Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 17.7, Problem 137RP

Helium expands in a nozzle from 220 psia, 740 R, and negligible velocity to 15 psia. Calculate the throat and exit areas for a mass flow rate of 0.2 lbm/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 137RP

The throat area of nozzle is 8.19×104ft2.

The exit area is 0.00164ft2.

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=740RP1=P01=220psia

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

T01=T02=704RP01=P02=220psia

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-1E, “Molar mass, gas constant, and critical-point properties”.

The gas constant (R) of helium is 2.6809psiaft3/lbmR or 0.4961Btu/lbmR.

Refer Table A-2E, “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 air is R.

Write formula for the velocity of air at exit.

V2=Ma2c2=Ma2kRT2 (IX)

Write the formula for mass flow rate of air 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 740R for T0, and 1.667 for k in Equation (I).

T=740R(21.667+1)=740R(0.7499)=554.93R

Substitute 220psia for P0, and 1.667 for k in Equation (II).

P=220psia(21.667+1)1.667/(1.6671)=220psia(0.7499)2.499=107.158psia107.2psia

Substitute 107.2psia for P, 2.6809psiaft3/lbmR for R, and 554.93R for T in Equation (III).

ρ=107.2psia(2.6809psiaft3/lbmR)(554.93R)=107.2psia1487.7118psiaft3/lbm=0.07205lbm/ft3

Substitute 1.667 for k, 0.4961Btu/lbmR for R, and 554.93R for T in Equation (IV).

V=(1.667)(0.4961Btu/lbmR)(554.93R)=458.9264Btu/lbm×25037ft2/s21Btu/lbm=3389.711ft/s3390ft/s

Substitute 3390ft/s for V, 0.2lbm/s for m˙, and 0.07205lbm/ft3 for ρ in

Equation (VI).

A=0.2lbm/s(0.07205lbm/ft3)(3390ft/s)=0.2lbm/s244.2495lbm/ft2s=8.1883×104ft28.19×104ft2

Thus, the throat area of nozzle is 8.19×104ft2.

Substitute 220psia for P02, 1.667 for k, and 15psia for P2 in Equation (VII).

220psia15psia=(1+1.66712Ma22)1.667/(1.6671)14.6667=(1+0.3335Ma22)2.499(14.6667)1/2.499=1+0.3335Ma220.3335Ma22=2.9291

Ma2=1.9290.3335Ma2=2.405

Here, the downstream Mach number (Ma2) is 2.405 that is greater than the Mach number of 1. Hence, the given nozzle must be a converging-diverging nozzle.

Substitute 740R for T02, and 1.667 for k, and 2.405 for Ma2 in Equation (VIII).

T2=740R(22+(1.6671)(2.405)2)=740R(0.3414)=252.648R252.6R

Substitute 2.405 for Ma2, 1.667 for k, 0.4961Btu/lbmR for R, and 252.6R for T2 in Equation (IX).

V2=(2.405)(1.667)(0.4961Btu/lbmR)(252.6R)=(2.405)208.9396Btu/lbm×25037ft2/s21Btu/lbm=2.405(2286.9687ft/s)=5500.15ft/s

5500ft/s

Substitute 0.2lbm/s for m˙, 15psia for P2, 5500ft/s for V2, 2.6809psiaft3/lbmR for R, and 252.6R for T2 in Equation (XI).

A2=(0.2lbm/s)(2.6809psiaft3/lbmR)(252.6R)(5500ft/s)(15psia)=135.4391psiaft3/s82500psiaft/s=1.6416×103ft2=0.0016416ft2

0.00164ft2

Thus, the exit area is 0.00164ft2.

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Chapter 17 Solutions

Thermodynamics: An Engineering Approach

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