Flash chamber feeding a steam turbine: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q2

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. Saturated liquid at 300 kPa is flashed to 150 kPa; the vapour fraction expands in a turbine ($\eta_T=0.90$) to 15 kPa. Feed $\dot m = 100$ kg/s. Find. the turbine power.

Figure 2 — Left: saturated liquid is throttled (flashed) from 300 to 150 kPa and the flash vapour expands through the turbine to 15 kPa. Right: the same processes on a T–s diagram, drawn to scale against the water saturation dome. The throttle 1 → (2,3) is at constant enthalpy, so entropy increases while the temperature falls from 133.5 to 111.4 °C; states 2 and 3 are the two ends of the 150 kPa tie line. The turbine line 3→4s is vertical (isentropic, dashed) and the actual η = 0.90 exit 4 lies just to its right at the same 15 kPa saturation temperature.

Approach. Treat the flash chamber as an adiabatic throttle (energy balance fixes the vapour fraction), then apply the isentropic-efficiency definition to the turbine on the flashed vapour stream.

Flash (throttling) balance across the chamber. With no work or heat and negligible KE/PE, enthalpy is conserved. Saturated liquid feed has $h_1 = h_f(300\ \text{kPa}) = 561.4$ kJ/kg. At 150 kPa, $h_f = 467.1$, $h_g = 2693.1$ kJ/kg, so the vapour fraction (quality) leaving is $$x = \frac{h_1 - h_f}{h_g - h_f} = \frac{561.4-467.1}{2693.1-467.1} = 0.0424$$
Vapour mass flow to the turbine. $$\dot m_v = x\,\dot m = (0.0424)(100) = 4.24\ \text{kg/s}$$ $\dot m_v = 4.24$ kg/s (the balance of 95.8 kg/s leaves as saturated liquid).
Ideal turbine exit at 15 kPa. The inlet is saturated vapour at 150 kPa: $h_3 = 2693.1$ kJ/kg, $s_3 = 7.223$ kJ/kg·K. For the isentropic exit, $s_{4s}=s_3$; at 15 kPa ($s_f=0.7549$, $s_g=8.0071$): $$x_{4s} = \frac{7.223-0.7549}{8.0071-0.7549} = 0.892,\qquad h_{4s} = 225.9 + (0.892)(2372.3) = 2341.8\ \text{kJ/kg}$$
Actual specific work via isentropic efficiency. $$w = \eta_T\,(h_3 - h_{4s}) = 0.90\,(2693.1 - 2341.8) = 316.2\ \text{kJ/kg}$$
Turbine power. $$\dot W = \dot m_v\,w = (4.24)(316.2) = 1339\ \text{kW}$$ $\dot W \approx 1.34\ \text{MW}$

Quantity	Result
Flash vapour fraction, $x$	0.0424
Vapour flow to turbine, $\dot m_v$	4.24 kg/s
Ideal / actual specific work	351.3 / 316.2 kJ/kg
Turbine power, $\dot W$	≈ 1.34 MW (1339 kW)

Concept & Theory

Topic. Single-stage flash separation followed by an adiabatic steam turbine with isentropic efficiency.

Key relations. Throttle: $h_\text{out}=h_\text{in}$; lever rule $x=(h-h_f)/h_{fg}$; isentropic exit $s_{4s}=s_3$; turbine efficiency $\eta_T = (h_3-h_4)/(h_3-h_{4s})$; power $\dot W = \dot m_v (h_3-h_4)$.

Why this works. The flash chamber does no shaft work and is insulated, so the feed enthalpy simply redistributes into saturated liquid and vapour at the lower pressure; the fraction that becomes vapour is set by how far the feed enthalpy sits above $h_f$. Only that vapour reaches the turbine, so the plant power is the small vapour flow times the (efficiency-derated) isentropic enthalpy drop.

Common pitfall. Using the full 100 kg/s in the turbine instead of only the flashed vapour, or applying $\eta_T$ to the enthalpies before finding the isentropic exit state.

Source. Çengel & Boles, Thermodynamics §5 (throttling and separators) and §7–§8 (isentropic efficiency of turbines).

Question 2 of 8: Flash chamber feeding a steam turbine

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