אאא اﻟﻤﺆﺳﺴﺔ اﻟﻌﺎﻣﺔ ﻟﻠﺘﻌﻠﻴﻢ اﻟﻔﻨﻲ واﻟﺘﺪرﻳﺐ اﻟﻤﻬﻨﻲ אאאא אא א א ٤ אאא ١٦٢ א א EFאא אא אא א Wאא Kאאאא Wאא Wאא J א אאאאאWאא o Jא א–אאא Kאאא Kאא Jא o JE Fאאא אo Kאאא Wאאא ٪٨٠אא Wאא K٦ Wאא Kאאאאאאא J١ Kאא J٢ Wאא אאאEאאאFאא א Kא - ٥٨ - אאא ١٦٢ א א EFאא W Kאאאאאאאאאאא אאאאאאאא א،אאא Kאאאאאאא Types of Fluid Flowאאאא אאאאאא אאא א אK אאא א אאא،א אEParticlesFאאא אאאאKאאא אא،E١٤FE Laminar FlowFא אK אאא KE١٤FETurbulent FlowFאאאאאא אאE١٤F - ٥٩ - אאE١٤F אאא ١٦٢ א א EFאא WFluids in Motionאא אא،אאא אאאאאאK אא،Steady Uniform FlowאאאאK אEConvergentF אאאKE٢٤F אא E٢ ٤FאאFאא אאאאאK KESteady Non-Uniform FlowFאא E٢٤F Eא٢٤F Rate of Flowא א אאא،E٣ ٤Fא KQאאאא אאאאE٣٤F - ٦٠ - אאא ١٦٢ א א EFאא Wאאאא Q A= . ×ν אאZ Q ،אZA،אZν . m3/secאאא ⋅ m אאא W،א א × אZ∴ א ∴ m = V × ρ Q = ρ × ⋅ . m א א א. 0.6 m/sec 15mmא Kאא Wא Q A = . Wאאא ν × . π = Q (0.015)2X0.6 = 106.02X10-6 m3/sec 4 ⋅ ⋅ Wאאא Q = ρ × m ⋅ = 1000 × (106.02×10-6) = 0.106 kg/sec m - ٦١ - אאא ١٦٢ א א EFאא Continuity EquationEאאFאא אE אFאאאא אאאאאאאאא W ⋅ m = ρ1 × A1 × ν1 = ρ2 × A2 ×ν2 Continuity EquationEא،אאFאאאא KLaw of Conservation of Massא אאאKאאא ρ1 = ρ2 = ρ Wא W A1 × ν1 = A2 × ν2 ⋅ ⋅ ∴Q = Q 1 2 W E٤ ٤Fא EConverging PipeF אא،0.15mא0.46mאאא Kאאאאא،0.3m3א אE٤٤F - ٦٢ - אאא ١٦٢ א א EFאא Wא Wאא . Q A 1 = × ν1 KE٤٤Fאאא ⋅ Q ν1Z A1 WA1אא (0.462) = 166.19×10-3 m2 π = d12 π 4 ∴ν1 = 0.3 1.66019 × 10 −3 4 A1 = = 1.8 m/sec Kאאא ⋅ ν2Z Q A2 17m/sec ≅ = 16.98 π 4 0.3 0.15 2 ( )= K אא אK אאא Kאאאאא A1× ν1 = A2 × ν2 A ×ν ∴ν2 = 1 1 = A2 π 4 × d12 ×ν 1 π × d 22 4 × 0.462 × 1.8 = 16.93 m ∴ν2 = π 2 sec × 0.15 4 π 4 - ٦٣ - אאא ١٦٢ א א EFאא kg 5cmאאאא998 W /m3אאא א3.75cm א7.5cm Wא،E٥٤Fא0.15 m3/secאאא Kאאאאאא o Kאאאאא אo אE٥٤F Wא Wאאאא ⋅ = ρ×Α1×ν1 m1 ⋅ ⋅ m1 = ρ × Q ⋅ ( ⎠ ∴ m1 = ⎛⎜ 998 kg m3 ⎞⎟ 0.015 m ⎝ 3 ⋅ ∴ m1 = 14.970 kg sec - ٦٤ - sec ) אאא ١٦٢ א א EFאא Wאא . ⋅ ⋅ m1 = 2 × m3 ⋅ ⋅ ∴ m 3 = × m 1 1 2 ⋅ m3 = 12 × 14.97 = 7.485 kg sec Wאאא Kאאאא KA1א A =1 π 4 × d12 = π4 × (0.05) 2 = 0.001963m 2 ⋅ Q1 = A1 ×ν 1 ⋅ ν 1 = Q1 A1 V1 = 0.015 m 3 sec = 7.6 m sec 0.001963m 2 אאא A3א A3 = π 4 ⋅ 2 ×d3 = π 4 × (0.0375) 2 = 0.001104m 2 3 m Q 3 0.0075 sec ν3 = = = 6.76 m sec A3 0.001104m 2 - ٦٥ - אאא ١٦٢ א א EFאא : Energy of a Liquid in Motionאא אא، אא "the capacity to do אאא א .work" אאאאאאא، אאKאאאאאא K??אאא Wאא Kא E١ Kא E٢ Kא E٣ Kאאא Potential EnergyWא Datum Z mאאא WאאאKE٦٤FאLine ⎛ ⎝ Potential Energy = א m × g × Z ⎜ kg × m ⎞ × m⎟ = 2 sec ⎠ א،?J?אא ''Joule'' ??אאאא Wאאאא - ٦٦ - אאא ١٦٢ א א EFאא אm אE٦٤F ⎛ J ⎞ ⎟⎟ ⎝ kg ⎠ g × z⎜⎜ Potential Energy per Unit Mass = Pressure Energy Wא אא א אאPאאAאאK אאאKאא P) × (A א W،Qא ⋅ Q = A ×ν ⋅ KE F V = Q אP) × (Aא A WאאEwork)א Work = Force × Distance Work = p×A×ν ⋅ Q work = p × A × A ⋅ work = p × Q - ٦٧ - אאא ١٦٢ א א EFאא ⋅ Wאאא m אא . . m = ρ ×Q WEFאאאאא ⋅ work done per unit mass = p×Q ⋅ ρ ×Q = p⎛J ⎞ ⎜ ⎟ ρ ⎝ kg ⎠ J kg ،אLאאאאאא ⎛ p⎞ ⎜⎜ ρ ⎟⎟ אאאEאFא ⎝ ⎠ p = Pressure Energy= א ρ Kinetic EnergyWא אא،ν mאא KE½ ×m×ν2 FEאF W Kinetic Energy = ½× m × ν2 Wאאאא 2⎛ J ⎞ ⎜⎜ ⎟⎟ ν × Kinetic Energy Per Unit MassZ ⎝ kg ⎠ 1 2 - ٦٨ - אאא ١٦٢ א א EFאא ، א، אאאאאא WKא אHאHאZאאאא g×z+ p ρ + 12 ×ν 2 Total Energy Per Kg = אEאאאFאאא WEheight or headFא KEPotential HeadFא o KEPressure HeadFא o KEVelocity HeadFא o א אאאאאאאאאאא WKא Total head Per kg = z + p ν2 + ρ × g 2g Wאאאא Total head Per kg = m + N m3 sec 2 m 2 sec 2 × × + × m m m 2 kg sec 2 W Total Head Per kg = m + m + m = m - ٦٩ - אאא ١٦٢ א א EFאא Bernoulli Equation אFאאא?Wא א، אEא W?אאא KאHאHא Wא p1 ρ + ν 12 2 p2 gZ1 = ρ + ν 22 2 + gZ 2 K אאא אאא Kאאא E٧٤F אאאE٧ ٤F אאא KCBAאKhאאv KC אאZB אאZA אא pA ρ + ν A2 2 + gZ A = pB ρ + ν B2 2 + gZ B = - ٧٠ - pc ρ + ν c2 2 W + gZ c אאא ١٦٢ א א EFאא א،אאאאאאאא א א Bא אA Kא Wא אBאאAא WאאKאC א p A + ν A + gZ A Z pB + ν B + gZ B 2 ρ 2 ρ 2 2 BAאא אא Kאא Wאאאאאא pA ρ + ν A2 2 + gZ A = pB ρ + ν B2 2 + gZ B + energy loss between A and B W 1.52 m/secאאאא אאאK9.14 m/secאא K2.339kPaאאאאאK15.2mאא ؟K993 kg/m3אא998 kg/m3אא Wא Wאאא - ٧١ - אאא ١٦٢ א א EFאא P1 ρ1 + ν 12 2 gZ1 = P2 ρ2 + 2.339 × 1000 + 1.52 998 2 2.344 + 1.155 + 0 = ν 22 2 2 gZ 2 + 9.81 × 0 = P2 9.14 2 + + 9.81 × 15.2 983 2 P2 + 41.77 + 149.112 983 P2 = 983(2.344 + 1.155 − 41.77 − 149.112) = −184.2 KPa K؟ אא W Wאאאאא אא אאאא htZ hνHhs Kא אKא hν אאhs אאht KאאאאאWא Kאאאאא אאאא WאאאאאאKאא hν = ν2 2g ν = 2 gh W Em/secFאν EmFאhν - ٧٢ - אאא ١٦٢ א א EFאא W אאאאאK ؟KאאKאא 21.6m22mא Wא Kאאאאאאא hν= ht - hs hν = 22-21.6 = 0.4 m ∴ν = 2 ghν = 2 × 9.81 × 0.4 = 2.8 m sec אא 13600kg/m3א1.2kg/m3אאאK20mm ؟Kאאא Wא Wאאאא Pν = ρ ghHg = 13600 × 9.81 × 0.02 = 2668.32 N/m2 W Pν ρ air = ν2 2 ν 2 = 2× ν = 2× Pν ρ air 266.32 = 66.7 m sec 1 .2 - ٧٣ - אאא ١٦٢ א א EFאא Wא א، אא א אאאK אא אאאאKאאא KאאאKא Pitot Tube E٨٤FאKא אאא א KאאאאKא WאאE٢FE١Fא P1 + ν 1 2 ρ 2 + gZ 1 = P2 ρ + ν 22 2 + gZ 2 אKאאאE٢Fאא W P1 ρ ν 12 2 + ν 12 = 2 = P2 ρ P 2 − P1 ρ ν2 = 2× P2 − P1 ρ אא א אאP2 אאP1 Kא אאא - ٧٤ - אאא ١٦٢ א א EFאא אאאאאא،א Kאאאא אE٨٤F WVenturi-Meterא אא אא KE٩ ٤FאאKא KאאאE٢FE١Fא אE٩٤F אא אאאאא אאאKא אאK אKאא א،אאאא KאאE٢FאאE١Fא - ٧٥ - אאא ١٦٢ א א EFאא P1 ρ1 ν 12 + 2 P2 + gZ 1 = ρ2 + ν 22 2 + gZ 2 Kאאאא ∴ P1 ρ1 + ν 12 ν 22 − ν 12 2 2 = = P2 ρ2 + gz2 P1 − P2 ρ Wאאא ν 1 = A2 ×ν 2 A1 WאאE٢Fאאא ν 2 = 2( P1 − P2) ⎛ ρ − ⎜⎜1 − ⎝ A22 A12 ⎞ ⎟⎟ ⎠ KאאE٢FE١FאאA2 A١ E אאFאאאK25Coא אא2.5cmE٢Fא7.5cmE١Fאאא42.1cm ؟Kאא Wא E٢Fאא - ٧٦ - אאא ١٦٢ א א EFאא ν2 = ∴ν 2 = ν2 = 2 × ∆Ρ = A22 ρ (1 − 2 ) A1 2 g × ∆h A2 (1 − 22 ) A1 2 × 9.81 × 0.421 ⎛ 0.025 2 1 − ⎜⎜ 2 ⎝ 0.075 ⎞ ⎟⎟ ⎠ 2 8.08 = 8.18 = 2.86 m sec 0.988 Wאאאא m3 0.0014 ⋅ m3 π 2 sec = 1.4 lit = 84 lit Q = A2 × v2 = 4 × (0.025) × 2.86 = 0.0014 sec = m3 sec min 0.001 lit WOrifice MeterEאFא אא،א אKא אKE١٠ ٤F אאא אאאאאאK Kאאאא ν 2 = 2(P1 − P2 ) ⎛ A22 ⎞ ⎜ ρ × ⎜1 − 2 ⎟⎟ A1 ⎠ ⎝ - ٧٧ - אאא ١٦٢ א א EFאא אE١٠٤F W אא0.3m/secE١٠٤Fאאאא K אא אאא، 0.06 mא0.3 m K Cv= 0.82אא Wא ν 2 = Cν 2 × ∆Ρ ⎛ A ⎞ ρ × ⎜⎜1 − 2 ⎟⎟ A1 ⎠ ⎝ 2 Wאא ν א ν 2 = ν 1 × A1 A2 ν2 = 7.5 m/sec א אאאא A22 ν × ρ × (1 − 2 ) A1 ∆P = = 73048.9 Pa ≈ 73KPa 2 2 × Cν 2 2 - ٧٨ - אאא ١٦٢ א א EFאא Wאאא Wאאא KאאאEF אא30 C0 א1.5barאאEF ،250mmB א، 400mm × 300mm A א אBאאאא،BA0.2barאאאא K18m/secAאאR=287 J/kg Kאאאא Wאאא אאאאאאאאא ⋅ Kא Q אאאKא אא - ٧٩ - אאא ١٦٢ א א EFאא Wאאא Wאאאא אאא Kאאאאאאא • ρ1 × A1 × v1 = ρ 2 × A2 × v2 WאאEFאא A1 × v1 = A2 × v2 אאא• א EBFאEAFאKאאא Kאאאא WE١Fא Aא ρ1 P1 RT1 1.5 × 105 Kg ρ1 = = 1.725 3 287 × (30 + 273) m WE٢FאBא ρ2 P2 RT2 (1.5 − 0.2) × 105 = 1.495 Kg 287 × (30 + 273) m3 E٢FEBFאאאאא ρ2 = ρ1 × A1 × v1 = ρ 2 × A2 × v2 - ٨٠ - אאא ١٦٢ א א EFאא אא،v2א אאא Wאא v2 = ρ1 × A1 × v1 1.725 × (0.4 × 0.3) × 18 m = = 39.87 2 ρ 2 × A2 sec 1.495 × (0.25) Wאאאא Kאאאא P1 + v1 + Z1 g = ρ 2 P2 ρ + v2 + Z2 g 2 E١Fאא K אאE٢FE١F א אKאאאאE٢Fא Wאא 2 v2 2 = Z1 g − Z 2 g = g (Z1 − Z 2 ) Wאא،4m(Z1-Z2) v2 = (2 × g × (Z1 − Z 2 )) v2 = 2 × 9.8 × 4 = 8.86 m sec ⋅ W Q אא ⋅ Q = A2 × v2 π ( ) π m3 Q = × d × v2 = (0.1) × 8.86 = 0.0695 4 4 sec ⋅ m ∴ Q = 0.0695 3 sec ⋅ 2 2 2 - ٨١ -
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