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04-BS-10 · December 2016

Question 8 of 9: Air Turbine with Significant Exit Kinetic Energy

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-10, Thermodynamics — December 2016. 3 hours, Closed-Book Exam (approved calculator and one double-sided 8.5x11-inch aid sheet permitted; property tables and charts supplied in an appendix, interpolation not required). Part A: answer 2 of Questions 1-3 (20 marks each). Part B: answer 4 of Questions 4-9 (15 marks each), for a 100-mark paper. Only the first two Part-A and first four Part-B questions as they appear in the answer book are marked. All nine questions (Part A complete, Part B complete) are solved below for completeness.

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed.; Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 8th ed. All state properties (water/steam, R-134a, air, N₂, CO₂) were computed from high-accuracy equations of state in place of printed property-table interpolation; every boxed numeric result.

Question 8: Air Turbine with Significant Exit Kinetic Energy (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. 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. $\dot m=10$ kg/s.

Find. (a) $\dot W$ [kW]; (b) $A_2$ [m$^2$].

Approach

Include the exit kinetic-energy term in the steady-flow energy balance since the inlet velocity is negligible but the exit is not. The exit area then follows from the mass-flow/continuity relation $\dot m=\rho_2A_2V_2$, using the ideal-gas specific volume at the exit state.

  1. Energy balance (part a). $h_1=h(900\ \text{K})=1059.40$ kJ/kg, $h_2=h(600\ \text{K})=733.42$ kJ/kg. $$w=(h_1-h_2)-\frac{V_2^2}{2}=(1059.40-733.42)-\frac{100^2}{2\times1000}=325.99-5.00 =\boxed{320.99\ \text{kJ/kg}}.$$ $$\dot W=\dot m\,w=10\times320.99=\boxed{3209.9\ \text{kW}}\quad\textbf{(a)}.$$
  2. Exit specific volume and area (part b). Ideal gas at the exit state: $$v_2=\frac{RT_2}{P_2}=\frac{0.287\times600}{100}=1.722\ \text{m}^3/\text{kg}.$$ $$A_2=\frac{\dot m\,v_2}{V_2}=\frac{10\times1.722}{100}=\boxed{0.1722\ \text{m}^2}\quad\textbf{(b)}.$$
QuantityResult
(a) $\dot W$3209.9 kW
(b) $A_2$0.1722 m²