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22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2019

Question 2 of 8: Flash chamber feeding a steam turbine

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

Notes on this paper

Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas processes and entropy generation, polytropic compression, rigid-vessel charging, the reciprocating air compressor, throttling/flash separation, the steam turbine, and vapour-compression refrigeration/heat-pump cycles; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — steady radial conduction through a cylindrical wall with convection, conduction with uniform internal generation, transient (lumped) cooling by combined convection and radiation, and the effectiveness–NTU method for s​hell-and-tube exchangers. Steam and Freon-12 (R-12) properties are evaluated, which reproduces the IAPWS steam tables and the standard R-12 property tables to graphing accuracy; enthalpy differences (the only quantities used) are datum-independent. Air and combustion gases are treated as ideal with constant specific heats ($\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005$, $c_v=0.718\ \text{kJ/kg}\cdot\text{K}$).

Question 2 — Flash chamber feeding a steam turbine (Part A, equal value)

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.

QuantityValue
Feed: saturated liquid water, 300 kPa$h_\text{in}=h_f(300\,\text{kPa})=561.4$ kJ/kg
Feed rate$\dot m=100$ kg/s
Flash pressure (sat. liq. + sat. vap. exit)150 kPa: $h_f=467.1,\ h_g=2693.1$ kJ/kg
Turbine isentropic efficiency$\eta_T=0.90$
Turbine exhaust pressure15 kPa

Find. Turbine shaft power.

Flashchamber feed 300 kPa100 kg/s (sat. liq.) vapour 150 kPa, 4.24 kg/s sat. liquid 150 kPa, 95.8 kg/s turbine 15 kPa W = 1.34 MW T–s (vapour path) sT 300 3 (150 kPa, sat. vap.) 4s 4
Only the flashed vapour (4.24 kg/s) reaches the turbine; the 95.8 kg/s of saturated liquid is drawn off the bottom and does no shaft work.

Approach. The flash chamber is an adiabatic throttle, so an energy balance on it fixes the vapour fraction; only that vapour enters the turbine, where the isentropic drop to 15 kPa is scaled by $\eta_T$ and multiplied by the vapour flow.

  1. Vapour fraction leaving the flash chamber (energy balance). The chamber is adiabatic with no work, so $h_\text{in}=x\,h_g+(1-x)h_f$ at 150 kPa: $$x=\frac{h_\text{in}-h_f}{h_g-h_f}=\frac{561.4-467.1}{2693.1-467.1}=0.0424$$
  2. Vapour mass flow to the turbine. $$\dot m_v=x\,\dot m=(0.0424)(100)=\boxed{4.236\ \text{kg/s}}\qquad(\text{liquid drawn off}=95.8\ \text{kg/s})$$
  3. Isentropic exhaust state at 15 kPa. The turbine inlet is saturated vapour at 150 kPa, $s_3=s_g=7.223\ \text{kJ/kg}\cdot\text{K}$. Expanding at constant $s$ to 15 kPa ($s_f=0.7549,\ s_g=8.0085$): $$x_{4s}=\frac{s_3-s_f}{s_g-s_f}=\frac{7.223-0.7549}{8.0085-0.7549}=0.892,\quad h_{4s}=h_f+x_{4s}h_{fg}=2341.8\ \text{kJ/kg}$$
  4. Actual enthalpy drop. $$\Delta h_s=h_3-h_{4s}=2693.1-2341.8=351.3\ \text{kJ/kg},\quad \Delta h=\eta_T\,\Delta h_s=0.90(351.3)=316.2\ \text{kJ/kg}$$
  5. Turbine power. $$\dot W=\dot m_v\,\Delta h=(4.236)(316.2)=\boxed{1339\ \text{kW}\approx1.34\ \text{MW}}$$
QuantityResult
Flash vapour fraction $x$0.0424
Vapour flow to turbine4.236 kg/s
Isentropic / actual enthalpy drop351.3 / 316.2 kJ/kg
Turbine power1339 kW (≈1.34 MW)