Question 1 of 9: Regenerative Rankine Cycle, One Closed Feedwater Heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations — December 2019, 04-BS-10, Thermodynamics. 3-hour
closed-book exam (one 8.5×11 in. double-sided note sheet allowed); property tables/charts
supplied, no interpolation required. Part A: two of three 20-mark questions; Part B: four of six
15-mark questions. Every question is answered in full.
Reference texts. Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed. (Vapor Power Cycles Ch. 10; Gas Power Cycles Ch. 9; Refrigeration Cycles
Ch. 11; Gas Mixtures Ch. 13; Psychrometrics Ch. 14; Exergy Ch. 8).
Questions 1, 3, 4, 6, 9 are fully legible and solved as printed. Question 2's specific
numbers (pressure ratio, reheat pressure, temperatures, efficiencies) were reconstructed from
surviving fragments that match a well-known textbook archetype (see Q2's callout). Questions 5, 7
and 8 have genuinely blank gaps in the source with no fully-confirming fingerprint; clearly-labelled
representative values are substituted so every question can still be solved in full (see each
question's own check callout).
Given. Turbine-1 inlet (state 1): 12.5 MPa, 550°C. Extraction/closed-FWH inlet
(state 2): 1 MPa, same entropy as state 1 (ideal turbine). Condenser (state 3): 5 kPa, same entropy
as state 1. Feedwater leaves the heater (state 6) at 12.5 MPa, 170°C; its drain (state 7) leaves
as saturated liquid at 1 MPa and is trapped (throttled, isenthalpic) into the condenser. All turbine
stages and both pumps (Pump I + Pump II, modelled together since no stream joins between them) are
isentropic. Source $T_H=1500$ K, sink $T_{sink}=300$ K, $T_0=300$ K.
State
Description
P
T
h (kJ/kg)
s (kJ/kg·K)
1
Turbine-1 inlet
12.5 MPa
550.0°C
3476.51
6.6317
2
Extraction / closed-FWH hot in
1 MPa
188.0°C
2798.44
6.6317
3
Turbine exit ($x=0.7774$)
5 kPa
32.87°C
2021.48
6.6317
4
Condenser exit, sat. liquid
5 kPa
32.87°C
137.75
0.4762
5
Pump exit (I+II combined)
12.5 MPa
—
150.31
0.4762
6
Closed-FWH cold out, to boiler
12.5 MPa
170.0°C
725.60
2.0270
7
Closed-FWH drain, sat. liquid
1 MPa
179.9°C
762.52
2.1381
8
Trap exit ($x=0.2579$)
5 kPa
32.87°C
762.52
2.5178
Find. (a) $w_{net}$ [kJ/kg]; (b) $\eta_{th}$; (c) $X_{dest}$ per process and total
[kJ/kg]; (d) $\eta_{II}$.
Fig. Q1 — T–s state points on the real
saturation dome. Full flow expands 1→2, then $(1-y)$ continues 2→3; the extracted fraction
$y$ condenses 2→7 in the closed heater and is throttled 7→8 into the condenser, where it
remixes with the turbine exhaust.
Approach
Fix states 1–3 from the constant entropy of the ideal turbine line, get the pump-exit and
FWH-exit states from the given pressures/temperature, close the closed-heater energy balance for the
extraction fraction $y$, then sum turbine/pump work and boiler heat per kg of steam entering the
first stage. Part (c) tracks entropy generation through every process (boiler heat addition from a
finite-temperature source, the adiabatic closed heater, the throttling trap, and the condenser's
mixing + heat rejection to the sink); part (d) compares the actual efficiency to the Carnot ceiling
set by the stated source/sink temperatures.
Closed feedwater heater energy balance (solve for $y$). The full feedwater flow
is heated from $h_5=150.31$ to $h_6=725.60$ kJ/kg by the extracted steam ($y$ kg) condensing from
$h_2=2798.44$ to its saturated-liquid drain $h_7=762.52$ kJ/kg:
$$y\,(h_2-h_7)=h_6-h_5\ \Rightarrow\
y=\frac{725.60-150.31}{2798.44-762.52}=\boxed{0.2826}.$$
Turbine work and pump work per kg entering the first stage. Full flow expands
1→2, and $(1-y)$ continues 2→3; the pump (I+II combined) moves the full 1 kg from 5 kPa to
12.5 MPa:
$$w_{turb}=(h_1-h_2)+(1-y)(h_2-h_3)=(3476.51-2798.44)+0.7174(2798.44-2021.48)
=678.07+557.42=\boxed{1235.49\ \text{kJ/kg}}$$
$$w_{pump}=h_5-h_4=150.31-137.75=\boxed{12.56\ \text{kJ/kg}}.$$
Net power output (part a) and boiler heat input.
$$w_{net}=w_{turb}-w_{pump}=1235.49-12.56=\boxed{1222.93\ \text{kJ/kg}}$$
$$q_{in}=h_1-h_6=3476.51-725.60=\boxed{2750.92\ \text{kJ/kg}}.$$
Exergy destruction, boiler (part c, 1 of 4). Heat $q_{in}$ is supplied by a source
at $T_H=1500$ K:
$$S_{gen,boiler}=(s_1-s_6)-\frac{q_{in}}{T_H}=(6.6317-2.0270)-\frac{2750.92}{1500}=4.6047-1.8339=2.7708\ \text{kJ/kg}\cdot\text{K}$$
$$X_{boiler}=T_0\,S_{gen,boiler}=300\times2.7708=\boxed{831.21\ \text{kJ/kg}}.$$
(Both turbine stages and the pump are ideal/isentropic, so $X_{turb}=X_{pump}=0$.)
Exergy destruction, closed FWH and trap (part c, 2 of 4). The heater is adiabatic
to the surroundings (only its two internal streams exchange heat), so its entropy generation is just
the sum of both sides' entropy change; the trap is an isenthalpic throttle:
$$S_{gen,fwh}=(s_6-s_5)+y\,(s_7-s_2)=(2.0270-0.4762)+0.2826(2.1381-6.6317)=1.5508-1.2698=0.2810\ \text{kJ/kg}\cdot\text{K}$$
$$X_{fwh}=T_0\,S_{gen,fwh}=300\times0.2810=\boxed{84.32\ \text{kJ/kg}}$$
$$S_{gen,trap}=y\,(s_8-s_7)=0.2826\times(2.5178-2.1381)=0.1073\ \text{kJ/kg}\cdot\text{K},\qquad
X_{trap}=300\times0.1073=\boxed{32.19\ \text{kJ/kg}}.$$
Exergy destruction, condenser (part c, 3 of 4). The turbine exhaust $(1-y)$ at
state 3 and the trap output $y$ at state 8 mix and reject $q_{out}$ to the sink at $T_{sink}=300$ K,
leaving as 1 kg of saturated liquid (state 4):
$$q_{out}=[(1-y)h_3+y\,h_8]-h_4=[0.7174(2021.48)+0.2826(762.52)]-137.75=1665.4-137.75=1527.7\ \text{kJ/kg}$$
$$S_{gen,cond}=s_4-[(1-y)s_3+y\,s_8]+\frac{q_{out}}{T_{sink}}=0.4762-6.6317+\frac{1527.7}{300}=0.1003\ \text{kJ/kg}\cdot\text{K}$$
$$X_{cond}=300\times0.1003=\boxed{30.08\ \text{kJ/kg}}.$$
Total exergy destruction (part c, 4 of 4).
$$X_{total}=X_{boiler}+X_{fwh}+X_{trap}+X_{cond}=831.21+84.32+32.19+30.08=\boxed{977.80\ \text{kJ/kg}}.$$
Check: $w_{net}+X_{total}=1222.93+977.80=2200.73$ kJ/kg, matching $q_{in}(1-T_0/T_H)=2750.92\times0.80=2200.73$
kJ/kg exactly — the reversible work available from $q_{in}$ splits cleanly between actual net
work and total destruction.
