Steam turbine on the Mollier chart: 22-Mec-B3 Energy Conversion and Power Generation, December 2017, Q2

Question 2: Steam turbine on the Mollier chart (10 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.

Given. A throttle-governed reheat turbine at part load, with the states named 1 (stop-valve inlet), 2 (throttled, HP inlet), 3 (HP exhaust), 4 (reheat outlet, LP inlet) and 5 (LP exhaust).

Point	Condition	Value
1	stop valve	16 MPa, 540 °C
2	after the throttle valve	8 MPa, h₂ = h₁
3	HP turbine exhaust	2 MPa, η_i = 0.90
4	after reheat	2 MPa, 500 °C
5	LP turbine exhaust	0.004 MPa, η_i = 0.80
—	steam flow through both cylinders	75 kg/s

Find. The expansion line on the h-s chart, the HP exhaust pressure and temperature, the LP exhaust moisture and temperature, and the total power at 75 kg/s.

Part (a): the process plotted on the Mollier chart. Throttling runs horizontally to the right (constant enthalpy), each expansion drops to its isentropic end point on the lower isobar and is then shortened by the internal efficiency, and the reheat line climbs back up the 2 MPa isobar.

Approach. Read h and s at the stop valve, carry the enthalpy horizontally across the throttle, drop vertically to each lower isobar to get the isentropic end point, shorten each drop by the internal efficiency, and multiply the sum of the two real drops by the mass flow. Chart readings are quoted to the nearest 10 kJ/kg, with the steam-table values alongside them, because the chart cannot be read more finely than that.

Part (a) — locate the stop-valve state and carry it across the throttle. At 16 MPa and 540 °C the chart reads about 3410 kJ/kg (tables: 3412.1) with $s_1 = 6.449$ kJ/kg·K. A throttle valve does no work and passes no heat, so the steady-flow energy equation collapses to $h_2 = h_1$: $$h_2 = h_1 = 3412\ \text{kJ/kg} \quad \text{at}\quad p_2 = 8\ \text{MPa}$$ Following the horizontal line to the 8 MPa isobar gives $t_2 \approx 505\ {}^\circ\text{C}$ and $s_2 = 6.743$ kJ/kg·K. The entropy has risen by 0.294 kJ/kg·K — that increase is the work the governing valve has destroyed before the steam ever reaches a blade.
Part (a) — the high pressure expansion. Dropping vertically from point 2 to the 2 MPa isobar gives the isentropic end point $h_{3s} \approx 3010$ kJ/kg, so $$\Delta h_{s,\text{HP}} = 3412 - 3010 = 402.5\ \text{kJ/kg}$$ The internal efficiency shortens that drop, $$\Delta h_{\text{HP}} = 0.90 \times 402.5 = \boxed{362.2\ \text{kJ/kg}}$$ and the real end state sits at $h_3 = 3412 - 362.2 = 3050$ kJ/kg on the 2 MPa line. The 40 kJ/kg by which point 3 lies above point 3s is the frictional reheating the question refers to.
Part (b) — the high pressure turbine outlet conditions. Reading along the 2 MPa isobar at $h_3 = 3050$ kJ/kg, the state is still well superheated: $$\boxed{p_3 = 2\ \text{MPa}, \qquad t_3 \approx 311\ {}^\circ\text{C}}$$ (saturation at 2 MPa is 212.4 °C, so the steam leaves the HP cylinder with about 99 K of superheat and the cold reheat pipework carries no moisture).
Part (a) — the reheat and the low pressure expansion. Reheating at constant pressure to 500 °C moves the state up the 2 MPa isobar to $h_4 = 3468$ kJ/kg, $s_4 = 7.434$ kJ/kg·K. Dropping vertically to the 0.004 MPa isobar reaches the wet region at $h_{5s} \approx 2240$ kJ/kg, i.e. a quality of 0.871: $$\Delta h_{s,\text{LP}} = 3468 - 2240 = 1228.6\ \text{kJ/kg}, \qquad \Delta h_{\text{LP}} = 0.80 \times 1228.6 = \boxed{982.9\ \text{kJ/kg}}$$ so the real exhaust enthalpy is $h_5 = 3468 - 982.9 = 2485$ kJ/kg.
Part (c) — the low pressure turbine outlet conditions. At 0.004 MPa the saturation values are $h_f = 121.4$ kJ/kg and $h_{fg} = 2432.3$ kJ/kg, so the exhaust quality and moisture are $$x_5 = \frac{2485.3 - 121.4}{2432.3} = 0.972, \qquad \text{moisture} = 1 - x_5 = \boxed{2.8\ \%}$$ and, because the exhaust is a saturated mixture, its temperature is simply the saturation temperature at 0.004 MPa: $$\boxed{t_5 = 28.96\ {}^\circ\text{C}}$$ The exhaust is unusually dry for a condensing turbine precisely because the LP cylinder is only 80 % efficient — the losses reappear as enthalpy in the steam. Had the machine been isentropic the moisture would have been 12.9 %, which is past the usual erosion limit of about 12 %.
Part (d) — the power output. With no extraction the same 75 kg/s passes through both cylinders, so the two real enthalpy drops simply add: $$P = \dot m\,(\Delta h_{\text{HP}} + \Delta h_{\text{LP}}) = 75 \times (362.2 + 982.9)$$ $$P = 75 \times 1345.1 = \boxed{100\,880\ \text{kW} = 100.9\ \text{MW}}$$ As a cross-check on the heat side, the reheater has to supply $\dot m (h_4 - h_3) = 75 \times 418.3 = 31\,370$ kW, which is 31 % of the output — the usual proportion for a single reheat at this pressure ratio.

Quantity	Symbol	Value
HP turbine outlet pressure	p₃	2 MPa
HP turbine outlet temperature	t₃	311 °C (superheated)
HP enthalpy drop, actual	Δh_HP	362.2 kJ/kg
LP turbine exhaust quality	x₅	0.972
LP turbine exhaust moisture	1 − x₅	2.8 %
LP turbine exhaust temperature	t₅	28.96 °C
LP enthalpy drop, actual	Δh_LP	982.9 kJ/kg
Total power at 75 kg/s	P	100.9 MW
Reheater duty	Q_rh	31.4 MW

Concept & Theory

Topic. Reading a throttle-governed reheat expansion off the Mollier (h-s) chart, and the difference between an isentropic end state and a real one.

Key relations. Throttling: $h_2 = h_1$ with $s_2 \gt s_1$. Internal (isentropic) efficiency $\eta_i = \Delta h_{\text{actual}} / \Delta h_{\text{isentropic}}$. Quality from the saturation line, $x = (h - h_f)/h_{fg}$. Power $P = \dot m \sum \Delta h$.

Why this works. The Mollier chart is drawn with enthalpy vertical and entropy horizontal precisely so that the two operations a steam plant performs become simple geometry: an adiabatic machine moves down the page by the work it delivers, and a throttle moves horizontally across it. The isentropic end state is found by a vertical drop to the lower isobar; multiplying that drop by the internal efficiency and stepping back up the vertical line gives the real end state, which always lies at higher enthalpy and higher entropy. Everything the question asks — outlet temperature, moisture, power — follows from where the real end state falls relative to the saturation line.

Common pitfall. Applying the efficiency to the enthalpy itself rather than to the enthalpy drop. A second trap is reading the exhaust temperature off a superheated-steam scale: once the state is inside the dome, the temperature is the saturation temperature and nothing else, so it is set by the condenser pressure alone.

Source. Granet & Bluestein, Thermodynamics and Heat Power, 6th ed., ch. 8 (vapour power cycles, Mollier diagram); El-Wakil, Powerplant Technology, ch. 2 (turbine expansion lines and internal efficiency).

Question 2 of 8: Steam turbine on the Mollier chart

Question 2: Steam turbine on the Mollier chart (10 marks)