Question

An ideal gas with \(k=1.4\) is flowing through a nozzle such that the Mach number is 1.8 where the flow area is \(36 \mathrm{cm}^{2} .\) Approximating the flow as isentropic, determine the flow area at the location where the Mach number is 0.9.

Short answer

Answer: The flow area at the location where the Mach number is 0.9 is approximately 25.45 cm².

Step-by-step solution

Recall the isentropic flow relation

For an isentropic flow of an ideal gas, the flow area (A) is related to the Mach number (M) by the following equation: \[A = A^* \left( \frac{1+[(k-1)/2]M^2}{(1+[k-1]/2)} \right)^\frac{k+1}{2(k-1)}\] where \(A^*\) is the critical (throat) area, and \(k\) is the specific heat ratio. In this case, we know that \(k = 1.4\), \(M_1 = 1.8\), and \(A_1 = 36\,\text{cm}^2\). We want to find \(A_2\) for \(M_2 = 0.9\).

Find the ratio \(A_1/A^*\)

First, let's find the ratio of the flow area at Mach number 1.8 (\(A_1\)) to the critical (throat) area (\(A^*\)): \[\frac{A_1}{A^*} = \left( \frac{1+[(1.4-1)/2]1.8^2}{(1+[1.4-1]/2)} \right)^\frac{1.4+1}{2(1.4-1)}\] Calculate this ratio: \[\frac{A_1}{A^*} \approx 1.755\]

Find the ratio \(A_2/A^*\)

Now, let's find the ratio of the flow area at Mach number 0.9 (\(A_2\)) to the critical (throat) area (\(A^*\)): \[\frac{A_2}{A^*} = \left( \frac{1+[(1.4-1)/2]0.9^2}{(1+[1.4-1]/2)} \right)^\frac{1.4+1}{2(1.4-1)}\] Calculate this ratio: \[\frac{A_2}{A^*} \approx 1.243\]

Use the critical area ratios to find the flow area at Mach 0.9

We can now use the area ratios we found to solve for \(A_2\): \[\frac{A_1}{A^*} = \frac{A_2}{A^*} \frac{A_1}{A_2}\] From the previous steps, we know that \(\frac{A_1}{A^*} \approx 1.755\), and \(\frac{A_2}{A^*} \approx 1.243\). We also know that \(A_1 = 36\,\text{cm}^2\). Solving for \(A_2\): \[A_2 = \frac{1.243}{1.755} \times 36\,\text{cm}^2\] Calculate the flow area at Mach 0.9: \[A_2 \approx 25.45\, \text{cm}^2\] So, the flow area at the location where the Mach number is 0.9 is approximately 25.45 cm².