Chapter 12, Problem 136P

To determine

The throat area of nozzle.

The exit areas of nozzle.

That the condition for the nozzle be converging-diverging.

Answer to Problem 136P

The throat area of nozzle is $6.46{\text{cm}}^{\text{2}}$.

The exit area of the nozzle is $10.8{\text{cm}}^{\text{2}}$.

The mach number at the exit is greater than $1$ that is why the nozzle must be converging-diverging.

Explanation of Solution

Given information:

The nozzle is isentropic, the pressure of helium gas in the nozzle is $1\text{MPa}$, the temperature of helium gas in the nozzle is $500\text{K}$, the exit pressure is $0.1\text{MPa}$, the velocity is negligible, and the mass flow rate is $0.46\text{kg}/\text{s}$.

Helium is an ideal gas with constant specific heats, flow through the nozzle is steady, one dimensional, isentropic and the specific heat constant is $1.667$.

Write the expression for the critical temperature.

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

Here, the stagnation temperature is $T_0$ and the specific heat ratio is $k$.

Write the expression for the critical pressure.

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

Here, the stagnation pressure is $P_0$.

Write the expression for the critical density.

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

Here, the critical pressure is $P^{\ast }$, the gas constant is $R,$ and the critical temperature is $T^{\ast }$.

Write the expression for the critical velocity.

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

Write the expression for throat area.

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

Here, the mass flow rate is $\dot{m}$, the critical density is ${\rho }^{\ast }$ and the critical velocity is $V^{\ast }$.

Write the expression for the mach number at the exit of the nozzle.

   $\frac{P_0}{P_2}={\left(1+\frac{k-1}{2}M{a}_2^2\right)}^{k/\left(k-1\right)}$    ...... (VI)

Here the exit pressure is $P_2$ and the mach number at the exit is $M{a}_2$.

Write the expression for the exit temperature.

   $T_2=T_0\left(\frac{2}{2+\left(k-1\right)M{a}_2^2}\right)$    ...... (VII)

Write the expression for the exit density.

   ${\rho }_2=\frac{P_2}{R{T}_2}$    ...... (VIII)

Here, the exit pressure is $P_2$ and the exit temperature is $T_2$.

Write the expression for the exit velocity.

   $V_2=M{a}_2\sqrt{kR{T}_2}$    ...... (IX)

Write the expression for the exit area of the nozzle.

   $A_2=\frac{\dot{m}}{{\rho }_2{V}_2}$    ...... (X)

Here, the exit density is ${\rho }_2$ and the exit velocity is $V_2$.

Calculation:

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

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

Substitute $1.667$ for $k$ and $1\text{MPa}$ for $P_0$ in the Equation (II).

   $\begin{array}{c}{P}^{\ast }=\left(1\text{MPa}\right){\left(\frac{2}{1.667+1}\right)}^{1.667/\left(1.667-1\right)}\\ =\left(1\text{MPa}\right){\left(0.7499\right)}^{2.4992}\\ =\left(1\text{MPa}\right)\left(0.48708\right)\\ =0.487\text{MPa}\end{array}$

Substitute $0.487\text{MPa}$ for $P^{\ast }$, $375\text{K}$ for $T^{\ast },$ and $2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$ in the Equation (III).

   $\begin{array}{c}{\rho }^{\ast }=\frac{0.487\text{MPa}}{\left(2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(375\text{K}\right)}\\ =\frac{0.487\text{MPa}\left(\frac{1000\text{kPa}}{1\text{MPa}}\right)}{\left(2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(375\text{K}\right)}\\ =\frac{0.487\text{kPa}}{\left(2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(375\text{K}\right)}\\ =0.625\text{kg}/{\text{m}}^{\text{3}}\end{array}$

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

   $\begin{array}{c}{V}^{\ast }=\sqrt{\left(1.667\right)\left(2.0769\text{kJ}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(375\text{K}\right)}\\ =\sqrt{\left(625.125\text{K}\right)\left(2.0769\text{kJ}/\text{kg}\left(\frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kJ}/\text{kg}}\right)\text{K}\right)}\\ =1139.4\text{m}/\text{s}\end{array}$

Substitute $1139.4\text{m}/\text{s}$ for $V^{\ast }$, $0.625\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }^{\ast }$ and $0.46\text{kg}/\text{s}$ for $\dot{m}$ in the Equation (V).

   $\begin{array}{c}{A}^{\ast }=\frac{0.46\text{kg}/\text{s}}{\left(0.625\text{kg}/{\text{m}}^{\text{3}}\right)\left(1139.4\text{m}/\text{s}\right)}\\ =\frac{0.46\text{kg}/\text{s}}{712.125\text{kg}/{\text{m}}^{\text{2}}\text{s}}\\ =6.460\times {10}^{-4}{\text{m}}^{\text{2}}\left(\frac{10000{\text{cm}}^{\text{2}}}{1{\text{m}}^{\text{2}}}\right)\\ =6.46{\text{cm}}^{\text{2}}\end{array}$

Substitute $1\text{MPa}$ for $P_0$, $0.1\text{MPa}$ for $P_2$ and $1.667$ for $k$ in the Equation (VI).

   $\begin{array}{l}\frac{1\text{MPa}}{0.1\text{MPa}}={\left(1+\frac{1.667-1}{2}M{a}_2^2\right)}^{1.667/\left(1.667-1\right)}\\ 10={\left(1+0.3335M{a}_2^2\right)}^{2.4992}\\ M{a}_2=2.130\end{array}$

Substitute $500\text{K}$ for $T_0$, $2.13$ for $Ma$ and $1.667$ for $k$ in the Equation (VII).

   $\begin{array}{c}{T}_2=\left(500\text{K}\right)\left(\frac{2}{2+\left(1.667-1\right){\left(2.13\right)}^2}\right)\\ =\left(500\text{K}\right)\left(0.39792\right)\\ =199\text{K}\end{array}$

Substitute $199\text{K}$ for $T_2$, $2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R,$ and $0.1\text{MPa}$ for $P_2$ in the Equation (VIII).

   $\begin{array}{c}{\rho }_2=\frac{0.1\text{MPa}}{\left(2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(199\text{K}\right)}\\ =\frac{0.1\text{MPa}\left(\frac{1000\text{kPa}}{1\text{MPa}}\right)}{\left(2.0769\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(199\text{K}\right)}\\ =\frac{100\text{kPa}}{\left(413.3031\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)}\\ =0.242\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $2.13$ for $Ma$, $1.667$ for $k$, $2.0769\text{kJ}/\text{kg}\cdot \text{K}$ for $R,$ and $199\text{K}$ for $T_2$ in the Equation (IX).

   $\begin{array}{c}{V}_2=2.13\sqrt{\left(1.667\right)\left(2.0769\text{kJ}/\text{kg}\cdot \text{K}\right)\left(199\text{K}\right)}\\ =2.13\sqrt{\left(331.733\text{K}\right)\left(2.0769\text{kJ}/\text{kg}\cdot \text{K}\left(\frac{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{kJ}/\text{kg}}\right)\right)}\\ =1768.0\text{m}/\text{s}\end{array}$

Substitute $1768.0\text{m}/\text{s}$ for $V_2$, $0.242\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_2$ and $0.46\text{kg}/\text{s}$ for $\dot{m}$ in the Equation (X).

   $\begin{array}{c}{A}_2=\frac{0.46\text{kg}/\text{s}}{\left(0.242\text{kg}/{\text{m}}^{\text{3}}\right)\left(1768.0\text{m}/\text{s}\right)}\\ =\frac{0.46\text{kg}/\text{s}}{\left(427.856\text{kg}/{\text{m}}^2\cdot \text{s}\right)}\\ =0.1075\times {10}^{-3}{m}^2\left(\frac{10000{\text{cm}}^{\text{2}}}{1{\text{m}}^{\text{2}}}\right)\\ =10.8{\text{cm}}^{\text{2}}\end{array}$

Conclusion:

The throat area of nozzle is $6.46{\text{cm}}^{\text{2}}$.

The exit area of the nozzle is $10.8{\text{cm}}^{\text{2}}$.

The mach number at the exit is greater than $1$ that is why the nozzle must be converging-diverging.