Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 12, Problem 127P
To determine

The throat and exit areas of the nozzle for given mass flow rates and also the reason the nozzle must be of converging-diverging type.

Expert Solution & Answer
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Answer to Problem 127P

The throat area of the nozzle is A*=5.96 cm2

The exit area of the nozzle is A2=8.97cm2 

The exit Mach number comes out to be greater than the 1, hence the nozzle must be of converging-diverging type.

Explanation of Solution

Given:

Ambient temperature at inlet T1= 500 K

Stagnation Pressure P01=P1=220 psia

Exit pressure P2 = 0.1 MPa

Mass flow rate m. = 0.34 kg/s

Calculation:

Properties of helium,

  R=2.0769×103J/kg.K

  cp=5.1926 KJ/kg.K

  k=1.667

R is gas constant, cp is specific heat, k is specific heat ratio.

Stagnation temperature of helium at inlet is given by

  T01=T1+V122cpThe ambient temperature at inlet is T1,inlet velocity of fluid is V1.T01=500+ ( 0 )22( 5.1926)×(1 1000 )T01=500.0K

Stagnation pressure of helium at inlet is given by

  P0=P1( T 01 T 1 )k/k-1Pressure of the fluid P1.P01=0.8( 500 500)1.667/1.667-1P01=0.8MPa

The flow is considered to be isentropic, so the stagnation temperature and stagnation pressure remain same throughout the nozzle.

  T02=T01= 500 K

  P02=P01=0.8 Mpa

Now the critical temperature is given by

  T*=T0(2 k+1)T*=500(2 1.667+1)T*=374.95 K

And critical pressure is given by

  P*=P0( 2 k+1)k/k1P*=0.8( 2 1.667+1)1.667/1.6671P*0.3896 MPa

The critical density is given by

  ρ*=P*RT*ρ*=0.3896×1062.0769×103×374.9531ρ*=0.50039 kg/m3

Inlet velocity is given by

  V*=c*=kRT*V*=1.667×2.0769( 374.95)( 1000 1 )V*=1139.36 m/s

Exit area of throat is given by

  A*=m.ρ*V*m. is mass flow rate,ρ*is critical density,V&*#x00A0;is inlet velocity.A*=0.341139.36×0.50039A*=5.96×10-4m2A*=5.96 cm2

Pressure at the exit of nozzle is given as P2 = 0.1 MPa.

  P0P2 = (1+ k1 2M a 2 2)k/k1exit pressure is P2.0.815 = (1+ 1.6671 2M a 2 2)1.667/1.6671Ma2=1.972806>1  nozzle is converging-diverging.

Temperature at nozzle exit is given by

  T2 =T0 (2 2+( k-1 )M a 2 2 )T2 =500 (2 2+( 1.667-1 )1.97280 6 2 )T2 =217.58 K

Density at exit of nozzle is given by

  ρ2=P2RT2ρ2=0.1×1062.0769×103×217.58ρ2=0.2212 kg/m3

Velocity at exit of nozzle is given by

  c2=kRT2R is gas constant,k specific heat ratio,T2 exit temperature.c2=1.667×0.4961( 217.58)( 1000 1 )c2=867.93 m/sV2=Ma2c2V2=(1.972806)867.93V2=1712.272 m/s

Area of exit at nozzle is given by

  A2=m.ρ2V2m. is mass flow rate,ρ2is exit density,V2 is exit velocity.A2=0.341712.272 ×0.2212 A2=8.97×104m2 A2=8.97 cm2

Conclusion:

Therefore, the throat area of the nozzle is A*=5.96 cm2.

The exit area of the nozzle is A2=8.97cm2 .

The exit Mach number comes out to be greater than 1, hence the nozzle must be of converging-diverging type.

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Chapter 12 Solutions

Fluid Mechanics: Fundamentals and Applications

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