Question

Air flows steadily through an adiabatic turbine, entering at 150 psia, \(900^{\circ} \mathrm{F}\), and \(350 \mathrm{ft} / \mathrm{s}\) and leaving at 20 psia, \(300^{\circ} \mathrm{F}\), and \(700 \mathrm{ft} / \mathrm{s}\). The inlet area of the turbine is \(0.1 \mathrm{ft}^{2} .\) Determine \((a)\) the mass flow rate of the air and \((b)\) the power output of the turbine.

Short answer

Answer: The mass flow rate of the air is 2.184 lb/s, and the power output of the turbine is 446.7 hp.

Step-by-step solution

Convert temperatures to Rankine

To convert temperatures from Fahrenheit to Rankine, we add 459.67: \(T_{1} = 900^{\circ} \mathrm{F} + 459.67 = 1359.67\, \mathrm{R}\) \(T_{2} = 300^{\circ} \mathrm{F} + 459.67 = 759.67\, \mathrm{R}\)

Calculate the mass flow rate

Using the continuity equation, which states that the mass flow rate is constant across the turbine, we get: \(\dot{m} = \rho AV\) We will assume air is an ideal gas, which allows us to use the equation of state: \(\rho = \frac{P} {RT}\) where \(\rho\): density of air \(P\): pressure \(R\): specific gas constant for air (\(53.35 \, \mathrm{ft.lb/(lb.R})\)) \(T\): temperature in Rankine At the inlet of the turbine, \(P_1 = 150\,\mathrm{psia}\) and \(T_1 = 1359.67\,\mathrm{R}\): $\rho_1 = \frac{P_1} {R T_1} = \frac{150 \times 144} {53.35 \times 1359.67} = 0.0624 \,\mathrm{lb/ft^3}$ (we multiply by 144 to convert from psia to lb/ft^2) The inlet area is given, \(A_1 = 0.1\,\mathrm{ft^2}\), and the inlet velocity is given, \(V_1 = 350\,\mathrm{ft/s}\). Now we can calculate the mass flow rate: \(\dot{m} = \rho_1 A_1 V_1 = (0.0624)(0.1)(350) = 2.184\, \mathrm{lb/s}\) (a) The mass flow rate of the air is \(2.184\, \mathrm{lb/s}\).

Calculate the power output

The power output of the adiabatic turbine can be calculated using the adiabatic equation: \(W_{out} = \dot{m}\left(h_1 - h_2 + \frac{1}{2}(V_1^2 - V_2^2)\right)\) We will use the ideal gas equation for enthalpy: \(h = c_p T\) where \(c_p = 0.240\, \mathrm{Btu/lb.R}\) is the specific heat capacity at constant pressure for air. For the inlet, \(h_1 = c_p T_1 = 0.240 \times 1359.67 = 326.32\, \mathrm{Btu/lb}\) For the outlet, \(h_2 = c_p T_2 = 0.240 \times 759.67 = 182.32\, \mathrm{Btu/lb}\) Additionally, the outlet velocity is given as \(V_2 = 700\,\mathrm{ft/s}\). Now we can calculate the power output: \(W_{out} = 2.184 \left((326.32 - 182.32) + \frac{1}{2}(350^2 - 700^2)\frac{1}{778}\right)\) (the 1/778 factor is used to convert ft.lb to Btu) \(W_{out} = 315.85\, \mathrm{Btu/s}\) To convert to horsepower, we use the conversion factor 1 Btu/s = 1.414 horsepower: \(W_{out} = 315.85 \times 1.414 = 446.7\, \mathrm{hp}\) (b) The power output of the turbine is \(446.7\, \mathrm{hp}\).