Compressible Fluid Flow, Oosthuizen ( Solution Manual ) PDF

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫ Kapten Eng. Mamoon Nazmi Al-Niser … An isothermal process, for example, is a process during which the temperature T remains cons...

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

An isothermal process, for example, is a process during which the temperature T remains constant; an isobaric process is a process during which the pressure P remains constant; and an isochoric (or isometric) process is a process during which the specific volume v remains constant. A process during which there is no heat transfer is called an adiabatic process

Solution: 2

2

V V C p T2 + 2 = C p T1 + 1 2 2 2 ( 200) (100) 2 1006 × T2 + = 1006 × T1 − 2 2 2 (100) ( 200) 2 1006 × T2 − 1006 × T1 = + 2 2 ∆T = −15 K

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution: 2

2

V V C p T2 + 2 = C p T1 + 1 2 2 (300) 2 (500) 2 = 1006 × ( 25 + 273) + 1006 × T2 + 2 2 T2 = 377.5 K

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution: 2

2

V2 V = C p T1 + 1 2 2 2 V 1006 × (15 + 273) + 2 = 1006 × (30 + 273) + 0 2 2 V2 = 304818 − 289728 2 V = 30180 = 173.7 m / s C p T2 +

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution: ℜ 8314 = = 2078.5 J / kg .K 4 m R = C p − Cv R=

2078.5 = C p − C v → C p = 2078.5 + C v

γ=

Cp Cv

→ 1.67 =

2078.5 + C v → 1.67C v = 2078.5 + C v Cv

1.67C v = 2078.5 + C v → C v = 3102.24 J / kg C p = 2078.5 + C v → C p = 5180.74 J / kg 2

2

V2 V = C p T1 + 1 2 2 2 V2 (180) 2 5180.74 × 263 + = 5180.74 × 283 + 2 2 2 V2 = 1482349 .42 − 1362534 .62 2 V2 = 239629 .6 = 489.52m / s C p T2 +

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution: dA dV dρ + ρA + VA =0 dx dx dx 150 × 10 3 P ρ= = = 1.7 kg / m 3 RT 287 × 308 πd 2 π (0.2) 2 = = 0.3m 2 A= 4 4 2 π (0.1) 2 dA πd = + A= = 0.31m 2 / m 4 4 dx dρ Assuming incompressible flow so =0 dx dA dV dρ ρV + ρA + VA =0 dx dx dx dV 1.7 × 250 × 0.31 + 1.7 × 0.3 =0 dx dV = −258.33( m / s ) / m dx dV dP = ρv = 1.7 × 250 × 258.33 = 109.8kPa / m dx dx dρ ⎡ V V ⎤ ρ dV ⎡ 250 250 ⎤ 1.7 kg / m 3 258 . 33 0 . 89 × = =⎢ − ⎥ =⎢ − dx ⎢⎣ C p R ⎥⎦ T dx ⎣1004 287 ⎥⎦ 308 m

ρV

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution: dA dV dρ + ρA + VA =0 dx dx dx P 120 × 10 3 ρ= = 1.4kg / m 3 = RT 287 × 298 dV = (0.3 × 200) = 60( m / s ) / m dx dP dV = ρv = 1.4 × 200 × 60 = 16.8kPa / m dx dx dρ ⎡ V V ⎤ ρ dV ⎡ 200 200 ⎤ 1.4 kg / m 3 =⎢ − ⎥ × −60 = 0.14 = − dx ⎣⎢ C p R ⎦⎥ T dx ⎢⎣1004 287 ⎥⎦ 298 m

ρV

ds dT 1 dP 1 1 = Cp × −R × = 0 − 287 (16.8)( ) = 0.04kJ / K dx dx T dx P 120 × 10 3

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Problem 2.7

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Problem 2.8

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

-9-

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 2 : The Equations of Steady One Dimensional Compressible Fluid Flow Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Problem 2.9

- 10 -

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution: γ hydrogen = 1.4

γ helium = 1.7

γ nitrogen = 1.1

a = TRγ helium = 288 × 287 × 1.7 = 340.17 m / s a = TRγ hydrogen = 288 × 287 × 1.4 = 374.9m / s a = TRγ nitrogen = 288 × 287 × 1.1 = 301.5m / s

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: γ Co = 1.3 2

a = TRγ = 293 × 287 × 1.3 = 330.6m / s a = TRγ = 873 × 287 × 1.3 = 570.7 m / s

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: a = TRγ = 303 × 287 × 1.4 = 348.9m / s P 101 × 10 3 = 1.2kg / m 3 = ρ= RT 287 × 303 dp 30 dv = = = 0.07 m / s ρa 1.2 × 348.9 v2 v2 C pT = ⇒ 1004 × 303 = ⇒ v = 24.6m / s 2 2 v 24.6 dT = dv = × 0.07 = 1.435 K 1. 2 ρ 1 .2 ρ dρ = dv = × 0.07 = 2.4 × 10 − 4 kg / m 3 a 348.9

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: a = TRγ = 213 × 287 × 1.4 = 292.55m / s M =

V = a

1000 3600 = 1.4 292.55

1500 ×

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: a = TRγ = 223 × 287 × 1.4 = 299.3m / s V M = = a

1000 3600 = 1.9 299.3

2000 ×

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: a = TRγ = 288 × 287 × 1.4 = 340.17 m / s 1000 V 3600 = 0.65 M = = 340.17 a a = TRγ = 229 × 287 × 1.4 = 303.3m / s 800 ×

V = Ma = 0.65 × 303.3 = 197.145m / s

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: A = (1.22) 2 = 1.5m 2 a = TRγ = 163 × 287 × 1.4 = 255.92 m / s V = Ma = 3.5 × 255.92 = 895.72m / s P 20 × 10 3 = = 0.43kg / m 3 RT 287 × 163 m& = ρVA = 0.43 × 895.72 × 1.5 = 577.7 kg / s

ρ=

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: T = 288.16 − 0.0065 H = 288.16 − 0.0065(11019) = 216.5 K a = TRγ = 216.5 × 287 × 1.4 = 295m / s V M = = a

1000 3600 = 0.11 295

120 ×

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: 1 1 = = 1 .7 sin α sin 35 a = TRγ = 318 × 287 × 1.4 = 257.5m / s M =

V = Ma = 1.7 × 257.5 = 607.7 m / s

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: 1 1 = = 1 .6 sin α sin 40 a = TRγ = 263 × 287 × 1.4 = 325.1m / s M=

V = Ma = 1.6 × 325.1 = 520.1m / s

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: ℜ 8314 = = 189 J / kg .K m 44 1 1 M = = = 1 .4 sin α sin 45 a = TRγ = 189 × 287 × 1.67 = 398.8m / s R=

V = Ma = 1.4 × 398.8 = 558.3m / s

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution:

F − 32 80 − 32 = = 26.7 1 .8 1 .8 a = TRγ = 299.7 × 287 × 1.4 = 347.02m / s

Co =

1 1 = sin −1 = 31.2 o M 1 .9 V = Ma = 1.9 × 347.02 = 659.34m / s

α = sin −1

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: a = TRγ = 283 × 287 × 1.4 = 346.03m / s V 180 = = 0 .5 a 346.03 V 800 M = = = 2. 3 a 346.03 1 1 α = sin −1 = sin −1 = 25.8 o M 2. 3 M =

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution:

7000m

13km

T = 288.16 − 0.0065 H ⇒ T = 288.16 − 0.0065(3500) ⇒ T = 268.65 K 7000 = 28.3 o 13000 a = TRγ = 268.65 × 287 × 1.4 = 328.5 / s

α = tan −1

1 1 = = 2 .1 sin α sin 28.3 V = Ma = 2.1 × 328.5 = 690 m / s M =

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Solution: 1 3

α = sin −1 = 19.5 o D=

H 6000 = = 16.943km tan α tan 19.5

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Solution:

2500m

6km

T = 288.16 − 0.0065 H ⇒ T = 288.16 − 0.0065(1250) ⇒ T = 280.04 K 2500 = 22.62 o 6000 a = TRγ = 280.04 × 287 × 1.4 = 335.4m / s

α = tan −1

1 1 = = 2 .6 sin α sin 22.62 V = Ma = 2.6 × 335.4 = 872.04m / s M =

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.18:

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.19:

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.20:

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.21:

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.22:

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.23:

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.24:

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Kapten Eng. Mamoon Nazmi Al-Niser …

Problem 3.25:

With Our Best Wishes Eng. “Mohammad Luay” Mahmoud Shaban ☺♪♫ Kapten Eng. Mamoon Nazmi Al-Nise ☺

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 3 : Some Fundamental Accepts of Compressible Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫

Problem 3.26:

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Kapten Eng. Mamoon Nazmi Al-Niser …

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 4 : One Dimentional Isentropic Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫ Kapten Eng. Mamoon Nazmi Al-Niser … -----------------------------------------------------------------------------------------------------------------------------------------------

Solution: ℜ 8314 = = 2078.5 J / kg .K m 4 a = TRγ = 283 × 2078.5 × 1.67 = 991.1m / s R=

V 200 = = 0 .2 a 991.1 λ −1 2 1.67 − 1 M1 1+ ( 0 .2 ) 2 1+ T2 1.67 − 1 1.0134 263 2 2 2 →⇒ 1 + M2 = →⇒ = = λ −1 2 1.67 − 1 2 T1 2 0.93 283 M2 M2 1+ 1+ 2 2 o.2 1.67 − 1 2 M 2 = 1 .1 → M 2 = = 1 .8 2 0.335 M =

a = TRγ = 263 × 2078.5 × 1.67 = 955.46m / s V = Ma = 1.8 × 955.46 = 1719.83m / s

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Solution: p1 680 × 10 3 = 7.1kg / m 3 = RT 287 × 333 m& = ρvA = 7.1 × 100 × 5 × 10 − 4 = 0.355kg / s

ρ=

a1 = TRγ = 333 × 287 × 1.4 = 365.8m / s M =

V 100 = = 0 .3 a 365.8 λ

1. 4

p 2 ⎡ λ − 1 2 ⎤ λ −1 p ⎤ 1.4−1 ⎡ 1 .4 − 1 →⇒ p 2 = 723.8kPa = ⎢1 + M ⎥ →⇒ 2 = ⎢1 + (0.3) 2 ⎥ p1 ⎣ 2 2 680 ⎣ ⎦ ⎦ λ −1

0.29

T2 ⎛ p 2 ⎞ λ T ⎛ 723.8 ⎞ = ⎜⎜ ⎟⎟ ⇒ 2 = ⎜ ⎟ ⇒ T2 = 339.7 K T1 ⎝ p1 ⎠ 333 ⎝ 680 ⎠ p 723.8 × 10 3 = 7.4kg / m 3 ρ2 = 2 = RT2 287 × 339.7 m& = ρ 2 vA ⇒ 0.355 = 7.4 × v × 3.8 × 10 − 4 v=

0.355 = 119.13m / s 7.4 × 3.8 × 10 − 4

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 4 : One Dimentional Isentropic Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫ Kapten Eng. Mamoon Nazmi Al-Niser … -----------------------------------------------------------------------------------------------------------------------------------------------

Solution: ℜ 8314 = = 259.8 J / kg .K m 32 a = TRγ = 1073 × 259.8 × 1.3 = 602 m / s R=

λ −1

⎡ 2 ⎤ M1 ⎥ 1+ P2 ⎢ 2 =⎢ ⎥ −1 λ 2 P1 ⎢ M2 ⎥ 1+ 2 ⎣ ⎦

λ λ −1

⎛ 80 ⎞ →⇒ ⎜ ⎟ ⎝ 1500 ⎠

1.3−1 1.3

⎡ 1. 3 − 1 2 ⎤ ⎢ 1 + 2 ( 0 .2 ) ⎥ =⎢ ⎥ ⎢ 1 + 1 .3 − 1 M 2 2 ⎥ 2 ⎣ ⎦

0.51 + 0.0765 M 2 = 1.006 ⇒ M 2 = 2.5 2

λ −1

1.3−1

T2 ⎛ p 2 ⎞ λ T ⎛ 80 ⎞ 1.3 = ⎜⎜ ⎟⎟ ⇒ 2 = ⎜ ⇒ T2 = 547.23K ⎟ 1073 ⎝ 1500 ⎠ T1 ⎝ p1 ⎠ p 1500 × 10 3 = 5.4kg / m 3 ρ= 1 = RT 259.8 × 1073 V1 = Ma = 0.2 × 602 = 120.4m / s m& = ρvA = 5.4 × 120.4 × 0.2 = 130.032kg / s

ρ2 =

p2 80 × 10 3 = = 0.56 kg / m 3 RT2 259.8 × 547.23

a 2 = TRγ = 547.23 × 259.8 × 1.3 = 429.7 m / s V2 = Ma = 2.5 × 429.7 = 1074.2m / s m& = ρvA ⇒ 130.032 = 5.4 × 1074 .20 × A ⇒ A = 0.22 m 2

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 4 : One Dimentional Isentropic Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫ Kapten Eng. Mamoon Nazmi Al-Niser … -----------------------------------------------------------------------------------------------------------------------------------------------

Solution:

ℜ 8314 = = 593.9 J / kg .K m 14 a1 = TRγ = 773 × 593.9 × 1.25 = 757.5m / s R=

V 100 = = 0.13 a 757.5 p1 1000 × 10 3 = = 2.21kg / m 3 ρ= RT 593.9 × 773 m& = ρvA = 2.2 × 100 × 0.7 = 154 kg / s M =

λ

1.25

p 2 ⎡ λ − 1 2 ⎤ λ −1 p ⎡ 1.25 − 1 ⎤ 1.25−1 ⇒ p 2 = 1010.6kPa = ⎢1 + M ⎥ ⇒ 2 = ⎢1 + (0.13) 2 ⎥ p1 ⎣ 2 1000 ⎣ 2 ⎦ ⎦ T2 ⎡ λ − 1 2 ⎤ T ⎡ 1.25 − 1 ⎤ = ⎢1 + M ⎥ ⇒ 2 = ⎢1 + (0.13) 2 ⎥ ⇒ T2 = 774.6 K T1 ⎣ 2 773 ⎣ 2 ⎦ ⎦ a 2 = TRγ = 774.6 × 593.9 × 1.25 = 758.32m / s V = Ma = 0.13 × 758.32 = 98.6m / s λ

⎡ λ − 1 2 ⎤ λ −1 ⎡ 1.25 − 1 ⎤ 1..25 −1 M1 ⎥ 1+ (0.13) 2 ⎥ 1+ ⎢ ⎢ 1 . 25 P2 ⎛ 1010.6 ⎞ 2 2 =⎢ →⇒ ⎜ =⎢ ⎟ ⎥ ⎥ 1.25 − 1 2 2 P1 ⎢ λ − 1 1000 ⎠ ⎝ ⎥ ⎢ M2 M2 ⎥ 1+ 1+ 2 2 ⎣ ⎦ ⎣ ⎦ 1.002 + 0.1253M 2 = 1.006 ⇒ M 2 = 0.2 2

T2 ⎛ p 2 =⎜ T1 ⎜⎝ p1

⎞ ⎟⎟ ⎠

λ −1 λ

T ⎛ 1010.6 ⎞ ⇒ 2 =⎜ ⎟ 773 ⎝ 1000 ⎠

1.25 −1 1.25

⇒ T2 = 774.55 K

a 2 = TRγ = 774.55 × 593.9 × 1.25 = 758.3m / s V2 = Ma = 0.2 × 758.3 = 151.7 m / s

‫ ﺍﻹﺗﺠﺎﻩ ﺍﻹﺳﻼﻣﻲ‬- ‫ﻟﺠﻨﺔ ﺍﻟﻤﻴﻜﺎﻧﻴﻚ‬

Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 4 : One Dimentional Isentropic Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫ Kapten Eng. Mamoon Nazmi Al-Niser … -----------------------------------------------------------------------------------------------------------------------------------------------

Solution: ℜ 8314 = = 189 J / kg .K m 44 a1 = TRγ = 288 × 189 × 1.3 = 266.01m / s R=

M =

V 100 = = 0.38 a 266.01 λ

1.3

p 2 ⎡ λ − 1 2 ⎤ λ −1 p ⎡ 1 .3 − 1 ⎤ 1.3−1 M ⎥ ⇒ 2 = ⎢1 + (0.38) 2 ⎥ = ⎢1 + ⇒ p 2 = 87.7 kPa p1 ⎣ 2 2 80 ⎣ ⎦ ⎦ T2 ⎛ p 2 ⎞ =⎜ ⎟ T1 ⎜⎝ p1 ⎟⎠

λ −1 λ

T ⎛ 87.7 ⎞ ⇒ 2 =⎜ ⎟ 288 ⎝ 80 ⎠

1.3−1 1.3

⇒ T2 = 294.2 K

a 2 = TRγ = 294.2 × 189 × 1.3 = 268.9m / s M =

V 300 = = 1. 1 a 268.9

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Compressible Fluid Flow, Oosthuizen ( Solution Manual ) Chapter 4 : One Dimentional Isentropic Flow. Eng. “Mohammad Luay” Mahmoud Shaban ♪♫ Kapten Eng. ...