Question

Air at \(200 \mathrm{kPa}, 100^{\circ} \mathrm{C},\) and Mach number \(\mathrm{Ma}=0.8\) flows through a duct. Calculate the velocity and the stagnation pressure, temperature, and density of the air

Short answer

Answer: The velocity of the air is 277.96 m/s, the stagnation pressure is 259743.77 Pa, the stagnation temperature is 439.84 K, and the stagnation density is 1.928 kg/m³.

Step-by-step solution

Calculate the speed of sound in the air

To calculate the speed of sound in the air (\(a\)), we use the following formula: $$ a = \sqrt{\gamma R T} $$ where \(\gamma\) is the ratio of specific heats (1.4 for air), \(R\) is the specific gas constant for air (\(287\, J/(kg \cdot K)\)), and \(T\) is the temperature in Kelvin (\(100^\circ C\) is equal to \(373.15\,K\)). $$ a = \sqrt{1.4 \cdot 287 \cdot 373.15} = 347.45\,m/s $$

Calculate the velocity of the air

Now that we have the speed of sound in the air, we can calculate the velocity of the air (\(V\)) using the given Mach number (Ma = 0.8): $$ V = \mathrm{Ma} \cdot a = 0.8 \cdot 347.45 = 277.96\,m/s $$

Calculate the stagnation pressure

Using the isentropic flow relations for perfect gases, we can find the stagnation pressure (\(P_0\)) given the Mach number and pressure. The formula for stagnation pressure is: $$ \frac{P_0}{P} = \left(1 + \frac{\gamma - 1}{2} \mathrm{Ma}^2 \right)^\frac{\gamma}{\gamma -1} $$ Rearranging the equation for \(P_0\): $$ P_0 = P \left(1 + \frac{\gamma - 1}{2} \mathrm{Ma}^2 \right)^\frac{\gamma}{\gamma -1} $$ Using the given pressure (\(P = 200\,kPa\) or \(200000\,Pa\)): $$ P_0 = 200000 \cdot \left(1 + \frac{1.4 - 1}{2} \cdot 0.8^2 \right)^\frac{1.4}{1.4 -1} = 259743.77\,Pa $$

Calculate the stagnation temperature

Similarly, we can calculate the stagnation temperature (\(T_0\)) using the isentropic flow relations for perfect gases and the given temperature: $$ \frac{T_0}{T} = 1 + \frac{\gamma - 1}{2} \mathrm{Ma}^2 $$ Rearranging the equation for \(T_0\): $$ T_0 = T \cdot \left(1 + \frac{\gamma - 1}{2} \mathrm{Ma}^2 \right) $$ Using the given temperature in Kelvin (\(T = 373.15\,K\)): $$ T_0 = 373.15 \cdot \left(1 + \frac{1.4 - 1}{2} \cdot 0.8^2 \right) = 439.84\,K $$

Calculate the density of the air

We can now find the density of the air (\(\rho\)) using the Ideal Gas Law: $$ \rho = \frac{P}{R T} $$ Using the given pressure and temperature: $$ \rho = \frac{200000}{287 \cdot 373.15} = 1.806\,kg/m^3 $$

Calculate the stagnation density

Lastly, we can calculate the stagnation density (\(\rho_0\)) using the isentropic flow relations for perfect gases: $$ \frac{\rho_0}{\rho} = \left(1 + \frac{\gamma - 1}{2} \mathrm{Ma}^2 \right)^\frac{1}{\gamma -1} $$ Rearranging the equation for \(\rho_0\): $$ \rho_0 = \rho \cdot \left(1 + \frac{\gamma - 1}{2} \mathrm{Ma}^2 \right)^\frac{1}{\gamma -1} $$ Using the calculated density of the air: $$ \rho_0 = 1.806 \cdot \left(1 + \frac{1.4 - 1}{2} \cdot 0.8^2 \right)^\frac{1}{1.4 -1} = 1.928\,kg/m^3 $$ So, the velocity of the air is \(277.96\,m/s\), the stagnation pressure is \(259743.77\,Pa\), the stagnation temperature is \(439.84\,K\), and the stagnation density is \(1.928\,kg/m^3\).