Flash-Chamber Separation and Steam Turbine Power: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2017, Q2

Question 2: Flash-Chamber Separation and Steam Turbine Power

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 water at 300 kPa is throttled (adiabatically, no work) in a flash chamber to 150 kPa, where it separates by gravity into a saturated-liquid stream (drained from the bottom) and a saturated-vapour stream (drawn off the top); the vapour stream alone feeds a steam turbine of isentropic efficiency 90% that exhausts at 15 kPa.

Given data
Quantity	Symbol	Value
Total mass flow into flash chamber	$\dot m_1$	100 kg/s
Flash-chamber inlet pressure	$P_1$	300 kPa (sat. liquid)
Flash-chamber exit pressure	$P_2$	150 kPa
Turbine isentropic efficiency	$\eta_t$	90%
Turbine exhaust pressure	$P_4$	15 kPa

Find. The turbine power output $\dot W_t$, with the processes sketched on a $T$–$s$ diagram.

Flash chamber fed by saturated liquid at 300 kPa; the saturated-vapour stream drawn off the top alone feeds the steam turbine.

Approach. The flash chamber is an adiabatic throttle (no work, negligible ΔKE/ΔPE), so its steady-flow energy equation collapses to $h_1=h_2$; splitting that mixed enthalpy between the saturated-liquid and saturated-vapour states at 150 kPa gives the vapour quality $x_2$, and hence the mass flow rate actually entering the turbine. The turbine is then a standard isentropic-efficiency expansion from saturated vapour at 150 kPa to 15 kPa.

Flash-chamber energy balance — vapour fraction. Sat. liquid at 300 kPa: $h_1=561.43\text{ kJ/kg}$. At 150 kPa: $h_f=467.13\text{ kJ/kg}$, $h_g=2693.11\text{ kJ/kg}$. Since the throttle conserves enthalpy, $h_1=h_2=x_2h_g+(1-x_2)h_f$, so $$x_2=\frac{h_1-h_f}{h_g-h_f}=\frac{561.43-467.13}{2693.11-467.13}$$ $$\boxed{x_2=0.0424}$$ The vapour stream feeding the turbine is therefore $\dot m_{vap}=x_2\dot m_1=0.0424\times100=4.236\text{ kg/s}$ (the remaining 95.76 kg/s leaves as saturated liquid and is not used further).
Turbine inlet and isentropic exit state. State 3 (turbine inlet) is saturated vapour at 150 kPa: $h_3=h_g=2693.11\text{ kJ/kg}$, $s_3=s_g=7.2230\text{ kJ/kg}\cdot\text{K}$. Expanding isentropically to 15 kPa ($s_{4s}=s_3$), with $s_f=0.7549$, $s_g=7.2536\text{ kJ/kg}\cdot\text{K}$ at 15 kPa: $$\begin{aligned} x_{4s}&=\frac{s_3-s_f}{s_g-s_f}=0.8919 \\ h_{4s}&=h_f+x_{4s}h_{fg}=225.99+0.8919\times2373.14=2341.79\text{ kJ/kg} \end{aligned}$$ so the isentropic specific work is $$w_s=h_3-h_{4s}=2693.11-2341.79=351.32\text{ kJ/kg}$$
Actual turbine work and power. With $\eta_t=90\%$, $$w_a=\eta_tw_s=0.90\times351.32=316.18\text{ kJ/kg}$$ $$\dot W_t=\dot m_{vap}\,w_a=4.236\times316.18$$ $$\boxed{\dot W_t=1339.5\text{ kW}\approx1.34\text{ MW}}$$ (the actual exit state is wet steam, $x_{4a}=0.907$, confirming the turbine exhausts safely inside the two-phase region rather than as superheated vapour.)

$T$–$s$ diagram: 1 (sat. liquid, 300 kPa) → 2 (throttling to 150 kPa, $x=0.042$) · the mixture splits into the saturated-vapour branch 2→3 (150 kPa) that feeds the turbine, expanding to 4s (isentropic) / 4a (actual, $\eta_t=90\%$) at 15 kPa.

Question 2 — results
Quantity	Value
Flash quality $x_2$	0.0424
Vapour mass flow to turbine $\dot m_{vap}$	4.236 kg/s
Isentropic specific work $w_s$	351.32 kJ/kg
Actual specific work $w_a$	316.18 kJ/kg
Turbine power $\dot W_t$	1339.5 kW (1.34 MW)

Topic. Adiabatic flash separation feeding a steam turbine with isentropic efficiency.

Key relations. Throttling (SFEE, no work/heat, negligible ΔKE/ΔPE): $h_{in}=h_{out}$; two-phase quality $x=(h-h_f)/h_{fg}$; turbine efficiency $\eta_t=(h_3-h_4)/(h_3-h_{4s})$.

Why this works. A flash chamber is physically just a throttle valve opening into a larger vessel: no shaft crosses its boundary and it is well insulated, so the SFEE forces $h_1=h_2$ regardless of the internal flow details. Gravity then does the separation for free — the lower-density vapour collects at the top while the liquid drains from the bottom — so each exit stream is a distinct SATURATED state at the new (lower) pressure, not a single mixed-quality stream leaving through one port.

Common pitfall. Feeding the turbine with the flash chamber's full 100 kg/s instead of only the vapour fraction $x_2\dot m_1$ that physically enters the turbine — this is the single most consequential slip in this problem, since the turbine power scales directly with that mass flow rate.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4–5 (throttling devices, flash chambers, steam turbines and isentropic efficiency).

Question 2 of 8: Flash-Chamber Separation and Steam Turbine Power

Question 2: Flash-Chamber Separation and Steam Turbine Power