17-Phys-B6 Applied Thermodynamics and Heat Transfer · December 2013
Question 2 of 8: Reheat Rankine Cycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B6 Applied Thermodynamics and Heat Transfer,
National Examination December 2013 — a three-hour open-book examination;
candidates are expected to bring both a thermodynamics text and a heat-transfer text to make use
of the property tables. A complete examination is five questions — either three from Part A
(Thermodynamics, Q1–Q4) and two from Part B (Heat Transfer, Q5–Q8), or two from Part A
and three from Part B — every question carrying equal value; all eight are solved below as a
complete study set. Candidates are invited to state any assumptions where a question is open to
interpretation; this licence is used explicitly in Question 1(b) (standard air properties, not
printed directly) and Question 7 (a printed convective "rate" read as a heat-transfer coefficient
with a temperature unit omitted in print — the same omission the paper shows in Question 6's specific-heat units).
Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An
Engineering Approach, 8th ed. (ideal-gas processes, vapour power cycles, gas-turbine/Brayton
cycles, vapour-compression refrigeration); F. P. Incropera and D. P. DeWitt, Fundamentals of
Heat and Mass Transfer, 7th ed. (conduction, natural convection, lumped-capacitance transient
conduction, radiation exchange between surfaces, heat-exchanger LMTD analysis).
Given. A reheat Rankine cycle: HP turbine inlet 3.5 MPa/350°C,
expansion to 0.5 MPa, reheat back to 350°C at 0.5 MPa, LP expansion to 7.5 kPa,
condenser exit liquid at 30°C pumped to 3.5 MPa. Both turbines 85% efficient, pump 80%
efficient, combined turbine power output 1000 kW.
Given data
State
Description
Condition
1
HP turbine inlet
3.5 MPa, $350\,{}^{\circ}\text{C}$
2
HP turbine exit / reheat inlet
0.5 MPa
3
LP turbine inlet (after reheat)
0.5 MPa, $350\,{}^{\circ}\text{C}$
4
LP turbine exit / condenser inlet
7.5 kPa
5
Condenser exit / pump inlet
$30\,{}^{\circ}\text{C}$, saturated liquid
6
Pump exit / boiler inlet
3.5 MPa
—
Turbine efficiency (each)
$\eta_t = 85\%$
—
Pump efficiency
$\eta_p = 80\%$
—
Total turbine power output
1000 kW
Find. (a) The steam mass flow rate $\dot m$; (b) the pump power; (c) the
cycle thermal efficiency $\eta_{th}$.
T–s diagram drawn to scale from steam properties: 1→2 HP turbine
(actual, dashed — irreversible) · 2→3 reheat at 0.5 MPa · 3→4 LP turbine
(dashed) · 4→5 condenser (condensing at 40.3°C, then subcooled to 30°C) ·
5→6 feed pump · 6→1 boiler. Both turbine exhausts lie just inside the dome
(state 2: $x\approx0.998$; state 4: $x\approx0.969$).
Approach. Fix all six state enthalpies from steam-table data:
isentropic exits for both turbines from the inlet entropy, actual exits via the turbine
efficiency, the condenser exit as saturated liquid at 30°C, and the pump exit via the pump
efficiency applied to the ideal pump work $v_5(P_6-P_5)$. The given 1000 kW total turbine power
then fixes $\dot m$ directly; pump power, boiler + reheater heat input and thermal efficiency
follow from energy balances on the remaining components.
HP turbine (1→2). At state 1 (3.5 MPa, $350\,{}^{\circ}\text{C}$),
$h_1 = 3104.84\text{ kJ/kg}$, $s_1 = 6.6601\text{ kJ/kg}\cdot\text{K}$. Expanding isentropically
to 0.5 MPa ($s_{2s}=s_1$) gives $h_{2s}=2679.86\text{ kJ/kg}$. The actual exit enthalpy uses the
turbine efficiency $\eta_t = (h_1-h_2)/(h_1-h_{2s})$:
$$h_2 = h_1 - \eta_t(h_1-h_{2s}) = 3104.84 - 0.85(3104.84-2679.86) = 2743.61\text{ kJ/kg}$$
(this lands almost exactly on the 0.5 MPa saturation temperature, $T_2\approx151.8\,{}^{\circ}\text{C}$
— the HP exhaust is wet or just-saturated steam.)
Reheat and LP turbine (3→4). Reheating at 0.5 MPa back to
$350\,{}^{\circ}\text{C}$ gives $h_3=3168.08\text{ kJ/kg}$, $s_3=7.6346\text{ kJ/kg}\cdot\text{K}$.
Isentropic expansion to 7.5 kPa gives $h_{4s}=2381.11\text{ kJ/kg}$, and
$$h_4 = h_3 - \eta_t(h_3-h_{4s}) = 3168.08 - 0.85(3168.08-2381.11) = 2499.15\text{ kJ/kg}$$
Total turbine work and mass flow rate — part (a). Per unit mass,
$$w_{\text{turb}} = (h_1-h_2)+(h_3-h_4) = 361.23 + 668.93 = 1030.16\text{ kJ/kg}$$
The stated 1000 kW is the combined power of both turbines, so
$$\dot m = \frac{\dot W_{\text{turb}}}{w_{\text{turb}}} = \frac{1000}{1030.16}$$
$$\boxed{\dot m = 0.9707\text{ kg/s}}$$
Condenser exit and pump work — part (b). Liquid leaves the condenser
at $30\,{}^{\circ}\text{C}$, taken as saturated liquid at that temperature (the condenser
pressure of 7.5 kPa corresponds to $T_{\text{sat}}\approx40\,{}^{\circ}\text{C}$, so the exit is
slightly subcooled — a standard idealisation since liquid properties barely depend on
pressure): $h_5=125.734\text{ kJ/kg}$, $v_5 = 0.001004\text{ m}^3/\text{kg}$ at
$P_5=P_{\text{sat}}(30\,{}^{\circ}\text{C})=4.247\text{ kPa}$. The ideal (isentropic) pump work is
$$w_{p,s} = v_5(P_6-P_5) = 0.001004\times(3500-4.247) = 3.511\text{ kJ/kg}$$
and with 80% pump efficiency the actual work is larger,
$$w_{p,a} = \frac{w_{p,s}}{\eta_p} = \frac{3.511}{0.80} = 4.389\text{ kJ/kg}, \qquad h_6=h_5+w_{p,a}=130.123\text{ kJ/kg}$$
$$\boxed{\dot W_{\text{pump}} = \dot m\, w_{p,a} = 0.9707\times4.389 = 4.260\text{ kW}}$$
Heat input and thermal efficiency — part (c). Heat is added in the
boiler (6→1) and reheater (2→3):
$$\dot Q_{in} = \dot m\big[(h_1-h_6)+(h_3-h_2)\big]
= 0.9707\times\big[(3104.84-130.123)+(3168.08-2743.61)\big] = 3299.66\text{ kW}$$
Net cycle power is the turbine output less the pump power,
$$\dot W_{net} = 1000 - 4.260 = 995.74\text{ kW}$$
$$\boxed{\eta_{th} = \frac{\dot W_{net}}{\dot Q_{in}} = \frac{995.74}{3299.66} = 30.18\%}$$
