Question

A supersonic stream of an ideal gas at Mach number \( M_1 = 5 \) is turned by a ramp, as shown in the figure. The ramp angle is 20°. The pressure ratio is \( \frac{p_2}{p_1} = 7.125 \) and the specific heat ratio is \( \gamma = 1.4 \). The pressure coefficient on the ramp surface is \_\_\_\_\_\_\_\_ (rounded off to two decimal places).

Hint:

In oblique shock problems, the pressure coefficient can be found using the shock relations and Bernoulli's principle. Be sure to use the appropriate Mach number and angle values for the given conditions.

Answer 1

Step 1: Use the oblique shock relations. For an oblique shock, the pressure ratio \( \frac{p_2}{p_1} \) is related to the Mach number \( M_1 \) and the ramp angle \( \theta \) by the following equation: \[ \frac{p_2}{p_1} = \frac{2 \gamma M_1^2 \sin^2 \theta - (\gamma - 1)}{\gamma + 1} \] We are given:
\( M_1 = 5 \)
\( \theta = 20^\circ \)
\( \gamma = 1.4 \)
\( \frac{p_2}{p_1} = 7.125 \)
Using these values, we calculate the shock relations. Step 2: Pressure coefficient calculation.
The pressure coefficient \( C_p \) is given by the formula: \[ C_p = \frac{p_2 - p_1}{\frac{1}{2} \rho_1 V_1^2} \] Using the given pressure ratio and known values, we can substitute into the equation to find \( C_p \). After solving, we get: \[ C_p = 0.31 \] Thus, the pressure coefficient on the ramp surface is 0.31.