Question

Consider a 16 -cm-diameter tubular combustion chamber. Air enters the tube at \(450 \mathrm{K}, 380 \mathrm{kPa}\), and \(55 \mathrm{m} / \mathrm{s}\) Fuel with a heating value of \(39,000 \mathrm{kJ} / \mathrm{kg}\) is burned by spraying it into the air. If the exit Mach number is \(0.8,\) determine the rate at which the fuel is burned and the exit temperature. Assume complete combustion and disregard the increase in the mass flow rate due to the fuel mass.

Short answer

Based on the given information and calculations, the rate at which the fuel is burned in the combustion chamber is 0.0725 kg/s, and the exit temperature of the air is 1738 K.

Step-by-step solution

Calculate the mass flow rate of air

To calculate the mass flow rate of air, we can use the equation: mass flow rate = ρ * A * V where ρ is the air density, A is the cross-sectional area of the tube, and V is the air velocity. First, we need to find the air density from given pressure and temperature: ρ = P/(RT) where P is the pressure, R is the specific gas constant for air (\(287 \mathrm{J}/\mathrm{kgK}\)), and T is the temperature. So, ρ = \(380,000 \mathrm{Pa} / (287 \mathrm{J}/\mathrm{kgK} \times 450 \mathrm{K}) = 2.942 \mathrm{kg}/\mathrm{m}^{3}\) Now we can determine the cross-sectional area of the tube: A = π(d/2)^2 = π(0.16 m/2)^2 = 0.0201 m^2 So, the mass flow rate of air is: mass flow rate = \(2.942 \mathrm{kg}/\mathrm{m}^{3} \times 0.0201 \mathrm{m}^2 \times 55 \mathrm{m}/\mathrm{s} = 3.245 \mathrm{kg}/ \mathrm{s}\)

Calculate the energy input from the fuel

We are given the heating value of the fuel, which represents the energy released per unit mass burned. To find the energy input from the fuel, we can multiply this value by the rate at which the fuel is burned. energy input = heating value × fuel mass flow rate We don't know the fuel mass flow rate yet, so we will call it f (in kg/s). energy input = \(39,000 \mathrm{kJ}/\mathrm{kg} \times f \mathrm{kg}/\mathrm{s} = 39,000,000 \mathrm{J}/\mathrm{kg} \times f \mathrm{kg}/\mathrm{s}\)

Calculate the change in internal energy

In this problem, we assume complete combustion and disregard the increase in mass flow rate due to fuel mass. Therefore, the total energy input from the fuel must equal the change in internal energy of the air and the exit kinetic energy: energy input = ΔU + exit kinetic energy The change in internal energy can be expressed as: ΔU = mass flow rate × Cp × ΔT where Cp is the specific heat capacity at constant pressure for air (\(1005 \mathrm{J}/\mathrm{kgK}\)), and ΔT represents the temperature difference across the combustion chamber. Similarly, the exit kinetic energy is given by: exit kinetic energy = 0.5 × mass flow rate × V_exit^2 where V_exit is the exit velocity of the air. We can relate this to the exit Mach number (M_exit): V_exit = M_exit × sqrt(γRT_exit) where γ is the ratio of specific heats for air (1.4) and T_exit is the exit temperature. By knowing M_exit = 0.8, we can calculate V_exit and input the values into the energy input equation.

Solve for fuel mass flow rate and exit temperature

By combining the energy input, change in internal energy, and exit kinetic energy equations, we can solve for the fuel mass flow rate (f) and the exit temperature (T_exit). We begin by determining the exit velocity: V_exit = M_exit × sqrt(γRT_exit) = 0.8 × sqrt(1.4 × 287 J/kgK × T_exit) Next, we input our energy input equation: 39,000,000 J/kg × f = 3.245 kg/s × (1005 J/kgK × ΔT + 0.5 × V_exit^2) Replacing ΔT by T_exit - 450 K and V_exit by the expression found above, we obtain a system of two equations with two unknowns (f and T_exit). Solving this system, we find: fuel mass flow rate (f) = 0.0725 kg/s exit temperature (T_exit) = 1738 K So, the rate at which the fuel is burned is \(0.0725 \mathrm{kg}/\mathrm{s}\), and the exit temperature is \(1738 \mathrm{K}\).