Air Turbine — Power and Exit Area: 04-BS-10, December 2018, Q8

Question 8: Air Turbine — Power and Exit Area (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\dot m=10$ kg/s. Inlet: $P_1=500$ kPa, $T_1=900$ K, $V_1\approx0$. Exit: $P_2=100$ kPa, $T_2=600$ K, $V_2=100$ m/s. Adiabatic, steady, negligible PE.

Find. (a) $\dot W$ [kW]; (b) $A_2$ [m²].

Approach

Apply the steady-flow energy equation with the exit kinetic-energy term retained (the inlet term is explicitly negligible) using real variable-specific-heat air enthalpies, then size the exit area from the ideal-gas continuity equation at the exit state.

Power developed (part a). Steady-flow energy balance, $\dot Q\approx0$, $\Delta pe\approx0$: $$\dot W=\dot m\left[(h_1-h_2)-\frac{V_2^2-V_1^2}{2}\right] =\dot m\left[(h_1-h_2)-\frac{V_2^2}{2}\right]$$ using real air enthalpies $h(900\ \text{K})=1059.40$, $h(600\ \text{K})=733.42$ kJ/kg: $$\dot W=10\left[(1059.40-733.42)-\frac{100^2}{2\times1000}\right]=10\,[325.99-5.00]=10\times320.99$$ $$=\boxed{3209.9\ \text{kW}}.$$
Exit specific volume and area (part b). Ideal gas at the exit state: $$v_2=\frac{RT_2}{P_2}=\frac{0.287\times600}{100}=1.7220\ \text{m}^3/\text{kg}.$$ Continuity, $\dot m=\rho_2 A_2V_2=A_2V_2/v_2$: $$A_2=\frac{\dot m\,v_2}{V_2}=\frac{10\times1.7220}{100}=\boxed{0.1722\ \text{m}^2}.$$

Quantity	Result
(a) $\dot W$	3209.9 kW
(b) $A_2$	0.1722 m²

Concept & Theory

Topic. Steady-flow turbine with a non-negligible exit kinetic-energy term, sized against real (variable-specific-heat) air properties.

Key relations. Steady-flow energy equation: $\dot W=\dot m[(h_1-h_2)+\tfrac12(V_1^2-V_2^2)]$. Ideal-gas continuity for exit area: $A_2=\dot m v_2/V_2$ with $v_2=RT_2/P_2$.

Why this works. The problem explicitly flags the inlet velocity as negligible relative to the exit velocity, not zero in an absolute sense — the exit kinetic-energy term is retained because a 100 m/s exit is a genuine few-percent correction (5.00 kJ/kg against a 326 kJ/kg enthalpy drop, about 1.5%), which the problem is testing the student not to drop by reflex.

Common pitfall. Sizing the exit area from the wrong state's properties (e.g. using $T_1,P_1$ instead of $T_2,P_2$) — the continuity equation must use the density and velocity at the SAME cross-section (the exit), never a mix of inlet and exit conditions.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 (Mass and Energy Analysis of Control Volumes).

Question 8 of 9: Air Turbine — Power and Exit Area

Question 8: Air Turbine — Power and Exit Area (15 marks)

Approach