Question

Air at \(0.9 \mathrm{MPa}\) and \(400 \mathrm{K}\) enters a converging nozzle with a velocity of \(180 \mathrm{m} / \mathrm{s}\). The throat area is \(10 \mathrm{cm}^{2} .\) Assuming isentropic flow, calculate and plot the mass flow rate through the nozzle, the exit velocity, the exit Mach number, and the exit pressure-stagnation pressure ratio versus the back pressure-stagnation pressure ratio for a back pressure range of \(0.9 \geq P_{b} \geq 0.1 \mathrm{MPa}\).

Short answer

Answer: The mass flow rate calculated for this converging nozzle is approximately \(2.555\:kg/s\).

Step-by-step solution

Calculate the stagnation pressure and temperature

As the flow is isentropic, we can use the isentropic flow relationships for an ideal gas. The relevant relationships are: $$ \frac{T_0}{T}=\left(1+\frac{\gamma - 1}{2} M^2\right) $$ $$ \frac{P_0}{P}=\left(1+\frac{\gamma - 1}{2} M^2\right)^\frac{\gamma}{\gamma-1} $$ Here, \(\gamma\) is the specific heat ratio (\(\gamma=1.4\) for air), \(M\) is the Mach number, and \(T_0\) and \(P_0\) are the stagnation temperature and pressure, respectively. - At the entrance of the nozzle, we have: $$ P_1=0.9\:MPa,\: T_1=400\:K,\: v_1=180\:m/s $$ - To find the Mach number at the entrance, we can use : $$ M_1 = \frac{v_1}{\sqrt{\gamma R T_1}} $$ where R is the specific gas constant for air, R=287 J/kg.K. Therefore, $$ M_1 \approx 0.7206. $$ - Now we can use the isentropic flow relationships to find \(P_{01}\) and \(T_{01}\). Here \(P_{01}=P_1\left(1+\frac{\gamma - 1}{2} M_1^2\right)^\frac{\gamma}{\gamma-1}\) and \(T_{01}=T_1\left(1+\frac{\gamma - 1}{2} M_1^2\right)\). So, $$ P_{01} \approx 1.409\:MPa,\: T_{01}\approx 479.95\:K. $$

Determine the mass flow rate through the nozzle

The mass flow rate through the nozzle can be determined using the following equation: $$ \dot{m} = \rho_1A_1v_1 = \frac{P_1A_1v_1}{RT_1}, $$ where \(A_1\) is the throat area and \(\rho_1\) is the air density at the inlet. Using the given values, we find $$ \dot{m} \approx 2.555\:kg/s. $$

Calculate the exit velocity and Mach number

- At the critical state (throat), the Mach number is 1 (M=1). We can use the isentropic flow relationships to find the stagnation pressure ratio (\(P_{02}/P_1\)) at this point. $$ \frac{P_{02}}{P_1}=\left(1+\frac{\gamma - 1}{2} \right)^\frac{\gamma}{\gamma-1}. $$ - Now the exit pressure, \(P_2\), will vary due to the range of back-pressure \(P_b\): $$ 0.9\:MPa\geq P_{b}\geq 0.1\:MPa. $$ As \(P_{2}\leq P_{02}\) and \(P_{02} = P_{01}\), the highest value of the exit pressure would be obtained when back pressure equals the highest limit possible for choked flow which is \(0.9\:MPa\). As \(P_{02}\approx 1.409\:MPa\), we need to find a range of pressure ratios \(\frac{P_{02}}{P_2}\) that would satisfy the given range of variation of back pressure. - For each value of \(P_b\), we can calculate the exit Mach number using the isentropic flow relationship: $$ M_2=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_{02}}{P_2}\right)^\frac{\gamma-1}{\gamma}-1\right]}. $$ - With the Mach number, we can find the isentropic exit velocity: $$ v_2 = M_2\sqrt{\gamma RT_2}. $$

Plot the requested ratios over the back pressure range

To plot the ratios of interest (\(\frac{\dot{m}}{A_2v_2}\), \(\frac{v_2}{v_{02}}\), \(\frac{P_2}{P_{02}}\)), we need to evaluate them over the specified range of back pressure (\(P_b\)) using: 1. Calculate the exit pressure (\(P_{2}\)) within the specified range. 2. Find the corresponding Mach number (\(M_2\)) using the isentropic flow relationships. 3. Calculate the required ratios. Finally, create plots for each of the ratios as a function of \(\frac{P_b}{P_{02}}\), where \(P_{02}\) is the exit stagnation pressure. Calculus and graph plotting software can be used to easily create these plots, once you have calculated the necessary ratios. This process will provide the flow properties of air through the converging nozzle under isentropic flow conditions. Keep in mind that factors such as heat transfer, friction, and air quality are not accounted for in this basic analysis.