Question 1 of 9: Regenerative Rankine Cycle, Closed + Open Feedwater Heaters
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — December 2016. 3 hours, Closed-Book Exam
(approved calculator and one double-sided 8.5x11-inch aid sheet permitted; property tables and
charts supplied in an appendix, interpolation not required). Part A: answer 2 of Questions 1-3 (20
marks each). Part B: answer 4 of Questions 4-9 (15 marks each), for a 100-mark paper. Only the
first two Part-A and first four Part-B questions as they appear in the answer book are marked. All
nine questions (Part A complete, Part B complete) are solved below for completeness.
Reference texts: Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed.; Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering
Thermodynamics, 8th ed. All state properties (water/steam, R-134a, air, N₂, CO₂)
were computed from high-accuracy equations of state
in place of printed property-table interpolation; every boxed numeric result.
Check: part (a) asks for "the net power output, in kJ/kg" — kJ/kg is a
specific-work unit, not a power unit, and the given 150 MW is already the net power. Read
part (a) as asking for the net specific work output $w_{net}$ (the standard quantity this
cycle archetype asks for), which then sizes the mass flow rate in part (b) via $\dot m=\dot
W_{net}/w_{net}$.
Given. Turbine inlet (state A): $P=12.5$ MPa, $T=550\ ^\circ$C. First extraction
at 0.8 MPa (closed FWH) with feedwater leaving at 170°C, 12.5 MPa; drain trapped as saturated
liquid at 0.8 MPa. Second extraction at 0.3 MPa (open FWH), saturated liquid leaves at 0.3 MPa.
Condenser at 10 kPa. $\dot W_{net}=150$ MW.
Fig. Q1 — T–s path of the main feedwater
loop (1→2 Pump I · 2→3 open FWH mixing · 3→4 Pump II · 4→5
closed FWH · 5→6 boiler · 6→7→8→9 isentropic turbine expansion,
constant $s$, with extractions at 7 and 8 · 9→1 condenser).
Approach
Fix the turbine-inlet and both extraction states isentropically from the boiler condition, then
solve the two extraction fractions $y$ and $z$ from energy balances on the closed and open
feedwater heaters (in that order, since the closed heater's drain feeds the open heater). Sum the
per-kg turbine and pump work terms into $w_{net}$, divide the given 150 MW by $w_{net}$ for the
mass flow rate, and divide $w_{net}$ by the boiler heat input for $\eta_{th}$.
Turbine states (isentropic expansion from A). At 12.5 MPa, 550°C:
$h_A=3476.51$ kJ/kg, $s_A=6.6317$ kJ/kg·K. Following the same entropy to 0.8 MPa and 0.3 MPa:
$h_{e1}=2755.03$ kJ/kg, $h_{e2}=2578.51$ kJ/kg. Continuing to the condenser at 10 kPa:
$h_{condin}=2099.96$ kJ/kg.
Condensate and Pump I (10→0.3 MPa). Saturated liquid at 10 kPa:
$h=191.81$ kJ/kg, $s=0.6492$ kJ/kg·K. Isentropic compression to 0.3 MPa gives $h_{pI,out}=
192.10$ kJ/kg (negligible pump work at this stage, as expected for a small liquid pressure rise).
Open FWH energy balance (solve $z$). The open FWH mixes the Pump-I stream
$(1-y-z)$ at $h_{pI,out}$ with the second extraction $z$ at $h_{e2}=2578.51$ kJ/kg and the closed
FWH's drain $y$ at $h_{drain}=720.86$ kJ/kg (all trapped forward into the open heater), leaving as
saturated liquid at 0.3 MPa, $h=561.43$ kJ/kg. This balance is solved together with Step 4 below
since $y$ appears in both.
Closed FWH energy balance (solve $y$). On the feedwater side, the closed FWH
raises the main flow $(1)$ from Pump II's exit $h_{pII,out}=574.48$ kJ/kg to $h_{fw,out}=725.60$
kJ/kg (170°C, 12.5 MPa); on the extraction side, fraction $y$ condenses from $h_{e1}=2755.03$
kJ/kg down to the saturated-liquid drain $h_{drain}=720.86$ kJ/kg. Equal duties give
$$y=\frac{h_{fw,out}-h_{pII,out}}{h_{e1}-h_{drain}}=\frac{725.60-574.48}{2755.03-720.86}
=\boxed{0.07429}.$$
Substituting into the open-FWH balance from Step 3 and solving for $z$:
$$z=\frac{h_{OFWH,out}-h_{pI,out}+y\,(h_{pI,out}-h_{drain})}{h_{e2}-h_{pI,out}}
=\frac{561.43-192.10+0.07429(192.10-720.86)}{2578.51-192.10}=\boxed{0.13830}.$$
Pump II (0.3→12.5 MPa). From the open-FWH exit (sat. liquid, 0.3 MPa,
$s=1.6717$ kJ/kg·K) isentropically to 12.5 MPa gives $h_{pII,out}=574.48$ kJ/kg, so
$w_{pII}=574.48-561.43=13.06$ kJ/kg.
Turbine and pump work per kg. Using the extraction fractions from Steps 3-4,
$$w_{turb}=(h_A-h_{e1})+(1-y)(h_{e1}-h_{e2})+(1-y-z)(h_{e2}-h_{condin})=1261.71\ \text{kJ/kg},$$
$$w_{pI}=(1-y-z)(h_{pI,out}-h_{condout})=0.231\ \text{kJ/kg},\qquad w_{pII}=13.055\ \text{kJ/kg}.$$
$$w_{net}=w_{turb}-w_{pI}-w_{pII}=1261.71-0.23-13.06=\boxed{1248.42\ \text{kJ/kg}}\quad\textbf{(a)}.$$
Mass flow rate (part b). With $\dot W_{net}=150\,000$ kW,
$$\dot m=\frac{\dot W_{net}}{w_{net}}=\frac{150{,}000}{1248.42}=120.15\ \text{kg/s}
=\boxed{432{,}547\ \text{kg/h}}\quad\textbf{(b)}.$$
Thermal efficiency (part c). Boiler heat input per kg of main flow:
$$q_{in}=h_A-h_{fw,out}=3476.51-725.60=2750.92\ \text{kJ/kg}.$$
$$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1248.42}{2750.92}=\boxed{0.4538\ (45.4\%)}\quad\textbf{(c)}.$$
