Steam Plant Turbomachinery: 22-Mec-B3 Energy Conversion and Power Generation, December 2013, Q3

Question 3: Steam Plant Turbomachinery (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.

Part I — Turbine expansion line (10 marks)

Given.

Quantity	Value
Supply steam to the control valve	6 MPa, dry saturated
Pressure after throttling (HP turbine inlet)	3 MPa
HP turbine exhaust pressure	0.6 MPa
Temperature after reheat, at 0.6 MPa	255 °C
LP turbine exhaust pressure	0.004 MPa
Internal efficiency, both turbines	0.80
Steam flow, HP turbine	600 kg/s
Steam flow, LP turbine	500 kg/s

Find. The enthalpy at every key point of the throttling and the two expansions, plotted on the Mollier chart with the ideal and actual lines distinguished; the temperature and moisture at the HP inlet, the HP exit and the LP exit; and the total power output of the machine.

Figure 3.1 — Mollier chart construction. 1→2 throttling at constant enthalpy, 2→3 the HP expansion (3s ideal), 3→4 reheat, 4→5 the LP expansion (5s ideal). Ideal paths dashed, actual paths solid.

Approach. Fix each state in turn on the \(h\)–\(s\) plane: the throttle is a constant-enthalpy horizontal move, each ideal expansion is a vertical drop to the exhaust isobar, and each actual expansion recovers only 80% of that drop, which places the real end state to the right of the ideal one.

Part (a), point 1 — fix the supply state. Dry saturated steam at 6 MPa lies on the saturation line, so from the steam tables \[ h_1 = h_g(6\ \text{MPa}) = 2784.6\ \text{kJ/kg}, \quad s_1 = 5.8901\ \text{kJ/kg}\cdot\text{K}, \quad T_1 = 275.6\ {}^{\circ}\text{C} \]
Point 2 — throttle the steam to 3 MPa. A throttling valve is an adiabatic device that does no work and has negligible change in kinetic energy, so the steady-flow energy equation reduces to constant enthalpy: \[ h_2 = h_1 = 2784.6\ \text{kJ/kg} \quad \text{at } 3\ \text{MPa} \] On the Mollier chart this is the horizontal move 1→2, and it is irreversible: entropy rises from 5.8901 to \(s_2 = 6.1490\) kJ/kg·K. Because \(h_g\) at 3 MPa is 2803.2 kJ/kg, the throttled steam is very slightly wet rather than dry saturated.
Point 3s — ideal HP expansion to 0.6 MPa. An isentropic expansion is a vertical drop at \(s = s_2 = 6.1490\) kJ/kg·K to the 0.6 MPa isobar. Interpolating in the wet region, \(x_{3s} = (6.1490-1.9310)/4.8282 = 0.8734\) and \[ h_{3s} = h_f + x_{3s}h_{fg} = 670.6 + 0.8734(2086.3) = 2492.5\ \text{kJ/kg} \] so the ideal enthalpy drop available to the HP turbine is \(2784.6 - 2492.5 = 291.8\) kJ/kg.
Point 3 — actual HP exit at 80% internal efficiency. Internal efficiency is actual work over isentropic work, so \[ h_3 = h_2 - \eta_t\,(h_2 - h_{3s}) = 2784.6 - 0.80(291.8) \] \[ \boxed{h_3 = 2550.9\ \text{kJ/kg}, \quad \Delta h_{HP} = 233.7\ \text{kJ/kg}} \] The lost 58.4 kJ/kg reappears as entropy, moving point 3 to \(s_3 = 6.284\) kJ/kg·K — to the right of 3s, exactly as drawn.
Point 4 — reheat at 0.6 MPa to 255 °C. Since \(T_{sat}\) at 0.6 MPa is 158.8 °C, the reheated steam is superheated by 96 K. Interpolating the 0.6 MPa superheat table between 250 and 300 °C, \[ h_4 = 2968.1\ \text{kJ/kg}, \quad s_4 = 7.2032\ \text{kJ/kg}\cdot\text{K} \] Reheating is the horizontal-ish rise 3→4 on the chart, and its purpose is visible in the geometry: it lifts the expansion line so the LP turbine can drop a long way before reaching a damaging moisture level.
Points 5s and 5 — the LP expansion to 0.004 MPa. Dropping vertically at \(s = 7.2032\) kJ/kg·K to the 0.004 MPa isobar gives \(x_{5s} = (7.2032-0.4226)/8.0520 = 0.8418\) and \(h_{5s} = 121.5 + 0.8418(2432.9) = 2169.9\) kJ/kg, an ideal drop of 798.2 kJ/kg. Applying the same internal efficiency, \[ h_5 = h_4 - 0.80(798.2) = 2968.1 - 638.5 \] \[ \boxed{h_5 = 2329.6\ \text{kJ/kg}, \quad \Delta h_{LP} = 638.5\ \text{kJ/kg}} \]
Part (b) — terminal conditions at the three requested points. Each of the three lies in the wet region, so the temperature is simply the saturation temperature at that pressure and the moisture is \(1-x\), with \(x\) obtained from \(x = (h-h_f)/h_{fg}\).
At the HP inlet (point 2, 3 MPa): \(x = (2784.6-1008.4)/1795.7 = 0.9897\), so \(T = 233.9\ {}^{\circ}\text{C}\) with 1.0% moisture. At the HP exit (point 3, 0.6 MPa): \(x = (2550.9-670.6)/2086.3 = 0.9016\), so \(T = 158.8\ {}^{\circ}\text{C}\) with 9.8% moisture. At the LP exit (point 5, 0.004 MPa): \(x = (2329.6-121.5)/2432.9 = 0.9079\), so \(T = 29.0\ {}^{\circ}\text{C}\) with 9.2% moisture.

Both exhaust moistures sit just below the 10–12% figure at which blade erosion becomes a serious concern, which is the engineering reason the reheat is there at all: without it the LP expansion would start from point 3 and end far wetter.
Part (c) — power output of the whole turbine. Each cylinder passes its own flow through its own actual enthalpy drop: \[ \dot W = \dot m_{HP}\,\Delta h_{HP} + \dot m_{LP}\,\Delta h_{LP} = 600(233.7) + 500(638.5) \] \[ \dot W = 140\,195 + 319\,243 \] \[ \boxed{\dot W = 459\,438\ \text{kW} \approx 459\ \text{MW}} \] The LP cylinder produces 70% of the output on five-sixths of the flow, because its available enthalpy drop is nearly three times the HP drop — a direct consequence of expanding into a deep vacuum.

Check: the 100 kg/s difference between the HP and LP flows is the extraction bled off between the cylinders (feedheating and moisture separation, typical of a CANDU secondary circuit). The question gives only the two flows, so the answer applies each drop to its stated flow and takes no credit for work done by the extracted steam beyond the HP exit. Enthalpies above are IAPWS values, which agree with the issued Granet & Bluestein tables to better than 0.1%.

Point	State	h (kJ/kg)	T (°C)	Moisture
1	6 MPa, dry saturated	2784.6	275.6	0%
2 (HP inlet)	3 MPa, after throttling	2784.6	233.9	1.0%
3s	0.6 MPa, ideal HP exit	2492.5	158.8	12.7%
3 (HP exit)	0.6 MPa, actual	2550.9	158.8	9.8%
4	0.6 MPa, reheated to 255 °C	2968.1	255.0	superheated
5s	0.004 MPa, ideal LP exit	2169.9	29.0	15.8%
5 (LP exit)	0.004 MPa, actual	2329.6	29.0	9.2%
HP turbine power	600 kg/s × 233.7 kJ/kg	140 195 kW
LP turbine power	500 kg/s × 638.5 kJ/kg	319 243 kW
(c) Total turbine power output	—	459 438 kW (459 MW)

Part II — Feedwater pump efficiency (5 marks)

Given. Feedwater enters the pump at 2.5 MPa and 210.0 °C and leaves at 20 MPa and 214.3 °C. The flow is approximately 250 kg/s but cannot be measured accurately.

Find. The internal (isentropic) efficiency of the pump.

Figure 3.2 — Feedwater pump on h–s coordinates. The measured 4.3 K temperature rise fixes point 2; the ratio of the two enthalpy rises is the efficiency.

Approach. The measured temperature rise fixes the actual outlet enthalpy, and the inlet entropy fixes the isentropic outlet enthalpy at the same discharge pressure; the ratio of the two enthalpy rises is the efficiency, and the unmeasurable flow rate cancels out of it.

Recognise why the temperature measurement is enough. Internal efficiency is \[ \eta_p = \frac{w_{ideal}}{w_{actual}} = \frac{h_{2s}-h_1}{h_2-h_1} \] Both numerator and denominator are per unit mass, so the flow rate cancels — which is exactly why the test was designed around temperatures rather than a flow measurement. This is the standard thermometric method for a high-head feed pump, where the temperature rise is a few kelvin and is measurable to far better accuracy than the flow.
Fix the inlet state. At 2.5 MPa the saturation temperature is 224.0 °C, so at 210.0 °C the water is compressed liquid. From the compressed-liquid table, \[ h_1 = 897.8\ \text{kJ/kg}, \qquad s_1 = 2.4235\ \text{kJ/kg}\cdot\text{K} \]
Fix the actual outlet state. At 20 MPa and the measured 214.3 °C, \[ h_2 = 923.6\ \text{kJ/kg} \] so the actual work input is \[ w_{actual} = h_2 - h_1 = 923.6 - 897.8 = 25.8\ \text{kJ/kg} \]
Fix the isentropic outlet state. Compressing reversibly to 20 MPa at \(s = s_1 = 2.4235\) kJ/kg·K gives \[ h_{2s} = 918.2\ \text{kJ/kg}, \qquad w_{ideal} = 918.2 - 897.8 = 20.4\ \text{kJ/kg} \] As a check against the incompressible approximation, \(w_{ideal} \approx v_1\,\Delta p = 0.001173 \times 17\,500 = 20.5\) kJ/kg, which confirms the table reading.
Form the efficiency. \[ \eta_p = \frac{918.2-897.8}{923.6-897.8} = \frac{20.4}{25.8} \] \[ \boxed{\eta_p = 0.790 \;\;\text{(79.0\%)}} \] This is a realistic value for a large multistage feed pump. Note how sensitive the result is: an error of only 0.1 K in either temperature moves the actual work by about 0.45 kJ/kg and the efficiency by roughly 1.4 percentage points, so the instrumentation must be matched pairs of platinum resistance thermometers, not ordinary thermocouples. Of the 25.8 kJ/kg supplied, 5.4 kJ/kg is dissipated internally and reappears as the extra temperature rise the measurement detects — the whole basis of the method.

Quantity	Result
Inlet enthalpy and entropy (2.5 MPa, 210.0 °C)	897.8 kJ/kg, 2.4235 kJ/kg·K
Actual outlet enthalpy (20 MPa, 214.3 °C)	923.6 kJ/kg
Isentropic outlet enthalpy (20 MPa, s = s₁)	918.2 kJ/kg
Ideal work / actual work	20.4 / 25.8 kJ/kg
Pump internal efficiency	0.790 (79.0%)

Topic. Expansion-line construction on the Mollier chart, and the thermometric (temperature-rise) determination of pump efficiency.

Key relations. Throttling: \(h_2 = h_1\). Isentropic expansion: \(s = \) constant. Internal efficiency of a turbine: \(\eta_t = (h_{in}-h_{out})/(h_{in}-h_{out,s})\), so \(h_{out} = h_{in} - \eta_t(h_{in}-h_{out,s})\). Quality: \(x = (h-h_f)/h_{fg}\). Pump internal efficiency: \(\eta_p = (h_{2s}-h_1)/(h_2-h_1)\), with the incompressible check \(w_{ideal} \approx v\,\Delta p\).

Why this works. The Mollier chart is drawn precisely so these processes are simple geometry: a throttle is horizontal, an isentropic expansion is vertical, and a real expansion is a vertical drop shortened by the efficiency and displaced to the right. The efficiency definition is what turns the vertical ideal drop into the actual one, and reading quality from the constant-\(x\) lines gives the moisture directly. Turbines and pumps are treated by the same definition but inverted — a turbine's actual work is less than ideal, so efficiency is actual over ideal, whereas a pump's actual work is more than ideal, so efficiency is ideal over actual. Getting that the right way round is the whole of Part II.

Common pitfall. Applying the 80% internal efficiency to the pressure drop across the throttle valve as well — the valve is a pure throttle and produces no work, so the entire 6 → 3 MPa drop is lost with no output, and the turbine's efficiency applies only from point 2 onward. The second frequent error is computing the LP expansion from point 3 rather than from the reheated point 4, which understates the power by about a third; the third is inverting the pump efficiency to get 25.8/20.4 = 1.27, a value above unity that should be recognised immediately as impossible.

Source. Granet & Bluestein, Thermodynamics and Heat Power 6th ed., Ch. 9 (vapour power cycles, Mollier chart and turbine efficiency) and the accompanying steam tables; El-Wakil, Powerplant Technology, Ch. 5 (steam turbines, expansion lines and internal efficiency).

Question 3 of 6: Steam Plant Turbomachinery

Question 3: Steam Plant Turbomachinery (15 marks)

Part I — Turbine expansion line (10 marks)

Part II — Feedwater pump efficiency (5 marks)