Question

Compressed air from the compressor of a gas turbine enters the combustion chamber at \(T_{1}=700 \mathrm{K}, P_{1}\) \(=600 \mathrm{kPa},\) and \(\mathrm{Ma}_{1}=0.2\) at a rate of \(0.3 \mathrm{kg} / \mathrm{s} .\) Via combustion, heat is transferred to the air at a rate of \(150 \mathrm{kJ} / \mathrm{s}\) as it flows through the duct with negligible friction. Determine the Mach number at the duct exit, and the drop in stagnation pressure \(P_{01}-P_{02}\) during this process.

Short answer

Answer: To find the Mach number at the exit of the combustion chamber and the drop in stagnation pressure, we use the energy equation, speed of sound equation, and isentropic relations. We first apply the energy equation with heat addition and find the initial velocity using the given Mach number. Next, we find the final temperature and pressure by solving the energy equation and isentropic relations using an iterative process. We then calculate the final velocity and the exit Mach number. Finally, we calculate the drop in stagnation pressure using the isentropic relations. This process gives us the Mach number at the exit of the combustion chamber and the drop in stagnation pressure during the combustion process in the gas turbine.

Step-by-step solution

Apply the energy equation for heat addition

The energy equation is given as $$ q = c_p(T_2 - T_1) + \frac{V_2^2 - V_1^2}{2} $$ Where \(q\) is the heat transfer per unit mass, \(c_p\) is the specific heat at constant pressure, \(T_1\) and \(T_2\) are the initial and final temperatures, and \(V_1\) and \(V_2\) are the initial and final velocities. Now, we convert the given heat transfer rate to heat transfer per unit mass $$ q = \frac{150 kJ/s}{0.3 kg/s} = 500 kJ/kg $$

Calculate the initial velocity \(V_1\)

To find \(V_1\), we will use the Mach number and the speed of sound equation $$ Ma_1 = \frac{V_1}{a_1} $$ Where \(a_1 = \sqrt{\gamma R T_1}\) is the speed of sound in the inlet, \(\gamma\) is the adiabatic constant, and \(R\) is the specific gas constant. For air, we have \(\gamma = 1.4\) and \(R = 287 J/(kg K)\). $$ a_1 = \sqrt{1.4 \times 287 \times 700} = 566.8 m/s $$ Then, $$ V_1 = Ma_1 \times a_1 = 0.2 \times 566.8 = 113.4 m/s $$

Find \(T_2\) and \(P_2\) using the energy equation

We can rewrite the energy equation as $$ T_2 = T_1 + \frac{q - \frac{V_2^2 - V_1^2}{2}}{c_p} $$ Where \(c_p = \frac{\gamma R}{\gamma - 1} = 1.006 kJ/(kg K)\) for air. We can simplify the equation if we define a new variable: \(Z = \frac{V_2^2 - V_1^2}{2}\), then $$ T_2 = T_1 + \frac{q - Z}{c_p} $$ Now we know that the ratio of specific heats (\(\gamma\)) is constant, so we can write the following equation for the final pressure \(P_2\): $$ \frac{P_1}{\rho_1^\gamma} = \frac{P_2}{\rho_2^\gamma} $$ Where \(\rho_1\) and \(\rho_2\) are the initial and final densities, which can be found using the ideal gas law \(P=\rho RT\). $$ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{1}{\gamma-1}} $$

Calculate \(V_2\) and the Mach number at the exit \(Ma_2\)

Since \(q, T_1, V_1, \gamma\), and \(c_p\) are known (from the problem statement and previous steps), we can use trial and error to find \(Z\) that gives us the right-hand side of the equation equal to \(T_2/T_1\). One way to do this is to use an iterative process, where we test different values of \(Z\) until we find the desired ratio. Once we have found \(Z\), we can then find \(V_2\): $$ V_2 = \sqrt{V_1^2 + 2Z} $$ Now, we can calculate the Mach number at the exit using: $$ Ma_2 = \frac{V_2}{a_2} $$ Where \(a_2 = \sqrt{\gamma R T_2}\) is the speed of sound at the exit.

Calculate the drop in stagnation pressure \(P_{01}-P_{02}\)

Finally, we can find the drop in stagnation pressure using the isentropic relations for \(\gamma\) being constant: $$ \frac{P_{01}}{P_1} = \left(\frac{\rho_{1}}{\rho_0}\right)^\gamma $$ and $$ \frac{P_{02}}{P_2} = \left(\frac{\rho_{2}}{\rho_0}\right)^\gamma $$ With \(\rho_{1} = \frac{P_1}{R T_1}\) and \(\rho_{2} = \frac{P_2}{R T_2}\), we can calculate the stagnation pressures \(P_{01}\) and \(P_{02}\), then find the drop in stagnation pressure: $$ P_{01}-P_{02} = P_{01} - P_{02} $$ With this, we have found the Mach number at the duct exit and the drop in stagnation pressure during the combustion process.