Two-Stage Steam Turbine with Extraction and Reheat: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, Undated paper, Q2

Question 2: Two-Stage Steam Turbine with Extraction and Reheat

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. A high-pressure (HP) stage and a low-pressure (LP) stage share one shaft; steam enters the HP stage at state 1, exits at the SAME state (nodes 2 and 3 are literally the same split point, both at $P=0.4$ MPa, $T=300\,{}^{\circ}\text{C}$): a fraction $\xi$ leaves the turbine's own control volume to be reheated in the boiler and returns at state 4, while the remaining $(1-\xi)$ continues directly into the LP stage; the two streams recombine inside the LP stage and leave together at state 5.

Given data
State	Pressure	Temperature	Role
1	4.0 MPa	$650\,{}^{\circ}\text{C}$	HP inlet
2	0.4 MPa	$300\,{}^{\circ}\text{C}$	HP exit / extraction to reheat ($\xi\dot m$ leaves)
3	0.4 MPa	$300\,{}^{\circ}\text{C}$	HP exit / bypass to LP ($(1-\xi)\dot m$, same state as 2)
4	0.4 MPa	$550\,{}^{\circ}\text{C}$	reheated steam re-enters LP ($\xi\dot m$ returns)
5	7.5 kPa	$x_5=0.96$	combined LP exit
$\dot m=5.0$ kg/s (mass entering HP at state 1); $\dot W=7500$ kW (total shaft power, both stages)

[Figure not reproduced: Two-stage turbine with extraction/reheat schematic, as printed in the source exam. See the official exam paper or the cited reference text.]

As printed in the source: steam $\dot m$ leaves the boiler at state 1 and enters the HP stage; at the HP/LP junction the flow splits into $\xi\dot m$ (state 2, routed back through the boiler's reheat coil) and $(1-\xi)\dot m$ (state 3, direct to LP); the reheated stream re-enters at state 4 and the combined flow leaves the LP stage at state 5.

Find. (a) the extraction fraction $\xi$; (b) the entropy generated by the turbine per unit mass entering at state 1; (c) the turbine's overall isentropic efficiency.

Approach. Take the WHOLE two-stage turbine (both blade paths, not the boiler) as one adiabatic control volume with mass/energy crossing its boundary at 1 (in), 2 (out, to reheat), 4 (in, reheated) and 5 (out): energy balance gives one equation in the one unknown $\xi$. The SAME control volume, being adiabatic, has an entropy balance with no heat term, so its entropy generation is the bracketed inflow/outflow sum directly. The isentropic efficiency compares the actual 7500 kW to the work of a turbine at the same states 1 and 4, same $\xi$, expanding each stream ISENTROPICALLY to its own exit pressure.

Steam properties at all five states.

State	$h$ (kJ/kg)	$s$ (kJ/kg·K)
1 — 4.0 MPa, $650\,{}^{\circ}\text{C}$	3790.08	7.4988
2 = 3 — 0.4 MPa, $300\,{}^{\circ}\text{C}$	3067.08	7.5677
4 — 0.4 MPa, $550\,{}^{\circ}\text{C}$	3593.59	8.3287
5 — 7.5 kPa, $x=0.96$	2477.82	7.9431

Part (a) — energy balance on the turbine control volume. Mass and energy crossing the boundary: in at 1 ($\dot m$) and 4 ($\xi\dot m$); out at 2 ($\xi\dot m$) and 5 ($\dot m$): $$\dot m h_1+\xi\dot m h_4=\xi\dot m h_2+\dot m h_5+\dot W$$ Solving for $\xi$: $$\xi=\frac{\dot W-\dot m(h_1-h_5)}{\dot m(h_4-h_2)} =\frac{7500-5.0\times(3790.08-2477.82)}{5.0\times(3593.59-3067.08)}$$ $$\boxed{\xi=0.357\ (35.7\%)}$$
Part (b) — entropy generation (adiabatic turbine, $\dot Q=0$). Per unit mass entering at state 1 (divide the same in/out bookkeeping by $\dot m$): $$\frac{\dot S_{gen}}{\dot m}=\xi(s_2-s_4)+(s_5-s_1) =0.357\times(7.5677-8.3287)+(7.9431-7.4988)$$ $$\boxed{\frac{\dot S_{gen}}{\dot m}=0.173\text{ kJ/kg}\cdot\text{K}}$$ Positive, as the second law requires for a real (irreversible) adiabatic turbine.
Part (c) — isentropic reference states. HP stage isentropic exit ($s_{2s}=s_1$, at $P_2$): $h_{2s}=3028.27$ kJ/kg. The bypass fraction, having left the HP stage at this same entropy, expands isentropically through the LP stage to $P_5$: $h_{5s,bypass}=2338.56$ kJ/kg. The reheated fraction enters the LP stage at state 4's own entropy $s_4$ and expands isentropically to $P_5$: $h_{5s,reheat}=2599.17$ kJ/kg.
Part (c) — isentropic work and efficiency. $$\dot W_s=\dot m(h_1-h_{2s})+(1-\xi)\dot m(h_{2s}-h_{5s,bypass})+\xi\dot m(h_4-h_{5s,reheat})$$ $$\dot W_s=5.0\times(3790.08-3028.27)+0.643\times5.0\times(3028.27-2338.56)+0.357\times5.0\times(3593.59-2599.17)$$ $$\boxed{\dot W_s=7801\text{ kW}}$$ $$\eta_T=\frac{\dot W}{\dot W_s}=\frac{7500}{7801}$$ $$\boxed{\eta_T=0.961\ (96.1\%)}$$

T–s state points: 1 (HP inlet, superheated) expands to the shared split point 2,3; the extracted fraction is reheated at constant 0.4 MPa up to 4, then both streams expand together through the LP stage down to the wet-steam exit at 5.

Question 2 — results
Quantity	Value
Extraction fraction $\xi$	0.357 (35.7%)
Entropy generation per unit mass at 1	0.173 kJ/kg·K
Isentropic work $\dot W_s$	7801 kW
Overall isentropic efficiency $\eta_T$	96.1%

Topic. Steam-turbine control-volume analysis with an internal extraction (feed to reheat) and a returning reheated stream — a first/second-law bookkeeping problem, not a simple single-inlet/single-outlet turbine.

Key relations. Steady-flow energy balance $\sum\dot m_{in}h_{in}=\sum\dot m_{out}h_{out}+\dot W$; adiabatic entropy balance $\dot S_{gen}=\sum\dot m_{out}s_{out}-\sum\dot m_{in}s_{in}\ge0$; isentropic efficiency $\eta_T=\dot W_{actual}/\dot W_s$.

Why this works. Drawing the control volume around the WHOLE turbine (both blade paths) rather than trying to track a single mass flow lets one energy equation solve directly for $\xi$, because the reheater itself sits outside the boundary and its duty never has to be computed. The same boundary choice makes the entropy balance trivial: with $\dot Q=0$ identically, whatever entropy imbalance appears between the four crossing streams IS the generation, by definition.

Common pitfall. Treating states 2 and 3 as physically different points (they are the same node, split only in mass fraction) and thereby double-counting a property lookup; and computing isentropic efficiency from a single $s_1\to s_5$ expansion, which ignores that the reheated fraction re-enters the LP stage at $s_4\ne s_1$.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 & 7 (turbine control volumes, entropy generation, isentropic efficiency).

Question 2 of 8: Two-Stage Steam Turbine with Extraction and Reheat

Question 2: Two-Stage Steam Turbine with Extraction and Reheat