Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 17.7, Problem 131RP

Helium expands in a nozzle from 0.8 MPa, 500 K, and negligible velocity to 0.1 MPa. Calculate the throat and exit areas for a mass flow rate of 0.34 kg/s, assuming the nozzle is isentropic. Why must this nozzle be converging–diverging?

Expert Solution & Answer
Check Mark
To determine

The throat and exit area of the nozzle.

Answer to Problem 131RP

The throat area of nozzle is 5.96cm2.

The exit area is 8.97cm2.

Explanation of Solution

It is given that the initial velocity is negligible. Hence, the inlet properties are equal to the stagnation properties at inlet.

T1=T01=500KP1=P01=0.8MPa

Consider the flow through the nozzle is isentropic. Hence, the stagnation properties at inlet and exit equal.

T01=T02=500KP01=P02=0.8MPa

Write the formula for the critical temperature of the mixture.

T=T0(2k+1) (I)

Here, the critical temperature of mixture is T, the ratio of specific heats is k, and stagnation temperature of mixture is T0.

Write the formula for the critical pressure of the mixture.

P=P0(2k+1)k/(k1) (II)

Here, the critical pressure of mixture is P, and the stagnation pressure of mixture is P0.

Write the formula for the critical density.

ρ=PRT (III)

Here, the critical density of mixture is ρ, and the gas constant of helium is R.

Write the formula for critical velocity of helium gas through the nozzle.

V=kRT (IV)

Here, the superscript indicates the critical properties that is the properties at throat region.

Write the formula for mass flow rate of helium at throat region.

m˙=ρAV (V)

Here, the cross sectional area of the throat is A, and the velocity at throat is V.

Rearrange the Equation (V) to obtain A.

A=m˙ρV (VI)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant (R) of helium is 2.0769kJ/kgK or 2.0769kpam3/kgK.

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat ratio (k) of helium is 1.667.

Write the formula of ratio of stagnation pressure to the static pressure at exit of the nozzle.

P02P2=(1+k12Ma22)k/(k1)P2=P02(22+(k1)Ma22)k/(k1) (VII)

Here, the actual (static) pressure at the exit of nozzle is P2, and Mach number of helium gas at the exit of nozzle is Ma2.

Write the formula of ratio of stagnation temperature to the static temperature at exit of the nozzle.

T02T2=1+k12Ma22T02T2=2+(k1)Ma222T2=T02(22+(k1)Ma22) (VIII)

Here, the actual (static) temperature at the exit of nozzle is T2.

Write the formula for velocity of sound at the exit conditions.

c2=kRT2

Here, speed of sound at the exit condition is c2, gas constant of helium is R.

Write formula for the velocity of helium at exit.

V2=Ma2c2=Ma2kRT2 (IX)

Write the formula for mass flow rate of helium at exit condition.

m˙=A2V2v2=A2V2P2RT2 (X)

Here, the exit cross sectional area is A2 and exit specific volume is v2.

Rearrange the Equation (X) to obtain A2.

A2=m˙RT2V2P2 (XI)

Conclusion:

Substitute 500K for T0, and 1.667 for k in Equation (I).

T=500K(21.667+1)=500K(0.7499)=374.95K=375K

Substitute 0.8MPa for P0, and 1.667 for k in Equation (II).

P=0.8MPa(21.667+1)1.667/(1.6671)=0.8MPa(0.7499)2.499=0.3897MPa

Substitute 0.3897MPa for P, 2.0769kpam3/kgK for R, and 375K for T in Equation (III).

ρ=0.3897MPa(2.0769kpam3/kgK)(375K)=0.3897MPa×103kPa1MPa778.8375kpam3/kg=389.7kPa778.8375kpam3/kg=0.5003kg/m3

Substitute 1.667 for k, 2.0769kJ/kgK for R, and 375K for T in Equation (IV).

V=(1.667)(2.0769kJ/kgK)(375K)=1298.3221kJ/kg×1000m2/s21kJ/kg=1139.4394m/s1139.4m/s

Substitute 1139.4m/s for V, 0.34kg/s for m˙, and 0.5003kg/m3 for ρ in

Equation (VI).

A=0.34kg/s(0.5003kg/m3)(1139.4m/s)=0.34kg/s570.0418kg/m2s=5.9645×104m2×1cm2104m25.96cm2

Thus, the throat area of nozzle is 5.96cm2.

Substitute 0.8MPa for P02, 1.667 for k, and 0.1MPa for P2 in Equation (VII).

0.8MPa0.1MPa=(1+1.66712Ma22)1.667/(1.6671)8=(1+0.3335Ma22)2.499(8)1/2.499=1+0.3335Ma220.3335Ma22=2.2981

Ma2=1.29820.3335Ma2=1.973

Here, the downstream Mach number (Ma2) is 1.973 that is greater than the Mach number of 1. Hence, the given nozzle must be a converging-diverging nozzle.

Substitute 500K for T02, and 1.667 for k, and 1.973 for Ma2 in Equation (VIII).

T2=500K(22+(1.6671)(1.973)2)=500K(0.4351)=217.5592K217.6K

Substitute 1.973 for Ma2, 1.667 for k, 2.0769kJ/kgK for R, and 217.6K for T2 in Equation (IX).

V2=(1.973)(1.667)(2.0769kJ/kgK)(217.6K)=(1.973)753.373kJ/kg×1000m2/s21kJ/kg=1.973(867.9706m/s)=1712.5061m/s

1712.5m/s

Substitute 0.34kg/s for m˙, 0.1MPa for P2, 1712.5m/s for V2, 2.0769kpam3/kgK for R, and 217.6K for T2 in Equation (XI).

A2=(0.34kg/s)(2.0769kpam3/kgK)(217.6K)(1712.5m/s)(0.1MPa×103kPa1MPa)=153.6574kPam3/s171250kPam/s=8.9726×104m2×1cm2104m28.97cm2

Thus, the exit area is 8.97cm2.

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Chapter 17 Solutions

Thermodynamics: An Engineering Approach

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