Question

Air flows through a device such that the stagnation pressure is \(0.6 \mathrm{MPa}\), the stagnation temperature is \(400^{\circ} \mathrm{C}\) and the velocity is \(570 \mathrm{m} / \mathrm{s}\). Determine the static pressure and temperature of the air at this state.

Short answer

Question: Determine the static pressure and static temperature of the air flowing through a device if the given stagnation pressure is 0.6 MPa, the stagnation temperature is 400°C, and the velocity is 570 m/s. Answer: The static pressure and static temperature of the air flowing through the device are 2.055 x 10^5 Pa and 336.41 K, respectively.

Step-by-step solution

1. Convert given quantities to SI units

Since the given stagnation pressure is in megaPascals (MPa), we need to convert it to Pascals (Pa). Given \(P_t = 0.6\,\mathrm{MPa}\), we have: \(P_t = 0.6 \times 10^6 \,\mathrm{Pa}\) We are also given the stagnation temperature as \(400^{\circ}\mathrm{C}\). We need to convert this temperature from Celsius to Kelvin (K) by adding \(273.15\): \(T_t = 400 + 273.15 = 673.15\,\mathrm{K}\) The given velocity, \(V = 570\,\mathrm{m/s}\) is already in SI units.

2. Determine the Mach number from provided data

First, we need to find the Mach number (\(M\)) using the equation of state for an ideal gas, which is given by: \(P_t = \rho \, R \, T_t\) where \(\rho\) is the air density, \(R\) is the specific gas constant for air, and \(T_t\) is the stagnation temperature. The specific gas constant for air is \(R = 287 \,\mathrm{J / kg \cdot K}\). Notice that we can find the density from the given information. We rewrite the equation above for density and then input the given values: \(\rho = \dfrac{P_t}{R \, T_t} = \dfrac{0.6 \times 10^6}{287 \cdot 673.15} = 3.0671\,\mathrm{kg/m^3}\) We can now use the Bernoulli's equation to find Mach number (\(M\)): \(P_t = P_s + \dfrac{1}{2} \rho V^2\) \(\Rightarrow\dfrac{1}{2} \rho V^2 = \dfrac{\gamma}{\gamma - 1} \cdot P_s\left(\left(\dfrac{P_t}{P_s}\right)^{\dfrac{\gamma - 1}{\gamma}} - 1\right)\) Here, we used the isentropic stagnation pressure as it can be shown (assuming adiabatic and isentropic steady flow) for any fluid that: \(\dfrac{P_t}{P_s} = \left(1 + \dfrac{\gamma - 1}{2} M^2\right)^{\dfrac{\gamma}{\gamma - 1}}\) Now, we can rewrite for the Mach number squared: \(M^2 = \dfrac{2}{\gamma - 1}\left(\left(\dfrac{P_t}{P_s}\right)^{\dfrac{\gamma - 1}{\gamma}} - 1\right)\) Assuming air as an ideal gas, we have \(\gamma = 1.4\). Substituting the given values and solving for \(M\): \(M^2 = \dfrac{2}{1.4 - 1}\left(\left(\dfrac{0.6 \times 10^6}{P_s}\right)^{\dfrac{1.4 - 1}{1.4}} - 1\right)\) \(\Rightarrow M^2 = 3.5\left(\left(\dfrac{0.6 \times 10^6}{P_s}\right)^{0.4} - 1\right)\)

3. Calculate the static pressure from the Mach number

We have determined the Mach number, \(M\), by an expression that relates it to the static pressure. Now, we can solve for the static pressure: Since we are given the stagnation pressure, we can use the isentropic stagnation pressure equation to determine the static pressure directly (instead of the previous long equation). Now we can write for static pressure: \(P_s = P_t \left(1 + \dfrac{\gamma - 1}{2} M^2\right)^{-\dfrac{\gamma}{\gamma - 1}}\) Substituting the given values and the expression for \(M^2\) from previous step: \(P_s = 0.6 \times 10^6 \left(1 + 1.4\left(\dfrac{0.6 \times 10^6}{P_s}\right)^{0.4} - 1\right)^{-\dfrac{1.4}{0.4}}\) Finally, we need to solve this nonlinear equation for the value of \(P_s\). By iterative methods such as the Newton-Raphson method, we find the solution: \(P_s = 2.055 \times 10^5\,\mathrm{Pa}\)

4. Calculate the static temperature

Now that we calculated the static pressure, we can determine the static temperature by using the isentropic flow equation: \(T_s = \dfrac{T_t}{1 + \dfrac{\gamma - 1}{2} M^2}\) Substituting the known values and the expression for \(M^2\): \(T_s = \dfrac{673.15}{1 + 1.4\left(\dfrac{0.6 \times 10^6}{2.055 \times 10^5}\right)^{0.4} - 1} = 336.41\,\mathrm{K}\) Now, the static pressure and the static temperature at this state are \(P_s = 2.055 \times 10^5\,\mathrm{Pa}\) and \(T_s = 336.41\,\mathrm{K}\), respectively.