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?

To determine

The throat and exit area of the nozzle.

Answer to Problem 131RP

The throat area of nozzle is $5.96{\text{cm}}^2$.

The exit area is $8.97{\text{cm}}^2$.

Explanation of Solution

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

$\begin{array}{l}{T}_1=T_{01}=500\text{K}\\ P_1=P_{01}=0.8\text{MPa}\end{array}$

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

$\begin{array}{l}{T}_{01}=T_{02}=500\text{K}\\ P_{01}=P_{02}=0.8\text{MPa}\end{array}$

Write the formula for the critical temperature of the mixture.

$T^{\ast }=T_0\left(\frac{2}{k+1}\right)$ (I)

Here, the critical temperature of mixture is $T^{\ast }$, the ratio of specific heats is k, and stagnation temperature of mixture is $T_0$.

Write the formula for the critical pressure of the mixture.

$P^{\ast }=P_0{\left(\frac{2}{k+1}\right)}^{k/\left(k-1\right)}$ (II)

Here, the critical pressure of mixture is $P^{\ast }$, and the stagnation pressure of mixture is $P_0$.

Write the formula for the critical density.

${\rho }^{\ast }=\frac{P^{\ast }}{R{T}^{\ast }}$ (III)

Here, the critical density of mixture is ${\rho }^{\ast }$, and the gas constant of helium is $R$.

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

$V^{\ast }=\sqrt{kR{T}^{\ast }}$ (IV)

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

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

$\dot{m}={\rho }^{\ast }{A}^{\ast }{V}^{\ast }$ (V)

Here, the cross sectional area of the throat is $A^{\ast }$, and the velocity at throat is $V^{\ast }$.

Rearrange the Equation (V) to obtain $A^{\ast }$.

$A^{\ast }=\frac{\dot{m}}{{\rho }^{\ast }{V}^{\ast }}$ (VI)

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

The gas constant $\left(R\right)$ of helium is $2.0769\text{kJ/kg}\cdot \text{K}$ or $2.0769\text{kpa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$.

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

The specific heat ratio $\left(k\right)$ of helium is $1.667$.

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

$\begin{array}{l}\frac{P_{02}}{P_2}={\left(1+\frac{k-1}{2}{\text{Ma}}_2^2\right)}^{k/\left(k-1\right)}\\ P_2=P_{02}{\left(\frac{2}{2+\left(k-1\right){\text{Ma}}_2^2}\right)}^{k/\left(k-1\right)}\end{array}$ (VII)

Here, the actual (static) pressure at the exit of nozzle is $P_2$, and Mach number of helium gas at the exit of nozzle is ${\text{Ma}}_2$.

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

$\begin{array}{l}\frac{T_{02}}{T_2}=1+\frac{k-1}{2}{\text{Ma}}_2^2\\ \frac{T_{02}}{T_2}=\frac{2+\left(k-1\right){\text{Ma}}_2^2}{2}\\ T_2=T_{02}\left(\frac{2}{2+\left(k-1\right){\text{Ma}}_2^2}\right)\end{array}$ (VIII)

Here, the actual (static) temperature at the exit of nozzle is $T_2$.

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

$c_2=\sqrt{kR{T}_2}$

Here, speed of sound at the exit condition is $c_2$, gas constant of helium is R.

Write formula for the velocity of helium at exit.

$\begin{array}{c}{V}_2={\text{Ma}}_2{c}_2\\ ={\text{Ma}}_2\sqrt{kR{T}_2}\end{array}$ (IX)

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

$\begin{array}{c}\dot{m}=\frac{A_2{V}_2}{v_2}\\ =\frac{A_2{V}_2{P}_2}{R{T}_2}\end{array}$ (X)

Here, the exit cross sectional area is $A_2$ and exit specific volume is $v_2$.

Rearrange the Equation (X) to obtain $A_2$.

$A_2=\frac{\dot{m}R{T}_2}{V_2{P}_2}$ (XI)

Conclusion:

Substitute $500\text{K}$ for $T_0$, and 1.667 for k in Equation (I).

$\begin{array}{c}{T}^{\ast }=500\text{K}\left(\frac{2}{1.667+1}\right)\\ =500\text{K}\left(0.7499\right)\\ =374.95\text{K}\\ =375\text{K}\end{array}$

Substitute $0.8\text{MPa}$ for $P_0$, and 1.667 for k in Equation (II).

$\begin{array}{c}{P}^{\ast }=0.8\text{MPa}{\left(\frac{2}{1.667+1}\right)}^{1.667/\left(1.667-1\right)}\\ =0.8\text{MPa}{\left(0.7499\right)}^{2.499}\\ =0.3897\text{MPa}\end{array}$

Substitute $0.3897\text{MPa}$ for $P^{\ast }$, $2.0769\text{kpa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for R, and $375\text{K}$ for $T^{\ast }$ in Equation (III).

$\begin{array}{c}{\rho }^{\ast }=\frac{0.3897\text{MPa}}{\left(2.0769\text{kpa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(375\text{K}\right)}\\ =\frac{0.3897\text{MPa}\times \frac{{10}^3\text{kPa}}{1\text{MPa}}}{778.8375\text{kpa}\cdot {\text{m}}^3\text{/kg}}\\ =\frac{389.7\text{kPa}}{778.8375\text{kpa}\cdot {\text{m}}^3\text{/kg}}\\ =0.5003{\text{kg/m}}^3\end{array}$

Substitute 1.667 for $k$, $2.0769\text{kJ/kg}\cdot \text{K}$ for R, and $375\text{K}$ for $T^{\ast }$ in Equation (IV).

$\begin{array}{c}{V}^{\ast }=\sqrt{\left(1.667\right)\left(2.0769\text{kJ/kg}\cdot \text{K}\right)\left(375\text{K}\right)}\\ =\sqrt{1298.3221\text{kJ/kg}\times \frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kJ}/\text{kg}}}\\ =1139.4394\text{m}/\text{s}\\ \simeq 1139.4\text{m}/\text{s}\end{array}$

Substitute $1139.4\text{m}/\text{s}$ for $V^{\ast }$, $0.34\text{kg}/\text{s}$ for $\dot{m}$, and $0.5003{\text{kg/m}}^3$ for ${\rho }^{\ast }$ in

Equation (VI).

$\begin{array}{c}{A}^{\ast }=\frac{0.34\text{kg}/\text{s}}{\left(0.5003{\text{kg/m}}^3\right)\left(1139.4\text{m}/\text{s}\right)}\\ =\frac{0.34\text{kg}/\text{s}}{570.0418{\text{kg/m}}^2\cdot \text{s}}\\ =5.9645\times {10}^{-4}{\text{m}}^2\times \frac{1{\text{cm}}^2}{{10}^{-4}{\text{m}}^2}\\ \simeq 5.96{\text{cm}}^2\end{array}$

Thus, the throat area of nozzle is $5.96{\text{cm}}^2$.

Substitute $0.8\text{MPa}$ for $P_{02}$, 1.667 for $k$, and $0.1\text{MPa}$ for $P_2$ in Equation (VII).

$\begin{array}{l}\frac{0.8\text{MPa}}{0.1\text{MPa}}={\left(1+\frac{1.667-1}{2}{\text{Ma}}_2^2\right)}^{1.667/\left(1.667-1\right)}\\ 8={\left(1+0.3335{\text{Ma}}_2^2\right)}^{2.499}\\ {\left(8\right)}^{1/2.499}=1+0.3335{\text{Ma}}_2^2\\ \text{0}\text{.3335}{\text{Ma}}_2^2=2.298-1\end{array}$

$\begin{array}{l}{\text{Ma}}_2=\sqrt{\frac{1.2982}{0.3335}}\\ {\text{Ma}}_2=1.973\end{array}$

Here, the downstream Mach number $\left({\text{Ma}}_2\right)$ is $1.973$ that is greater than the Mach number of 1. Hence, the given nozzle must be a converging-diverging nozzle.

Substitute $500\text{K}$ for $T_{02}$, and 1.667 for $k$, and $1.973$ for ${\text{Ma}}_2$ in Equation (VIII).

$\begin{array}{c}{T}_2=500\text{K}\left(\frac{2}{2+\left(1.667-1\right){\left(1.973\right)}^2}\right)\\ =500\text{K}\left(0.4351\right)\\ =217.5592\text{K}\\ \simeq \text{217}\text{.6}\text{K}\end{array}$

Substitute $1.973$ for ${\text{Ma}}_2$, 1.667 for $k$, $2.0769\text{kJ/kg}\cdot \text{K}$ for R, and $\text{217}\text{.6}\text{K}$ for $T_2$ in Equation (IX).

$\begin{array}{c}{V}_2=\left(1.973\right)\sqrt{\left(1.667\right)\left(2.0769\text{kJ/kg}\cdot \text{K}\right)\left(\text{217}\text{.6}\text{K}\right)}\\ =\left(1.973\right)\sqrt{753.373\text{kJ/kg}\times \frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kJ/kg}}}\\ =1.973\left(867.9706\text{m}/\text{s}\right)\\ =1712.5061\text{m}/\text{s}\end{array}$

$\simeq 1712.5\text{m}/\text{s}$

Substitute $0.34\text{kg}/\text{s}$ for $\dot{m}$, $0.1\text{MPa}$ for $P_2$, $1712.5\text{m}/\text{s}$ for $V_2$, $2.0769\text{kpa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for R, and $\text{217}\text{.6}\text{K}$ for $T_2$ in Equation (XI).

$\begin{array}{c}{A}_2=\frac{\left(0.34\text{kg}/\text{s}\right)\left(2.0769\text{kpa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(\text{217}\text{.6}\text{K}\right)}{\left(1712.5\text{m}/\text{s}\right)\left(0.1\text{MPa}\times \frac{{10}^3\text{kPa}}{1\text{MPa}}\right)}\\ =\frac{153.6574\text{kPa}\cdot {\text{m}}^3\text{/s}}{171250\text{kPa}\cdot \text{m}/\text{s}}\\ =8.9726\times {10}^{-4}{\text{m}}^2\times \frac{1{\text{cm}}^2}{{10}^{-4}{\text{m}}^2}\\ \simeq 8.97{\text{cm}}^2\end{array}$

Thus, the exit area is $8.97{\text{cm}}^2$.