Solution to Exercise Problem 16.1 parts (b),
(a) & (c)
I have attached solutions in that order to show you what the impact is on the heat exchanger area when the dilute feed is at
(b) this is a typical dilute feed condition coming from a storage tank at ambient temperature, e.g. at 21.1 °C
(a) as if the dilute feed is coming from an economiser of a furnace where it has been warmed up to almost its boiling point (no sensible heat needed)
& (c) the dilute feed is coming superheated from a source on a plant where plenty of waste heat is available
Solution to Exercise Problem 16.1 part (b)
Q16.1 part (b)
20,000 kg/h of 8 % organic colloid solution is concentrated to 45% in a single effect evaporator operating at a pressure of 102
mm Hg absolute. Saturated steam is available at 1.03 atmospheres gauge (120.5°C). The overall heat transfer coefficient, U is
2800 W/ m2 °C. Assume that the solution has a specific heat of 3.77 J/g °C and negligible elevation in boiling point and
negligible heat of dilution. (SO IN THIS CASE, IT IS EASIER TO USE ENTHALPY ABSORPTION METHOD).
Calculate: Steam consumption, economy and heating surface (area) of the evaporator at feed temperatures of part b) 21.1 °C
Solution:
(Note that specific heat of 3.77 J/g °C is same as specific heat of 3.77 kJ/kg. °C or specific heat of 3.77 kJ/kg. K)
Draw a diagram showing all the given data.
EVAPORATOR
U= 2800 W/m2 °C
Sp. Heat, Cp = 3.77 kJ/kg °C
A = ? m2
NEGLIGIBLE BPE & HEAT OF
DILUTION
FEED
mF, = 20,000 kg/h
Tf, = b) 21.1 °C (294.2K)
Cf, = 0.08 (8%)
Sp. Heat, CpF = 3.77 kJ/kg °C
VAPOUR
PV, = 102 mm Hg absolute,
mV, = ? kg/h
Cv, = 0.0
THICK LIQUOR
ml = ?
kg/h
Tl , = ?
Cl , = 0.45 (45%)
Sp. Heat, Cpl = 3.77 kJ/kg °C
Since most of the given data is in SI units, convert all of the units to SI units:
Steam : a) Steam Tables are based on absolute Pressure basis. So, we have to convert steam pressure of 1.03 Atmosphere gauge to
absolute by adding 1 Atmosphere, giving the absolute steam pressure of 2.03 Atmosphere absolute. Since the pressure
in SI units Steam Table is given in kPa, we have to covert 2.03 Atmospheres to kPa. From conversions given in slide 6, 1 Atmosphere is 101.3 kPa. The steam pressure
therefore is, 2.03*101.3 kPa = 205.6 kPa
b) From the SI unit Steam Table on slide 4, (using the pressure of 200 kPa which is close to 205.6 kPa) the temperature of
the steam is 393.4 K and latent heat λs is 2201.5 kJ/kg. (Note that the latent heat Hlg in SI steam tables is same as λv in the equations)
Evaporator Pressure : a) Since we are dealing with SI units, we have to convert evaporator pressure of 102 mm Hg absolute to
SI units in kPa. From slide 6, 760 mm Hg is equal to 101.3 kPa. Evaporator pressure is therefore (102/760)*101.3 kPa = 13.6 kPa
b) From the same SI unit Steam Table, (using the pressure of 14 kPa which is close to 13.6 kPa), the temperature of
the boiling liquid (Tl) and the vapour (Tv) is 325.7 K and latent heat of vaporization λv is 2375.8 kJ/kg.
Equations to be used are:
Mass Balance,
mF = m V + m L
Solute Balance,
cf*mf = cV*mV + cL*mL
Enthalpy Absorption Method is used, (since specific heat of solution, Cp, is given and question states there is negligible elevation in boiling point (B.P.E.))
q = λS*mS = mf *Cp *(Tb-Tf ) + λv*mV
Steam required,
m S = q/λS = {mf *Cp *(Tb-Tf ) + λv*mV }/ λS
Exchanger sizing, heat transferred by steam
q = λS*mS = U*A*ΔT = U*A*(TS-Tb) which can be rearranged to give are, A = q/{ U*(Ts-Tb)}
U = q/ A *(TS-Tb)
Economy
Economy = mV / mS
Watt
Watt (W) = 1 Joule/second = (1000/3600) kJ/h
Saturated Steam Table in SI Unit
by Pressure
Press.
(kPa)
Temp
(K)
Enthalpy (kJ/kg)
Hℓg (Latent Heat, λ)
Volume (m3/kg)
Vℓ
Vg
Hℓ
Hg
0.80
276.92
0.001000
159.7
15.4
2492.4
2507.8
1
280.13
0.001000
129.2
28.6
2485.1
2513.7
1.2
282.81
0.001000
108.7
39.7
2479.0
2518.7
1.4
285.13
0.001001
93.92
49.3
2473.6
2522.9
1.6
287.17
0.001001
82.76
57.8
2468.9
2526.7
1.8
288.99
0.001001
74.03
65.5
2464.5
2530.0
2.0
290.65
0.001002
67.00
72.4
2460.6
2533.0
2.5
294.23
0.001002
54.25
87.5
2452.1
2539.6
3.0
297.23
0.001003
45.67
100.1
2445.0
2545.1
4.0
302.12
0.001004
34.80
120.7
2433.2
2553.9
5.0
306.03
0.001005
28.19
137.2
2423.8
2561.0
6.0
309.31
0.001007
23.74
151.0
2415.9
2566.9
8.0
314.66
0.001009
18.10
173.7
2402.8
2576.5
10
318.96
0.001010
14.67
191.8
2392.4
2584.2
12
322.57
0.001012
12.36
207.1
2383.5
2590.6
14
325.70
0.001013
10.69
220.3
2375.8
2596.1
16
328.47
0.001015
9.433
231.9
2369.1
2601.0
18
330.96
0.001016
8.445
242.4
2362.9
2605.3
20
333.22
0.001017
7.649
251.9
2357.4
2609.3
25
338.12
0.001020
6.204
272.6
2345.1
2617.7
30
342.26
0.001022
5.229
289.9
2334.9
2624.8
40
349.02
0.001026
3.993
318.3
2318.0
2636.3
50
354.48
0.001030
3.240
341.3
2304.1
2645.4
60
359.09
0.001033
2.732
360.6
2292.4
2653.0
80
366.65
0.001038
2.087
392.3
2273.0
2665.3
100
372.78
0.001043
1.694
418.0
2257.0
2675.0
101.33
373.14
0.001043
1.673
419.5
2256.1
2675.6
120
377.96
0.001047
1.428
439.7
2243.4
2683.1
140
382.46
0.001051
1.237
458.6
2231.4
2690.0
160
386.47
0.001054
1.091
475.5
2220.5
2696.0
180
390.09
0.001058
0.9775
490.8
2210.6
2701.4
200
393.38
0.001061
0.8857
504.7
2201.5
2706.2
250
400.59
0.001067
0.7187
535.2
2181.3
2716.5
300
406.70
0.001073
0.6058
561.2
2163.7
2724.9
400
416.78
0.001084
0.4625
604.3
2133.8
2738.1
Calculations:
1) Mass Flows:
Solute Balance,
cf*mf = cV*mV + cL*mL ; 0.08*20,000 = 0*mV + 0.45*mL ; mL = 0.08*20,000/ 0.45;
Mass Balance,
m f = mV + mL ;
20,000 = mV + 3555.6 ; mV = 20,000 – 3,555.6 ;
mL = 3,555.6 kg/h
mV = 16,444.4 kg/h
2) Enthalpy Balances:
Heat provided by steam as it condenses, q = λS*mS = mf *Cp *(Tb-Tf ) + λV*mV ; (this heat, q, provides the sensible heat to raise feed from Tf
to Tb at the evaporator pressure plus latent heat to convert vapour portion of the boiling solution to vapour)
q = 2,201.5*mS =20,000 *3.77 *(325.7-294.2) + 2375.8*16,444.4 = 41,443,705.52 kJ/h
(where, 2,201.5 kJ/kg. is the latent heat of steam at 200 kPa and 2375.8 kJ/kg. is latent heat of vapour at 14kPa.
Tb = 325.7 K, the boiling temperature of liquid at 14 kPa and Tf is the given feed temperature 294.2 K (21.1 ° C) . Mass flow
rate of feed, mf , is 20,000kg/h and mass flow rate of vapour, mV , is 16,444.4 kg./h)
Steam required,
m S = q/λS = 41,443,705.52 kJ/h/2201.5 kJ/kg.
mS = 18,825.21 kg/h
3) Economy:
Economy = mV/ms = 16,444.4/18,825.21 = 87.35 %
4) Heat Exchanger Area:
A = q/{U*(Ts-Tb)} = 41,443,705.52 /{2800*(3600/1000)*(393.4-325.7)}
Area, A = 60.73 m2
Evaporator Related Units and Conversion Factors
Pressure:
1 Atmosphere = 14.696 psi
= 101.3 kPa
= 760 mm Hg
Pressure gauges typically measure gauge pressure not absolute pressure. (unless the pressure gauge states absolute)
To convert gauge pressure given in Atmospheres (guage) to absolute pressure you add 1 Atmosphere,
e,g. 5 atmospheres guage is 6 atmospheres absolute.
To convert gauge pressure given in psig to absolute pressure you add 14.696 psi,
e,g. 50 psig is 64.696 psia. (g denotes gauge, a denotes absolute)
To convert gauge pressure given in kPag to absolute pressure you add 101.3 kPa,
e,g. 600 kPag is 701.3.kPa. (if g it is not stated it is understood that the pressure is absolute)
IN EVAPORATOR CALCULATIONS IN THIS COURSE THE STEAM PRESSURE WILL BE GIVEN AS GAUGE PRESSURE
Temperature:
32 °F= 0 °C =273 K
Conversion from °F to °C, °C = (°F – 32)/1.8
Conversion from °C to °F, °F = (°C *1.8) +32
Mass:
1 pound (Lb) = 0.45359237 kilograms (kg.)
1 kilogram (kg) = 2.20462 pounds (Lb)
Energy:
1 Btu = 1.0555 kJ
1 Watt (W) = 1 J/s
Area:
1 ft2 = 0.0929 m2
Evaporator Diagram with Given Data &
Calculated Information
VAPOUR
PV, = 102 mm Hg absolute, (13.6 kPa)
SATURATD STEAM
Ps , = 1.03 Atm. Gauge (205.6 kPa)
Ms, = 18,825 kg/h
Ts, = 120.5 °C (393.5 K)
λs
= 2201.5 kJ/kg
FEED
mF, = 20,000 kg/h
TF, = b) 21.1 °C
CF, = 0.08 (8%)
Sp. Heat, Cpf = 3.77 kJ/kg °C
EVAPORATOR
U= 2800 W/m2 °C
Sp. Heat, Cp = 3.77 kJ/kg °C
Area =60.73 m2
mV, = 16,444.4 kg/h
Cv, = 0.0
Ts, = 52.5 °C (325.7 K)
λs = 2375.6 kJ/kg
NEGLIGIBLE BPE & HEAT OF
DILUTION
THICK LIQUOR
ml = 3,555.56 kg/h
Tl , = 52.5 °C, (325.7 K)
Cl , = 0.45
Sp. Heat, Cpl = 3.77 kJ/kg °C
Solution to Exercise Problem 16.1 part (a)
Calculate: Steam consumption, economy and heating surface (area) of the evaporator at feed temperatures of 51.7 °C (324.9K)
Calculations:
1) Mass Flows:
Solute Balance,
cf*mf = cV*mV + cL*mL ; 0.08*20,000 = 0*mV + 0.45*mL ; mL = 0.08*20,000/ 0.45;
Mass Balance,
mf = mV + mL ;
20,000 = mV + 3555.6 ; mV = 20,000 – 3,555.6 ;
mL = 3,555.6 kg/h
mV = 16,444.4 kg/h
2) Enthalpy Balances:
Heat provided by steam as it condenses, q = λS*mS = mf *Cp *(Tb-Tf ) + λV*mV ; (this heat, q, provides the sensible heat to raise feed from Tf (324.9 K)
to Tb (325.7 K) at the evaporator pressure plus latent heat to convert vapour portion of the boiling solution to vapour)
q = 2,201.5*mS =20,000 *3.77 *(325.7-324.9) + 2375.8*16,444.4 = 39,128,925.52 kJ/h
(where, 2,201.5 kJ/kg. is the latent heat of steam at 200 kPa and 2375.8 kJ/kg. is latent heat of vapour at 14kPa.
Tb = 325.7 K, the boiling temperature of liquid at 14 kPa and Tf is the given feed temperature 324.9 K (51.7 ° C) .
Mass flow rate of feed, mf , is 20,000kg/h and mass flow rate of vapour, mV , is 16,444.4 kg./h)
Steam required,
mS = q/λS = 39,128,925.52 kJ/h/2201.5 kJ/kg.
mS = 17,773.76 kg/h
3) Economy:
Economy = mV/ms = 16,444.4/18,801.24 = 92.52 %
4) Heat Exchanger Area:
A = q/{U*(Ts-Tb)} = 39,128,925.52 /{2800*(3600/1000)*(393.4-325.7)}
Area, A = 57.34 m2
Solution to Exercise Problem 16.1 part (c)
Calculate: Steam consumption, economy and heating surface (area) of the evaporator at feed temperatures of 93.3 °C (366.5 K)
Calculations:
1) Mass Flows:
Solute Balance,
cf*mf = cV*mV + cL*mL ; 0.08*20,000 = 0*mV + 0.45*mL ; mL = 0.08*20,000/ 0.45;
Mass Balance,
mf = mV + mL ;
20,000 = mV + 3555.6 ; mV = 20,000 – 3,555.6 ;
mL = 3,555.6 kg/h
mV = 16,444.4 kg/h
2) Enthalpy Balances:
Heat provided by steam as it condenses, q = λS*mS = mf *Cp *(Tb-Tf ) + λV*mV ; (this heat, q, provides the sensible heat to raise feed from Tf (366.5 K)
to Tb (325.7 K) at the evaporator pressure plus latent heat to convert vapour portion of the boiling solution to vapour)
q = 2,201.5*mS =20,000 *3.77 *(325.7-366.5) + 2375.8*16,444.4 = 35,992,285.52 kJ/h
(where, 2,201.5 kJ/kg. is the latent heat of steam at 200 kPa and 2375.8 kJ/kg. is latent heat of vapour at 14kPa.
Tb = 325.7 K, the boiling temperature of liquid at 14 kPa and Tf is the given feed temperature 366.5 K (93.3 ° C) .
Mass flow rate of feed, mf , is 20,000kg/h and mass flow rate of vapour, mV , is 16,444.4 kg./h)
Steam required,
mS = q/λS = 35,992,285.52 kJ/h/2201.5 kJ/kg.
mS = 16,348.98 kg/h
3) Economy:
Economy = mV/ms = 16,444.4/16,348.98 = 100.58 %
4) Heat Exchanger Area:
A = q/{U*(Ts-Tb)} = 35,992,285.52 /{2800*(3600/1000)*(393.4-325.7)}
Area, A = 52.74 m2
Summary of Results:
a)
Feed Temp, °C
51.7
Steam Consumption, kg./h
17773.76
Economy, %
92.52
Heat Exchanger Area, m2
57.34
b)
21.1
18825.21
87.35
60.73
c)
93.3
16348.98
100.58
52.74
More steam and exchanger area are required when feed is colder and the corresponding economy is lower