Chapter 2, Problem 63P

Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s sand leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach number (a) at the inlet and (b) at the exit of the nozzle. Assess the accuracy of the constant specific heat approximation.

To determine

(a)

The Mach at the inlet of the nozzle.

Answer to Problem 63P

The Mach at the inlet of the nozzle is $0.0925$.

Explanation of Solution

Given information:

The nozzle is adiabatic, the inlet temperature carbon dioxide at inlet is $1200\text{K}$, the velocity of carbon dioxide through the nozzle is $50\text{m}/\text{s}$, the outlet temperature of carbon dioxide is $400\text{K}$.

Write the expression for the speed of sound.

  $c=\sqrt{kRT}$...... (I)

Here, the gas constant is $R$, specific heat ratio is $k$ and the temperature is $T$.

Write the expression for the Mach number.

  $M=\frac{v}{c}$...... (II)

Here the velocity of the flow is $v$ and the velocity of the sound is $c$.

Calculation:

Refer to the table A1, "molar mass gas constant and ideal gas specific heat of some substances" to obtain the gas constant for carbon dioxide as $188.9\text{J}/\text{kg}\cdot \text{K}$.

Refer to the table A1, "molar mass gas constant and ideal gas specific heat of some substances" to obtain the specific heat ratio for carbon dioxide as $1.288$.

Substitute $1200\text{K}$ for $T$, $1.288$ for $k$ and $188.9\text{J}/\text{kg}\cdot \text{K}$ for $R$ in Equation (I).

  $\begin{array}{c}c=\sqrt{\left(1.288\right)\left(188.9\text{J}/\text{kg}\cdot \text{K}\right)\left(1200\text{K}\right)}\\ =\sqrt{291963.84\text{J}/\text{kg}}\\ =540.336\text{m}/\text{s}\end{array}$

Substitute $540.336\text{m}/\text{s}$ for $c$ and $50\text{m}/\text{s}$ for $v$ in Equation (II)

  $\begin{array}{c}M=\frac{50\text{m}/\text{s}}{540.336\text{m}/\text{s}}\\ =0.0925\end{array}$

Conclusion:

The Mach at the inlet of the nozzle is $0.0925$.

To determine

(b)

The Mach at the exit of the nozzle.

Answer to Problem 63P

The Mach at the exit of the nozzle is $3.728$.

Explanation of Solution

Write the expression for the energy balance Equation applied at inlet and exit of the nozzle.

  $h_1+\frac{v_1^2}{2}=h_2+\frac{v_2^2}{2}$...... (III)

Here, the velocity at inlet is $v_1$, the enthalpy at inlet is $h_1$, the velocity at outlet is $v_2$ and the enthalpy at outlet is $h_2$

Write the expression for the enthalpy at inlet.

  $h_1=c_p{T}_1$..... (IV)

Here, the specific heat at constant pressure is $c_p$ and the temperature at the inlet is $T_1$.

Write the expression for the enthalpy at Outlet.

  $h_2=c_p{T}_2$..... (V)

Here, the specific heat at constant pressure is $c_p$ and the temperature at the outlet is $T_2$.

Substitute $c_p{T}_1$ for $h_1$ and $c_p{T}_2$ for $h_2$ in Equation (III)

  $\begin{array}{l}\left(c_p{T}_1\right)+\frac{v_1^2}{2}=\left(c_p{T}_2\right)+\frac{v_2^2}{2}\\ \frac{v_2^2}{2}=c_p\left(T_1-T_2\right)+\frac{v_1^2}{2}\end{array}$..... (VI)

Calculation:

Refer to the table A1 "molar mass gas constant and ideal gas specific heat of some substances" to obtain the specific heat for constant pressure for carbon dioxide as $0.8439\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $1200\text{K}$ for $T_1$, $400\text{K}$ for $T_2$, $0.8439\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ and $50\text{m}/\text{s}$ for $v_1$ in Equation (VI).

  $\begin{array}{c}\frac{v_2^2}{2}=\left(0.8439\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(1200\text{K}\right)-\left(400\text{K}\right)\right)+\frac{{\left(50\text{m}/\text{s}\right)}^2}{2}\\ \frac{v_2^2}{2}=675.12\text{kJ}/\text{kg}\left(\frac{1000\text{J}}{1\text{kJ}}\right)+\frac{{\left(50\text{m}/\text{s}\right)}^2}{2}\\ \frac{v_2^2}{2}=676370{\left(\text{m}/\text{s}\right)}^2\\ v_2=1163.073\text{m}/\text{s}\end{array}$

Substitute $400\text{K}$ for $T$, $1.288$ for $k$ and $188.9\text{J}/\text{kg}\cdot \text{K}$ for $R$ in Equation (I).

  $\begin{array}{c}c=\sqrt{\left(1.288\right)\left(188.9\text{J}/\text{kg}\cdot \text{K}\right)\left(400\text{K}\right)}\\ =\sqrt{97321.28\text{J}/\text{kg}}\\ =311.96\text{m}/\text{s}\end{array}$

Substitute $311.96\text{m}/\text{s}$ for $c$ and $1163.073\text{m}/\text{s}$ for $v$ in Equation (II)

  $\begin{array}{c}M=\frac{1163.073\text{m}/\text{s}}{311.96\text{m}/\text{s}}\\ =3.728\end{array}$

Conclusion:

The Mach at the exit of the nozzle is $3.728$.