 , ṁ, and that β3 = −β2 . 46 Find: Ẇ , rate of loss of energy by irreversiblities. Solution: The components of the velocities are Vu2 = V2 sin α2 = 400.0 sin(74◦ ) = 384.5 m/s Vx2 = V2 cos α2 = 400.0 cos(74◦ ) = 110.3, m/s Also Wx2 = Vx2 Vu3 = V3 sin α3 = 94.0 sin(−8.2◦ ) = −13.4 m/s Vx3 = V3 cos α2 = 94.0 cos(−8.2◦ ) = 93.0, m/s Wx3 = Vx3 From Wu2 = W2 sin β2 = Vu2 − U Wu3 = W3 sin β3 = Vu3 − U so that W2 sin β3 − W3 sin β2 = Vu2 − Vu3 from which W2 (1 + cv ) sin β2 = Vu2 − Vu3 Since the velocity coefficient is the same in the nozzles and the rotor, it can be written as W3 Wx3 Vx3 93.0 cv = = = = = 0.843 W2 Wx2 Vx2 110.3 From the previous equation W2 sin β2 = Vu2 − Vu3 1 + cv In addition W2 cos β2 = Vx2 the ratio of these gives tan β2 = 384.5 + 13.4 Vu2 − Vu3 = = 1.957 Vx2 (1 + cR ) 110.3(1 + 0.843) Thus β2 = 62.93◦ and W2 = 110.3 Vx2 + = 242.4 m/s W3 = cv W2 = 0.843·242.4 = 204.5 m/s cos β2 cos(62.94◦ 47 From these Wu2 = W2 sin β2 = 242.4 sin(62.94◦ ) = 215.8 m/s Wu3 = W3 sin β3 = 242.4 sin(−62.94◦ ) = −182.1 m/s and the blade speed is U = Vu2 − Wu2 = 384.5 − 215.8 = 168.7 m/s The specific work is w = U (Vu2 − V u3) = U (Wu2 − Wu3 ) = 168.7(215.8 + 182.8) = 67.13 kJ/kg so that Ẇ = ṁw = 7.3 · 67.13 = 490.0 kW The loss is 1 1 KEloss = (W22 − W32 ) = (242.42 − 204.62 ) = 8.454 kJ/kg 2 2 and the power loss is Ẇloss = ṁKEloss = 7.3 · 8.454 = 61.7 kW Exercise 5.5 Carry out the steps in the development of the expression for ratio of the optimum blade speed to the steam velocity for a single stage impulse turbine with equiangular blades. Note that this expression is independent of the velocity coefficient. Carry out the algebra to obtain the expression for the rotor efficiency at this condition. (a) Find numerical value for the velocity ratio when the nozzle angle is 76◦ . (b) Find the rotor efficiency at this condition assuming that cR = 0.9. (c) Find the flow angle of the relative velocity entering the blades at the optimum condition. Given: α2 , optimum conditions. Find: ηR and β3 . Solution: At optimum conditions 1 1 U = sin α2 = sin(76◦ ) = 0.485 V2 2 2 48 ⇐ (a) then sin2 α2 1 ηR = (1 + cv C) = (1 + 0.9) sin(76◦ ) = 0.894 2 2 ⇐ (b) From W2 sin β2 = V2 sin β2 − U W2 cos β2 = V2 cos β2 by dividing gives tan β2 = sin α2 − 12 sin α2 1 sin α2 − U/V2 = = tan α2 cos α2 cos α2 2 ( so that −1 β2 = tan ) 1 ◦ tan(76 ) = 63.50◦ 2 ⇐ (c) Exercise 5.6 Steam flows from a set of nozzles of a single stage impulse turbine at α2 = 78◦ with velocity V2 = 305 m/s and the blade speed is U = 146 m/s. The outlet flow angle of the relative velocity is 3◦ greater than its inlet angle and the velocity coefficient is cR = 0.84. The nozzle velocity coefficient is cN = 1. The power delivered by the wheel is 1000 kW. Draw the velocity diagrams at the inlet and outlet of the blades. Calculate the mass flow rate of steam. Given: V2 , U , α2 , β3 = −β2 − 3◦ , cR = 0.84, CN = 1.0 and Ẇ . Find: ṁ. Solution: The components of the velocities are Vu2 = V2 sin α2 = 305 sin(78◦ ) = 290.1 m/s Vx2 = V2 cos α2 = 305 cos(78◦ ) = 94.3, m/s Wx2 = Vx2 Wu2 = Vu2 − U = 290.1 − 146 = 144.1 m/s so that W2 sin β3 − W3 sin β2 = Vu2 − Vu3 √ √ 2 2 = 94.32 + 144.12 = 172.7 m/s W2 = Wx2 + Wu2 and the flow angle of the relative velocity is ) ( ( ) Wu2 144.1 −1 −1 β2 = tan = tan = 56.80◦ Wx2 94.3 49 At the exit of the rotor W3 = CR W2 = 0.84 · 172.2 = 144.6 m/s and β3 = −(56.8◦ + 3◦ ) = −59.8◦ . Then Wu3 = W3 sin β3 = 144.6 sin(−59.8◦ ) = −125.0 m/s Wx3 = W3 cos β3 = 144.6 cos(−59.8◦ ) = 72.7 m/s Then Vu3 = Wu3 + U = −125.0 + 146 = 21.0 m/s ( therefore α3 = tan −1 Vu3 Vx3 ) ( −1 = tan 21.0 72.7 Vx3 = 72.7 m/s ) = 16.11◦ The specific work is w = U (Vu2 − V u3) = 146(290.1 − 21.0) = 39.29 kJ/kg so that ṁ = Ẇ 1000 = = 25.5 kg/s w 39.29 Exercise 5.7 Steam flows from a set of nozzles of a single stage impulse turbine at angle α2 = 70◦ . (a). Find the maximum total-to-static efficiency if the velocity coefficients are cR = 0.83 and cN = 0.98. (b). If the rotor efficiency is 90 % of its maximum value, find the possible outlet flow angles for the relative velocity. Given: α2 , cR = 0.83, CN = 0.83 and speed ratio U/V2 = 0.9(U/V2 )opt . Find: ηts and ṁ. Solution: At optimal condition U sin α2 sin(70◦ ) = = = 0.470 V2 2 2 and 1 1.83 2 ◦ ηRmax = (1 + CR ) sin2 α2 = sin (70 ) = 0.808 2 2 Also 0.982 (1 + 0.83) 2 ◦ 1 ηts = c2N (1 + cR ) sin2 α2 = sin (70 ) = 0.776 2 2 50 ⇐ (a) The actual rotor efficiency is ηR = f ηRmax = 0.9 · 0.808 = 0.727 and therefore ηR = 2(1 + cR )λ(sin α2 − λ) = which reduces to f (1 + cR ) sin2 α2 2 1 λ2 − sin α2 λ + f sin2 α2 = 0 4 The solution of this is One root is λa = √ 1 λ sin α2 (1 ± (1 − f )) 2 √ U sin(70◦ ) (1 + (1 − 0.9) = 0.6184 = 2 V2 and the other root is √ sin(70◦ ) U λa = (1 − (1 − 0.9) = 0.3123 = 2 V2 The flow angles are ( −1 β2a = tan ( −1 β2b = tan sin α2 − λa cos α2 sin α2 − λb cos α2 ) = 43.21◦ ) = 61.06◦ Exercise 5.8 The nozzles of a single stage impulse turbine have a wall thickness t = 0.3 cm and height b = 15 cm. The mean diameter of the wheel is D = 116 cm and the nozzle angle is α2 = 72◦ . The number of nozzles in a ring is 72. The specific volume of steam at the exit of the nozzles is 15.3 m3 /kg and the velocity there is V2 = 366m/s. (a) Find the mass flow rate of steam through the steam nozzle ring. (b) Find the power developed by the blades for an impulse wheel of equiangular blades, given that the velocity coefficient cR = 0.86 and cN = 1.0. The shaft turns at 3000 rpm. Given: t, b, D, α2 , Z, v1 , cR , cN , and V2 . Find: ṁ and Ẇ . 51 Solution: Vu2 = V2 sin α2 = 366 sin(72◦ ) = 348.1 m/s Vx2 = V2 cos α2 = 366 cos(72◦ ) = 113.1, m/s The exit area is A = (2πr cos α2 − Zt)b = (2π · 0.58 cos(72◦ ) − 72 · 0.003)0.15 = 0.1365, m2 Mass flow rate is ṁ = 366 · 0.1365 V2 A = = 3.266 kg/s ⇐ (a) v2 15.3 Next Wu2 = Vu2 − U = 348.14 − 182.2 = 165.9 m/s Wx2 = 113.1 m/s and the magnitude of the relative velocity leaving the nozzles is √ √ 2 2 W2 = Wx2 + Wu2 = 113.12 + 165.92 = 200.8 m/s and its flow angle is β2 = tan ( −1 Wu2 Wx2 ) ( −1 = tan 165.9 113.1 ) = 55.71◦ Therefore after the rotor W3 = cR W2 = 0.86 · 200.8 = 172.7 m/s Then β3 = −55.71◦ Wu3 = W3 sin β3 = 172.7 sin(−55.71◦ ) = −142.7 m/s Wx3 = W3 cos β3 = 172.7 cos(−55.71◦ ) = 97.3, m/s The tangential component of the absolute velocity at the exit of the rotor is Vu3 = U + Wu3 = 182.2 − 142.7 = 39.5 m/s and the specific work is w = U (Vu2 − Vu3 ) = 182.2(348.1 − 39.5) = 56.23 kJ/kg and the power is Ẇ = ṁw = 3.266 · 56.23 = 183.6 kW 52 ⇐ (b) Exercise 5.9 The isentropic static enthalpy change across a stage of a single stage impulse turbine is ∆hs = 22 kJ/kg. The nozzle exit angle is α2 = 74◦ . The mean diameter of the wheel is 148 cm and the shaft turns as 1500 rpm. The blades are equiangular with a velocity coefficient cR = 0.87. The nozzle velocity coefficient is cN = 0.98. (a) Find the steam velocity at the exit from the nozzles. (b) Find the flow angles of the relative velocity at the inlet and exit of the wheel. (c) Find the overall efficiency of the stage. Given: r2 , α2 , Ω, cR , cN , and ∆hs . Find: V2 , β2 , β3 and ηtt . Solution: The blade speed is U = r2 Ω = 0.74 · 1500 · π = 116.2 m/s 30 For isentropic flow assuming that the exit velocity is the same as at the inlet to the stage ws = ∆hs = 2U (V2s sin α2 − U ) therefore ( V2s = ∆hs +U 2U ) 1 = sin α2 ( ) 22, 000 1 + 116.2 = 219.4 m/s 2 · 116.2 sin(74◦ ) Then from CN = V2 V2s V2 = CN V2s = 0.98 · 219.4 = 215.0 m/s ⇐ (a) and U 116.2 = = 0.540 V2 215.0 To find the flow angles, first calculate λ= Vu2 = V2 sin α2 = 215 sin(74◦ ) = 206.7 m/s Vx2 = V2 cos α2 = 215 cos(74◦ ) = 59.3, m/s then Wu2 = Vu2 − U = 206.7 − 116.2 = 90.5 m/s 53 Wx2 = 59.3 m/s and the angle at which the relative velocity enters the rotor is ( ( ) ) 90.5 Wu2 −1 −1 β2 = tan = tan = 56.76◦ ⇐ Wx2 59.3 (b) and the exit angle from the rotor is β3 = −56.76◦ . The efficiency is ηts = 2λCn2 (1 + CR )(sin α2 − λ) = 2 · 0.540 · 0.982 (1 + 0.85)(sin(74◦ ) − 0.534) = 0.808 ⇐ (c) Exercise 5.10 An impulse turbine has a nozzle angle α2 = 72◦ and steam velocity V2 = 244 m/s. The velocity coefficient for the rotor blades is cR = 0.85 and the nozzle efficiency is ηn = 0.92. The output power generated by the wheel is Ẇ = 562 kW when the mass flow rate is ṁ = 23 kg/s. Find the efficiency of the turbine. Given: α2 , V2 , cR , cN , ṁ, and Ẇ . Find: ηts . Solution: The specific work is w= Ẇ 562 = = 24.43 kJ/kg ṁ 23 and from w = U (Vu2 − Vu3 ) = U (Wu2 − Wu3 ) = U (1 + cR )WU 2 But Wu2 = V2 sin α2 − U so that (1 + cR )U 2 − (1 + cR )V2 sin α2 U + The solution of this is 1 U = V2 sin α2 ± 2 and 244 sin(72◦ ) U1 = + 2 √ √ w =0 1 + cR w 1 2 2 V2 sin α2 − 4 1 + cR 2442 sin2 (72◦ ) 24, 430 − = 132 m/s 4 1.85 54 and the second root is 244 sin(72◦ ) U2 = − 2 √ 2442 sin2 (72◦ ) 24, 430 − = 100.1 m/s 4 1.85 From these U2 U1 = 0.541 lambda2 = = 0.410 V2 V2 The efficiency using either root becomes λ1 = ηts = 2λc2N (1 + cR )(sin α2 − λ) ηts = 2 · 0.541 · 0.922 (1 + 0.85)(sin(72◦ ) − 0.541) = 0.755 ⇐ Exercise 5.11 A two row velocity compounded impulse wheel is part of a steam turbine with many other stages. The steam velocity from the nozzles is V2 = 580 m/s and the means speed of the blades is U = 116 m/s. The flow angle leaving the nozzle is α2 = 74◦ and the flow angle of the relative velocity leaving the first set of rotor blades is β3 = −72◦ . The absolute velocity of the flow as it leaves the stator vanes between the two rotors is α4 = 68◦ and the outlet angle of the relative velocity leaving the second rotor is β5 = −54◦ . The steam flow rate is ṁ = 2.4 kg/s. The velocity coefficient is cv = 0.84 for both the stator and rotor rows. (a) Find the axial thrust from each wheel. (b) Find the tangential thrust from each wheel. (c) Find the efficiency of the rotors defined as the work out divided by the kinetic energy available from the nozzles. Given: V2 , U , α2 , β3 , α4 , β5 , cN = CR , and ṁ. Find: Fx , Fu , ηR . Solution: The velocity components are Vu2 = V2 sin α2 = 580 sin(74◦ ) = 557.5 m/s Vx2 = V2 cos α2 = 580 cos(74◦ ) = 159.9 m/s and Wu2 = Vu2 − U = 557.5 − 116 = 441.5 m/s Wx2 = 159.9 m/s and the magnitude of the relative velocity leaving the nozzles is √ 2 2 + Wu2 = 469.6 m/s W3 = 0.84 · 469.6 = 394.4 m/s W2 = Wx2 55 Next Wu3 = W3 sin β3 = 394.4 sin(−72◦ ) = −375.1 m/s Wx3 = W3 cos β3 = 394.4 cos(−72◦ ) = 121.9 m/s The components of the absolute velocity are Vu3 = Wu3 + U = −375.1 + 116 = −259.1 m/s and √ 2 2 V3 = Vx3 + Vu3 = 286.3 m/s Next Vx3 = 121.9 m/s V4 = 0.84 · 286.3 = 240.5 m/s Vu4 = V4 sin α4 = 240.5 sin(68◦ ) = 223.0 m/s Vx4 = V4 cos α4 = 240.5 cos(68◦ ) = 90.1 m/s and the components of the relative velocity become Wu4 = Vu4 − U = 223.0 − 116 = 107.0 m/s and its magnitude is √ 2 2 W4 = Wx4 + Wu4 = 139.9 m/s Wx4 = V x4 = 90.1 m/s W5 = 0.84 · 139.9 = 117.5 m/s The components after the second rotor become Wu5 = W5 sin β5 = 117.5 sin(−54◦ ) = −95.1 m/s Wx5 = W5 cos β5 = 117.5 cos(−54◦ ) = 69.1 m/s and finally Vu5 = Wu5 + U = −95.1 + 116 = 20.9 m/s Vx5 = Wx5 = 69.1 m/s The force components are then Fx1 = ṁ(Vx2 − Vx3 ) = 2.4(159.9 − 121.9) = 91.2 N Fu1 = ṁ(Vu2 − Vu3 ) = 2.4(557.5 − 259.1) = 1960.0 N For the second rotor Fx2 = ṁ(Vx4 − Vx5 ) = 2.4(90.1 − 69.1) = 50.4 N 56 Fu2 = ṁ(Vu4 − Vu5 ) = 2.4(223 − 20.9) = 485.0 N The power output is Ẇ1 = Fu1 U = 1960 · 116 = 227.4 kW Ẇ2 = Fu2 U = 485 · 116 = 52.26 kW an the efficiency is η= w1 + w2 2U = 2 (Vu2 − Vu3 + Vu4 − Vu5 ) 2 V2 /2 V2 with a numerical value η= 2 · 116 (555705259.1 + 223 − 20.9) = 0.703 5802 ⇐ Exercise 5.12 A velocity-compounded impulse wheel has two rows of moving blades with a mean diameter of D = 72 cm. The shaft rotates at 3000 rpm. Steam issues from the nozzles at an angle α2 = 74◦ with velocity V2 = 555 m/s. The mass flow rate is ṁ = 5.1 kg/s. The energy loss through each of the moving blades is 24 percent of the kinetic energy entering the blades based on the relative velocity. Steam leaves the first set of moving blades at β3 = −72◦ the guide vanes between the rows at α4 = 68◦ and the second set of moving blades at β5 = −52◦ . (a) Draw the velocity diagrams and find the flow angles at the blade inlets both for absolute and relative velocities. (b) Find the power developed by each row of blades. (c) Find the rotor efficiency as a whole. Given: V2 , D, Ω, α2 , β3 , α4 , β5 , cN = CR , and ṁ. Find: Fx , Fu , ηR . Solution: The blade speed is 0.36 · 3000 · π = 113.1 m/s 30 √ and the velocity coefficients are cR = cN = 0.76 = 0.872. The velocity components are U = r2 Ω = Vu2 = V2 sin α2 = 555 sin(74◦ ) = 533.5 m/s Vx2 = V2 cos α2 = 555 cos(74◦ ) = 153.0 m/s 57 and Wu2 = Vu2 − U = 533.5 − 113.1 = 420.4 m/s Wx2 = 153.0 m/s and the magnitude of the relative velocity leaving the nozzles is √ 2 2 W2 = Wx2 + Wu2 = 447.4 m/s W3 = 0.872 · 447.4 = 390.0 m/s Next Wu3 = W3 sin β3 = 390.0 sin(−72◦ ) = −370.9 m/s Wx3 = W3 cos β3 = 390.0 cos(−72◦ ) = 120.5 m/s The components of the absolute velocity are Vu3 = Wu3 + U = −370.9 + 113.1 = −257.8 m/s and √ V3 = Next Vx3 = 120.5 m/s V4 = 0.872 · 284.6 = 248.1 m/s 2 2 Vx3 + Vu3 = 284.6 m/s Vu4 = V4 sin α4 = 248.1 sin(68◦ ) = 230.0 m/s Vx4 = V4 cos α4 = 248.1 cos(68◦ ) = 92.9 m/s and the components of the relative velocity become Wu4 = Vu4 − U = 230.0 − 113.1 = 116.9 m/s and its magnitude is √ 2 2 W4 = Wx4 + Wu4 = 149.3 m/s Wx4 = V x4 = 92.9 m/s W5 = 0.872 · 149.3 = 130.2 m/s The components after the second rotor become Wu5 = W5 sin β5 = 130.2 sin(−52◦ ) = −102.6 m/s Wx5 = W5 cos β5 = 130.2 cos(−52◦ ) = 80.2 m/s and finally Vu5 = Wu5 + U = −102.6 + 113.1 = 10.5 m/s 58 Vx5 = Wx5 = 80.2 m/s The work done is w1 = U (Vu2 − Vu3 ) = 113.1(533.5 + 257.8) = 89, 496 J/kg w2 = U (Vu4 − Vu5 ) = 113.1(230.0 − 10.5) = 24, 825 J/kg and the power is Ẇ1 = ṁw1 = 5.1·89.496 = 456.4 kW Ẇ2 = ṁw2 = 5.1·24.825 = 126.6 kW so the efficiency is η= w1 + w2 2(89496 + 24825) = = 0.742 2 V2 /2 5552 ⇐ (c) The flow angles are ( ) ( ) Wu2 420.4 −1 β2 = tan = tan = 70◦ Wx2 153.0 ( ) ( ) Vu3 −257.8 −1 −1 α3 = tan = tan = −64.9◦ Vx3 120.5 ) ( ) ( 116.9 Wu4 −1 −1 = tan = 51.5◦ β4 = tan Wx4 92.9 −1 Exercise 5.13 Steam flows from the nozzles of a zero percent repeating stage at an angle α2 = 69◦ and speed V2 = 450 m/s and enters the rotor with blade speed moving at U = 200 m/s. (a) Find its efficiency when the loss coefficients are calculated from Soderberg’s correlation. (b) Find the work delivered by the stage. Given: V2 , U , and α2 . Find: η, w. Solution: The velocity components are Vu2 = V2 sin α2 = 450 sin(69◦ ) = 420.1 m/s Vx2 = V2 cos α2 = 450 cos(69◦ ) = 161.3 m/s so that Wu2 = Vu2 − U = 420.1 − 200 = 220.1 m/s 59 Wx2 = Vx2 = 161.3 m/s √ 2 2 Wx2 + Wu2 = 272.9 m/s. Next ( ) ( ) 220.1 Wu2 −1 −1 = tan β2 = tan = 53.76◦ Wx2 161.3 and W2 = β3 = −53.73◦ The amount of turning is ∆α = 68◦ ∆β = 107/5◦ The loss coefficients are ( ζN = 0.04 + 0.06 ( ζR = 0.04 + 0.06 69 100 )2 107.5 100 = 0.0686 )2 = 0.1094 The velocity coefficients are cN = √ 1 = 0.9674 1 + ζN cR = √ 1 = 0.9494 1 + ζR Now W3 = CR W2 = 0/9494 · 272.9 = 259.1 m/s and Wu3 = W3 sin β3 = 259.1 sin(−53.76◦ ) = −209.0 m/s Wx3 = W3 cos β3 = 2559.1 cos ∗(−53.76◦ ) = 153.1 m/s and the components of the absolute velocity after the rotor are Vu3]=U +Wu3 =300−209.0=−9.0 m/s V[ x3 = 153.1 m/s so that the flow angle is ( α3 = tan and V3 = −1 Vu3 Vx3 ) ( −1 = tan −9.0 153.1 ) = −3.35◦ √ 2 2 Vx3 + Vu3 = 153.3 m/s. The value of the speed ratio is λ= 200 U = = 0.444 V2 450 60 and the word delivered is w = U (Vu2 − Vu3 ) = 200(420.1 + 9.0) = 85.818 kJ/kg ⇐ (b) ⇐ (a) The efficiency is ηtt = (ζR − 4)λ2 4λ(sin α2 − λ) = 0.888 + (4 − 2ζR ) sin α2 + ζR + ζN Exercise 5.14 Show that for a repeating stage the efficiency of a zero percent reaction is 4λ(sin α2 − λ) ηtt = 2 (ζR − 4)λ + (4 − 2ζR )λ sin α2 + ζR + ζN and this maximum is at the condition λ = U/V2 given by √ ζR + ζS − (ζR + ζN )(ζR + ζS − ζR sin2 α2 ) λ= ζR sin α2 Given: A stage with zero reaction. Find: ηtt , and λ for maximum efficiency. Solution: Examination of the figure for a stage with zero reaction shows that ηtt = h01 − h03 h01 − h03ss which can be written as 1 h01 − h03ss h03 − h03ss −1= −1= ηtt h01 − h03 w which can be written as 2 h3 − h3ss + 12 (V32 − V3ss ) 1 −1= ηtt w Since T2 T3s = T2s T3ss then h3s − h3ss = T3ss T3s (h2 − h2s ) = (h2 − h2s ) T3s T2 61 and the expression for efficiency becomes h3 − h3s + 1 −1= ηtt T3s (h2 T2 − h2s ) + (1 − T3ss )V32 T3 w or ζR W32 + TT3s2 ζN V22 + (1 − 1 −1= ηtt 2w Neglecting the temperature factors leads to T3ss )V33 T3 1 ζR W32 + ζN V22 −1= ηtt 2w Since W3 = W2 and W22 = V22 − 2U V2 sin α2 + U 2 w = 2U W2 sin β2 = 2U (V2 sin α2 − U ) Substituting these into the expression for efficiency and introducing the parameter λ = U/V2 gives 1 ζR (λ2 − 2λ sin α2 + 1) + ζN −1= ηtt 4λ(sin α2 − λ) This can be now written as ηtt = (ζR − 4)λ2 4λ(sin α2 − λ) + (4 − 2ζR )λ sin α2 + ζR + ζN ⇐ (a) Differentiating this with respect to λ and setting the derivative to zero gives ζr sin α2 λ2 − 2(ζR + ζN )λ + (ζR + ζN ) sin α2 = 0 and the solution of this is √ ζR + ζS − (ζR + ζN )(ζR + ζS − ζR sin2 α2 ) λ= ζR sin α2 62 ⇐ (b) Chapter 6 Exercise 6.1 At inlet to the rotor in a single stage axial-flow turbine the magnitude of the absolute velocity of fluid is 610 m/s. Its direction is 61◦ as measured from the cascade front in the direction of the blade motion. At exit of this rotor the absolute velocity of the fluid is 305 m/s directed such that its tangential component is negative. The axial velocity is constant, the blade speed is 305 m/s, and the flow rate through the rotor is 5 kg/s. (a) Construct the rotor inlet and exit velocity diagrams showing the axial and tangential components of the absolute velocities. (b) Evaluate the change in total enthalpy across the rotor. (c) Evaluate the power delivered by the rotor. (d) Evaluate the average driving force exerted on the blades. (e) Evaluate the change in static and stagnation temperature of the fluid across the rotor, assuming the fluid to be a perfect gas with cp = 1148 J/kg K. (f) Calculate the flow coefficient and the blade loading coefficient. Are they reasonable? Given: V2 , α2 , V1 , U , and ṁ. Find: ∆T , ∆T0 , ϕ, ψ. Solution: For a The axial and tangential velocity components at the inlet to the rotor are Vx = V2 cos α2 = 610 cos(61◦ ) = 295.7 m/s Vu2 = V2 sin α2 = 610 sin(61◦ ) = 533.5 m/s At the exit of the rotor cos α1 = Vx /V3 so that ( ) 295.7 −1 α3 = cos = −14.16◦ 305 and Vu3 = V3 sin α2 = 610 sin(−14.16◦ ) = −74.6 m/s To construct the velocity diagrams the relative flow angles are needed. First the relative velocity components are Wu2 = Vu2 − U = 533.5 − 305 = 228.5 m/s ( so that −1 β2 = tan Wu2 Wx2 ) ( −1 = tan 228.5 295.7 Wx2 = Vx2 ) = 37.7◦ The relative velocity component after the rotor are Wu3 = Vu3 − U = −74.5 − 305 = −379.5 m/s 63 Wx3 = Vx3 = so that ( −1 β3 = tan Wu3 Wx3 ( ) −1 = tan −379.5 295.7 ) = −52.08◦ ⇐ (a) The specific work delivered is w = U (vu2 − Vu3 ) = 305(533.5 + 74.6) = 185, 480 J/kg which gives a drop in stagnation enthalpy of h02 − h03 = 185.480 kJ/kg ⇐ (b) Ẇ = ṁ = w = 5 · 185.48 = 92.7 MW ⇐ The power delivered is (c) and the force on the blades is Fu = Ẇ 185, 480 = = 3040 N U 305 ⇐ (d) The drop in the stagnation temperature of T02 − T03 = w 185, 4801 = = 161.6 K cp 1148 (e) and the drop in static temperature is T2 − T3 = T02 − T03 − 1 6102 − 3052 (V22 − V32 ) = 161.6 + = 40.0 K ⇐ (e) 2cp 2 · 1148 The flow coefficient and the blade loading coefficients are (reasonable) ϕ= Vx 295.7 = = 0.969 ψ = 2.00 U 305 ⇐ (f ) Exercise 6.2 A small axial-flow turbine must have an output power of 37 kW when handling one half of a kilogram of combustion gases per second with an inlet total temperature of 410 K. The value of the gas constant is 287 J/kg K and γ = 4/3. The total-to-total efficiency of the turbine is 80%. The rotor operates at 50, 000 rpm and the mean blade diameter is 10 cm. Evaluate (a) the average driving force on the turbine blades, (b) the change in the tangential component 64 of the absolute velocity across the rotor, and (c) the required total pressure ratio across the turbine. Given: Ẇ , T01 , Ω, D, ηtt and ṁ. Find: Fu , Vu2 − Vu3 , p01 /p03 . Solution: The blade speed is U = rΩ = 0.05 · 50, 000 · π = 261.8 m/s 30 and the average blade force is Fu = Ẇ 37, 000 = = 141.3 N U 261.8 ⇐ (a) The specific work is w= Ẇ 37, 000 = = 74 kJ/kg ṁ 0.5 and therefore the change in the tangential velocity components is Vu2 − Vu3 = w 74000 = = 282.7 m/s U 261.8 ⇐ (b) The isentropic work is ws = 74 w = = 92.5 kJ/kg ηtt 0.8 so that the exit stagnation temperature for an isentropic expansion is T03s = T01 − ws 925, 000 = 410 − = 329.4 K cp 1148 and the pressure ratio is p01 = p03 ( T01 T03s ( )γ/(γ−1) = 410 329.4 )4 = 2.4 ⇐ (c) Exercise 6.3 A turbine stage of a multi-stage axial turbine is shown in Figure ??. The value of the gas constant is 287 J/kg K and γ = 4/3. The inlet gas angle to the stator is α1 = −36.8◦ and the outlet angle from the stator is α2 = 60.3◦ . 65 The flow angle of the relative velocity at the inlet to the rotor is β2 = 36.8◦ and the flow leaves at β3 = −60.3◦ . (a) If the blade speed is U = 220 m/s, find the axial velocity, which is assumed constant throughout the turbine. (b) Find the work done by the fluid on the rotor blades for one stage. (c) The inlet stagnation temperature to the turbine is 950 K and the mass flow rate is ṁ = 400 kg/s. If this turbine produces a power output of 145 MW, find the number of stages. (d) Find the overall stagnation pressure ratio if its isentropic adiabatic efficiency is ηt = 0.85. (e) Why does the static pressure fall across the stator and the rotor? Given: α1 , α2 , β2 , β3 , ηtt , U , Ẇ , ṁ, and T01 . Find: Vx , w, number of stages, p01 /p0e . Solution: From Vu2 = Wu2 + U follows Vx tan α2 = Vx tan β2 + U so that Vx = U 220 = = 218.9 m/s tan α2 − tan β2 tan(60.3◦ ) − tan(36.8◦ ) ⇐ (a) Work delivered by a repeating stage is w = U (Vu2 − Vu3 ) = U Vx (tan α2 − tan α1 ) w = 220 · 218.9(tan(60.3◦ ) − tan(−36.8◦ )) = 120.5 kJ/kg ⇐ (b) ⇐ (c). To work done by all the stages is wt = Ẇ 145 · 106 = = 362.5 kJ/kg ṁ 400 Hence the number of stages is N = wt /w = 362.5/120.5 = 3.00 The temperature drop per stage is T01 − T03 = w 120, 500 = = 104.9 K cp 1148 and the pressure drop across the entire machine is T01] − T0e = 3 · 104.9 = 314.7 K and the exit temperature for an isentropic process is T03s = T01 − 314.7 T01 − T0e = 950 − = 579.7 K ηtt 0.85 66 and the pressure ratio is ( )γ/(γ−1) ( )4 p01 T01 950 = 7.21 = = p0e T0es 579.7 ⇐ (d) The static pressure drops because velocity increases. Static pressure drops across the rotor because the relative stagnation enthalpy remains constant and the relative velocity increases across the rotor. Exercise 6.4 A single stage axial turbine has a total pressure ratio of 1.5 to 1, with an inlet total pressure 300 kPa and temperature of 600 K. The absolute velocity at the inlet to the stator row is in the axial direction. The adiabatic total-to-total efficiency is 80%. The relative velocity is at an angle of 30◦ at the inlet of the rotor and it exits at −35◦ . If the flow coefficient is ϕ = 0.9, find the blade velocity. Use compressible flow analysis with cp = 1148 J/kg K, γ = 4/3, and R = 287 J/kg K. Given: p01 /p03 , p01 , T01 , α1 , β2 , β3 , ηtt , ϕ, Find: U . Solution: From w = U (Vu2 − Vu3 ) = U (Wu2 − Wu3 ) = U Vx (tan β2 − tan β3 ) Dividing by U 2 gives ψ = ϕ(tan β2 − tan β3 ) = 0.9(tan(30◦ ) − tan(−35◦ )) = 1.15 Then from T01 = T03s ( p01 p03 )γ/(γ−1) = 1.50.25 = 1.1067 so that T03s = 600/1.1067 = 542.2 K, and T03 = T01 − ηtt (T01 − T03s ) = 600 − 0.8(600 − 542.2) = 553.7 K The work is now w = cp (T01 − T03 ) = 1148(600 − 553.7) = 53.15 kJ/kg From w = ψU 2 the blade speed is √ √ w 53150 U= = = 215 m/s ψ 1.15 67 ⇐ Exercise 6.5 An axial turbine has a total pressure ratio of 4 to 1, with an inlet total pressure 650 kPa and total temperature of 800 K. The combustion gases that pass through the turbine have γ = 4/3, and R = 287 J/kg K. (a) Justify the choice of two stages for this turbine? Each stage is normal stage and they are designed the same way, with the blade loading coefficient equal to 1.1 and the flow coefficient equal to 0.6. The absolute velocity at the inlet to the stator row is at angle 5◦ from the axial direction. The overall total-to-total efficiency of the turbine is 91.0%. Find: (b) The angle at which the the absolute velocity leaves the stator,(c) the angle of the relative velocity at the inlet of the rotor, (d) the angle at which the relative velocity leaves the rotor. (e) Draw the velocity diagrams at the inlet and outlet of the rotor. (f) What are the blade speed and the axial velocity? (g) A consequence of the design is that each stage has the same work output and efficiency. Find (g) the stage efficiency, and (h) the pressure ratio for each stage. Given: p01 /p0e , p01 , T01 , psi, ϕ, and ηtt . Find: Justify that two stages is appropriate, α2 , β2 , β3 , U , Vx , ηs , p01 /p03 . Solution: The stagnation temperature at the exit of the turbine is ( T0es = T01 p0e p01 )(γ−1)/γ = 800 · 0.250.25 = 565.7 K and the specific work delivered is w = ηtt cp (T01 − T0es ) = 0.91 · 1148(800 − 565.7) = 244.8 kJ/kg The temperature drop is T02 − T0e = w 244, 800 = = 213.2 K cp 1148 ⇐ (a) A typical temperature drop across a stage is about 120 K. Hence two stages are justified and ws = w/2 = 122.39 kJ/kg. Each stage has the same flow angles. Hence their efficiency is the same. Solving from 1 − R − ψ/2 tan α1 = ϕ for the reaction R gives R=1− ψ − ϕ tan α1 = 1 − 0.555 − 0.6 tan(5◦ ) = 0.3975 2 68 and tan β1 = − 0.3975 + 0.55 R + ψ/2 =− = −1.5792 β1 = −57.65◦ ϕ 0.6 tan α2 = ⇐ (c) 1 − R + ψ/2 1 − 0.3975 + 0.55 = = 1.9208 α2 = 62.50◦ ϕ 0.6 R − ψ/2 0.3975 − 0.55 =− = 02542 β2 = 14.26◦ ϕ 0.6 The blade speed is √ √ w 122, 390 = = 333.6 m/s ⇐ (e) U= ψ 1.1 tan β2 = − ⇐ (b) and the axial velocity is Vx = ϕU = 0.6 · 333.6 = 200.1 m/s ⇐ (d) The stage efficiency is obtained as follows. For each stage so that ηs = T01 − T02 1 − T02 /T 01 = T01 − T02s 1 − (p02 /p01 )γ/(γ−1 ) ηs = T02 − T0e 1 − T0e /T 02 = T02 − T0es 1 − (p0e /p02 )γ/(γ−1 ) [ p02 = 1− p01 [ p0e = 1− p02 1 ηs 1 ηs ( ( T02 1− T01 T0e 1− T02 )]γ/(γ−1) )]γ/(γ−1) and [ ( )]γ/(γ−1) [ ( )]γ/(γ−1) p0e p02 p0e 1 T02 1 T0e = = 1− 1− 1− 1− p01 p01 p02 ηs T01 ηs T02 Hence ( )( ) ( ) ( )(γ−1)/γ T02 T0e 1 T02 T0e 1 p0e 1− 1− − 1− +1− +1− T01 T02 ηs T01 T02 ηs p01 69 Which when solved for the stage efficiency 1/ηs gives 1 2 − T02 /T01 − T0e /T02 = + ηs 2(1 − T02 /T01 )(1 − T0e /T02 ) √ (2 − T02 /T01 − T0e /T02 )2 − 4(1 − T02 /T01 )(1 − T0e /T02 )(1 − p0e /p01 )(γ−1)/γ 2(1 − T02 /T01 )(1 − T0e /T02 ) and and T02 693.4 = = 0.827 T01 800 0.287 − 1 = ηs √ 586.8 T0e = = 0.846 T02 693.4 0.0824 − 4 · 0.133 · 0.154 · 0.293 = 1.108 2 · 0.133 · 0.154 so that ηs = 0.9024 and ηs = 1 − T02 /T01 1 − (p02 /p01 )(γ−1)/γ so that [ ( )]γ/(γ−1) ( )4 p02 1 T02 0.133 = 1− 1− = 1− = 0.528 p01 ηs T01 0.9024 and p02 = 0.528 · 0.528 = 343 kPa. Similarly ηs = 1 − T0e /T02 1 − (p0e /p02 )(γ−1)/γ [ ( )]γ/(γ−1) ( )4 p0e 1 T02 0.154 = 1− 1− = 1− = 0.474 p02 ηs T01 0.9024 or p01 = 1.895 p02 p02 = 2.111 p0e ⇐ (h) Exercise 6.6 For a steam turbine rotor the blade speed at the casing is U = 300 m/s and at the hub its speed is 240 m/s. The absolute velocity at the casing section at the inlet to the rotor is V2c = 540 m/s and at the hub section it is V2h = 667 m/s. The angle of the absolute and relative velocities at the inlet and exit of the casing and hub sections are αc2 = 65◦ , βc3 = −60◦ , αh2 = 70◦ , and 70 βh3 = −50◦ . The exit relative velocity at the casing is Wc3 = 456 m/s and at the hub it is Wh3 = 355 m/s. Evaluate for the tip section: (a) The axial velocity at the inlet and exit; (b) the change in total enthalpy of the steam across the rotor; (c) the outlet total and static temperatures at the hub and casing sections, if the inlet static temperature is 540 C and inlet total pressure is 7 MPa and they are the same at all radii. Assume that the process is adiabatic and steam can be considered a perfect gas with γ = 1.3. The static pressure at the exit of the rotor is the same for all radii and is equal to the static pressure at inlet of the hub section. Repeat the calculations for the hub section. (d) Find the stagnation pressure at the outlet at the casing and the hub. Given: Uc , Uh , V2c , V2h , α2c , α2h , β3c , β3h , W3c , W3h , T2 , p02 . Find: Vx3c , Vx2c , ∆h0 , T03 , T3 , and similarly for the hub. Solution: The axial velocities are Vx2c Vx2h Vx3c Vx3h V2c cos α2c = 540 cos(65◦ ) = 228.2 m/s ⇐ (a) ◦ V2h cos α2h = 667 cos(70 ) = 228.1 m/s ⇐ (a) ◦ V3c cos α3c = 456 cos(−60 ) = 228.0 m/s ⇐ (a) V3h cos α3h = 355 cos(−50◦ ) = 228.2 m/s ⇐ (a) = = = = and the flow coefficients are ϕc = Vx 228.1 = = 0.76 Uc 300 ϕh = Vx 228.0 = = 0.95 Uh 240 The tangential components of the velocities are Vu2c = V2c sin α2c = 540 sin(65◦ ) = 489.4 m/s Vu2h = V2h sin α2h = 667 sin(70◦ ) = 626.8 m/s and Vu3c = Wu1c + Uc = W3c sin β3c + Uc = 456 sin(−60◦ ) + 300 = −94.6 m/s Vu3h = Wu1h + Uh = W3h sin β3h + Uh = 667 sin(−50◦ ) + 240 = −31.9 m/s The stagnation enthalpy change is therefore ∆h0c = wc = Uc (Vu2c − Vu3c ) = 300(489.34 + 94.9) = 175, 290 J/kg ⇐ (b) ∆h0h = wh = Uh (Vu2h − Vu3h ) = 240(626.8 + 31.9) = 158, 088 J/kg ⇐ (b) 71 Note that if rVu = C then w is independent of the radius. Here rc Vu2c 489.4 = 1.25 = 0.976 rh Vu2h 626.8 so that the condition for a free vortex distribution does not quite hold. Take 1 1 w = (wc + wh ) = (175, 290 + 158, 088) = 166, 690 J/kg 2 2 The blade loading coefficients are ψc = wc 175, 290 = = 1.947 Uc2 3002 ψh = wh 158, 088 = = 2.744 2 Uh 2402 Using the average value U = 0.5(Uc + Uh ) = 270 m/s, the average value for the blade loading coefficient is ψ = w/U 2 = 166, 690/2702 = 2.286. The average force on the blades is therefore Fu = ψU = 2.286 · 270 = 617.2 N The reactions at the casing and the hub are Rc = −ϕc tan β3c − ψ = −0.76 tan(−60◦ )−0.5·1.947 = 0.343 2 Rh = −ϕh tan β3h − The negative reaction at the hub is undesirable. The static temperature at the in let to the rotor is uniform T2 = 813.15 K. and the gas constant is R = R̄/M = 8314/18 = 461.9 J/kg K. The specific heat is cp = γR/(γ −1) = 1.3·416.9/0.3 = 2001.2001.5 J/kg K, and the speed of sound √ is c2 = sqrtγRT2 = 1.3 · 461.9 · 813.15 = 698.8 m/s. So the Mach number is M2c = V2c 540 = = 0.773 c2 689.8 and the stagnation temperature at the casing is T2c = T2 (1 + γ−1 2 M2c ) = 813.15(1 + 0.15 · 0.7732 ) = 886.0 K 2 From the work done on the casing end of the blade the exit stagnation temperature is wc = cp (T02c − T03c ) T03c = T02c − 72 175, 290 wc = 886.0 − = 798.4 K cp 2001.6 ψ = −0.95 tan 2 and the absolute velocity is √ √ 2 2 V3c = Vx3c + Vu3c = 228.22 + 94.62 = 247.0 m/s so that T3c = T03c − V3c2 2472 = 798.4 − = 783.2 K 2cp 2 · 2001.6 At the hub V2h 667 = = 0.995 c2 689.8 and the stagnation temperature at the hub is M2h = T2h = T2 (1 + γ−1 2 M2h ) = 813.15(1 + 0.15 · 0.9952 ) = 924.3 K 2 The exit stagnation temperature is T03h = T02h − wh 158, 088 = 924.3 − = 845.3 K cp 2001.6 and the absolute velocity is √ √ 2 2 V3h = Vx3h + Vu3h = 228.22 + 31.92 = 230.4 m/s so that T3h = T03h − 2 230.4 V3h = 845.3 − = 832.0 K 2cp 2 · 2001.6 The stagnation pressure is a uniform p02 = 7 MPa at the inlet to the rotor. so that the static pressure at the casing is ( )−(γ−1)/γ γ−1 p2c = p02 1 + = 7000(1 + 0.15 · 0.7732 )−1.3/0.3 = 4826 kPa 2 and a similar calculation at the hub give ( )−(γ−1)/γ γ−1 = 7000(1 + 0.15 · 0.9952 )−1.3/0.3 = 4018 kPa p2h = p02 1 + 2 At the exit p3 = 4016 kPa and it is uniform. The mach number at the casing is M3c = V3c V3c 247 =√ =√ = 0.360 c3c γRT3c 1.3 · 2001.6 · 783.2 73 and ( p03c = p3c γ−1 2 1+ M3c 2 )γ/(γ−1) = (1 + 0.15 · 0.3602 )1.3/0.3 = 4368 kPa At the hub M3h = V3h V3h 230.2 =√ =√ = 0.326 c3h γRT3h 1.3 · 2001.6 · 832.0 and ( p03h = p3h γ−1 2 1+ M3h 2 )γ/(γ−1) = (1 + 0.15 · 0.3262 )1.3/0.3 = 4303 kPa Exercise 6.7 Combustion gases, with γ = 4/3 and R = 287 kJ/kg K, flow thorough a turbine stage. The inlet flow angle for a normal stage is α1 = 0◦ . The flow coefficient is ϕ = 0.52 and the blade loading coefficient is ψ = 1.4. (a) Draw the velocity diagrams for the stage. (b) Determine the angle at which relative velocity leaves the rotor. (c) Find the flow angle at the exit of the stator. (d) A two-stage turbine has an inlet stagnation temperature of T01 = 1250 K and blade speed U = 320 m/s. If the adiabatic efficiency of the turbine is ηt = 0.89, find the stagnation temperature of the gas at the exit of the turbine and the stagnation pressure ratio for the turbine. (e) Assuming that the density ratio across the turbine based on static temperature and pressure ratios is the same as that based on the stagnation temperature and stagnation pressure ratios, find the ratio of the cross-sectional areas across the two-stage turbine. Given: ϕ, ψ, α1 , T01 , U , ηtt and two stages. Find: β3 , α2 , T03 , Ae /A1 . Solution: With the flow coefficient and blade loading coefficient known, as well as the inlet flow angle α1 = 0, the reaction can be calculated from tan α1 = 1 − R − ψ/2 ϕ R=1− ψ 1.4 =1− = 0.3 2 2 and the flow angles are 1 − R + ψ/2 tan α2 = ϕ ( −1 α2 = tan 74 1 − 0.3 + 0.7 0.52 ) = 69.62◦ ⇐ (c) ( ) 0.3 + 0.7 β1 = tan − = −62.53◦ ⇐ (b) 0.52 ( ) R − ψ/2 0.3 − 0.7 −1 tan β2 = − α2 = tan − = 37.59◦ ⇐ (b) ϕ 0.53 R + ψ/2 tan β1 = − ϕ −1 With the flow angles fixed and axial velocity equal for both stages, the work is the same. Hence w = ψU 2 − 1.4 · 3202 = 143, 360 J/kg therefore wts = wt = 2w = 286, 720 K/kg 286, 720 wt = = 322.2 kJ/kg ηtt 0.89 and T0e = T01 − wt 286, 720 = 1250 − = 1000.0 K cp 1148 T0es = T01 − 322, 200 wts = 1250 − = 969.4 K cp 1148 Therefore p01 = p0e ( T01 T0es With p1 = ρ1 RT1 ] )γ/(γ−1) ( = 1250 969.4 )4 = 2.76 ⇐ (d) p e = ρ e T Te , ρ1 p1 p01 T05 1000 = |f racTe T1 ≈ = 2.76 = 2.208 ρe pe p0e T01 1250 From ρ1 Vx1 A1 = ρe Ve Ae the area ratio is Ae /A1 = 2.208 The detailed calculations show that Vx = ϕU = 0.52 · 320 = 166.4 m/s and Ve5 166.42 Te = T0e − = 1000 − = 988.2 K 2cp 2 · 1148 so that pe = p0e ( Te T0e ( )γ/(γ−1) = 75 988.2 1000 )4 = 0.9526 and T1 = T01 − so that p1 = p01 Therefore ( 166.42 V15 = 1450 − = 1237.9 K 2cp 2 · 1148 T1 T01 )γ/(γ−1) ( = 1237.9 1250 )4 = 0.962 p5 p5 p05 p01 0.9526 = = = 0.3582 p1 p05 p01 p1 2.76 · 0.962 so that ρ1 p1 Te 988.2 = = = 2.229 ρe pe T1 0.362 · 1250 and the approximation is reasonably good. ⇐ (e) Exercise 6.8 Steam enters a 10-stage fifty percent reaction turbine at pressure 0.8 MPa and 200 C and leaves at pressure 5 kPa and with quality equal to 0.86. (a) If the steam flow rate is 7 kg/s, find the power output and the overall efficiency of the turbine. (b) The steam enters each stator stage axially with velocity of 75 m/s. The mean rotor diameter for all stages is 1.4 m and the axial velocity is constant through the machine. Find the rotational speed of the shaft. (c) Find the inlet and exit flow angles at the mean blade height assuming equal enthalpy drops for each stage. Given: R, p01 , T01 , α1 , p0e , x = 0.86, Vx , D, ṁ and the number of stages is 10 and equal stagnation enthalpy drops across each stage. Find: Ẇ , ηtt , α2 , and β1 . Solution: From the steam tables with p01 = 80, 00 kPa and T01 , enthalpy is h01 = 2839.3 kJ/kg and entropy is s1 = 6.8158 kJ/kg K. For an isentropic process to exit pressure of p0e = 5 kPa, entropy is ses = s1 , the quality is xe = ses − sf 6.1858 − 0.4764 = = 0.808 sg − sf 7.9187 and h0es = hf + xes hf g = 137.82 + 0.808 · 2424.4 = 2077.3 kJ/kg h0e = hf + xe hf g = 137.82 + 0.86 · 2424.4 = 2222.8 kJ/kg Therefore ηtt = 2839.3 − 2222.8 h01 − h0e = 2839.3 − 2077.3 = 0.811 h01 − h0es | 76 ⇐ (a) Since equal work is done by each stage w= 2839.3 − 2222.8 = 61.74 kJ/kg 10 Since the inlet and exit are axial and the reaction is 50 percent, the velocity triangles are symmetric and β2 = 0 so that Vu2 = u. Then the expression for work can be solved for the blade speed √ w = U (Vu2 − Vu3 ) = U Vu2 = U 2 U = sqrtw = 61740 = 248.5 m/s Therefore U 248.5 · 30 = = 3390 rpm ⇐ (b) r 0.7 · π The flow angle to the rotor is ) ( ) ( U 248.5 −1 −1 α2 = −β1 = tan = tan = 73.2◦ Vx 75 Ω= ⇐ (c) Exercise 6.9 Combustion gases enter axially into a normal stage at stagnation temperature T01 = 1200 K and stagnation pressure p01 = 1500 kPa. The flow coefficient is ϕ = 0.8 and the reaction is R = 0.4 and the inlet Mach number to the stator is M1 = 0.4. Find; (a) the blade speed, (b) the Mach number leaving the stator and the relative Mach number leaving the rotor. (c) Using the Soderberg loss coefficients find the efficiency of the stage. (d) Repeat the calculations with inlet Mach number M1 = 0.52. Given: ϕ, α1 , R, T01 and p01 . Find: α2 , β1 , R, and plot results for α1 = 10◦ to 30◦ . Solution: With α1 = 0 and R = 0.4, then ψ = 2(1 − R) = 1.2. As a consequence tan α2 = 1 − R + ψ/2 0.6 + 0.6 = = 1.5 α2 = 56.31◦ ϕ 0.8 −(R + ψ/2) 0.4 + 0.6 =− = 1.25 β1 = −51.34◦ ϕ 0.8 0.4 − 0.6 −(R − ψ/2) =− = 0.25 β2 = 14.04◦ tan β2 = ϕ 0.8 With M1 = 0.4 and the inlet flow axial ( )−1 γ−1 2 0.42 T1 = T01 1 + = 1200(1 + M1 ) = 1168.8 K 2 6 tan β1 = 77 and V1 = Vx = M1 √ √ γRT1 = 0.4 1.333 · 287 · 1168.8 = 267.5 m/s Then U = Vx /ϕ = 267.5/0.8 = 334.4 m/s and the work done is w = U Vx (tan α2 − tan α3 ) = 334.4 · 267.5 tan(56.31◦ ) = 134.2 kJ/kg The velocity after the stator is V2 = Vx 267.5 = = 482.2 m/s cos α2 cos(56.31◦ ) The static temperature is T2 = T02 − V22 482.22 = 1200 − = 1098.7 K 2cp 2 · 1.148 and V2 482.2 =√ = 0.744 γRT2 1.333 · 287 · 1098.7 The stagnation temperature after the rotor is M2 = √ T03 = T02 − ⇐ (a) w 134.2 = 1200 − = 1083.1 K cp 1.148 Next the relative velocity is W3 = Wx 267.5 = = 428.3 m/s cos β3 cos(−51.34◦ ) and T3 = T03 − and V32 267.52 = 1083.1 − = 1051.9 K 2cp 2 · 1148 W3 428.3 =√ = 0.675 γRT3 1.333 · 287 · 1051.9 The deflections are ∆α = α2 − α1 = 56.31◦ M3R = √ ∆β = β2 − β3 = 14.0451.34 = 65.38◦ 78 ⇐ (b) so that ( ζS = 0.04 + 0.06 ( ζR = 0.04 + 0.06 ∆α 100 ∆β 100 )2 ( = 0.04 + 0.06 )2 ( = 0.04 + 0.06 56.31 100 65.38 100 )2 = 0.0590 )2 = 0.0656 ( ) ϕ2 ζS ζR 1 −1= + ηtt 2ψ cos2 β3 cos2 α2 ( ) 0.82 0.0590 0.0656 2 ◦ = (51.34 ) + = 0.09596 2 · 1.2 cos cos2 (56.312 ) so that and the total-to-total efficiency is ηtt = 0.9124 ⇐ (c). If the inlet Mach number is increased to M1 = 0.52, then M2 = 0.9889 and M3R = 0.9131. The efficiency remains the same. Exercise 6.10 For a normal turbine stage fluid enters the stator at the angle 10◦ . The relative velocity has an angle −40◦ as it leaves the rotor. The blade loading factor is 1.6. (a) Determine the exit angle of the flow leaving the stator and the angle of the relative velocity as it enters the rotor. Determine the degree of reaction. (b) For the conditions of part (a) calculate the flow exit angle and the angle of the relative velocity entering the rotor, as well as the degree of reaction and plot them as functions of α1 , from α1 = 10◦ to 30◦ . Given: ψ, α1 , β3 . Find: α2 , β1 , R, and plot results for α1 = 10◦ to 30◦ . Solution: 1 = tan α1 − tan β1 = tan(10◦ ) − tan(−40◦ ) = 1.015 ϕ = 0.985 ϕ Reaction is R=1− ψ − ϕ tan α1 = 1 − 0.8 − 0.985 tan(10◦ ) = 0.026 2 ⇐ (b) α2 = tan−1 (ψ/ϕ + tan α1 ) = tan−1 (1.6/0985 + tan(10◦ )) = 60.96◦ ⇐ (a) From β2 = tan−1 (tan α2 − 1/ϕ) = tan−1 (tan(60.96◦ ) − 1.015) = 38.15◦ ⇐ (a) 79 %Hw 6.10 alpha3deg=10; alpha3=alpha3deg*pi/180; beta3deg=-40; beta3=beta3deg*pi/180; phi=1/(tan(alpha3)-tan(beta3)); psi=1.6 alpha2=atan(psi/phi+tan(alpha3)); alpha2deg=alpha2*180/pi; R=1-0.5*phi*(tan(alpha2)+tan(alpha3)) % Computer part alpha3adeg=linspace(-20,10); alpha3a=alpha3adeg*pi/180; n=length(alpha3a); for i=1:n phia(i)=1/(tan(alpha3a(i))-tan(beta3)); alpha2a(i)=atan(psi/phia(i)+tan(alpha3a(i))); beta2a(i)=atan(tan(alpha2a(i)-1/phia(i))); Ra(i)=1-0.5*phia(i)*(tan(alpha2a(i))+tan(alpha3a(i))); end plot(alpha3adeg,Ra); grid; Exercise 6.11 For Example 6.6 plot the variation of the reaction from the hub to the casing. %HW=6.11 clear all Rgas=287; cp=1148; k=4/3; T01=1100; p01=420000; rho01=p01/(Rgas*T01); c01=sqrt(k*Rgas*T01); rm=0.17; kappa2=0.7; rc2=2*rm/(1+kappa2); rh2=kappa2*rc2; kappa3=0.65; rc3=2*rm/(1+kappa3); rh3=kappa3*rc3 ; alpha2mdeg=60.08; alpha2m=alpha2mdeg*pi/180; alpha3mdeg=-21.20; alpha3m=alpha3mdeg*pi/180; beta2mdeg=25.98; beta2m=beta2mdeg*pi/180; beta3mdeg=-58.59; beta3m=beta3mdeg*pi/180; Rinv=1-(tan(alpha2m)+tan(alpha3m))/(tan(beta2m)+tan(beta3m)); Rm=1/Rinv; phi1=-2*Rm/(tan(beta2m)+tan(beta3m)) phi=0.8; psim=phi*(tan(beta2m)-tan(beta3m)); Rm=-0.5*phi*(tan(beta3m)+tan(beta2m)); psi=2*(1-Rm-phi*tan(alpha3m)); kp=k/(k-1); Vx=231; U=Vx/phi; Vum2=Vx*tan(alpha2m); Vum3=Vx*tan(alpha3m); 80 n=100; for i=1:n+1; zeta(i)=(i-1)*0.01; r2(i)=2*((1-kappa2)/(1+kappa2))*zeta(i)+2*kappa2/(1+kappa2); r3(i)=2*((1-kappa3)/(1+kappa3))*zeta(i)+2*kappa3/(1+kappa3); alpha2(i)=atan(tan(alpha2m)/r2(i)); alpha3(i)=atan(tan(alpha3m)/r3(i)); beta2(i)=atan(tan(alpha2m)/r2(i)-r2(i)/phi); beta3(i)=atan(tan(alpha3m)/r3(i)-r3(i)/phi); R(i)=1-(1-Rm)/r2(i)^2; end alpha2deg=alpha2*180/pi; alpha3deg=alpha3*180/pi; beta2deg=beta2*180/pi; beta3deg=beta3*180/pi; plot(zeta,alpha2deg,zeta,alpha3deg,zeta,beta2deg,zeta,beta3deg) figure(2); plot(zeta,R) m=100; %Numerical check a=(2*pi*Vx*rho01); sum=0; dr=(rc2-rh2)/m; Vxr=Vx^2/(2*cp*T01); Vur=Vum2^2/(2*cp*T01); rhom=rho01*(1-Vxr-Vur)^(1/(k-1)); b=(2*pi*Vx*rhom); for i=1:m+1 r(i)=rh2+(i-1)*dr; rho(i)=(1-Vxr-Vur*rm^2/r(i)^2)^(1/(k-1)); f(i)=r(i)*rho(i); end mdota=0.5*a*(rc2^2-rh2^2) mdotb=0.5*b*(rc2^2-rh2^2) mdot=a*trapz(r,f) Rh = 0.204 Rc = 0.610 81 Exercise 6.12 For a normal turbine stage the exit blade angle of the stator at 70◦ and relative velocity has an angle −60◦ as it leaves the rotor. For a range of flow coefficients ϕ = 0.2 − 0.8 calculate and plot the gas exit angle from the rotor, the angle the relative velocity makes as it leaves the stator, the blade loading coefficient and the degree of reaction. Comment on what is a good operating range and what are the deleterious effects in the flow over the blades if the mass flow rate is reduced too much or if it is increased far beyond this range. Given: ψ, α1 , β3 . Find: α2 , β1 , R, and plot results for α1 = 10◦ to 30◦ . Solution: %HW6.12 clear all phi=0.2:0.05:0.8; alpha2d=70; alpha2=alpha2d*pi/180; beta1d=-60; beta1=beta1d*pi/180; psi=-1+phi.*(tan(alpha2)-tan(beta1)); alpha1=atan(tan(alpha2)-psi./phi); alpha1d=alpha1*180/pi; R=1-0.5.*phi.*(tan(alpha2)+tan(alpha1)); beta2=atan(tan(alpha2)-1./phi); beta2d=beta2*180/pi; subplot(2,2,1); plot(phi,alpha1d); grid; xlabel(’Flow Coefficient \phi’); ylabel(’Air rotor outlet angle \alpha_1’); subplot(2,2,2); plot(phi,beta2d); grid; xlabel(’Flow Coefficient, \phi’); ylabel(’Flow angle, \beta_2’); subplot(2,2,3); plot(phi, psi); grid; xlabel(’Flow Coefficient \phi’); ylabel(’Blade Loading Factor, \psi’); subplot(2,2,4); plot(phi,R); grid; xlabel(’Flow Coefficient \phi’); ylabel(’Degree of Reaction, R’); Exercise 6.13 Given: T01 , p01 , α1 , β3 , ψ, and V2 . Find: α2 , β2 , R, and DpLS . 82 Solution: The flow coefficient is ϕ= 1 1 = = 0.985 ◦ tan α1 − tan β1 tan(10 ) − tan(−40◦ ) The flow angles after the stator are α2 = tan−1 (ψ/ϕ + tan α1 ) = tan−1 (1.6/0.985 + tan(10◦ )) = 60.96◦ ⇐ (a) β2 = tan−1 (tan α2 − 1/ϕ) = tan−1 (tan(60.96◦ ) − 1/0.985) = 38.15◦ ⇐ The reaction is 1 1 R = 1 − ϕ(tan α2 + tan α1 ) = 1 − (tan(60.96◦ ) + tan(10◦ )) = 0.026 2 2 The Ainley-Mathieson correlations give ]( ) [ (α ) ( α )2 s 2 2 2 + 0.821 − 0.129 Ypa = −0.627 100 100 c ]( ) [ ( α )2 (α ) s 2 2 + 0.242 + 1.489 − 1.676 100 100 c ( α )2 (α ) 2 2 − 0.356 + 0.399 + 0.007 = 0.0268 100 100 Ype [ ]( ) ( α )2 s 2 α2 2 = −1.56 + 1.554 − 0.0639 100 100 c [ ]( ) ( α )2 s α2 2 + 3.73 − 3.435 + 0.289 100 100 c ( α )2 α 2 2 + −0.82 + 0.7806 + 0.078 = 0.1058 100 100 and then [ ] ( )|α1 /α2 | ( α )2 t/c 1 Yp = Ypa + (Ype − Ypa ) = 0.029 α2 0.2 The static temperature after the stator is T2 = T02 − 4202 V2 = 700 − = 699.8 K 2cp 2 · 1148 83 (b) so that the Mach number is M2 = √ 420 V2 = = 0.8116 1.333 · 287 · 699.38 γRT2 and the ratio of the stagnation pressure to the static pressure is p02 = p2 ( γ−1 2 1+ M2 2 )γ/(γ−1) = (1 + 0.792 4 ) = 1.517 6 and from the definition of the stagnation pressure loss coefficient ( ) p01 p2 = 1 + Yp 1 − = 1.010 p02 p02 Therefore p02 = p01 380 p02 = = 376.3 kPa p01 1.01 and 376.3 p2 = = 248.0 kPa p02 1.517 and the loss of stagnation pressure is p2 = p02 dp0LS = p01 − p02 = 380.0 − 376.2 = 3.8 kPa The static enthalpy loss coefficient is obtained as follows. First ( T2s = T02 p2 p02 )(γ−1)/γ ( = 700 248.0 376.3 )0.25 The stagnation density is ρ02 = and therefore ζN = ( p02 = 1.873 kg/m3 RT02 2dp0LS /ρ02 ) = 0.0205 1 + γ−1 M22 V22 2 84 = 630.8 K Chapter 7 Exercise 7.1 The inlet and exit total pressures of a air flowing through a compressor are 100 kPa and 1000 kPa. The inlet total temperature is 281 K. What is the work of compression if the adiabatic total-to-total efficiency is 0.75 Given: Inlet p01 and T01 , and exit p0e as well as ηtt . Find: Work to compress the air. Solution: From ηtt = ws h02s − h01 = w h02 − h01 the work done is w= so that 1 cp T01 (h02s − h01 ) = ηtt ηtt 1004.5 · 281 w= 0.75 [( 1000 100 [( )1/3.5 p02 p01 ] )(γ−1)/γ −1 ] − 1 = 350 kJ/kg Exercise 7.2 Air flows through an axial fan rotor at mean radius of 15 cm. The tangential component of the absolute velocity is increased by 15 m/s through the rotor. (a) Evaluate the torque exerted on the air by the rotor, if the flow rate is 0.472 m3 s and the pressure and temperature of the air are 100 kPa and 300 K. (b) What is the rate of energy transfer to the air, if the rotational speed is 3000 rpm. Given: Inlet p1 , T1 , and the change in the swirl velocity; the flow rate Q, rotational speed Ω, and r. Find: Torque and power. Solution: Density at the inlet is ρ1 = p1 100 = = 1.61 kg/m3 RT1 0.287 · 300 Mass flow rate is ṁ = ρ1 Q = 1.161 · 0.472 = 0.5575 kg/s 85 The blade speed is U = rΩ = 0.15 · 3000 · π = 47.12 kg/s 30 The power needed to compress the air is Ẇ = ṁU (Vu2 − vu1 ) = 0.5575 · 47.12 · 15 = 394 W ⇐ (b) and the torque is T = Ẇ 394 · 30 = = 1.23 N m Ω 3000· ⇐ (a) Exercise 7.3 The blade speed of a compressor rotor is U = 280 m/s and the total enthalpy change across a normal stage is 31.6 kJ/kg. If the flow coefficient ϕ = 0.5 and the inlet to the rotor is axial, find the gas angle leaving the rotor. Given: ϕ and w, and U . Find: Find the value of the flow angle at the exit of the rotor. Solution: Since the flow at the inlet is axial, the equation for work reduces to w = U (Vu2 − Vu1 ) = U Vu2 = U Vx tan α2 or ψ = ϕ tan α2 With 31600 w = = 0.403 U2 2802 ( ) ( ) ψ 0.403 −1 −1 α2 = tan = tan = 38.9◦ ϕ 0.5 ψ= then ⇐ Also Vx = ϕU = 0.5 · 280 = 140 m/s and Wu2 = Vu2 − U = Vx tan α2 − U = 140 tan(38.9◦ ) − 280 = −167.1 m/s ( and β2 = tan −1 Wu2 Vx ) ( −1 = tan 86 −167.1 140 ) − 50.05◦ ⇐ Exercise 7.4 Air flows through an axial flow fan, with an axial velocity of 40 m/s. The absolute velocities at the inlet and the outlet of the stator are at angles of 60◦ and 30◦ , respectively. The relative velocity at the outlet of the rotor is at an angle −25◦ . Assume reversible adiabatic flow and a normal stage. (a) Draw the velocity diagrams at the inlet and outlet of the rotor. (b) Determine the flow coefficient. (c) Determine the blade loading coefficient. (d) Determine at what angle the relative velocity enters the rotor. (e) Determine the static pressure increase across the rotor in Pascals. The inlet total temperature is 300 K and the inlet total pressure is 101.3 kPa. (f) Determine the degree of reaction. Given: Inlet p01 , T01 , Vx , α1 , β2 , α2 . Reversible adiabatic flow. Find: ϕ, ψ, R and p2 − p1 . Solution: The blade speed U = Vu2 −Wu2 = Vx (tan α2 −tan β2 ) = 40(tan(60◦ )−tan(−15◦ )) = 87.93 m/s and the flow coefficient is ϕ= Vx 40 = = 0.455 U 87.93 ⇐ (b) The work done is w = U Vx (tan α2 − tan α1 ) = 87.63 · 40(tan(60◦ ) − tan(30◦ )) = 4061.5 J/kg so that w 4061.5 = = 0.525 U2 87.932 Writing the equation for work as ψ= w = U (Wu2 − Wu1 ) = Vx (tan β1 − tan β2 ) so that ( −1 β1 = tan ψ tan β2 − ϕ ) ( −1 = tan ⇐ (c) ψ = ϕ(tan β1 − tan β2 ) 0.525 tan(−25 ) − 0.455 ◦ ) = −58.3◦ ⇐ (d) The reaction is 0.455 1 [tan(60◦ ) + tan(30◦ )] = 0.475 ⇐ (f ) R = 1− ϕ(tan α2 +tan α1 ) = 1− 2 2 87 Since the velocities are low, the flow may be assumed incompressible and since it is reversible and adiabatic ∆p0 w= ρ The density is ρ01 = p01 101.3 = = 1.1765 kg/m3 RT01 0.287 · 300 and it may be assumed that ρ1 = ρ01 . Thus p02 = p01 + ρw = 101.3 + 1.1765 · 4.0615 = 106.1 kPa The velocities are V1 = 40 Vx = = 46.19 m/s cos α1 cos(30◦ ) V2 = Vx 40 = = 80.00 m/s cos α2 cos(60◦ ) The static pressure at the inlet is 1 1.1765 · 46.192 p1 = p01 − ρV12 = 101.3 − = 100.05kPa 2 2 · 1000 1 1.1765 · 80.02 p2 = p02 − ρV22 = 106.1 − = 102.31kPa 2 2 · 1000 so that p2 − p1 = 102.31 − 100.05 = 2.26 kPa ⇐ (e) Exercise 7.5 Air flows through an axial flow compressor. The axial velocity is 60 percent of the blade speed at the mean radius. The reaction ratio is 0.4. The absolute velocity enters the stator at an angle of 55 deg from the axial direction. Assume a normal stage. (a) Draw the velocity diagrams at the inlet and outlet of the rotor. (b) Determine the flow coefficient. (c) Determine the blade loading coefficient. (d) Determine at what angle the relative velocity enters the rotor. (e) Determine at what angle the relative velocity leaves the rotor. (f) Determine at what angle the absolute velocity leaves the stator. Given: Vx = 0.U , R = 04, and α2 = 55◦ . Find: α1 , β1 , β2 and ψ. 88 Solution: Since Vx = 0.6U , the flow coefficient is ϕ = 0.6. The flow angle α1 can be obtained from ( ) ( ) 2 · 0.6 2(1 − R) −1 −1 ◦ α1 = tan − tan α2 = tan − tan(55 ) = 29.76◦ ϕ 0.6 Then the blade loading coefficient is ψ = 2(1 − R tan α1 ) = 2(1 − 0.4 − 0.6 tan(29.76◦ )) = 0.514 and then R + ψ/2 tan β1 = − ϕ R − ψ/2 tan β2 = − ϕ ( −1 β1 = tan ( −1 β2 = tan 0.5 + 0.514/2 − 0.6 0.5 − 0.514/2 − 0.6 ) = −47.59◦ ⇐ (e) = −13.42◦ ⇐ (e) ) Exercise 7.6 The angle at which the absolute velocity enters the rotor of single stage axial compressor is α1 = 40◦ and the relative velocity at the inlet of the rotor is β1 = −60◦ . These angles at the inlet of the stator are α2 = 60◦ and β2 = −40◦ . The mean radius of the rotor is 30 cm and the hub to tip radius is 0.8. The axial velocity is constant and has a value Vx = 125 m/s. The inlet air is atmospheric at pressure 101.325 kPa and temperature 293 K. (a) Find the mass flow rate. (b) What is the rotational speed of the shaft under these conditions? (c) What is the power requirement of the compressor? Given: α1 , α2 , β1 , r, rh /rt and Vx . The inlet conditions are p01 and T01 . Find: ṁ, Ω, Ẇ . Solution: To calculate the mass flow rate from ṁ = ρ1 AVx the area and density need to be determined. The area can be written as A = π(rc2 − rh2 ) = πrc2 (1 − κ2 ) in which κ = rh /rc = 0.8 and from 1 rm = (rh + rc ) 2 89 rc (1 + κ) 2 thus rc = 2rm 2 · 0.3 1 = = m 1+κ 1 + 0.8 3 A= π (1 − 0.64) = 0.1257 m2 9 and The velocity at the inlet is V1 = Vx 125 = = 163.18 m/s cos α1 cos(40◦ ) The stagnation Mach number is V1 163.18 =√ = 0.4756 γRT01 (1.4 · 287 · 293 M01 = and the Mach number is 0.4756 M01 =√ M1 = √ = 0.4876 2 1 − 0.2 · 0.27562 M 1 − γ−1 01 2 Then T1 = and ( p1 = p01 T01 1+ T1 T01 γ−1 M12 2 = 293 = 279.75 K 1 − 0.2 · 0.48762 )γ/(γ−1) ( = 101.325 279.75 293 )3.5 = 86.17 kPa and the density is ρ1 = p1 81.17 = = 1.073 kg/m3 RT1 0.287 · 279.75 The mass flow rate is ṁ = rho1 AVX = 1.073 · 0.1257 · 125 = 16.86 kg/s ⇐ (a) From U = Vu1 −Wu1 = Vx (tan α1 −tan β1 ) = 125 [tan(40◦ ) − tan(60◦ )] = 321.4 m/s so that Ω= U 321.4 · 30 = = 10230 rpm rm 0.3 · π 90 ⇐ (b) The work done is w = U Vx (tan α2 − tan α1 ) = 321.4 · 125 [tan(60◦ ) − tan(40◦ )] = 35.87 kJ/kg and the power is Ẇ = ṁw = 16.86 · 35.87 = 604.8 kW ⇐ (c) Exercise 7.7 The angle at which the absolute velocity enters the rotor a compressor stage is α1 = 35◦ and the relative velocity makes an angle β1 = −60◦ . The corresponding angles at the inlet to the stator are α2 = 60◦ and β2 = −35◦ . The stage is normal and the axial velocity is constant through the compressor. (a) Why does the static pressure rise across both the rotor and the stator. (b) Draw the velocity triangles. (c) If the blade speed is U = 290ms, find the axial velocity. (d) Find the work done per unit mass flow for a stage and the increase in stagnation temperature across it. (e) The stagnation temperature at the inlet is 300 K. The overall adiabatic efficiency of the compressor is ηc = 0.9 and the overall stagnation pressure ratio is 17.5. Determine the number of stages in the compressor. (f) How many axial turbine stages will it take to power this compressor? Given: Inlet T01 , U , α1 , α2 , β1 , β2 and p0e /p01 as well as ηc . Find: Vx , w and ∆T0 across as stage, the number of stages in the compressor and suggest how many turbine stages are needed to power the compressor. Solution: The first part asks why pressure increases across the stator and the rotor. Because the flow is turned toward the axis in the stator and the axil velocity remains constant, its kinetic energy decreases. This means that the static enthalpy increases, since the stagnation remains constant. With an increase in static enthalpy pressure increases, as seen from T ds = dh − vdp for isentropic flow and there is no qualitative change as irreversibilities are introduced. The same arguments hold for the rotor, but now the relative velocity is turned toward the axis and since the relative stagnation enthalpy remains constant, the static enthalpy and thus also the pressure increased across the rotor. Part (b) asks for the velocity diagrams. Since the angles are given, and the velocity U is also given, they can be drawn. From 1 = tan α1 − tan β1 = tan(35◦ ) − tan(−60◦ ) = 2.4323 ϕ 91 so that phi = 0.411 and Vx = ϕU = 0.4112̇90 = 119.2ms ⇐ (c) Next ϕ (tan α2 + tan α1 ) = 0.5 2 as was expected as the velocity diagrams are symmetric. Then R=1− ψ = 2(1 − R − ϕ tan α1 ) = 1 − 2 · 0.411 tan(35◦ ) = 0.424 and w = ψU 2 = 0.424 · 2902 = 35, 678 K/kg DT0 = w = 35.5 K cp ⇐ (d) With the pressure ratio known, the exit temperature can be determined from ( )(γ−1)/γ 1 p0e 1 f racT0e T01 = 1 + =1+ (17.51/3.5 ) = 2.401 ηc p01 0.9 so that T0e = 2.401·300 = 721.8 K. The number of stages can now be determined from T0e − T01 421.8 N= = = 11.88 N = 12 ⇐ (e) T03 − T01 35.5 Typically one turbine stage is needed to power 5 or 6 compressor stages, so 2 turbine stages are needed. This is the answer to part (f). Exercise 7.8 The blade speed of a rotor of an axial air compressor is U = 150 m/s. The axial velocity is constant and equal to Vx = 75 m/s. The tangential component of the relative velocity leaving the rotor is Wu2 = −30 m/s, the tangential component of the absolute velocity entering the rotor is Vu1 = 55m/s. The stagnation temperature and pressure at the inlet to the rotor are 340 K and 185 kPa. The stage efficiency is 0.9 and one-half of the loss in stagnation pressure takes place through the rotor. (a) Draw the velocity diagrams at the inlet and exit of the rotor. (b) Find the work done per unit mass flow through the compressor. (c) Draw the states in an hs-diagram. (d) The stagnation and static temperatures between the rotor and the stator. (e) The stagnation pressure between the rotor and the stator. Given: Inlet p01 and T01 , Vu1 , Vx , Wu2 and ηtt . Find: Work to compress the air, T02 and T2 , p02 . 92 Solution: The remaining velocity components are Vu2 = U + Wu2 = 150 − 30 = 20mm/s Wu1 = Vu1 − U = 65 − 150 = 95 m/s and the work done is w = U (Vu2 − Vu1 ) = 150(120 − 55) = 9750 J/kg and T02 = T01 + w 9750 = 340 + = 349.7 K cp 1004.5 ⇐ ⇐ (b) (d) The isentropic work is ws = ηtt w = 0.9 · 9750 = 8775 J/kg and T03s = T01 + so that ( p03 = p01 T03s T01 8775 = 348.7 K 1004.5 )γ/(γ−1) ( = 185 + 349.7 340 )3.5 = 202.18 kPa The ideal pressure reached after the rotor is there were no losses and the same amount of work is done as in the actual case is ( )γ/(γ−1) ( )3.5 T02 349.7 p02i = p01 = 185 = 204.15 kPa T01 340 and the stagnation pressure loss is Dp0L = p02i − p03 = 204.15 − 202.18 = 1.97 kPa The stagnation pressure after the rotor is ⇐ p02 = p03 + 0.5Dp0L = 202.18 + 0.985 = 203.17 kPa (e) The velocity after the rotor is √ √ 2 = 752 + 1202 = 141.51 m/s V2 = Vx2 + Vu2 and the static temperature is T2 = T02 − 141.512 V22 = 349.7 − = 339.7 K 2cp 2 · 1004.5 93 ⇐ (d) Exercise 7.9 Air from ambient at 101.325 kPa and temperature 20 C enters the first stage of multistage axial flow compressor with velocity 61 m/s. The blade tip radius is 60 cm and the hub radius is is 42 cm. The shaft speed is 1800 rpm. The air enters a stage axially and leaves it axially at the same speed. The rotor turns the relative velocity 18.7◦ toward the direction of the blade movement. The total-tototal stage efficiency is 0.87. (a) Draw the inlet and exit velocity diagrams for the rotor. (b) Draw the blade shapes. (c) Determine the flow coefficient and the blade loading factor. (d) Determine the mass flow rate. (e) What is the work required per unit mass. (f) What is the total pressure ratio for the stage? (g) Determine the degree of reaction. Given: p01 , T01 , Vx , Dβ, rc , rh , Ω, and ηtt . Find: ψ, ϕ, R, ṁ, w and p0e /p01 . Solution: Since the inlet velocity is axial and it is quite small the density at the inlet is the same as the stagnation density ρ01 = p01 101.325 = = 1.205 kg/m3 RT01 287 · 293 and the mass flow rate is ṁ = ρAVx = ρπ(rc2 − rh2 )Vx = 1.205 · π(0.62 − 0.422 ) = 42.40 kg/s ⇐ (d) The mean radius and the blade speed are 1 0.6 + 0.42 0.51 · 1800 · π r = (rc + rh ) = 0.51 m U = rm Ω = = 96.13 m/s 2 2 30 Now the flow angles can be determined. The relative velocity is at the angle β1 entering the rotor. It is ( ) ( ) −U −96.13 −1 −1 β1 = tan = tan = −57.6◦ Vx 61 and β2 = β1 + 18.7◦ = −57.6◦ + 18.7◦ = −38.9◦ The tangential component of the velocity entering the stator is Vu2 = Wu2 + U = Vx tan β2 + U = 62 tan(−38.9◦ ) + 96.13◦ = 46.91 m/s 94 ( and −1 α2 = tan Vu2 Vx ( ) = tan −1 46.91 61 ) = 37.56◦ The specific work done is w = U (Vu2 − Vu1 ) = 96.1(46.91 − 0) = 4509 J/kg ⇐ (e) and ws = ηtt w = 0.87 · 4509 = 3922.8 J/kg so that ψ= 4509 Vx w 61 = = 0.488 ϕ = = = 0.635 2 2 U 96.13 U 96.13 ⇐ (c) The stagnation temperature after the rotor is T02 = T01 + w 4509 = 293 + = 297.5 K cp 1004.5 T02s = T01 + ws 3922.8 = 293 + = 296.9 K cp 1004.5 and The pressure ratio is therefore p02 = p01 ( T02s T01 )γ/(γ−1) ( = 296.9 293 )3.5 = 1.047 The reaction is R=1− ϕ 0.635 (α2 + tan α1 ) = 1 − (tan(37.56◦ ) − 0) = 0.756 2 2 ⇐ (g) Exercise 7.10 Air from ambient at 101.325 kPa and temperature 300 K enters an axial flow compressor stage axially with velocity 122 m/s. The blade tip radius is 35 cm and the hub radius is 30 cm. The shaft speed is 6000 rpm. At the exit relative velocity is at an angle −45◦ . The total-to-total stage efficiency is 0.86. (a) Draw the inlet and exit velocity diagrams for the rotor. (b) Draw the blade shapes. (c) Determine the flow coefficient and the blade loading factor. (d) Determine the mass flow rate. (e) What is the total pressure ratio for the stage? (f) Determine the degree of reaction. 95 Given: Inlet p01 , T01 , Vx , rc , rh , Ω β1 and ηtt . Find: ψ, ϕ, R, and p03 /p01 . Solution: The mean radius and the blade speed are 1 0.35 + 0.30 0.325 · 6000 · π r = (rc +rh ) = 0.325 m U = rm Ω = = 204.2 m/s 2 2 30 Now the flow angles can be determined. The relative velocity is at the angle β1 entering the rotor. It is ( ) ( ) −U −204.2 −1 −1 = tan = −59.1◦ β1 = tan Vx 122 From the expressoion for the tangential component of the velocity Vx tan α2 = U + Vx tan β2 which gives ( α2 = tan −1 U + tan β2 Vx ) ( −1 = tan ) 204.2 ◦ + tan(−45 ) = 34.0◦ 122 The specific work done is w = U (Vu2 − Vu1 ) = 204.2(82.3 − 0) = 16, 806 J/kg and ws = ηtt w = 0.86 · 16806 = 14, 453 J/kg so that ψ= w 16806 Vx 122 = = 0.40 ϕ = = = 0.60 2 2 U 204.2 U 204.2 The stagnation temperature after the rotor is T02 = T01 + w 16, 806 = 300 + = 316.7 K cp 1004.5 T02s = T01 + ws 14, 453 = 300 + = 314.4 K cp 1004.5 and 96 ⇐ (c) The pressure ratio is therefore ( )γ/(γ−1) ( )3.5 p02 T02s 314.4 = = = 1.178 p01 T01 300 ⇐ (e) The reaction is R=1− ϕ 0.60 (α2 + tan α1 ) = 1 − (tan(34.0◦ ) − 0) = 0.80 2 2 ⇐ (f ) The stagnation density at the inlet is ρ01 = p01 101.325 = = 1.205 kg/m3 RT01 287 · 300 The static temperature at the inlet is T1 = T01 − V12 1222 = 300 − = 292.6 K 2cp 2 · 1004.5 and the static pressure is ( )γ/(γ−1) ( )3.5 T1 292.6 p1 = p01 = 101.325 = 92.83 kPa T01 300 and the static density is ρ1 = p1 92.83 = = 1.1055 kg/m3 RT1 287 · 292.6 and the mass flow rate is ṁ = ρAVx = ρπ(rc2 − rh2 )Vx = 1.1055 · π(0.352 − 0.302 ) = 13.77 kg/s ⇐ (d) Exercise 7.11 Carry out design calculations for a compressor stage with blade loading factor in the range ψ = 0.25 to ψ = 0.55, flow coefficient ϕ = 0.7 and reaction R = 0.6. Keep the diffusion factor equal to 0.45. Calculate and plot 1−η, solidity, and the static pressure rise 1 − V22 /V12 for the rotor and stator (including the de Haller criterion). What are the stagnation losses across the stator and rotor? Given: The range of blade loading factors and R = 0.6 Find: Solidity, diffusion factor, loss, diffusion. Solution: 97 % Compressor calculations clear all; clf; % Specified conditions R=0.4; phi=0.6; DF=0.45; % Blade angles alpha1=atan((1-R-0.5*psi)./phi); alpha2=atan((1-R+0.5*psi)./phi); beta1=-atan((R+0.5*psi)./phi); beta2=-atan((R-0.5*psi)./phi); % Solidities using Lieblein’s diffusion factor sigmar=0.5*sin(beta2-beta1)./(cos(beta1)-(1-DF).*cos(beta2)); sigmas=0.5*sin(alpha2-alpha1)./(cos(alpha2)-(1-DF).*cos(alpha1)); % Loss coefficients omegar=0.014*sigmar./cos(beta2); omegas=0.014*sigmas./cos(alpha1); % Pressure rise and de Haller Conditions w1sq=phi.^2+(0.5*psi+R).^2; w2sq=phi.^2.*(1+tan(beta2).^2); v2sq=phi.^2+(0.5*psi+1-R).^2; v1sq=phi^2.*(1+tan(alpha1).^2); hallerr=1-w2sq./w1sq; hallers=1-v1sq./v2sq; % Stagantion pressure loss and efficiency Dp0LR=(0.5*phi^2./psi).*omegar./cos(beta1).^2; Dp0LS=(0.5*phi^2./psi).*omegas./cos(alpha2).^2; Dp0L=Dp0LR+Dp0LS; % Plots n=length(hallers); dehaller=0.44*ones(1,n); subplot(2,2,1), plot(psi,hallerr,’.-’,psi,hallers,psi,dehaller); grid on; axis([0.25 0.52 0.2 0.7]); xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12) ylabel(’1-V_2^2/V_1^2’,’Fontsize’,12) text(0.26, 0.47,’de Haller’) legend(’Rotor’,’Stator’) set(gca,’Fontsize’,12); hold on; subplot(2,2,2), plot(psi,Dp0LR,’.-’,psi,Dp0LS,psi,Dp0L); grid on; axis([0.25 0.52 0.00 0.08]); xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12) 98 ylabel(’Loss, 1-\eta’,’Fontsize’,12) legend(’Rotor’,’Stator’,’Total’) subplot(2,2,3), plot(psi,sigmar,’.-’,psi,sigmas); grid on; axis([0.25 0.52 0.0 3.0]); xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12) ylabel(’Solidity’,’Fontsize’,12) legend(’Rotor’,’Stator’) s=[1.0 1.4 2.0]; subplot(2,2,4); for i=1:3 ddf(i,:)=1-cos(beta1)./cos(beta2) + ... 0.5*cos(beta1).*(tan(beta2)-tan(beta1))./s(i); plot(psi,ddf(i,:)); hold on; end axis([0.25 0.52 0.2 0.6]); grid; xlabel(’Blade Loading, \psi=\Delta h_0/U^2’,’Fontsize’,12) ylabel(’Diffusion Factor’,’Fontsize’,12) text(0.405,0.55,’\sigma=1.0’), text(0.405,0.35,’\sigma=2.0’) text(0.405,0.482,’\sigma=1.4’); % End of script Exercise 7.12 A compressor stage with reaction ratio R = 0.54, and stator outlet metal angle χ3 = 14.5◦ . The camber angle is θ = 32◦ , pitch chord ratio is s/c = 0.82, and the position of maximum camber a/c = 0.45. The ratio of the blade height to the chord is b/c = 1.7. (a) Find the deviation of the flow leaving the stator. (b) Find the deflection. (c) Find the flow coefficient and blade loading coefficient if the inflow is at zero incidence. Given: R, χ2 , θ, s/c, a/c, b/c. Find: Deviation, deflection, ϕ, ψ. Solution: The flow angle at the exit of the stator is √ ( a )2 √ s ∗ α s 3 θ + θ α3∗ = χ3 + δ ∗ = χ3 + 0.902 c c 500 c and solving this for α3∗ gives α3∗ √ χ3 + 0.92(a/c)2 θ s/c √ = = 21.12◦ θ 1 − 500 s/c 99 and the deviation at the exit is δ ∗ = α3∗ − χ3 = 21.12◦ − 14.5◦ = 6.62◦ ⇐ (a) From the tangent difference formula ( ∗ −1 α2 = tan tan α3∗ + ) 1.55 1 + 1.5s/c ) ( 1.55 −1 ◦ = 47.24◦ = tan tan(21.12 ) + 1 + 1.5 · 0.82 and the deflection is ϵ∗ = α2∗ − α3∗ = 47.24◦ − 21.12◦ = 26.12◦ ⇐ (b) The flow coefficient is 1 = tan α3∗ + tan α2∗ = tan(21.12◦ ) + tan(47.24◦ ) = 1.429 ϕ ϕ = 0.7 ⇐ (c) and ψ = ϕ(tan α2∗ − tan α3∗ ) = 0.7[tan(47.24◦ ) − tan(21.12◦ )] = 0.474 ⇐ (c) Exercise 7.13 The circular arc blades of a compressor cascade have camber θ = 30◦ and the maximum thickness at a/c = 0.5. The spacing to chord ratio is s/c = 1.0. The nominal outflow angle is α∗ = 25◦ . (a) Determine the nominal incidence. (b) Determine that lift coefficient at the nominal incidence if the drag coefficient is CD = 0.016. Given: θ, s/c, a/c, α2∗ , and CD . Find: The nominal incidence and the lift coefficient. Solution: From the tangent difference formula ( ) 1.55 ∗ −1 ∗ α2 = tan tan α3 + 1 + 1.5s/c ( ) 1.55 −1 ◦ ∗ tan(25 ) + = 47.37◦ α2 = tan 2.5 100 The deviation is obtained from √ δ ∗ = mθ in which m = 0.92 so that ( a )2 c + s c α3∗ 25 = 0.92 · 0.25 + = 0.28 500 500 δ ∗ = 0.28 · 30◦ · 1 = 8.4◦ ⇐ (a) The metal angle at the exit is χ2 = χ3 + θ = 16.6◦ + 30◦ = 46.6◦ and the incidence at the inlet is i∗ = α2∗ − δ ∗ = 47.37◦ − 46.6◦ = 0.77◦ The lift coefficient is CL = 2 (s) c cos αm (tan α2∗ − tan∗3 ) − CD tan αm the mean flow direction is ( ) [ ] 1 −1 ∗ ∗ −1 1 ◦ ◦ ∗ αm = tan (tan α2 + tan α3 ) = tan (tan(47.37 ) + tan(35 )3 ) = 37.83◦ 2 2 so that CL = 2·1·cos(37.82◦ )(tan(47.37◦ )−tan(25◦ ))−0.018 tan(37.82◦ ) = 0.966 ⇐ (b) Exercise 7.14 Air with density 1.21 kg/m3 flows into a compressor stator with velocity V2 = 120 m/s and leaves at the angle α3 = 30◦ . If the Leiblein’s diffusion factor is to be held at 0.5 and the the stagnation pressure loss across a compressor stator is 0.165 kPa, find the static pressure increase across the stage assuming that the flow is incompressible. Given: ρ1 , V2 , Dp0 LS, and DF . Find: p3 − p2 . 101 Solution: The definition of the diffusion factor DF = V2 − V3 Vu2 − Vu3 s + V2 2V2 c can be written as DF = 1 − cos α2 Vx s + (tan α2 tan α3 ) cos α3 2V2 c or cos α2 cos α2 s + (tan α2 tan α3 ) cos α3 2 c Letting x = cosα2 then inserting the numerical values gives (√ ) 1 2 1 4x 1 − x2 =1− √ x+ −√ 2 10 x 3 3 DF = 1 − which reduces to √ 9 13x2 − 5 3x + =0 16 and the solution of this is √ √ 5 3 + 75 − 13 · 9/4 x= 26 x = 0.593236 α2 = 53.61◦ Assuming incompressible flow the difference is the stagnation pressures is 1 p02 − p03 = p2 − p3 + ρ(V22 − V32 ) 2 and [ ] 1 2 cos2 α2 p3 − p2 = ρV2 1 − − ∆p0 2 cos2 α3 ( ) 1.21 cos2 (53.61◦ ) 2 = · 120 1 − − 125, 000 = 4.46 kPa 2 cos2 (30◦ ) Exercise 7.15 Air at temperature 288 K and at pressure 101.325 kPa flows into a compressor with 10 stages. The efficiency of the first stage is 0.920 and the second stage 0.916. For the rest of the stages the stage efficiency drops by 0.004 successively so that the last stage has an efficiency 0.884. The axial velocity is constant and the flow angles are the same at the inlet and exit of each of the stages. Hence the work done by each stage is the same. (a) Find the overall efficiency of 102 the compressor. (b) Find the overall efficiency by using the theory for a polytropic compression with small stage efficiency ηp = 0.902. Given: Inlet p01 and T01 , and the stage efficiencies of ten stages. Find: Find the overall efficiency. Solution: The following Matlab script gives the overall effieciency. %HW 7.15 clear all p0(1)=101325; T0(1)=288; T0s(1)=T0(1); n=10; r=4.3; k=1.4; kp=k/(k-1); etapg=0.902; etag=(r^(1/kp)-1)/(r^(1/(kp*etapg))-1); T0eg=T0(1)*r^(1/kp) T0e=T0(1)+(T0eg-T0(1))/etag; dT0a=(T0e-T0(1))/n dT0=16.8323 etap=[0.92, 0.916, 0.912, 0.908, 0.904, ... 0.900, 0.896, 0.892, 0.888, 0.884] for i=2:n+1 T0s(i)=T0(i-1)+etap(i-1)*dT0; T0(i)=T0(i-1)+dT0 p0(i)=p0(i-1)*(T0s(i)/T0(i-1))^kp end rt=p0(n+1)/p0(1) T0ss=T0(1)*r^(1/kp) eta=(T0ss-T0(1))/(T0(n+1)-T0(1)) etaa=(r^(1/kp)-1)/(r^(1/(etapg*kp))-1) The exact value for the overall efficiency is ηc = 0.8846 and the approximate value is ηc = 0.8804. Hence the use of the average small stage efficiency does not change the overall efficiency by much. Exercise 7.16 Air flows in a axial flow fan of free vortex design, with hub radius rh = 7.5 cm and casing radius 17 cm. The fan operates at 2400 rpm. The volumetric flow rate Q = 1.1 m3 /s and the stagnation pressure rise is 3 cm H2 O. The fan efficiency is ηtt = 0.86. (a) Find the axial velocity. (b) Find the work done on 103 the fluid. (c) Find the absolute and relative flow angles at the inlet and exit of the rotor when the inlet is axial. Given: rh and rc , Ω, Q, ∆p, ηtt . Find: w α1 , α − 2, β1 , β2 . Solution: Free vortex design. Then mean radius is 1 1 rm = (rh + rc ) = (7.5 + 17.0) = 12.25 cm 2 2 and the blade speed at the mean radius is Um = rm Ω = 0.1225 · 2400 · π = 30.79 m/s 30 The blade speed at the hub and the casing are Uh = rh Um rm Uc = rc Um rm The density is ρ01 = p01 101325 = = 1.20 kgm2 RT01 287 · 293 and the flow area is A = π(rc2 − rh2 ) = π(172 − 7.52 ) = 731.2 cm2 and the axial velocity is Vx = Q 1.1 = = 15.0m/s A 0.07312 The ideal work is ws = ∆p0 0.03 cdot9.81 · 998 = = 243.8 J/kg ρ 1.2 and the actual work is w = ws /eta = 243.8/0.86 = 283.4 J/kg. Thus since the inlet is axial w 243.8 Vu2 = = = 9.21 m/s Um 30.79 Now ( ) ( ) Um 30.79 −1 −1 β1m = tan − = tan − = −64◦ ⇐ (a) Vx 15 104 and ( −1 α2m = tan Vu2m Vx ) ( −1 = tan 9.217 − 15 ) = 31.6◦ ⇐ (a) From Vu2m = Um + Wu2m which is also Vx tan α2m = Um + Vx tan β2m from which ) ( ) ( 30.79 Um −1 ◦ −1 = tan tan(31.6 ) − = −55.2◦ ⇐ (a) β2m = tan tan α2m − Vx 15 Similarly at the hub and the casing α1h = 0◦ α2h = 45◦ α1c = 0◦ α2c = 23.8◦ β1h = −51.4◦ β2h = −14.2◦ β1c = −70.6◦ β2c = −67.4◦ Exercise 7.17 Consider an axial flow compressor in which flow leaves the stator with a tangential velocity distributed as a free vortex. The hub radius is 20 cm and the static pressure at the hub is 94 kPa and the static temperature there is 292 K. The radius of the casing is 30 cm and the static pressure at the casing is 97 kPa. The total pressure at this location is 101.3 kPa. Find the exit flow angles at the hub and the casing. Given: pc , p01h , ph Th , rh , rc . Find: αc and αh . Solution: The density at the hub is ρh = From ph 94 = = 1.12 kg/m3 RTh 0.287 · 292 1 p0h = ph + ρh Vh2 2 the velocity at the hub is √ √ 2(p0h − ph ) 2(101.3 − 94)1000 = = 114.1 m/s Vh = ρh 1.122 so that ∫ pt − p h = rc rh Vu ρ dr = ρK 2 r ∫ 105 rc rh dr ρK 2 = r3 2 ( 1 1 − 2 2 rh rc ) so that √ K= 2(pc − ph ) = ρ(1/rh2 − 1/rc2 ) √ 2(97000 − 94000) = 9.81 m2 /s 1.12(1/0.102 − 1/0.152 ) Also, assuming that the density is the same 1 1 pc + ρVc2 = ph + ρVh2 2 2 The velocity at the casing is √ √ 2(p0c − pc ) 2(101300 − 97000) Vc = = = 87.56 m/s ρ 1.12 and Vuc = K 9.81 = = 65.4 m/s rc 0.15 Vuh = 9.81 = 98.1 m/s 0.1 and Vx = and √ Vc2 − Vuc2 √ √ 2 = 87.552 − 65.42 = 58.21 m/s = Vh2 − Vuh ( ) 65.4 αc = tan = tan = 48.3◦ 58.21 ) ( ) ( 98.1 Vuh −1 −1 = tan = 59.3◦ αh = tan Vx 58.21 −1 Vuc Vx ) ( −1 106 Chapter 8 Exercise 8.1 An industrial air compressor has 29 backward swept blades with blade angle −21◦ . The tip speed of the blades is 440 m/s and the radial component of the velocity is 110 m/s. Air is inducted from atmospheric conditions at 101.3 kPa and 298 K with an axial velocity equal to 95 m/s. The hub to tip ratio at the inlet is 0.4. The total-to-total efficiency of the compressor is 0.83, and the mass flow rate is 2.4 kg/s. (a) Find the total pressure ratio using the Stodola slip factor. (b) Find the tip radius of the impeller. Given: p01 , T01 , V1 , r1h /r1s , ṁ, Z , χ2 , U2 , Vr2 . Find: p03 /p01 and U2 . Solution: The flow coefficient is ϕ= Vr2 110 = = .35 U2 44) The slip coefficient is σ =1− π cos χ2 π cos(−21◦ ) =1− = 0.899 Z 29 and the tangential velocity is Vu2 = σU + Vr2 tan χ2 = 0.899 · 440 + 110 tan(−21◦ ) = 353.3 ms The work done is w = U Vu2 = 440 · 353.3 = 155.44 kJ/kg and the isentropic work is ws = ηw = 0.83 · 155.44 = 129.02 kJ/kg The total temperature at the end of the isentropic process is T02s = T01 + ws 129.02 = 298 + = 426.4 K cp 1.0045 and the pressure ratio is ( )γ/(γ−1) ( )3.5 426.4 p02 T02s = = = 3.505 p01 T01 298 107 ⇐ (a) At the inlet T1 = T01 − M1 = √ 952 V12 = 298 − = 293.5 K 2cp 2 · 1004.5 V1 95 =√ = 0.277 γRT1 1.4 · 287 · 293.5 The stagnation density is ρ01 = p01 101300 = = 1.184 kg/m3 RT01 287 · 298 and the static density is ( ρ1 = ρ01 T1 T01 )1/(γ−1) ( = 1.184 293/5 298 )2.5 = 1.140 kg/m3 From the mass flow rate ṁ = ρ1 A1 V1 A1 = ṁ 2.4 = = 0.02216 m2 ρ1 V1 1.1.4 · 95 and from the expression for the area A1 = π(rs2 − rh2 ) = πrs2 (1 − κ2 ) the shroud radius is √ rs = A1 = π(1 − κ2 ) √ 0.002216 = 0.0916 m ⇐ (b) π(1 − 0.16) Exercise 8.2 A centrifugal compressor has 23 radial vanes and an exit area equal to 0.12 m2 . The radial velocity is 27 m/s and the tip speed of the impeller is 350 m/s. The total-to-total efficiency is 0.83. (a) Find the mass flow rate of air, if at the inlet the total pressure and temperature are 101.3 kPa and 298 K. (b) Find the exit Mach number. (c) If the blade height at the exit is b = 3 cm and there is no leakage flow, find the tip radius of the impeller. (d) Find the rotational speed of the compressor wheel. (e) Find the required power neglecting mechanical losses. Given: p01 , T01 , Z , χ2 , U2 , Vr2 , and ηtt . Find: ṁ, M2 , r2 , and Ω. 108 Solution: The slip factor is σ =1− π cos χ2 π =1− = 0.863 Z 23 The tangential velocity is Vu2 = σU = 0.863 · 350 = 302.2 ms and the specific work is w = U2 Vu2 = 302.2 · 350 = 105.8 kJ/kg and the temperature at the exit is T02 = T01 + w 105.8 = 298 + = 403.3 K cp 1.0045 The isentropic process leads to T02s )] [ ( )] [ ( 403.3 T02 T02s = T01 1 + ηtt −1 = 298 1 + 0.83 −1 = 385.4 K T01 298 The exit pressure is ( p02 = p01 T02s T01 )γ/(γ−1) ( = 101.3 )3.5 385.4 −1 = 249.2 kPa 298 The velocity at the exit is √ √ 2 V2 = Vr22 + Vu2 = 272 + 302.22 = 303.3 m/s and the static temperature at the exit is T2 = T02 − V22 303.32 = 403.3 − = 357.5 K 2cp 2 · 1004.5 The Mach number at the exit is √ ) √ ) ( ( 2 T02 2 403.3 −1 = − 1 = 0.8 M2 = γ − 1 T2 0.4 357.5 109 The static pressure at the exit is ( p2 = p02 T2 T02 )γ/(γ−1) ( = 249.2 357.5 403.3 )3.5 = 163.4 kPa The static density is ρ2 = 163.4 p2 = = 1.593 kg/m3 RT2 0.287 · 357.5 The mass flow rate is ṁ = ρ2 A2 Vr2 = 1.593 · 0.12 · 27 = 5.16 kg/s The exit radius is therefore r2 = A2 0.12 = = 0.637 m 2πb2 2π · 0.03 Ω= U 35) · 30 = = 5250 rpm r2 0.637 · π and the shaft speed is The power to the compressor is Ẇ = ṁw = 5.16 · 105.7 = 545.7 kW ⇐ (e) Exercise 8.3 A centrifugal compressor has an axial inlet and the outlet blades at an angle such that the tangential component of the exit velocity has a value equal to 0.9 times the blade speed. The outlet radius is 30 cm and the desired pressure ratio is 3.5. The inlet stagnation temperature is T0 = 298 K. If the total-tototal efficiency of the compressor is 0.8, at what angular speed does it need to be operated. Given: Axial inlet with T01 . At the exit Vu2 = 0.9U2 , r2 = 0.3 m, p03 /p01 = 3.5, ηtt = 0.8. Find: Ω. Solution: Work done is w = Vu2 U2 = 0.9U22 110 so that the isentropic work is [( ws = ηtt0.9U22 Solving this for U2 gives √ U2 = and Ω= = cp (T02s − T01 ) = cp T01 )] p03 −1 p01 1004.5 · 298(3.51/3.5 − 1) = 423.0m/s 0.8 · 0.9 423.0 · 30 U2 = = 13, 464 rpm Lef tarrow r2 0.3 · π Exercise 8.4 A small centrifugal compressor as a part of a turbocharger operates at 55,000 rpm. It draws air from atmosphere at temperature 288 K and pressure 101.325 kPa. The inlet Mach number is M1 = 0.4 and the flow angle of the relative velocity is β1s = −60◦ at the shroud. The radius ratio at the inlet is κ = r1h /r1s = 0.43. Find, (a) the tip blade speed at the inlet, (b) the mass flow rate and (c) the throat are if the inducer is choked. Given: Axial inlet T01 , p01 , M1 , β1s , and r1s /r2 . The shaft speed is Ω. Find: U1s , ṁ, At . Solution: The inlet static temperature is )−1 ( γ−1 2 T1 = T01 1 + M1 = 288(1 + 0.2 · 0.42 )−1 = 279.1 K 2 and the stagnation density is ρ01 = 101325 p01 = = 1.226 kg/m3 RT01 287 · 288 The static density is therefore ( )1/(γ−1) ( )1/(γ−1) 279.1 T1 = 1.226 = 1.333 kg/m3 ρ1 = ρ01 T01 288 The relative Mach number at the inlet shroud is M1Rs = M1 0.4 = = 0.8 cos β1s cos(−60◦ ) 111 The sound speed and stagnation sound speed at the inlet are √ √ c1 = γRT1 = 1.4 · 287 · 279.1 = 334.9 m/s √ √ c01 = γRT01 = 1.4 · 287 · 288.0 = 340.2 m/s and the blade speed at the shroud inlet is √ √ √ 2 2 − M12 = 334.9 0.82 − 0.42 = 232.0 m/s ⇐ (a) − V12 = c1s M1Rs U1s = W1s and the inlet radius of the shroud is r1s = U1s 232.0 · 30 = = 0.0403 m Ω 55, 000 · π The mean radius at the inlet is 1 r1s 0.0403 r1 = (r1h + r1s ) = (1 + κ) = (1 + 0.62) = 0.0288 m 2 2 2 The inlet area is 2 2 2 A1 = π(r1s − r1h ) = pir1s (1 − κ2 ) = π · 0.4032 (1 − 0.622 ) = 0.004155 m2 The inlet velocity is V1 = M1 c1 = 0.4 · 334.9 = 133.96 m/s and the mass flow rate is ṁ = ρ1 A1 V1 = 1.133 · 0.00415 · 133.9 = 0.6305 kg/s ⇐ (b) and the mass balance gives ( A1 cos β1s M1Rs γ−1 2 M1Rs 1+ 2 )−(γ+1)/2(γ−1) ( = At Mt γ−1 2 1+ Mt 2 )−(γ+1)/2(γ−1) and when the throat is choked At = A1 cos1s ( 2 γ+1 M1RS 2 )−(γ+1)/2(γ−1) = 0.0020 m 2 + γ−1 M1Rs γ+1 112 ⇐ (c) Exercise 8.5 A centrifugal compressor in a turbocharger operates at 40,000 rpm and inlet Mach number M1 = 0.35. It draws air from atmosphere at temperature 293 K and pressure 101.325 kPa. The radius ratio is r1s /r2 = 0.71 and the diffusion ratio is W1s /W2 = 1.8. The inlet angle of the relative velocity at the shroud is β1s = −63◦ . The slip factor is σ = 0.85 and the flow angle at the exit is α2 = 69◦ . Find, (a) the tip speed of the blade at the inlet, (b) the tip speed of the blade at the outlet, (c) the loading coefficient, and (d) the metal angle at the exit. Given: Axial inlet and T01 , p01 , M1 , β1s , r1s /r2 , W1s /W2 , σ and α2 . Find: U1s , U2 , ψ, χ2 . Solution: The inlet static temperature is ( )−1 γ−1 2 T1 = T01 1 + M1 = 293(1 + 0.2 · 0.352 )−1 = 286.0 K 2 and the stagnation density is ρ01 = p01 101325 = = 1.205 kg/m3 RT01 287 · 293 The static density is therefore ( ρ1 = ρ01 T1 T01 )1/(γ−1) ( = 1.205 286.0 293 )1/(γ−1) = 1.134 kg/m3 The sound speed at the inlet is √ √ c1 = γRT1 = 1.4 · 287 · 286 = 339.0 m/s and the velocity is V1 = M1 c1 = 0.35 · 339 = 118.6 m/s The relative Mach number at the inlet shroud is M1Rs = 0.35 M1 = = 0.771 cos β1s cos(−63◦ ) and the blade speed at the shroud inlet is √ √ √ 2 2 U1s = W1s − V12 = c1s M1Rs − M12 = 339.0 0.7712 − 0.352 = 232.8 m/s ⇐ (a) 113 The radius of the inlet shroud is r1s = U1s 232.8 · 30 = = 0.0556 m Ω 40, 000 · π and the exit radius is r2 = r1s 0.0556 r2 = = 0.0783 m r1s 0.71 ⇐ (b) and the blade speed there is U2 = U1s With 0.0783 r2 = 232.8 = 328.0 m/s r1s 0.0556 W1s r1s /r2 √ =− W2 sin β1s 1 − 2ψ + ψ 2 / sin2 α2 which can be written as 1 − 2ψ + where (r1s /r2 )2 c= = (W1s /W2 )2 sin2 β1s ψ2 =c sin2 α2 ( 0.71 1.8 sin(−65◦ ) )2 = 0.196 Hence the loading factor can be found by solving ψ 2 − 2 sin2 α2 ψ + (1 − c) sin2 α2 = 0 or √ ψ = sin α2 − sin4 α2 − (1 − c) sin2 α2 = 0.629 2 ⇐ (c) The choice of the minus sign gives backward swept blades. Then ( ) σ tan χ2 = 1 − tan α2 ψ gives χ2 = tan−1 [tan(60◦ )(1 − 0.85/0.629)] = −42.48◦ 114 ⇐ (d) Exercise 8.6 A small centrifugal compressor draws atmospheric air at 293 K and 101.325 kPa, At the inlet r1h = 3.2 cm and r1s = 5 cm. The total-to-total efficiency of the compressor is 0.88. The relative Mach number at the inlet shroud is 0.9 and the corresponding relative flow angle is β1s = −62◦ . At the outlet the absolute velocity is at the angle α2 = 69◦ . The diffusion ratio is W1s /W2 = 1.8 and the radius ratio is r1s /r2 = 0.72. (a) Find the rotational speed of the shaft. (b). Find the blade loading coefficient w/U22 . (c) Find the flow coefficient ϕ = Vr2 /U2 (d) Find the blade width at the exit. Given: Axial inlet and T01 , p01 , Ω, ηtt , M1Rs , β1s , r1s /r2 , W1s /W2 , and α2 . Find: Ω, ψ = w/U22 , ϕ = Vr2 /U2 , b2 . Solution: The inlet static temperature is ( )−1 γ−1 2 T1 = T01 1 + M1 = 293(1 + 0.2 · 0.52 )−1 = 279.0 K 2 and the sound speed at the inlet is √ √ c1 = γRT1 = 1.4 · 287 · 279 = 334.8 m/s and the velocity is V1 = M1 c1 = 0.5 · 334.8 = 167.7 m/s The blade speed at the shroud inlet is √ √ √ 2 2 2 U1s = W1s − V1 = c1s M1Rs − M12 = 334.8 0.812 − 0.52 = 213.1 m/s The shaft speed is Ω= U1s 213.1 · 30 = = 40, 700 rpm r1s 0.05 · π ⇐ and the exit radius is r2 = r1s r2 0.05 = = 0.0694 m r1s 0.72 and the blade speed there is U2 = U1s 213.1 r2 = = 296.0 m/s r1s 0.72 115 (a) By squaring and adding the equations V2 cos α2 = W2 cos β2 V2 sin α2 − U2 = W2 sin β2 leads to V22 − 2U2 V2 sin α2 + U22 − W22 = 0 which when solved for V2 gives √ V2 = U2 sin α2 − U22 sin2 α2 − U22 + W22 = 296 cos(69◦ ) − and √ 2962 cos2 (69◦ ) − 2962 + 150.672 = 169.35 m/s Vu2 = V2 sin α2 = 169.35 sin(69◦ ) = 158.1 m/s The work done is w = U2 Vu2 = 158.1 · 296 = 46.795 kJ/kg and the blade loading factor is ψ= w Vu2 158.1 = = = 0.534 2 U2 U2 296 ⇐ (b) The radial velocity is Vr2 = V2 cos α2 = 169.35 cos(69◦ ) = 60.69 m/s and the flow coefficient is ϕ= Vr2 60.69 = = 0.205 U2 296 ⇐ (c) The stagnation temperature at the exit is T02 = T01 + 46, 795 w = 293 + = 339.6 K cp 1004.5 The isentropic work is ws = ηtt w = 0.88 · 46, 795 = 41, 180 J/kg 116 and the stagnation temperature for an isentropic process is T02s = T01 + w2 41, 180 = 334.0 K = 293 + cp 1004.5 The stagnation pressure at the exit is ( )γ/(γ−1) ( )3.5 T02s 334.0 p02 = p01 = 101325 = 160.24 kPa T01 293 The flow function at the inlet is ( )−(γ+1)/2(γ−1) γ γ−1 2 1.4 F1 = √ M1 1 + M1 = √ ·0.5(1+0.2·0.52 )−3 = 0.957 2 γ−1 0.4 The stagnation Mach number at the exit is M02 = 169.35 V2 =√ = 0.458 c02 1.4 · 287 · 339.6 and the Mach number is M2 = √ M02 1− γ−1 2 M02 2 =√ 0.458 1 = 0.2 · 0.4582 = 0.468 The flow function is )−(γ+1)/2(γ−1) ( γ γ−1 2 F2 = √ M2 1 + M2 2 γ−1 From 1.4 =√ · 0.468(1 + 0.2 · 0.4682 )−3 = 0.9115 0.4 √ √ A2 T02 p01 F1 339.6 101325 0.957 = = = 0.7149 A1 T01 p02 F2 293 160240 0.9115 With 2 2 A1 = π(r1s − r1h ) = π(0.052 − 0.0322 ) = 0.004637 m2 then A2 = 0.7149 · 0.004637 = 0.003315 m2 and the blade width is A2 0.003315 b2 = = = 0.0076 m = 7.6 mm 2πr2 2]pi · 0.0694 117 ⇐ (d) Exercise 8.7 Show that in the incompressible limit the angle of the relative velocity at the inlet is optimum at 54.7◦ . Given: The expression which yields the correct angle. Find: β1s in the incompressible limit. Solution: The general equation is ( cos2 β1s = 2 3 + γM1R 2 2M1R √ )( 1− 1− 2 4M1R 2 2 (2 + γM1R ) ) For M1R ≪ 1, the radical becomes √ √ 2 2 2 4M1R 4 1 1− = 1 − M1R = 1 − M1R 2 2 (2 + γM1R ) 9 9 Therefore ( 2 cos β1s = So that cos β1s = 2 3 + γM1R 2 2M1R √1 3 ) 2 2 2 2(3 + γM1R 1 ) (1 − 1 + M1R )= ≈= 9 2·9 3 and thus β1s = 54.7◦ . Exercise 8.8 Show that the expression for the dimensionless mass flow rate for a compressor with pre-swirl at angle α1 is 3 M1R (tan α1 − tan β1s )2 cos3 β1s ( ) Φf = γ − 1 2 cos2 β1s 1+ M1R 2 sin2 α1 Plot the results for α1 = 30◦ , with β1s on the abscissa and Φf on the ordinate, for relative Mach numbers 0.6, 0.7, 0.8. For a given mass flow rate does the pre-swirl increase or decrease the allowable relative Mach number and does the absolute value of the relative flow angle increase or decrease with pre-swirl. Solution: The mass balance can be written as 2 (1 − κ2 )W1s cos β1s ṁ = ρ1 A1 V1 = ρ1 A1 W1s cos β1s = ρ1 πr1s which in the non-dimensional form is Φ= 2 2 W1s ρ1 U1s W1s ρ1 πr1s 2 (1 − κ ) cos β = (1 − κ2 ) cos β1s 1s 2 2 ρ01 πr2 c01 ρ01 U2 c01 118 Since U1s = Vu1 − Wu1 Vx = W1s cos β1s this can be written as Φ= ρ1 Vx2 c31 W1s c201 (1 − κ2 )(tan α1 − tan β1s )2 cos β1s ρ01 c21 c301 c01 U22 or as ρ1 Φ= ρ01 With Φf = so that ( T1 T01 )3 2 ΦM0u 1−κ 3 M1R (1 − κ2 )(tan α1 − tan β1s )2 cos3 β1s 3 M0u M12 = 2 W1s cos2 β1 s V12 Vx2 = = c21 c21 cos2 α1 c21 cos2 α1 M 2 (tan α1 − tan β1s )2 cos3 β1s ϕf = ( 1R )(3γ−1)/(2γ−2) 2 cos2 β1 s 1 + γ−1 M 1R cos2 α1 2 Exercise 8.9 Water with density 998 kg/m3 flows through the inlet pipe of a centrifugal pump at a velocity of 6 m/s. The inlet shroud radius is 6.5 cm and water flow has an axial entry. The relative velocity at the exit of the impeller is 15 m/s and is directed by backward curved impeller blades such that the exit angle of the absolute velocity is α2 = 65◦ . The impeller rotates at 1800 rpm and has a tip radius of 15 cm. Assume that the adiabatic efficiency of the pump is 75 %. Evaluate (a) the power into the pump, (b) the increase in total pressure pressure of the water across the impeller, (c) the change in static pressure of the water between the inlet and outlet of the impeller. (d) If the inlet mean inlet radius is 5.0 cm, what is the ratio of the change in kinetic energy of the water across the impeller to the total enthalpy of the water across the pump. (d) What is the ratio of the change in the absolute kinetic energy, the change in the relative kinetic energy and the change in the kinetic energy owing to the centrifugal effect as a fraction of work done. (f) If the velocity at the exit of the volute is 6 m/s, what is the ratio of the change in static pressure across the rotor to the change in static pressure across the entire pump? , Given: V1 , r1s , R1h , Ω, W2 , α2 , r2 . Find: Ẇ , ∆p0 , ∆p, ∆KE/∆h0 , ∆KEU /∆h0 , ∆KEW /∆h0 , (∆p)R /(∆p)p . Solution: The inlet area is 2 2 ) = 0.00942 m2 − r1h A1 = π(r1s 119 and the mass flow rate is ṁ = ρ1 A1 V1 = 998 · 0.00942 · 6 = 56.4 kg/s The blade speed is U2 = r2 Ω = 0.15 · 1800 · π = 28.2 m/s 30 By squaring and adding the equations V2 cos α2 = W2 cos β2 V2 sin α2 − U2 = W2 sin β2 leads to V22 − 2U2 V2 sin α2 + U22 − W22 = 0 which when solved for V2 gives √ V2 = U2 sin α2 − U22 sin2 α2 − U22 + W22 √ = 28.27 cos(65 ) − 28.272 sin2 (65◦ ) − 28.272 + 15.02 = 16.56 m/s ◦ in which the minus sign was chosen for backward swept blades. Next Vu2 = V2 sin α2 = 16.56 sin(65◦ ) = 15.00 m/s Vr2 = V2 cos α2 = 16.56 cos(65◦ ) = 7.00 m/s so that Wu2 = Vu2 − U2 = 15.00 − 28.27 = 13.29 m/s Wr2 = 7.00 m/s and the flow angle of the relative velocity is ( ) ( ) Wu2 13.27 −1 −1 β2 = tan = tan − = −62.19◦ Wr2 7.00 and the specific work is w = Vu2 U2 = 15.0 · 28.27 = 318.2 J/kg and the power is Ẇ = ṁw = 56.4 · 318.2 = 23.95 kW 120 ⇐ (a) The isentropic work is ws = ηtt w = 0.75 · 424.3 = 318.2 J/kg The stagnation pressure increase is therefore ∆p0 = ρws = 998 · 317.6 = J/kg and the static pressure increase is 998 ρ p2 − p1 = p02 − p01 − (v22 − V12 ) = 317.6 − (16.562 − 62 ) = 198.75 kPa 2 2 The kinetic energies are obtained from 1 1 1800 · π U1 = r1 Ω = (r1h + r1s )Ω = (0.065 + 0.035) = 9.425 m/s 2 2 30 √ and W1 = so that V12 + U12 = √ 62 + 9.4252 = 11.17 m/s 1 1 ∆KEV = (V22 − V12 ) = (16.562 − 62 ) = 119.1 J/kg 2 2 1 1 ∆KEU = (U22 − U12 ) = (28.272 − 9.4252 ) = 355.3 J/kg 2 2 1 1 ∆KEW = (W12 − W22 ) = (11.172 − 192 ) = 50.17 J/kg 2 2 so that ∆KEV 119.1 = = 0.281 w 424.3 ∆KEU 355.3 = = 0.827 w 424.3 ∆KEW 50.1 =− = −0.118 w 424.3 For the stator ∆p0 = 0, so that p3 − p2 = so that ) 988 ( ) ρ( 2 V2 − V32 = 16.562 − 62 = 118.8 kPa 2 2 ∆pR 198.75 = = 0.626 ∆pR + ∆pV 198.75 + 118.8 121 ⇐ (d) Exercise 8.10 Centrifugal pump handling water operates with backward curving blades. The angle between the relative velocity and the tip section is 45◦ . The radial velocity at the tip section is 4.5 m/s, the flow at the inlet is axial, and the impeller rotational speed is 1800 rpm. Assume that there is no leakage and that the mechanical friction may be neglected, and the total-to-total efficiency is 70%. (a) Construct the velocity diagram at the impeller exit. (b). Evaluate the required tip radius for a water pressure rise of 600 kPa, (c) For the total pressure rise of 600 kPa, evaluate the difference between the total and static pressure of water at the impeller tip section. Given: Vr2 , Wu2 , β2 , p3 − p1 . Find: r2 , p03 − p3 . Solution: From w = Vu2 U2 = U2 (U2 + Wu 2) and ∆p0 = ρηtt w = ρηtt U2 (U2 + Wu2 ) and ∆p0 600, 000 = = 858.86 ρηtt 998 · 0.7 so that U22 + Wu2 U2 − ∆p0 =0 ρηtt the solution of which is √ √ 1 2 1 ∆p0 U2 = − Wu2 + Wu2 + = 2.25 + 2.252 + 858.86 = 31.64mm/s 2 4 ρηtt Hence r2 = U2 31.64 · 30 = = 0.168 m Ω 1800 · π ⇐ (b) and Vu2 = U2 + Wu2 = 31.64 − 4.5 = 27.145 m/s and V2 = sqrtVu2 + Vr2 = √ 4.52 + 27.142 = 27.51 m/s so that the difference in the stagnation and static pressure is 998 1 p02 − p2 = ρV22 = · 27.512 = 377.6 kPa 2 2 122 ⇐ (c) Exercise 8.11 A centrifugal water pump has an impeller diameter D2 = 27 cm and when its shaft speed is 1750 rpm it produces a head H = 33 m. (a) Find the volumetric flow rate. (b) Find the blade height at the exit of the impeller. (c) Find the blade angle at the exit of the impeller if it has 11 blades. Given: D2 , Ω, H, Z. Find: Q, b2 , χ2 . Solution: The blade tip speed is U2 = r2 Ω = 0.135 · 1750 · π = 24.74 m/s 30 The blade loading coefficient is ψs = gH 9.81 · 33 = = 0.5289 2 U 24.742 From the expression ψs = 0.383 1/3 Ωs the specific speed becomes ( )3 ( )3 0.383 0.383 Ωs = = = 0.380 ψs 0.5289 √ Ω Q Ωs = (gh)3/4 Then from ( )2 ( )2 Ωs (gh)3/4 0.380 · 30(9.81 · 33)3/4 Q= = = 0.0250 m3 /s Ω 1750 · π The flow coefficient is √ √ ϕ = 0.1715 Ωs = 0.1715 0.380 = 0.1057 so that the radial component of the velocity is Vr2 = ϕU2 = 0.1057 · 24.74 = 2.615 m/s and the blade width is b2 = 0.1057 ϕ = = 0.0113 m 2πr2 Vr2 2π · 0.135 · 2.615 123 ⇐ (b) ⇐ (a) The hydraulic efficiency is ηh = 1 − 0.4 0.4 = 1/4 = 0.821 1/4 Q 25 and the work done is w= ws gH 9.81 · 33 = = = 394.3 J/kg ηh ηh 0.821 The tangential component of velocity is w 394.3 = = 15.94 m/s U2 24.74 Vu2 = The flow angle is therefore ( α2 = tan −1 Vu2 Vr2 ) ( −1 = tan 15.94 2.615 ) = 80.7◦ and Wu2 = Vu2 − U2 = 15.94 − 24.74 = −8.80 m/s ( so that −1 β2 = tan Wu2 Wr2 ) ( −1 = tan 8.80 − 2.615 ) = −73.46◦ The slip coefficient is σ =1− 0.63 · π 0.63π =1− = 0.82 Z 11 so that the metal angle can be obtained by solving Vu2 = σU2 − Vr2 tan χ2 for χ2 , which is ( ) ( ) Vu2 − σU2 15.94 − 0.82 · 24.74 −1 −1 χ2 = tan = tan − = −59.0◦ Vr2 2.615 ⇐ (c) Exercise 8.12 A centrifugal water pump has an impeller diameter of D2 = 25 cm and when its shaft speed is 1750 rpm, it delivers 20 liters per second of water. (a) Find the head of water delivered by the pump. (b) Find the power to drive the 124 pump. (c) The impeller has 9 blades. Use the Stanitz slip factor to find the blade angle at the exit of the impeller. (d) Use the Wiesner slip factor to find the blade exit blade angle. Given: D2 , Ω, Q, Z. Find: H, ṁ, χ2 . Solution: The blade speed is U2 = r2 Ω = The specific speed is 0.125 · 1750 · π = 22.91 m/s 30 √ Ω Q Ωs = (gH)3/4 and the blade loading coefficient is ϕs = gH U22 In addition, the blade loading is related to the specific speed by the empirical equation 0.383 ψs = 1/3 Ωs Equating these expressions for ψs and solving for gH gives ( gH = 0.383U22 √ (Ω Q)1/3 )4/3 Then H= ( = 0.383 · 22.912 √ (1750 · π 0.020/30)1/3 gH 277.1 = = 28.24 m g 9.81 ⇐ )4/3 = 277.1 J/kg (a) and the mass flow rate is ṁ = ρQ = 998 · 0.020 = 19.96 kg/m3 The hydraulic efficiency is ηh = 1 − 0.4 0.4 = 1 − = 0.811 Q1/4 201/4 125 and the actual work is w= ws gH 277.1 = = = 341.7 J/kg ηh ηh 0.811 and the power to the pump is Ẇ = ṁw = 19.96 · 341.7 = 6.82 kW ⇐ (b) The specific speed is √ √ 1750 · π 0.02 Ω Q = = 0.3816 Ωs = (gh)3/4 30 277.13/4 and the flow coefficient is √ √ ϕ = 0.1715 Ωs = 0.1715 0.3816 = 0.1059 and the radial component of the exit velocity is Vr2 = ϕU2 = 0.1059 · 22.91 = 2.43 m/s The tangential component is Vu2 = w 341.7 = = 14.9 m/s U2 22.91 and the flow angle at the exit is ( ) ( ) Vu2 14.9 −1 −1 α2 = tan = tan = 80.76◦ Vr2 2.43 The tangential component of the relative velocity is Wu2 = Vu2 − U2 = 14.9 − 22.91 = −7.99 m/s ( so that −1 β2 = tan Wu2 Wr2 ) ( = tan −1 7.99 − 2.43 ) = −73.1◦ The first guess of the slip coefficient is obtained from the Stanitz formula σ =1− 0.63π 0.63 · π =1− = 0.78 Z 9 126 and this is used to obtain the metal angle ( ) ( ) Vu2 − σU2 14.9 − 0.78 · 22.91 −1 −1 χ2 = tan = tan = −50.71◦ Vr2 2.43 Using this in the Wiesner’s formula √ √ cos χ2 cos(50.71◦ ) σ =1− = = 0.829 Z 0.7 90.7 Iterating gives ( −1 χ2 = tan Vu2 − σU2 W Vr2 ) ( = tan −1 14.9 − 0.829 · 22.91 2.43 ) = −59.3◦ and one more iteration leads to χ2 = −62.4◦ ⇐ (c) Exercise 8.13 A centrifugal pump delivers water at 0.075 m3 /s with a head of 20 m while operating at 880 rpm. The hub to shroud radius ratio at the inlet is 0.35 and the relative velocity makes an angle −52◦ at the inlet. (a) Find the reversible work done by the pump. (b) What is the work done by the impeller. (c) Find the radius of impeller and the inlet radius ratio of the shroud to the hub. (d) Determine the blade width at the exit of the impeller. (e) Assume a reasonable number of blades and calculate the blade angle at the exit. Use the Pfleiderer equation to determine more accurately the number of blades and recalculate the blade angle at the exit if needed. (f) What is the power to drive the pump? Given: Q, Ω, H, β1s , κ. Find: ws , w, r2 , r1s /r2 , b2 , χ2 . Solution: The isentropic work is ws = gH = 9.81 · 20 = 196.2 J/kg ⇐ (a) Since the inlet flow is axis, the flow rate can be written as Q = A1 V1 = 2 (1 − κ2 )U1s πr1s πr3 (1 − κ2 )U Ω = 1s tan(−β1s ) tan(−β1s ) from which ( )1/3 ( )1/3 75 tan(52◦ )30 Q tan(−β1s ) = = 0.0723 m r1s = π(1 − κ2 )Ω 1000π(1 − 0.352 ) · 880 · π 127 ⇐ (c) and the tip speed of the blade at the inlet shroud is U1s = r1s Ω = 0.0723 · 880 · π = 6.66 m/s 30 The hydraulic efficiency is ηh = 1 − 0.4 0.4 = 1 − 1/4 = 0.864 1/4 Q 75 so that the specific work done by the rotor is w = f racws ηh = 196.2 = 227.1 J/kg 0.864 ⇐ (b) The specific speed is √ √ Ω Q 880π 0.075 Ωs = = = 0.481 (gH)3/4 30196/23/4 1/3 (Note that ψs = 0.383/(Ωs ) = 0.489, which is close). The blade speed is √ √ w 227.1 U2 = = = 21.56 m/s ψs 0.481 The flow coefficient is √ √ ϕ = 0.1715 Ωs = 0.1715 0.481 = 0.119 Next the velocity components at the exit are Vu2 = w 227.1 = = 10.53 m/s U2 21.56 Vr2 = ϕU2 = 0.119 · 21.56 = 2.565 m/s The volumetric efficiency is obtained by using the correlation in the text. It is ηv = 1 − Thus QR = 0.0988 C =1− = 0.977 n Q 0.0750.3387 Q 0.075 = = 0.0768 m3 /s ηv 0.977 128 and the blade width is b2 = QR 0.0768 = = 0.0204 m 2πr2 V r2 2π · 0.234 · 2.565 ⇐ (d) The overall efficiency can be obtained from the correlation in the text. It is η = 0.8393 The exit flow angle is ( α2 = tan −1 Vu2 Vr2 ) ( −1 = tan 10.53 2.565 ) = 76.3◦ The tangential component of the relative velocity is Wu2 = Vu2 − U2 = 10.53 − 21.56 = −11.02 m/s ( so that −1 β2 = tan Wu2 Wr2 ) ( −1 = tan 11.02 − 2.565 ) = −76.9◦ Assume that Z = 6 and assume that the metal angle is χ2 = −70◦ , then √ √ χ2 cos(−70◦ ) σ = 1 − 0.7 = 1 − = 0.833 Z 60.7 and this is used to obtain the metal angle ( ) ( ) Vu2 − σU2 10.53 − 0.833 · 21.56 −1 −1 χ2 = tan = tan = −70.94◦ Vr2 2.565 After one iteration the metal angle is χ2 = −71.19◦ . Using the Pfleider’s formula gives ( ) ( ) 1 + r1s /r2 1 Z = 6.5 cos (β1s + χ2 ) 1 − r1s /r2 2 ) ( ) ( 1 + 0.3091 1 ◦ ◦ (−52.0 − 71.9 ) = 6.4 ⇐ (e) = 6.5 cos 1 − 0.3091 2 Thus Z = 6 is an appropriate number. Finally the power to the pup is Ẇ = ρQ 998 · 75 · 96.2 ws = = 17.5 kW ηh 1000 · 0.839 129 ⇐ (f ) Exercise 8.14 A fan draws in atmospheric air at 0.4 m3 /s at pressure 101.32, kPa and temperature 288 K. The total pressure rise across the fan, which has 30 radial blades, is 2.8 cm of water. The inner radius is 14.8 cm and outer radius is 17.0 cm. The rotational speed of the fan is 980 rpm. Use the Stanitz slip factor and a fan efficiency of 78 %. (a) If the velocity into the fan is radially outward, find the angle of the relative velocity at the inlet. (b) Determine the power to the fan if 4 % is lost to mechanical friction. (c) Find the angle blade angle at the exit. Given: p01 , T01 , Ω, Q, r1s , r2 , b2 , η, ηm , total pressure rise in h of water. Find: β1 , Ẇ , β2 . Solution: The stagnation density at the inlet is ρ01 = p01 101325 = = 1.226 kg/m3 RT01 287 · 288 The pressure rise is small and this can be taken to be incompressible flow. The stagnation pressure rise is ∆p0 = ρH2 O gh = 998 · 9.81 · 0.028 = 274.1 Pa and the isentropic work is ws = ∆p0 274.1 = = 223.6 J/kg ρ 1.226 and the actual work is w= ws 223.6 = = 286.7 J/kg ηh 0.78 The power to the fan is therefore Ẇ = ρQw 1.226 · 0.4 · 286.7 = = 146.4 W ηm 0.96 The blade speed at the inlet is U1 = r2 Ω = 0.148 · 980 · π = 15.19 m/s 30 and at the exit it is U2 = U1 17 r2 = 15.19 = 17.45 m/s r1 14.8 130 and the inlet radial velocity is Vr1 = Q 0.4 = = 2.76 m/s 2πr1 b 2π · 0.148 · 0.156 The tangential component of the relative velocity at the inlet is wu1 = Vu1 − U1 = 0 − 15.19 = −15.19 m/s The flow angle is therefore ( ) ) ( Wu1 15.19 −1 −1 β1 = tan = −79.7◦ = tan − Wr11 2.76 At the exit 14.8 r1 = 2.76 = 2.40 m/s r2 17.0 and the tangential component of the velocity is Vr2 = Vr1 Vu2 = W 286.7 = = 16.43 m/s U2 17.45 The slip factor is σ =1− 0.63π 0.63π =1− = 0.934 Z 30 so that the metal angle is ( ) ( ) Vu2 − σU2 16.43 − 0.934 · 17.45 −1 −1 χ2 = tan = tan = 3.3◦ Vr2 2.4 Exercise 8.15 A pump draws water at the rate of 75 liters per second from a large tank with the air pressure above the free surface at 98.00 kPa. The pump is z = 2 m above the water level in the tank. The pipe diameter is 14.0 cm and the suction pipe is 20 m in length. The entrance loss coefficient is Ki = 0.2 and the loss coefficient of the elbow is Ke = 0.6 and the pipe roughness is 55 µ m. Find the suction specific speed if the shaft speed is 1800 rpm. The viscosity of water is 1.08 · 10−3 kg/m s. Given: Q, pa , z1 , Dp , Ki , Ke , ϵ, Ω. Find: Ωss . 131 Solution: From the Bernoulli equation [ ( ) ] 2 Vp pa 1 2 pt L + V1 + gz1 = + f Ki + Ke ρ 2 ρ D + 2 The velocity in the pipe is Vp = 4Q 4 · 0.075 Q = = = 4.87 m/s 2 Ap πD π · 0.142 Reynolds number is Re = ρVp Dp 998 · 4.87 · 0.14 = = 630, 300 µ 1080 · 10−6 The friction factor is obtained from the Colebrook formula in Chapter 3. For a pipe with roughness ϵ = 55 µm, it gives f¯ = 0.01677. Thus the head loss is [ ( ) ] 20 4.872 hl = 0.01677 + 0.2 + 0.6 = 3.867 m 0.14 2 · 9.81 The vapor pressure is pv = 37.8 kPa and the pressure at the free surface is pa = 98 kPa. Thus the NPSH is NPSH = pa pv 98000 3780 − z1 − hl − = − 2 − 3.863 − = 3.75 m ρg ρg 998 · 9.81 998 · 9.81 The suction specific speed is therefore √ √ Ω Q 1800 · π 0.075 Ωss = = = 3.455 (gNPSH)3/4 30 · (9.81 · 3.75)3/4 So cavitation is likely to take place. Exercise 8.16 Consider a volute consisting only a circular section. (a) If the tangential velocity varies as Vu = K/r, show that the value of K in terms of the volumetric flow rate Q, the radius of the section R and the radius to the center of the section a is given by K= Q √ 2πR(λ − λ2 − 1) 132 in which λ = a/R. (b) For R = 0.5 m, a = 2 m and Q = 1.5 m3 /s, find the pressure difference p2 − p1 at the centerline, between the outside and inside edge of the section. Given: R, a, Q. Find: K and p2 − p1 . Solution: With the coordinate origin at the √ center of the pipe its circumference is 2 2 2 given by x + y = a . From this y = a2 − x2 for the upper half of the pipe. Then flow rate is given by ∫ ∫ K Q= V dA = dA A A r but dA = 2y(x)dx and r = R + x, so that ∫ a √ 2 K a − x2 Q=2 dx R+x a Let z = x/a and λ = R/a, and then dx = adz and the equation becomes ∫ 1 √ 1 − z2 Q = 2aK dz −1 λ + z Since R > a it follows that λ > 1. It remains to evaluate the integral, which is an exercise in calculus. First write ∫ 1 √ 1 − z2 I= dz −1 λ + z as ∫ 1 I= −1 (√ 1 − z2 λ−z −√ λ+z 1 − z2 ∫ This is I = (1 − λ ) 1 2 −1 ) ∫ dz + −1 dz √ + (λ + z) 1 − z 2 or considering the two integrals separately ∫ 1 dz 2 √ I1 = (1 − λ ) 2 −1 (λ + z) 1 − z 133 1 ∫ λ−z √ dz 1 − z2 1 −1 ∫ I2 = λ−z √ dz 1 − z2 1 −1 λ−z √ dz 1 − z2 In evaluating I1 let z = tanh u so that dz = du/ cosh2 u and ∫ ∫ du du 2 2 I1 = (1 − λ ) = (1 − λ ) 2 λ cosh u + sinh u cosh u(λ + tanh u) But λ ( + −u ) 1 ( − −u ) λ + 1 u λ − 1 −u e e + e e = e + e 2 2 2 2 λ cosh u + sinh u = Thus ∫ du + 2(1 − λ2 ) u (λ + 1)e + (λ − 1)e−u I1 = 2(1 − λ ) 2 Next let v = eu and dv = eu du, then ∫ eu du (λ + 1)e2u + λ − 1 ∫ dv (λ + 1)v 2 + λ − 1 √ √ √ Now write λ + 1v = λ − 1 tan θ, so thatdv = (λ − 1)/(λ + 1) tan θdθ. Hence (√ ) ∫ √ √ λ + 1 dθ I1 = −2 λ2 − 1 = −2 λ2 − 1 tan−1 eu cos2 θ(tan2 θ + 1) λ−1 I1 = 2(1 − λ ) 2 Now from z = tanh u = (eu − e−u )/(eu + e−u ) it follows that ] √ 1+z u e = 1−z and √ I1 = −2 λ2 − 1 tan−1 (√ so that λ+1 λ−1 √ 1+z 1−z ) (π ) √ |1−1 = −2 λ2 − 1 −0 2 √ I1 = −π λ2 − 1 The second integral ∫ I2 = 1 −1 λ−z √ dz = 1 − z2 134 ∫ λ sin θ cos θdθ cos θ or ∫ I2 = √ π π (λ−sin θ)dθ = λθ−cos θ = z sin−1 z|1−1 + 1 − z 2 |1−1 = λ −λ(− ) = λπ 2 2 Hence I = π(λ − and K= √ λ2 − 1 Q √ 2πR(λ − λ2 − 1) 135 Chapter 9 Exercise 9.1 Combustion gases with γ = 4/3 and cp = 1148 J/kg K, at T01 = 1050 K and p01 = 310 kPa enter a radial inflow turbine. At the exit of the stator M2 = 0.9. As the flow leaves the turbine it is diffused to atmospheric pressure at p4 = 101.325 kPa. The total-to-total efficiency of the turbine is ηtt = 0.89. Find the stator exit angle. Given: T01 , p01 , M2 , p4 , ηts . Find: α2 Solution: Since T02 = T01 , the temperature at the inlet the the rotor can be obtained from γ − 1 2 −1 1050 T2 = T 02(1 + M2 ) = = 925.1 K 2 1 + 0.81/6 and the velocity is V2 = M2 √ √ γRT2 = 0.8 1.333 · 287 · 925.1 = 535.49 m/s From T01 − T03 T01 − T4s The stagnation temperature at the exit is [ ( ( )(γ−1)/γ )] p4 T03 = T01 1 − ηts 1 − p01 ))] [ ( ( 101.325 = 822.1 K T03 = 1050 1 − 0.89 1 − 310 The word delivered is ηts = w = cp (T01 − T03 ) = 1148(1050 − 822.1) = 261.65 kJ/kg Since w = U 2 , the blade speed is √ √ U2 = w = 261, 640 = 511.5 m/s and the flow angle is α2 = sin ( −1 U2 V2 ) ( −1 = sin 136 511.5 535.49 ) = 72.8◦ ⇐ Exercise 9.2 During a test air runs through a radial inflow turbine at the rate of ṁ = 0.323 kg/s when the shaft speed is 55, 000 rpm. The inlet stagnation temperature is T01 = 1000 K and the pressure ratio is p01 /p3 = 2.1. The blade radius at the inlet is r2 = 6.35 cm. The relative velocity entering the blade is radial and the flow leaves the blade without swirl. Find, (a) the spouting velocity, (b) the total-to-static efficiency, and (c) the power delivered. Given: T01 , p01 /p3 , ṁ, Ω, and r2 . Find: V0 , ηts and Ẇ . Solution: The blase speed is U2 = r2 Ω = 6.35 · 55, 000 · π = 365.73 m/s 100 · 30 [ and ( ws = cp (T01 − T3ss ) = cp T01 1 − [ so that ( V0 = 2cp T01 V0 = ( 1− p3 p01 p3 p01 )(γ−1)/γ ] 1 = V02 2 )(γ−1)/γ )] √ 2 · 1148 · 980 (1 − (1/2.1)0 .35) = 617.20 m/s ⇐ (a) and the work and isentropic work delivered are w = U22 = 365.752 = 133.7 kJ/kg so that ηts = 1 617.22 ws = V02 = = 190.5 kJ/kg 2 2 w 133.7 = = 0.702 ws 90.5 ⇐ (b) and the power delivered is Ẇ = ṁw = 0.35 · 133.7 = 46.8 kW ⇐ (c) Exercise 9.3 A radial turbine delivers Ẇ = 80 kW as its shaft turns at 44, 000 rpm. Combustion gases with γ = 4/3 and cp = 1148 J/kg K enter the rotor with relative velocity radially inward at radius r2 = 8.10 cm. At the exit the shroud radius is r3s = 6.00 cm and at this location M3Rs = 0.59. The exit pressure is 137 p3 = 101.325 kPa and exit temperature is T3 = 650 K. Find the hub to shroud ratio κ = r3h /r3s at the exit. Given: T3 , p3 , Ẇ , Ω, r2 , and r3s . Find: b3 Solution: The blade speed is 0.081 · 44, 000 · π = 373.22 m/s 30 0.060 · 44, 000 · π U3s = r3s Ω = = 276.46 m/s 30 and the relative velocity at the shroud is √ √ W3s = M3s γRT3 = 0.59 1.333 · 287 · 650 = 294.25 m, /s U2 = r2 Ω = The absolute velocity at the exit is √ √ 2 2 V3 = W3s − U3s = 294.252 − 276.462 = 100.8 m/s The specific work is w = U22 = 373.222 = 139.3 kJ/kg and the mass flow rate is therefore Ẇ 80, 000 = = 0.574 kg/s w 139.3 The static density at the exit is ṁ = ρ3 = p3 101325 = = 0.543 kg/m3 RT3 287 · 650 and the exit area is A3 = ṁ 0.574 = = 0.01049 m2 ρ3 V 3 0.543 · 100.8 √ so that r3h = 2 r3s − A3 = 0.0161 m π and the radius ratio is κ= 0.0161 r3h = = 0.268 r3s 0.060 138 ⇐ Exercise 9.4 A radial inflow turbine rotor, with rotor inlet radius r2 = 9.3 cm and blade height b2 = 1.8 cm, turns at 42, 000 rpm. Its working fluid is a gas mixture with cp = 1148 J/kg K and γ = 4/3. The exhaust pressure is p3 = 101.325 kPa and the total-to-static efficiency is ηts = 0.82. The nozzle (stator) angle is α2 = 67◦ and the velocity coefficient for the flow through the stator is cv = 0.96 and the Mach number at the inlet to the rotor is M2 = 0.9. Find, (a) the inlet stagnation pressure to the stator, and (b) the stagnation pressure loss across the stator. Given: r2 , b2 , Ω, p3 , α2 , and ηts . Find: T01 and ∆p0LS . Solution: The blade speed is U2 = r2 Ω = 9.3 · 42, 000 · π = 409.03 m/s 100 · 30 and the absolute velocity is V2 = U2 409.03 = = 444.36 m/s sin α2 sin(67◦ ) The radial component is Vr2 = Wr2 = V2 cos α2 = 444.36 cos(67◦ ) = 173.62 m/s Since the M2 is known, the inlet static temperature is T2 = V22 444.362 = = 637.0 K γRT M22 1.333 · 287 · 0.92 and the stagnation temperature T02 = T01 is T02 = T2 (1 + γ−1 2 0.81 M2 ) = 637(1 + ) = 723.0 K 2 6 The velocity for isentropic flow is V2s = V2 444.36 = = 462.88 m/s cv 0.96 and the static temperature for the isentropic process is T2s = T02 − V2s2 462.882 = 637 − = 629.7 K 2cp 2 · 1148 139 The specific work is w = U 2 = 409.032 = 167.31 kJ/kg The exit stagnation temperature is therefore T03 = T02 − w 167, 310 = 723 − = 577.3 K cp 1148 From the total-to-static efficiency ηts = T01 − T03 T01 − T3ss the static temperature T3ss is T3ss = T01 − 1 1 (T01 − T03 ) = 723 − (723 − 573.3)545.3 K ηts 0.822 The stagnation pressure at the inlet is therefore )γ/(γ−1) ( )4 ( 723 T01 p01 = p3 = 101.325 = 313.25 kPa T3ss 545.3 ⇐ (a) The stagnation pressure p02 is obtained from ( )4 ( )γ/(γ−1) 629.7 T2s p02 = p01 = 313.2 = 299.1 kPa T2 637 and the stagnation pressure loss is ∆p0LS = p01 − p02 = 313.2 − 299.1 = 14.1 kPa ⇐ (b) Exercise 9.5 Gas with γ = 4/3 and cp = 1148 K/kg K flows in a radial inflow turbine, in which the inlet stagnation temperature is T01 = 980 K and the inlet stagnation pressure is p01 = 205.00 kPa. The exit pressure is p3 = 101.325 kPa and the exit temperature is T3 = 831.5 K. The stagnation temperature at the exit is T03 = 836.7 K. The pressure at the inlet to the rotor is T2 = 901.6 K and the pressure is p2 = 142.340 kPa. The shaft speed is 160,000 rpm and the radius ratio is r3 /r2 = 0.57. Assume that the relative velocity is radial at the inlet and that there is no exit swirl. Find, (a) the total-to-static efficiency, (b) the flow angles α2 and β3 , and (c) ηS and ηR . 140 Given: T01 , p01 , p3 , T3 , p2 , T2 , Ω, r3 /r2 . Find: ηts , β3 , ηN , and ηR . Solution: The velocity at the inlet to the rotor is √ √ V2 = 2cp (T02 − T2 ) = 2 · 1148(980 − 901.6) = 424.3 m/s and the Mach number is M2 = √ 424.3 V2 =√ = 0.722 γRT2 1.333 · 287 · 901.6 The stagnation pressure at the inlet to the rotor is ( )γ/(γ−1) ( )4 γ−1 2 0./722 p02 = p2 1 + M2 = 142, 340 1 + = 198.69 kPa 2 6 The temperature T2s can be calculated from ( )(γ−1)/γ ( )0.25 p02 198.69 T2s = T2 = 901.6 = 894.6 K p01 205.00 The static enthalpy loss coefficient for the stator is ζS = 2cp (T2 − T2s ) 2 · 1148(901.6 − 894.6) = = 0.0895 ⇐ (c) 2 V2 424.32 and the total-to-static efficiency is ηts = 1 − T03 /T01 1 − 836.7/908 = = 0.905 (γ−1)/γ 1 − (p3 /p01 ) 1 − (101.325/205.00)0.25 ⇐ The specific work delivered is w = cp (T01 − T03 ) = 1148(980 − 836.7) = 164.51 kJ/kg The blade speed is U2 = √ w= √ 164, 510 = 405.6 m/s The flow angle at the inlet to the rotor is ( ) ( ) U2 405.6 −1 −1 α2 = sin = sin = 72.94◦ V2 421.90 141 ⇐ (b) (a) The blade velocity at the exit at the mean radius is U3 = U2 r3 = 405.6 · 0.57 = 231.2 m/s r2 The velocity at the exit is √ √ V3 = 2cp (T03 − T3 ) = 2 · 1148(836.7 − 831.5) = 109.2 m/s and the flow angle of the relative velocity is ( ) ( ) U3 231.2 −1 −1 = − tan = −64.7◦ β3 = − tan V3 109.2 ⇐ (c) The calculation of the static enthalpy loss coefficient of the rotor flow is carried out by ] [ ( )2 r3 2 sin2 β3 1 1 ζS Tr ζR = −1− − ηts 2 tan2 β3 r2 2 sin2 α2 (r3 /r2 )2 [ ] 1 0.572 0.0895 2 sin2 (64.72 ) = −1− − = 0.0974 0.905 2 tan2 (64.72 ) 2 sin2 (73◦ ) 0.325 The temperature ratio Tr = T2s /T2 was taken to be unity in the previous forgoing calculation. It can now determined to be [( ) ] 2 r3 T2s γ−1 2 2 1 + ζS 1 =1= M2 sin α2 = 0.919 −1 +1− 2 T2 2 r2 tan2 α2 sin β3 and when this is used for the value of Tr the static enthalpy loss coefficient for the rotor can be recalculated. The result is ζR = 0.117 ⇐ (c) Exercise 9.6 For an exit with no swirl show that √ W3 (r) sin(α2 − β2 ) r3 r2 = + cot2 β3 2 W2 cos α2 r2 r3 in which r is the radius at an arbitrary location of the exit plane of the blade and r3 and β3 are the mean values of the exit radius and angle. Show further that at the mean radius sin(α2 − β2 ) r3 W3 =− W2 cos α2 sin β3 r2 142 and plot the angle β3 for the range 0.53 < r3 /r2 < 0.65 when W3 /W2 = 2 and α2 = 70◦ and β2 = −40◦ . Given: A range of r3 /r2 , and W3 /W2 = 2 as well as α2 = 70c irc and β2 = −40◦ . Find: β3 Solution: From the exit velocity diagram W32 (r) = U32 (r) + V32 and U3 (r) = U3 r/r3 in which U3 is the blade speed at the mean radius r3 , this expression can be written as W32 (r) = U32 ( or W32 (r) = U32 r2 + cot2 β3 U32 r32 r2 + cos2 β3 r32 ) = r2 U22 32 r2 ( r2 + cos2 β3 r32 The tangential component of the velocity give U2 = V2 sin α2 − W2 sin β2 and dividing by W2 leads to U2 V2 = sin α2 − sin β2 W2 W2 From the radial components V2 cos α2 = W2 cos β2 the equation f racV2 W2 = is obtained. Thus or cos β2 cos α2 U2 = tan α2 cos β2 − sin β2 W2 U2 sin α2 cos β2 − cos α − 2 sin β2 = W2 cos α2 143 ) which is also sin(α2 − β2 U2 = W2 cos α2 Therefore sin(α2 − β2 ) r3 W3 (r) = W2 cos α2 r2 At the shroud W3s sin(α2 − β2 ) r3 = W2 cos α2 r2 and at the hub W3h sin(α2 − β2 ) r3 = W2 cos α2 r2 √ so that W3s = W3h √ r2 + cot2 β3 r32 √ r2 + cot2 β3 2 r3s √ r2 + cot2 β3 2 r3h 2 r3s + r32 cos2 β3 2 r3h + r32 cot2 β3 at the mean radius sin(α2 − β2 ) r3 √ W3 = 1 + cot2 β3 W2 cos α2 r2 or W3 sin(α2 − β2 ) r3 =− W2 cos α2 sin β3 r2 For β2 − 40◦ and α2 = 70◦ the relationship − r3 sin(110◦ ) sin β3 = = 1.373739 r2 2 cos(70◦ ) so that sin β3 = −1.373739 r3 r2 and the results are shown in the table below. r3 /r2 β3 0.53 −46.7◦ 0.56 −50.3◦ 0.59 −54.1◦ 144 0.62 −58.4◦ 0.65 −63.2◦ Exercise 9.7 An inexpensive radial inflow turbine has flat radial blades both at the inlet and the exit of the rotor. The shaft speed is 20,000 rpm. The radius of the inlet to the rotor is 10 cm and the mean radius at the exit is 6 cm. The ratio of blade widths is b3 /b2 = 1.8. The inlet stagnation temperature is T01 = 420 K and the exhaust flows into the atmospheric pressure 101.325 kPa. Assume that the gases which flow through the turbine have γ = 4/3 and cp = 1148 J/kg K. If the power delivered by the turbine is 10 kW, find (a) the mass flow rate, (b) the static temperature at the exit of the stator, (c) the static temperature at the exit of the turbine, (d) the blade height at the inlet and the exit of the turbine, (e) the total-to-total efficiency, and the total-to-static efficiency. Given: r3 , r2 , α2 , cN , ζR , T01 , Ω, p3 , and b3 /b2 . Find: ṁ, T3 , p01 , b3 , b2 , ηtt and ηts . Solution: The blade speeds are U2 = r2 Ω = and U3 = U2 0.1 · 20, 000 · π = 209.4 m/s 30 r3 0.06 = 209.4 = 125.7 m/s r2 0.1 The specific work delivered is w = U2 − U3 = 209.42 − 125.72 = 28.07 kJ/kg and the mass flow rate is therefore ṁ = Ẇ 10, 000 = = 0.356 kg/s w 28, 070 ⇐ (a) The inlet velocity to the rotor is V2 = U2 209.4 = = 216.8 m/s sin α2 sin(75◦ ) and the radial component of the velocity is Vr2 = V2 cos α2 = 216.8 cos(75◦ ) = 56.12 m/s W2 = 56.12 m/s The static temperature leaving the stator is T2 = T02 − 216.82 V22 = 420 − = 399.5 K 2cp 2 · 1148 145 ⇐ (b) The isentropic velocity is V2 216.8 = = 223.5 m/s cN 0.97 V2s = and the isentropic static temperature is therefore T2s = T02 − 223.52 V2s2 = 420 − = 398.2 K 2cp 2 · 1148 The ratio of stagnation pressures at the inlet and exit of the stator are therefore p01 = p02 ( T2 T2s )γ/(γ−1) ( = 399.5 398.2 )4 = 1.013 The stagnation temperature after the rotor is T03 = T02 − w 28.07 = 420 − = 395.5 K cp 1.148 Mass balance gives ρ2 r2 b2 W2 = ρ3 r3 b3 W3 This equation contains three unknowns ρ2 , ρ3 and W3 . One may proceed by trial by assuming a trial value for the flow angle at the exit. As the first attempt the value α3 = 50◦ is chosen. Then V3 = U3 125.7 = = 164.0 m/s sin α3 sin(50◦ ) and the axial velocity component at the exit is Vx3 = V3 cos α3 = 164.0 cos(50◦ ) = 105.4 m/s W3 = 105.4 m/s The static temperature is T3 = T03 − V32 164.02 = 395.5 − = 383.3 K 2cp 2 · 1148 Next the temperature for an isentropic expansion is calculated from T3s = T3 − ζR 105.42 W32 = 383.8 − 0.5 = 381.4 K 2cp 2 · 1148 146 The static pressure at the inlet can now be obtained from ( )γ/(γ−1) ( )4 T2 399.5 p 2 = p3 = 122.0 kPa = 101.325 T3s 381.4 The densities can now be calculated. They are ρ2 = p2 122, 000 = = 1.064 kg/m3 RT2 287 · 399.5 and p3 101, 325 = = 0.920 kg/m3 RT3 287 · 383.8 The value of density ρ̄3 may now be calculated from ρ3 = ρ̄3 = ρ2 0.1 · 56.12 r2 W2 b2 = 0.920 = 0.524 kg/m3 r3 W3 b3 0.06 · 105.4 · 1.8 This is in error by the amount r = ρ3 − ρ̄3 = 0.396 kg/m3 . By writing a Matlab scrip to carry out the calculations, and changing the trial value of α3 , the error is made nearly zero when α3 = 65.414◦ . The calculations with this value give V3 = U3 125.7 = = 138.2 m/s sin α3 sin(65.414◦ ) and the axial velocity component at the exit is Vx3 = V3 cos α3 = 138.2 cos(65.414◦ ) = 57.5 m/s W3 = 57.5 m/s The static temperature is T3 = T03 − V32 138.22 = 395.5 − = 387.2 K 2cp 2 · 1148 ⇐ (c) Next the temperature for an isentropic expansion is calculated from T3s = T3 − ζR W32 57.52 = 387.2 − 0.5 = 386.5 K 2cp 2 · 1148 The static pressure at the inlet can now be obtained from ( ( )γ/(γ−1) )4 399.5 T2 = 101.325 p2 = p 3 ) = 115.68 kPa T3s 386.5 147 The densities can now be calculated. They are ρ2 = p2 115, 680 = = 1.0088 kg/m3 RT2 287 · 399.5 and p3 101, 325 = = 0.9117, kg/m3 RT3 287 · 387.2 The value of density ρ̄3 may now be calculated from ρ3 = ρ̄3 = ρ2 0.1 · 56.12 r2 W2 b2 = 0.9117 = 0.9117 kg/m3 r3 W3 b3 0.06 · 57.50 · 1.8 and the error is now only e = −0.000014 kg/m3 . The static temperature )0.25 ( )(γ−1)/γ ( p02 1 = 385.3 K T3ss = T3s = 386.5 p01 1.013 and the stagnation pressure at the inlet to the stator is )γ/(γ−1) ( )4 ( 420 T01 p01 = p3 = 101.325 = 143.11 kPa T3ss 385.26 The blade widths can be now determined. They are b2 = ṁ 0.3562 = = 0.010 m 2πr2 ρ2 W2 2 · π · 0.10 · 1.0088 · 56.12 ⇐ (d) b3 = 0.3562 ṁ = = 0.018 m 2πr3 ρ3 W3 2 · π · 0.06 · 0.9117 · 57.50 ⇐ (d) and The total-to-static efficiency is ηts = T01 − T03 420 − 395.5 = = 0.704 T01 − T3ss 420 − 385.3 ⇐ (e) To determine the total-to-total efficiency, the stagnation temperature T03ss is needed. It is ( ( )(γ−1)/γ )0.25 110.320 p03 = 385.3 = 393.5 K T03ss = T3ss p3 101.325 so that 420 − 395.5 T01 − T03 = = 0.924 ⇐ (d) ηtt = T01 − T03ss 420 − 393.5 148 Exercise 9.8 Combustion gases with γ = 4/3 and cp = 1148 J/kg K enter a stator of a radial flow turbine with T01 = 1150 K, p01 = 1300 kPa, and M1 = 0.5, and with a flow rate of ṁ = 5.2 kg/s. The radius of the inlet is r1 = 17.4 cm, the exit from the stator is at r2 = 15.8 cm and the inlet to the rotor is at r2 = 15.2 cm. The chord of the stator is ct = 4.8 cm and the width of the channel is b = 1.6 cm. The rotational speed of the rotor is 31, 000 rpm. The exit static pressure is p3 = 320 kPa. The trailing edge thickness of the 17 stator vanes can be ignored. Find, (a) the total-to-static efficiency of the turbine, (b) the stagnation pressure loss across the stator, and (c) the stagnation pressure loss across the gap. Given: T3 , p3 , Ẇ , Ω, r2 , and r3s . Find: b3 Solution: : (a) 0.811, (b) 10.6 kPa, (c) 3.67 kPa. Exercise 9.9 Combustion gases enter the stator of a radial inflow turbine at the stagnation pressure p01 = 346 kPa and stagnation temperature T01 = 980 K. They enter the rotor at the speed V2 = 481.4 m/s with the relative flow making an angle β2 = −35◦ and exhaust into the atmosphere at 101.3325 kPa. The total-to-static efficiency of the turbine is ηts = 0.83. Find, (a) the angle at which the flow enters the rotor and (b) the relative Mach number at the inlet. Given: T3 , p3 , Ẇ , Ω, r2 , and r3s . Find: b3 Solution: : α2 = 61.41◦ ,(b) M2R = 0.67. 149 Chapter 10 Exercise 10.1 A Pelton wheel operates from an effective head of He = 300 m and at a flow rate of 4.2 m3 /s. The wheel radius is r2 = 0.75 m and its rotational speed is 450 rpm. The water which leaves the penstock is divided into 5 streams. The nozzle coefficient is cN = 0.98 for each of the nozzles. Impulse blades turn the flow into the direction β3 = −65◦ and as a result of friction the relative velocity reduces by an amount which gives a velocity coefficient cv = 0.90. Find, (a) the efficiency of the turbine, (b) the power specific speed, and (c) the nozzle diameter and the number of buckets in the wheel. Given: He , Q, Ω, r2 , cN , cv , β3 Find: ηts , Ωsp , d, Z Solution: The exit velocity from the nozzle is √ √ V2 = cN 2gHe = 0.98 2 · 9.81 · 300 = 75.19 m/s and the blade speed is U = r2 Ω = 0.75 · 450 · π = 35.34 m/s 30 The relative velocity entering the wheel is W2 = V2 − U = 75.19 − 35.34 = 39.85 m/s and the relative velocity at the exit is W3 = cv W2 = 0.90 · 39.85 = 35.5, m/s The tangential component of the relative velocity at the exit is Vu3 = W3 sin β3 = 35.86 sin(−65◦ ) = −32.50 m/s The tangential component of the absolute velocity at the exit is Vu3 = U + Wu3 = 35.34 − 32.50 = 2.83 m/s and the specific work is w = U (Vu2 − Vu3 ) = 35.34(39.85 − 2.83) = 2556.8 J/kg 150 or w = U (Vu2 − U )(1 − cv sin β3 ) = 2556.8J/kg The power is therefore Ẇ = ρQw = 998 · 4.2 · 255.8 = 10.7 MW and the efficiency is w Ẇ = = 0.869 ρQgHe ηHe η= ⇐ (a) The specific speed is √ √ Ω Q 450 · π 4.2 Ωs = = = 0.24 (gHe )3/4 30(9.81 · 400)3/4 and the power specific speed is √ √ Ωsp = ηΩs = 0.869 0.24 = 0.22 ⇐ (b) The flow rate in each jet is Qj = Q 4.2 = = 0.84 m3 /s N 5 and the jet area is Aj = Qj 0.84 = = 0.0111 m2 V2 75.19 so that the jet diameter is √ d= 4Aj = π √ 4 · 0.0111 = 0.119 m π The ratio of the wheel diameter to the jet diameter is 2 · 0.75 D = = 12.6 d 0.119 ⇐ (c) and the number of buckets is Z= D + 15 = 21.3 2d 151 Z = 21 ⇐ (c) Exercise 10.2 The flow rate through a small Francis turbine is 4.5 m3 /s, its head is 150 m and the the rotational speed is 450 rpm. The inlet radius is r2 = 0.6 m and the water leaves the guide vanes at the angle α2 = 72◦ and velocity V2 = 53.3 m/s. It leaves the turbine without swirl. (a) Find the velocity coefficient of the stator (inlet spiral and gates). (b) Find the inlet angle of the relative velocity, β2 . (c) What is the output power? (d) What is the torque on the shaft? (d) Determine the power specific speed and comment if the runner shape in the figure is appropriate. Given: He , Q, V2 , r2 , α2 , cv , β3 Find: cN , β2 Ẇ , torque, Ωs , Ωsp . Solution: The nozzle coefficient is cN = √ V2 53.7 =√ = 0.9825 m/s 2gHe 2 · 9.81 · 150 ⇐ (a) The tangential component of the velocity is Vu2 = V2 sin α2 = 53.3 sin(72◦ ) = 50.69 m/s and the radial component is Wr2 = V2 cos α2 = 53.3 cos(72◦ ) = 16.44 m/s The blade speed is U2 = r2 Ω = 0.6 · 450 · π = 28.27 m/s 30 and the tangential component of the relative velocity is Wu2 = Vu2 − U2 = 50.69 − 28.27 = 22.42 ms The angle of the relative velocity is therefore β2 = Wu2 22.42 = = 54.6◦ Wr2 16.44 ⇐ (b) The specific work is w = U2 Vu2 = 28.27 · 50.69 = 1433.2 J/kg and the power is Ẇ = ρQw = 998 · 4.5 · 1433.3 = 6437 kW 152 ⇐ (c) and the efficiency is η= w 1433.3 = = 0.974 gHe 9.81 · 150 The shaft torque is T = Ẇ 6, 437, 000 · 30 = = 136.6 kN Ω 450 · π ⇐ The specific speed and power specific speed are √ √ 450 · π 4.5 Ω Q = = 0.421 Ωs = (gHe )3/4 30(9.81 · 150)3/4 and Ωsp = √ ηΩs = √ 0.9740.421 = 0.415 ⇐ (d) ⇐ (e) (e) Exercise 10.3 The pressure at the entrance of a Francis turbine runner is 189.5 kPa and at the exit it is 22.6 kPa. The shaft turns at 210 rpm. At the exit the flow leaves without swirl. The inlet radius is r2 = 910 mm, and the exit radius is r3 = 760 mm. The relative velocity entering the runner is W2 = 10.2 m/s, and the flow angle of the relative velocity leaving the runner is β3 = −72◦ . The blade height at the inlet is b2 = 600 mm. (a) Compute the stagnation pressure loss in the runner. (b) Find the power delivered by the turbine. Given: p2 , p3 , Ω, r2 , r3 , β3 , W2 . Find: dp0LR and Ẇ . Solution: The blade speed at the inlet is U2 = r2 Ω = 0.91 · 210 · π = 20.0 m/s 30 and at the exit it is U3 = U2 0.76 r3 = 20.0 = 16.7 m/s r2 0.91 The radial velocity at the exit is Vr3 = − 16.7 U3 =− = 5.43 m/s tan β3 tan(−72◦ ) 153 The mass balance gives Vr2 = r3 0.76 5.43 = 4.54 m/s Vr3 = r2 0.91 and thus the relative flow angle into the rotor is ) ) ( ( Vr2 5.45 −1 −1 β2 = cos = cos = 63.6◦ W2 10.2 and the tangential component of the velocity is Vu2 = U2 + Vr2 tan β2 = 20.0 + 4.54 tan(63.6◦ ) = 29.15 m/s The specific work is w = U2 Vu2 = 20/0 · 29.15 = 583.31 J/kg The isentropic work is ws = (p02 − p03 )/ρ. The stagnation pressure at the inlet is p02 = p2 + ρ V22 29.52 = 189, 600 + 998 = 623.8 kPa 2 2 and at the exit it is p03 = P3 + ρ V32 5.432 = 226, 000 + 998 = 37.2 kPa 2 2 and 623, 800 − 373, 000 p02 − p03 = = 587.68 J/kg ρ 998 and the stagnation pressure loss is ws = dp0LR = ρ(ws − w) = 998(587.68 − 583.31) = 4.36 kPa ⇐ The flow rate is Q = 2πr2 b2 Vr2 = 2π · 0.19 · 0.6 · 4.54 = 15.56 m3 /kg and the power delivered is Ẇ = ṁw = 998 · 15.56 · 583.3 = 9.06 MW 154 ⇐ (b) (a) Exercise 10.4 A Francis turbine has an inlet radius r2 = 1450 mm and outlet radius r3 = 1220 mm. The blade width is constant b = 370 mm. The shaft speed is 360 rpm and the volumetric flow rate is Q = 16.7 m3 /s. The flow enters the runner at α2 = 78◦ . Water leaves the the turbine without swirl and the outlet pressure p3 = 35 kPa. The loss through the runner is 0.20W23 /2g m. Find the pressure p2 at the inlet and the head loss through the runner. Given: r2 , r3 , b, Q, α2 , p3 and given the head loss across the runner as 0.02W32 /2g. Find: p2 . Solution: The blade speed is U2 = r2 Ω = 1.45 · 360 · π = 54.66 m/s 30 The flow area at the inlet to the runner is A2 = 2πr2 b = 2π · 1.11 · 0.37 = 3.37 m2 The radial component of the velocity is Vr2 = Q 16.7 = = 4.95 m/s A2 3.37 and the tangential component is Vu2 = Vr2 tan α2 = 4.95 tan(78◦ ) = 23.31 m/s and the absolute velocity is √ √ 2 +V2 = V2 = Vrw 4.952 + 23.312 = 23.83 m/s′ u2 Hence the specific work is w = U2 Vu2 = 54.66 · 23.31 = 1374 J/kg The blade speed at the exit is U3 = U2 r3 1.22 = 54.66 = 46.00 m/s r2 1.45 The exit area is A3 = 2πrr b = 2π · 1.22 · 0.37 = 2.384 m2 155 and the relative velocity at the exit is √ √ W3 = U32 + Vr32 = 46.002 + 5.892 = 46.37 m/s Hence the head loss is HR = 0.2 W32 0.2 · 46.372 = = 21.92 m 2g 2 · 9.81 The stagnation pressure at the exit is p03 = p3 + ρ V32 5.892 = 35000 + 998 = 52.30 kPa 2 2 The stagnation pressure loss is ∆p0LR = ρgHr = 998 · 9.81 · 21.92 = 214.6 kPa so that the stagnation pressure at the inlet is p02 = p03 + ∆p0LR + ρw = 52.3 + 214.6 + 998 · 1274 = 1538.3 kPa 1000 and the static pressure is V22 998 · 23.832 p2 = p02 − ρ = 1, 538, 300 − = 1255, kPa 2 2 Exercise 10.5 The relative velocities at the inlet and the exit of a Francis turbine are W2 = 10.0 m/s and W3 = 33.7 m/s. The shaft speed is Ω = 200 rpm. The inlet radius of the runner is r2 = 1880 mm and its outlet radius is r3 = 1500 mm. The runner blade width is constant b = 855 mm. Find, (a) the flow rate through the turbine and (b) the torque assuming the flow leaves without exit swirl. Given: W2 , W3 , Ω, r2 , r3 , b. Find: Q and torque. Solution: The blade speeds are U2 = r2 Ω = and U3 = U2 1.88 · 200 · π = 39.37 m/s 30 r3 1.88 = 39.37 = 31.42 m/s r2 1.50 156 and the redial velocity at the exit is √ √ Vr3 = W32 − U32 = 33.72 − 31.422 = 12.20 m/s The flow rate is Q = 2πr3 bV r3 = 2π · 1.50 · 0.855 · 12.20 = 98.3 m3 /s ⇐ (a) and the radial velocity at the inlet is Vr2 = Vr3 r3 1.50 = 12.20 = 9.74 m/s r2 1.88 The angle of the relative flow at the inlet is ) ( ) ( Vr2 9.734 −1 −1 β2 = cos = cos = 13.3◦ V2 10.0 Now the tangential component of the relative velocity is Wu2 = Wr2 tan β2 = 9.74 tan(1.3◦ ) = 2.31 m/s and the tangential component of the absolute velocity is therefore Vu2 = U2 + Wu2 = 39.37 + 2.31 = 41.68 m/s and the specific work is w = U2 Vu2 = 39.37 · 41.68 = 1641 J/kg and the power delivered is Ẇ = ρQw = 998 · 98.3 · 1641 = 161 MW The torque is therefore T = Ẇ 161 · 106 · 30 = = 7, 685, 000 N m Ω 200 · π ⇐ (b) Exercise 10.6 A Francis runner is to be designed for effective head of He = 140 m and flow rate Q = 20 m3 /s. Assume that the efficiency is η = 0.9 and that there is no exit swirl. Use the formula of Lugaresi and Massa to calculate 157 the specific speed. Use other the formulas in the text to obtain br and U2 /V0 from Ωsp . Assume that the mechanical and volumetric losses are negligible. Find the specific diameter on this basis. (a) Find the diameter at the inlet. (b) Find the blade speed at the inlet. (c) Find the flow angles of the absolute and relative velocities at the inlet. Given: He , Q, η. Find: D2 , U2 , α2 and β2 . Solution: The power delivered is Ẇ = ηρQgHe = 0.9 · 998 · 20 · 9.81 · 140 = 24, 672 kW and the specific work is w= Ẇ 24, 670, 000 = = 1236 J/kg ρQ 998 · 20 The specific speed is ( Ωs = 1.14 and Ωsp = Hr He √ )0.512 ηΩs = ( = 1.14 √ 100 140 )0.512 = 0.9596 0.90 · 0.9596 = 0.9104 The shaft speed can now be determined from Ωs (gHe )3/4 0.9596 · 30 · (9.81 · 140)3/4 √ √ Ω= = = 462 rpm Q 20 · π The blade width to diameter ratio is br = 0.0505Ω2sp + 0.26Ωsp + 0.018 = 0.213 and the U2 /V0 ratio is U2 = 0.724 = 0.74Ω0.238 sp V0 Thus √ 8U2 Ds = = Ωs V0 √ 8 · 0.724 = 2.13 0.9596 158 Hence √ √ Ds Q 2.13 20 D2 = = = 1.567 m (gHe )1/4 (9.81 · 140)1/4 and b2 br D2 = 0.213 · 1.567 = 0.334 m and D2 1.567 · 462 · π Ω= = 37.93 m/s 2 2 · 30 Now the radial velocity at the inlet is U2 = Vr2 = Q 20 = = 12.12 m/s 2πr2 b 2π0.7835 · 0.335 and the tangential velocity component can be obtained from Vu2 = so that ( α2 = tan −1 Vr2 Vu2 w 1236.1 = = 32.59 m/s U2 37.93 ) ( −1 = tan 12.12 32.59 ) = 69.5◦ ⇐ (a) and the tangential component of the relative flow is Wu2 = Vu2 − U2 = 32.59 − 37.93 = −5.33 ms and it makes the angle ( ) ( ) Wr2 12.12 −1 −1 β2 = tan = tan = −23.6◦ Wu2 −5.33 ⇐ (b) Exercise 10.7 Water enters the runner of a Francis turbine with a relative velocity at angle −12◦ . The inlet radius is 2.29 m and the mean radius at the exit is 1.37 m. The rotational speed is 200 rpm. The blade height at the inlet is b2 = 1.22 m and at the exit the inclined width of the blade is b3 = 1.55 m. The radial velocity at the runner inlet is 10.0 m/s and the flow leaves the runner without swirl. Evaluate (a) the change in total enthalpy of the water across the runner, (b) the torque exerted by the water on the runner normalized to a metric tonne per second, (c) the power developed, (d) the flow rate of water, (e) the change in total pressure across the runner if the total-to-total efficiency is 95%, and the volumetric and mechanical 159 losses can be neglected. (f) What is the change in static pressure across the runner, and (h) the total-to-static efficiency. Given: r2 , r3m , Ω, β2 , b2 , η, Vr2 , b3 . Find: h02 − h03 , T , Ẇ , Q, p02 − p03 , p2 − p3 , ηts . Solution: The blade speeds are U2 = r2 Ω = 2.29 · 200 · π = 47.96 m/s 30 and the blade speed at the mean radius at the exit is U3m = U2 1.37 r3m = 47.96 = 28.69 m/s r2 2.29 The flow areas are A2 = 2πr2 b2 = 2π · 2.29 · 1.22 = 17.55 m/s A3 = 2πr3m b3 = 2π · 1.37 · 1.55 = 13.34 m/s and the flow rate is Q = Vr2 A2 = 10 · 17.55 = 175.5 m3 /s ⇐ (d) and the mass flow rate is ṁ = ρQ = 998 · 175.5 = 175.2 tonne/s The mean radial velocity leaving the runner is V3m = Q 175.54 = = 13.16 m/s A3 12.342 The angle of the relative flow at the exit is ) ( ( ) U3 28.69 −1 −1 β3 = − tan = − tan = −65.4◦ Wm3 13.16 The tangential component of the relative velocity at the inlet is Wu2 = Vr2 tan β2 = 10 tan(−12◦ ) = −2.12 m/s 160 and the corresponding value of the absolute velocity is Vu2 = U2 + Wu2 = 47.96 − 2.12 = 45.84 m/s so that V2 = √ √ 2 Vr22 + Vu2 = 102 + 45.842 = 46.91 m/s and the flow angle is ( −1 α2 = tan Vu2 Vr2 ) ( −1 = − tan 45.84 10.0 ) = 77.7◦ The work done is w = U2 Vu2 = 47.96 · 45.384 = 2198.4 J/kg so this is also the total enthalpy difference across the rotor h02 − h03 = 2198.4 J/kg ⇐ (a) The power delivered is Ẇ = ρQw = 998 · 175.54 · 2198.4 = 385.1 MW ⇐ (c) and the torque is therefore T = Ẇ 385, 100, 000 · 30 = = 18, 39, 000 N m Ω 200 · pi Hence per ton of water flow it is Ts = T 1, 839, 100 = = 105 kN m 175.2 ṁ ⇐ (b) The isentropic work is ηHe = w 2198.4 = = 2314.1 J/kg η 0.95 The total pressure loss is ∆p0LR = ρ(gHe − w) = 998(2314.1 − 2198.4) = 115.5 kPa 161 and the change in stagnation pressure is p02 − p03 = ρgHe − ∆p0LR = 998 · 2314.1 − 115.5 = 2194 kPa ⇐ 1000 (e) The change in static pressure is 998 1 2 ) = 2, 194, 000− p2 −p3 = p02 −p03 − ρ(V22 −V3m (46.912 −13.162 ) = 1182 kPa ⇐ 2 2 : Exercise 10.8 The Otari number 2 power plant in Japan delivers 89.5 MW of power when the flow rate is 207 m3 /s and the head is 48.1 m. The diameter of the Kaplan turbine is D2t = 5.1 m and hub-to-tip ratio κ = 0.56. The generator has 36 poles and delivers power at line frequency is 50 Hz. (a) Find the efficiency of the turbine. (b) Calculate and flow angles entering and leaving the rotor and construct a graph to show their variation across the span. Solution: (a) The efficiency is η= Ẇ0 89.5 · 106 = = 0.918 ρQgHe 998 · 207 · 9.81 · 48.1 (b) With a 50 Hz line frequency and 36 poles, the shaft speed is Ω = 120·50/36 = 166.67 rpm. The tip speed of the runner blade is D2t 2.55 · 166.67 · π Ω= = 44.5 m/s 2 30 The axial velocity is uniform and it is given by U2t = Vx2 = 4Q 4 · 207 = = 14.76 m/s 2 −κ ) π · 5.12 (1 − 0.562 ) 2 πD2t (1 Since each blade element delivers the same amount of work, the tangential velocity at the tip can be determined from Vu2t = 0.91 · 9.81 · 51 ηgHe = = 9.73 m/s U2t 39.9 The flow angles for the absolute and relative velocities can now be determined. The calculations, at the mean radius r2m = 1.99 m the absolute velocity makes an angle ( ) ( ) Vu2t rt 9.73 · 2.55 −1 −1 α2m = tan = tan = 40.2◦ Vx2 rm 14.76 · 1.99 162 (f ) The tangential component of the relative velocity at this location is Wu2m = Vx2 tan α2m − U2t rm 44.5 · 1.99 = −22.25 m/s = 14.76 tan(40.2◦ ) − rt 2.55 and the flow angle is ( −1 β2m = tan Wu2m Wx2 ) ( −1 = tan −22.25 14.76 ) = −56.4◦ The flow leaves the runner axially. Therefore the tangential component of the relative velocity is Wu3m = −U2t rm /rt and with the axial velocity constant, the flow angle is ) ( ) ( Wu3m −44.5 · 1.99 −1 −1 β3m = tan = tan = −67.0◦ Wx3 14.76 · 2.55 163 Chapter 11 Exercise 11.1 A fluid coupling operates with oil flowing in a closed circuit. The device consists of two elements, the primary and secondary, each making up onehalf of a torus, as shown in the figure in the text. The input power is 100 hp and input rotational speed is 1800 rpm. The output rotational speed is 1200 rpm. (a) Evaluate the efficiency of and output power of this device. (b) At what rate must energy as heat be transferred to the cooling system, to prevent a temperature rise of the oil in the coupling? Given: Ẇp , Ωp , Ωs . Find: η, Q̇ = Ẇp − Ẇs . Solution: Since the torque in the primary and the secondary are the same Ẇp = T Ωp and Ẇs = Ẇp Ẇp = T Ωp Ωs 1200 = 100 = 66.7 HP Ωp 1800 The efficiency is η= Ẇs 66.7 = = 0.667 100 Ẇp and the rate of heat transfer from the converter fluid to the hardware and the surroundings is Q̇ = Ẇp − Ẇs = 33.3 · 745.7 = 24.86 kW After the hardware has reached its steady state temperature all the heat is transferred to the surroundings. Exercise 11.2 (a) Carry out the algebraic details to show that the expression for the flow rate through a fluid coupling is given by √ √( )( ) Ω2s r12 Dh Q = AΩp r2 1− 2 1− 2 (1) fL Ωp r2 and if for a low values of slip the friction factor is related to the flow rate by an expression cµA c = f= Re ρQD 164 express the flow rate in terms of the slip for small values of s. (b) Carry out the algebraic details to show that the expression for the torque of a fluid coupling is given by √ √( )( )( ) Ω2s r12 r12 Ωs Dh 2 3 T = ρAΩp r2 1− 2 1− 2 1− 2 (2) fL ΩP r2 r2 Ωp (c) What is the appropriate form for this equation for low values of slip. Given: The form of the equation for the flow rate and torque. Find: Derive the equations. Solution: The details are shown in the text. For small values of slip 1− Ω2s Ωs Ωs )(1 + ) = s(2 − s) = (1 − 2 Ωp Ωp Ωp and with c cµA D ReD = so that = Re ρQD fL cL and the flow rate becomes √ √ ρD2 Q Q = AΩp r2 s(2 − s)(1 − r12 /r22 ) cµAL f= Squaring and solving for Q gives ( Q= AΩ2p r22 s(2 which can be reduces to Q= ( 2AΩ2p r22 s r2 − s) 1 − 12 r2 r12 1− 2 r2 ) ρD2 cµL ) ρD2 cµL ⇐ (b) or the flow rate is directly proportional to slip. The expression for torque coefficient is obtained from ) ( r12 ρD2 2 2 2 2 T = ρQ(Ωp r2 − Ωs r1 ) = ρQΩp r2 s(2 − s) 1 − 2 r2 cµL 165 and ( )( ) r12 r12 ρAD2 Ωp T CT = = s(2 − s) 1 − 2 1 − (1 − s) 2 ρΩ2p r25 r2 r2 cµLr2 and for a small s this becomes ( r2 CT = 2s 1 − 12 r2 )2 ρAD2 Ωp cµLr2 and this is also linear in s for small values. Exercise 11.3 A fluid coupling operates with an input power of 200 hp, 5% slip and a circulatory flow rate of 1500 l/s. (a) What is the rate at which energy as heat must be transferred from the coupling in order for its temperature remain constant? (b) What would be the temperature rise of the coupling over a period of one-half hour assuming that no heat is transferred form the device and that it has a mass of 45 kg, consisting of 70% metal with a specific heat 840 J/kg K, and 30% oil with a specific heat 2000 J/kg K. Given: Ẇp , s and a time duration of 30 minutes. Find: η, Q̇ = Ẇp − Ẇs . Solution: The output power is Ẇs = ˙˙Wp ΩS = (1 − s)Ẇp = 0.95 · 200 = 190 hp Ωp The rate at which heat is transferred out from the coupling oil is Q̇ = Ẇp − Ẇs = 10 · 745.7 = 7457 W ⇐ (a) The change in internal energy is U2 − U1 = Q̇∆t = 7457 · 30 · 60 = 13, 422, 600 J The change in internal energy is also U2 − U1 = ms cs (T2 − T1 ) + mf cf (T2 − T1 ) = (ms cs + mf cf )(T2 − T1 ) which can be written as U2 − U1 = m(fs cs + ff cf )(T2 − T1 ) so that T2 − T1 = 13, 422, 600 = 251 C 45(0.7 · 840 + 0.3 · 2000) 166 ⇐ (b) Exercise 11.4 In a fluid coupling the input and output shafts rotate at 2000 and 1800 rpm, respectively. The fluid is an oil having a specific gravity of 0.88 and viscosity 0.25 kg/m s. The outer mean radius of the torus is r2 = 15 cm the inner mean radius is r1 = 7.5 cm. The radial height is b = 2r2 /15. The axial flow area around the torus is the same as the flow area at the outer clearance between the primary and secondary rotors. Given that the relative roughness of the flow conduit is 0.01, find the volumetric flow rate and the axial velocity. Given: Ωp , Ωs , r2 , r1 , b, SG, µ, ϵ Find: Q Solution: The cross sectional area of the flow is [ ] A = π (r2 = b/2)2 − (r2 − b/2)2 = 2πr2 b and the circumference is C = 2π(r2 − b/2) + 2π(r1 + b/2) + 4πr2 so that the hydraulic diameter is D= The flow rate is 4A 4 · 2πr2 b = = 2b C 4πr2 √ Q = AΩp r2 Dh fL √( √ or Q= 2πr22 bΩp b 2πf Ω2 1 − 2s Ωp √( )( Ω2 1 − 2s Ωp r2 1 − 12 r2 )( With r1 = 7.5 cm r2 = 15 cm b= ) r2 1 − 12 r2 ) 2 R2 = 2 cm 15 and the flow area is A = 2πr2 b = 2π0.15 · 0.02 = 0.0118 m2 Assume that the flow is laminar and f = 64/Re. If Reynolds number is Re = 3200, then f = 64/3200 = 0.02 and Q = 0.3256 m3 /s 167 so that the axial velocity is Vx = Q 0.3256 = = 17.28 m/s A 0.0118 and the Reynolds number is Re = ρVx D 0.88 · 998 · 17.28 · 0.04 = = 2428 µ 0.25 And the friction factor is f = 64/Re = 0.0264. Iterating gives finally f = 0.0346 and the flow rate is Q = 0.2476 m3 /s = 247.6 liters/s ⇐ (a) and the axial flow rate is Vx = Q 0.02476 = = 13.l1 m/s A 2π · 0.15 · 0.02 ⇐ (b) Exercise 11.5 Show that the kinetic energy loss model at the inlet to the turbine given by 1 2 r2 (Ωp − Ωs )2 2 is based on the conversion of the change in the one half of the tangential component of the velocity squared, irreversibly into internal energy. To show this, note that the incidence of the relative velocity at the inlet to the turbine is β2 since the blades are radial. This leads to a leading edge separation, after which the flow reattaches to the blade. After this reattachement the radial component of the relative velocity is the same as in the flow incident on the blade. Given: The velocity triangles Find: The loss Solution: Inspection of the velocity diagrams shows that Vus2 = Us2 + Wus2 ′ Vus2 = Us2 and therefore the expression ′ = Wus2 = Up2 − Us2 = r2 (Ωp − Ω) Vus2 − Vus2 so that 1 1 ′ )2 = r22 (Ωp − Ωs )2 (Vus2 − Vus2 2 2 168 Exercise 11.6 For a fluid coupling for which r1 /r2 = 0.7, find the value of the slip at which the power is maximum. Given: The expressions for power delivered by the secondary Find: For what value of the ratio ω = Ωs /Ωp is the power maximum. Solution: From ) ( √ Ẇs r12 2 = ω 1 − ω 1 − 2ω Tm Ωp r2 in which ω = ΩS /Ωp . Let w̄ = Ẇs /Tm Ωp , and k = r12 /r22 the following equation is obtained √ w̄ = ω 1 − ω 2 (1 − kω) Differentiating with respect to ω and setting the derivative to zero gives √ ω 1ω 2 (1 − 2kω) − √ (ω − kω 2 ) = 0 2 1−ω which simplifies to 2kω 3 − 2ω 2 − 2kω + 1 = 0 With k = 0.72 = 0.49 the correct root of this cubic equation gives ω = 0.6053 and the slip is s = 1 − ω = 1 − 0.6053 = 0.3947 ⇐ Exercise 11.7 A torque converter operates with oil flowing in a closed circuit. It consists of a torus is made of a pump, a turbine, and a stator. The input and output rotational speeds are 4000 rpm and 1200 rpm, respectively. At this operating condition the torque exerted on the stator is twice that exerted on the pump. Evaluate (a) the output to input torque ratio and (b) the efficiency. Given: Ωp , Ωs , and Tf = 2Tp . Find: Ts /Tp , Ẇs /Ẇp . Solution: From Tp = ṁ(r1 Vu2 − r1 Vu1 ) Ts = ṁ(r2 Vu2 − r3 Vu3 ) Tf = ṁ(r1 Vu1 − r3 Vu3 ) 169 it is seen that Ts = Tp + Tf . Therefore Ts = Tp + 2Tp = 2Tp and or Ts =3 Tp 3 · 1200 Ts Ωs Ẇs = = = 0.9 Tp Ωp 4000 Ẇp ⇐ ⇐ (a) (b) Exercise 11.8 A torque converter multiplies the torque by is designed to have to provide a torque multiplication ratio of 3.3 to 1. The circulating oil flow rate is 500 kg/s. The oil enters the fixed vanes in the axial direction at 10 m/s, and leaves at an angle 60 deg in the direction of the blade motion. The axial flow area is constant. Find the torque which the primary exerts on the fluid and the torque by the fluid on the blades of the secondary. The inlet and outlet radii of fixed vanes are 15 cm. Given: r2 , α1 = 0, α2 = 60c irc, Vx1 = Vx2 = 10 m/s, Ts = 3.3Tp , and ṁ = 500 kg/s Find: Tf , Tp , Ts . Solution: The tangential component of the velocity leaving the fixed member is Vu2 = Vx2 tan α2 = 10 tan(60◦ ) = 17.32 m/s and V u1 = 0. Thus Tf = ṁr2 Vu2 Tp = ṁ(r3 Vu3 − r2 Vu2 ) Tf = ṁr3 Vu3 From Tp + Tf = Ts Ts = 3.3Tp Tf = 2.3Tp and Tf = 500 · 0.15 · 17.32 = 1299 N m and thus Tp = 565 N m and Ts = 1864 N m. 170 ⇐ Exercise 11.9 Develop the equations for the torque ratio and efficiency for a torque converter given in the text. (a) At what ratio of the rotational speeds is the efficiency maximum. From this and the experimental curves shown in the text estimate (b) the ratio r2 /r3 , and the value of Qr1 tan α1 /Ar32 Ωp . Given: The graphs in the text. Find: r2 /r3 , and Qr1 tan α1 /Ar32 Ωp Solution: From Ts = ρQ(r22 Ωp − r32 Ωs ) and Tp = ρQ(r22 Ωp − r1 Vu1 ) Defining the ratios T = Ts Tp ω= Ωs Ωp R= r2 r3 a= r1Vu1 r32 Ωp the ratio of the first two equations can be written as T = The efficiency is η= R2 − ω R2 − a Ts Ωs ω(R2 − ω) = Tp Ωp R2 − a The graph shown that efficiency and torque ratio are zero when ω = R2 = 0.85. The maximum efficiency is obtained by differentiating dη R2 − 2ω = 2 =0 dω R −a so that ωmax = R2 /2 = 0.425 ⇐ (a). At this value the graph shows that T = 1.6. Then from the equation for the torque ratio may be written as (R2 − a)T = R2 − ω and solving this for a gives a = R2 − 0.85 − 0.425 R2 − ω = 0.85 − = 0.584 T 1.6 171 ⇐ (b) Chapter 12 Exercise 12.1 Reconsider the ducted windmill but now let the duct be a cylindrical control volume Ae in cross sectional area. Show that the axial force on the blades is Fd = ρAb Vb (V − Vb ) which is in agreement with the equation for wind turbine drag. Given: The flow variables and the control volume. Find: The drag force. Solution: From a cylindrical control volume ρAe V = ṁe + ρ(Ae − Ab ) + ρAb Vb in which Ae is the cross sectional area of the duct, Ab is the cross sectional area of the slipstream where the velocity is Vb and ṁe is the rate at which mass leaves the control volume across its lateral boundary. Solving this for ṁe gives ṁe = ρAb (V − Vb ) The x-momentum balance gives ρAb Vb2 + ρ(Ae − Ab )V 2 + ṁV − ρAe V 2 = −Fd which reduces to ρAb (V − Vb )[V − (V + Vb )] = −Fd so that Fd = ρAb Vb (V − Vb ) ⇐ Exercise 12.2 The non-dimensional pressure difference can be expressed as p + − pa 1 ρV 2 2 in which the pressure difference is between is that just before the disc and the free stream. (a) Use the momentum theory for a wind turbine to express this in terms of the interference factor a. For which value of a is this maximum? Interpret this physically. (b) Develop the expression for p a − p− 1 ρV 2 2 172 For which value of a is this the maximum? Given: The formulas for the pressure difference across the disk. Find: Express the pressure difference in terms of the interference factor a and for which value of a is the pressure difference the maximum. Solution: From p0+ p a 1 2 p+ 1 = + ρV = + (1 − a)V 2 ρ ρ 2 ρ 2 so that p+ − p a 1 = ρV 2 = 1 − (1 − a)2 = a(2 − a) ρ 2 Differentiating to obtain the maximum of this gives 2 − 2a = 0 a=1 which means that p+ is the stagnation pressure. Downstream p0− p− 1 pa 1 = + (1 − a)2 ρV 2 = + (1 − 2a)V 2 ρ ρ 2 ρ 2 Therefore p a − p− = (1 − a)2 − (1 − 2a)2 = a(2 − 3a) 1 2 ρV 2 Differentiating this with respect to a and setting the derivative to zero gives a= 1 3 This value of a corresponds to a maximum power from the turbine. Hence also maximum for the difference p+ − p− . or the lowest value of p− . Exercise 12.3 A wind turbine operates at wind speed of V = 12 m/s. Its blade radius is R = 20 m and its tip speed radius RΩ/V = 4. It operates at the condition Cp = 0.3. a. Find the rate of rotation of the blades. b. Find the power developed by the turbine. c. Find the value of the interference factor a based on the momentum theory. d. Find the pressure on the front of the actuator disc, if the free stream pressure is 101.30 kPa. Given: V , ΩR/V , Cp , and R. 173 Find: Ω, Ẇ , a, p+ . Solution: The shaft speed is Ω= 5V 5 · 12 · 30 = = 28.6 rpm R 20 · π From the expression for the power coefficient Cp = 4a(1 − a)2 and expanding the right hand side and setting it equal to 0.4 gives 4a3 − 8a2 + 4a − 0.4 = 0 This has the solutions a = (0.1330, 0.5874, 1.2749). The value in the range 0 < a < 1/2 is the appropriate one, or a = 0.1339 Lef tarrow (c). The power delivered is 1 1 Ẇ Cp AV 3 = 0.4 · 1.2 · π202 · 123 = 521 kW 2 2 ⇐ (b) The velocity at the actuator disk is Vd = (1 − a)V = 0.867 · 12 = 10.4 m/s ⇐ (c) The pressure upstream of the disk is 1 1 p+ = pa + V 2 − ρ(1 − a)2 − V 2 2 2 or 1 p+ = pa + ρa(2−a)V 2 = 101, 325+0.5·1.2·0.1330·1.867·122 = 101, 325+21 = 101, 346 Pa 2 Exercise 12.4 Using the axial momentum theory calculate the ratio of the slipstream radius to that of the disc radius in terms of the interference factor a. If the wind turbine blades are 80 m long, what is the radius of the slip stream far downstream. What is the radius of the streamtube far upstream. Given: a = 1/3, R. Find: Rb , R Ra . 174 ⇐ ( Solution: From continuity πRa2 V = πR2 (1 − a)2 V = πRb2 (1 − 2a)V it follows that Ra √ = 1−a R so that for a = 1/3 these give Ra √ = 2 R √ Rb 2 = R 3 Rb = R Ra = √ √ Ra = √ 1−a 1 − 2a 280 = 113 m 2 80 = 65.3 m 3 ⇐ ⇐ Exercise 12.5 In a wind with speed V = 8.7 m/s and air density ρ = 1.2 kg/m3 , a wind turbine operates at a condition with Cp = 0.31. Find the blade length, if the power delivered to the turbine is to be Ẇ = 250 kW. Given: V , Cp , and Ẇ . Find: R. Solution: From Cp = 4a(1 − a)2 with Cp = 0.31 the value of a can be obtained from 4a3 − 8a2 + 4a − 0.31 = 0 The roos are a = (1.2491, 0.6564, 0.0945). The only one in the the range of wind turbines is a = 0.0945 ⇐ (a). From 1 Ẇ = ρAV 3 2 the area is A= 2Ẇ 2 · 250, 000 = = 2041.1 m2 3 3 ρV Cp 1.2 · 8.7 · 0.31 so that the radius is R= A 2041.1 = = 25.5 m π π 175 Lef tarrow (b) Exercise 12.6 Consider a 3-bladed wind turbine with blade radius of R = 35 m and constant cord of c = 80 cm, which operates with rotational speed of Ω = 10 rpm. The wind speed is V = 12 m/s. Find the axial and tangential induction factors at r = 10 m assuming that angle of attack is 6◦ and CD = 0.01CL . Given: Z, R, r, c, V , and Ω. Find: a′ and a. Solution: Using the Matlab script on page 425 of the text gives a′ = 0.09390423 a = 0.06017493 ] Exercise 12.7 For the Example 12.4 calculate the force Fx and torque T by using the mean values at r = 0.6 and compare with the numerical solution given in the text. Given: Z, R, X, c, α, ϵ. Find: Fx and T . Solution: At r/R = 0.6 the values for a′ and a are a′ = 0.0055 and a = 0.0341, therefore Fx = 2a(1−a)πρV 2 R2 (1−0.22 ) = 2·0.046·0.954·π·1.2·122 ·402 ·(1−0.22 ) = 14, 638 N and the torque is T = πρΩV a′ (1 − a)R4 (1 − 0.24 ) = π · 1.2 · 10π · 354 · (1 − 0.24 ) = 23, 8670 N m 30 %HW 12.6 clear all; R=35; V=12; c=0.80; alphad=6; omega=10*pi/30; X=R*omega/V; alpha=alphad*pi/180; CL=2*pi*sin(alpha); Z=3; rho=1.2; dr=0.01; Fx=0; T=0; T2=0; r=[0.2:dr:1.00]*R; x=r*X/R; 176 n=length(r); kmax=20; fid=fopen(’span’,’w’); for i=1:n k=0; a(i) = 0; ap(i) = 0; temp=a(i); phi(i) = atan((1-a(i))/(x(i)*(1+ap(i)))); theta(i)=phi(i)-alpha; sigma(i)=Z*c/(2*pi*r(i)); sg(i)=sigma(i)/4; Cx(i)=CL*cos(phi(i)); Cy(i)=CL*sin(phi(i)); while k < kmax a(i)=sg(i)*Cx(i)/(sin(phi(i))^2+sg(i)*Cx(i)); ap(i)=sg(i)*Cy(i)/(cos(phi(i))*sin(phi(i))-sg(i)*Cy(i)); phi(i)=atan((1-a(i))/(x(i)*(1+ap(i)))); Cx(i)=CL*cos(phi(i)); Cy(i)=CL*sin(phi(i)); theta(i)=phi(i)-alpha; res=abs(1-temp/a(i)); temp= a(i); if res < 0.0001 break end k=k+1; end thetad(i)=theta(i)*180/pi; phid(i)=phi(i)*180/pi; fprintf(fid,’%12.4f%12.4f%12.4f%12.4f%12.4f\n’,r(i)/R, ... phid(i),thetad(i),a(i),ap(i)); g(i)=4*pi*rho*R*V^2*a(i)*(1-a(i))*r(i)*dr; gg(i)=4*pi*rho*X*V^2*ap(i)*(1-a(i))*r(i)^3*dr; Fx=Fx+g(i); T=T+gg(i); end Fx=Fx-0.5*(g(1)+g(n)); T=T-0.5*(gg(1)+gg(n)); % Put these !!!! P=T*X*V/R; fclose(fid); Fx1=2*rho*pi*V^2*a(41)*(1-a(41))*R^2*(1-0.2^2); T1=rho*pi*omega*V*ap(41)*(1-a(41))*R^4*(1-0.2^4); 177