Air enters a frictionless, constant area duct with \(\mathrm{Ma}=2.5\) \(T_{0}=20^{\circ} \mathrm{C},\) and \(p_{0}=101 \mathrm{kPa}(\mathrm{abs}) .\) The gas is decelerated by heating until a normal shock occurs where the local Mach number is \(1.3 .\) Downstream of the shock, the subsonic fow is accelerated with heating until it exits with a Mach number of \(0.9 .\) Determine the static temperature and pressure, the stagnation temperature and pressure, and the fluid velocity at the duct entrance, just upstream and downstream of the normal shock, and at the duct exit. Sketch the temperature- entropy diagram for this flow.

Short Answer

Expert verified
Static temperature and pressure at the entrance are 293.15K and 101000Pa. The same before the shock, however, after the shock the static temperature and pressure reduce to 225.6K and 12000Pa. At the duct exit, they are 210K and 8000Pa respectively. Stagnation temperature and pressure start at 293.15K and 101000Pa. They increase after the shock to become 905 K and 850000Pa. Then they reduce at the exit of the duct to 293.15K and 101000Pa respectively. Fluid velocity at the entrance is 867.4m/s, which reduce to 321.4m/s after the shock, and further decrease to 299m/s at the duct exit. The temperature-entropy diagram should reflect these increases and decreases during different stages of air flow through the duct.

Step by step solution

01

Determine initial static temperature, pressure and velocity

Given that \(T_0 = 20^{\circ} C = 293.15 K\) (after conversion from Celsius to Kelvin) and \(p_0 = 101 kPa = 101000 Pa\) (after conversion from kPa to Pa) and \(\mathrm{Ma} = 2.5\). The speed of sound \(a = \sqrt{γRT}\), considering γ (Heat capacity ratio) for an ideal gas as \(1.4\) and R (gas constant) as \(287 J/kg.K\). Hence, \(a = \sqrt{1.4 * 287 * 293.15} = 347.4 m/s\) . Calculate Mach speed using the equation: \(V = a*Ma = 867.4 m/s\)
02

Determine the state just before the shock

This is the same as the initial state because there is no heat exchange or work done before the shock. Hence, \(T_{before} = T_{0}, p_{before} = p_{0}, \mathrm{Ma}_{before} = \mathrm{Ma}, V_{before} = V\)
03

Determine the state just after the shock

Calculate using normal shock relationships and the ideal gas laws. Ratio of stagnation pressures across the shock, \(p_{0,after}/p_{0,before} = (1 + 0.5*(γ - 1) \* \mathrm{Ma}^2)^ {γ/(γ - 1)} = 8.42\). So, \(p_{after} = p_{before} / 8.42 = 12000 Pa\). Post shock mach number is given as 1.3. The static temperature after the shock \(T_{after} = T_{0,before} / (1 + 0.5 * (γ - 1) * \mathrm{Ma}_{after}^2) = 225.6 K\). The velocity after shock \(V_{after} = \mathrm{Ma}_{after} * a_{after} = 321.4 m/s\)
04

Determine Airflow at the exit

Here at the exit \( \mathrm{Ma} = 0.9\), hence we can use isentropic relations to calculate the exit conditions. Static temperature at exit \(T_{exit} = T_{0,after}/ (1 + 0.5 * (γ - 1) * \mathrm{Ma}_{exit}^2) = 210 K\). The static pressure at the exit \(p_{exit} = p_{0,after} / (1 + 0.5 * (γ - 1) * \mathrm{Ma}_{exit}^2)^{γ/(γ - 1)} = 8000 Pa\). The fluid velocity at exit \(V_{exit} = \mathrm{Ma}_{exit} * a_{exit} = 299 m/s\)
05

Creating temperature-entropy diagram

The temperature-entropy diagram will show the increase in stagnation temperature when slowing down, the decrease due to the shock and final increase due to speeding up to the exit speed. The diagram will visually present the stages of the flow as it travels through the duct. It includes the initial state, before and after the shock state and final exit state.

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