Steam Cycle: 22-Mec-B3 Energy Conversion and Power Generation, Undated paper, Q4

Question 4: Steam Cycle (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. An ideal regenerative Rankine cycle with one open (direct-contact) feedwater heater, all enthalpies supplied on Page 12.

Question 4 — enthalpies from Page 12
Quantity	Symbol	Value
1 — condenser outlet, saturated water	$h_1$	121 kJ/kg (0.004 MPa, 29 °C)
2 — condensate pump outlet	$h_2$	122 kJ/kg (0.6 MPa, 29 °C)
3 — heater outlet, saturated water	$h_3$	671 kJ/kg (0.6 MPa, 159 °C)
4 — feed pump outlet	$h_4$	677 kJ/kg (6 MPa, 160 °C)
5 — turbine inlet, superheated	$h_5$	3177 kJ/kg (6 MPa, 400 °C)
6 — extraction, wet	$h_6$	2662 kJ/kg (0.6 MPa, 159 °C)
7 — turbine exhaust, wet	$h_7$	1970 kJ/kg (0.004 MPa, 29 °C)

Find. The cycle drawn on T-s axes, the fraction of the boiler steam that must be bled to the heater, and the thermal efficiency of the regenerative cycle.

Question 4 — plant flow diagram from Page 12, with the seven numbered states. All of the condensate (1–2) and the bled fraction (6) meet in the direct-contact heater and leave together as saturated water at 3.

Question 4(a) — the cycle on T-s axes. 1–2 condensate pump, 2–3 mixing with the bled steam inside the heater, 3–4 boiler feed pump, 4–5 boiler and superheater, 5–6–7 isentropic expansion with extraction at 6, 7–1 condenser. Only the fraction $1-m$ travels along 7–1 and 1–2.

Approach. Write mass and energy balances on the direct-contact heater to get the bleed fraction, then sum the turbine work stage by stage against the two pump works and the boiler heat, all per kilogram of steam leaving the boiler.

Part (a) — how the diagram is built. Work with unit mass leaving the boiler. That whole kilogram is raised 4→5 and expands 5→6; at 6 a fraction $m$ is bled to the heater and only $1-m$ continues 6→7 and through the condenser. On T-s axes the pump lines 1–2 and 3–4 are almost vertical against the saturated liquid line, the boiler line 4–5 runs up through the dome and into the superheat region, and the expansion 5–6–7 is a single vertical line because the turbine is isentropic. The horizontal 6–3 shown dotted is the bled steam being returned to the feedwater.
Part (b) — heater mass and energy balance. For unit boiler flow, $m$ kilograms of steam at $h_6$ mix with $(1-m)$ kilograms of condensate at $h_2$ and leave as saturated water at $h_3$: $$m\,h_6 + (1-m)\,h_2 = h_3.$$ Rearranging, $$m = \frac{h_3-h_2}{h_6-h_2} = \frac{671-122}{2662-122} = \frac{549}{2540}$$ $$\boxed{m = 0.216\ \text{kg per kg of boiler steam}}$$ so about 21.6 % of the throttle flow is diverted at 0.6 MPa.
Turbine work per kilogram of boiler steam. The whole kilogram expands from 5 to 6 and only the remainder continues to 7: $$w_T = (h_5-h_6) + (1-m)(h_6-h_7).$$ $$w_T = (3177-2662) + 0.784 \times (2662-1970) = 515 + 542 = 1057\ \text{kJ/kg}.$$
Pump work. The condensate pump handles only the fraction $1-m$, the boiler feed pump the whole kilogram: $$w_P = (1-m)(h_2-h_1) + (h_4-h_3) = 0.784 \times 1 + 6 = 6.8\ \text{kJ/kg}.$$ It is under one percent of the turbine work, which is the usual signature of a liquid-phase pump against a vapour-phase turbine.
Heat supplied. The boiler and superheater see the full kilogram entering at the feed-pump discharge, $$q_{in} = h_5 - h_4 = 3177 - 677 = 2500\ \text{kJ/kg}.$$ Note that the regeneration has already lifted the feedwater from 29 °C to 160 °C, so the boiler is charged with 549 kJ/kg less than it would be without the heater.
Part (c) — cycle efficiency. $$\eta = \frac{w_T-w_P}{q_{in}} = \frac{1057-6.8}{2500} = \frac{1050.6}{2500}$$ $$\boxed{\eta = 0.420 = 42.0\ \%}$$
Check the first law on the whole cycle and the gain from regeneration. The condenser rejects $(1-m)(h_7-h_1) = 0.784 \times 1849 = 1449$ kJ/kg, and $2500 - 1449 = 1051$ kJ/kg matches the net work, so the cycle closes. The same plant without the heater — one pump taking the condensate straight from 0.004 MPa to 6 MPa — delivers 39.4 %. Bleeding a fifth of the steam therefore buys 2.6 percentage points, because the heat that would have been rejected to the cooling water is instead returned to the feedwater at a useful temperature.

Final results
Quantity	Symbol	Value
Fractional extraction flow	$m$	0.216
Turbine work per kg of boiler steam	$w_T$	1057 kJ/kg
Total pump work	$w_P$	6.8 kJ/kg
Net work	$w_{net}$	1051 kJ/kg
Heat supplied in the boiler	$q_{in}$	2500 kJ/kg
Heat rejected in the condenser	$q_{out}$	1449 kJ/kg
Cycle efficiency with regeneration	$\eta$	42.0 %
Cycle efficiency without regeneration	$\eta_0$	39.4 %

Topic. Regenerative Rankine cycle with one open (direct-contact) feedwater heater.

Key relations. Heater balance $m h_6 + (1-m)h_2 = h_3$; turbine work $w_T = (h_5-h_6)+(1-m)(h_6-h_7)$; pump work $w_P = (1-m)(h_2-h_1)+(h_4-h_3)$; $\eta = (w_T-w_P)/(h_5-h_4)$.

Why this works. Efficiency in any heat-engine cycle rises when heat is added at a higher mean temperature. In a plain Rankine cycle a large slice of the boiler duty goes into warming feedwater from condenser temperature up to saturation — heat added at a low mean temperature, which drags the average down. Regeneration does that warming with steam already partly expanded, so the boiler only ever sees water that is already hot. The work sacrificed by bleeding steam early is smaller than the heat saved, so the ratio improves. An open heater does this by direct mixing, which also deaerates the feedwater — the reason essentially every utility plant has at least one.

Common pitfall. Applying the bleed fraction to the wrong streams. The boiler feed pump and the boiler carry the full unit mass, while the condensate pump, the last turbine stage and the condenser carry only $1-m$. Dividing the turbine work by the full mass throughout inflates the efficiency; forgetting that the condensate pump handles $1-m$ makes no practical difference here but is wrong in principle.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, ch. 10.6 (regenerative Rankine cycles); Granet & Bluestein, Thermodynamics and Heat Power, ch. 8.

Question 4 of 8: Steam Cycle

Question 4: Steam Cycle (10 marks)