Question 2 of 6: Combined Cycle Plant — Brayton topping cycle with a regenerative Rankine bottoming cycle

Question 2: Combined Cycle Plant — Brayton topping cycle with a regenerative Rankine bottoming cycle (15 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. The system diagram and state table of page 11, a cold air-standard gas cycle with $k=1.4$ and $c_{p}=1.005\ \text{kJ}\cdot\text{kg}^{-1}\text{K}^{-1}$ (paper p. 14), and a gas mass flow of $\dot{M}_{g}=100\ \text{kg}\cdot\text{s}^{-1}$.

Find. Combustor duty, main and bled steam flows, net gas-turbine and steam-turbine power, the isentropic efficiencies of the compressor, gas turbine and steam turbine, the pump work, and the overall plant efficiency.

[Figure not reproduced: The combined cycle of page 11 redrawn. Gas states 1 to 5 form the open Brayton cycle; water and steam states 6 to 12 form the closed regenerative Rankine cycle, with the bleed taken at state 11 into the direct-contact heater. See the official exam paper.]

Approach. Treat the gas side as an ideal gas with constant specific heat so every duty is $\dot{M}c_{p}\Delta T$, then use the heat-recovery steam generator as the coupling: an energy balance across it fixes the steam flow, an energy balance on the direct-contact heater fixes the bleed, and the isentropic states supplied in the table give the component efficiencies directly.

1. Part (a) — combustor duty. Between the compressor discharge and the turbine inlet the gas is heated at essentially constant pressure, so $$\dot{Q}_{cc}=\dot{M}_{g}c_{p}(T_{3}-T_{2})=100\times1.005\times(1000-422)=\boxed{58\,089\ \text{kW}}$$ This is the heat input against which the whole plant will be assessed in part (j).
2. Part (b) — main steam flow. The heat recovery steam generator transfers the gas enthalpy drop from state 4 to state 5 into the water between states 9 and 10, with no external heat addition: $$\dot{M}_{g}c_{p}(T_{4}-T_{5})=\dot{M}_{s}(h_{10}-h_{9})$$ The gas side gives $100\times1.005\times(418-159)=26\,030\ \text{kW}$, and the water side rise is $3196-610=2586\ \text{kJ}\cdot\text{kg}^{-1}$, so $$\dot{M}_{s}=\frac{26\,030}{2586}=\boxed{10.07\ \text{kg}\cdot\text{s}^{-1}}$$ about one kilogram of steam for every ten kilograms of gas, which is the usual order for an unfired single-pressure boiler.
3. Part (c) — bled steam flow. The direct-contact heater mixes the bleed $\dot{m}$ at state 11 with the condensate $(\dot{M}_{s}-\dot{m})$ at state 7 and delivers saturated water at state 8. An adiabatic energy balance on the heater gives $$\dot{m}h_{11}+(\dot{M}_{s}-\dot{m})h_{7}=\dot{M}_{s}h_{8}\quad\Longrightarrow\quad\dot{m}=\dot{M}_{s}\frac{h_{8}-h_{7}}{h_{11}-h_{7}}$$ Substituting, $\dot{m}=10.07\times(605-136)/(2719-136)=\boxed{1.828\ \text{kg}\cdot\text{s}^{-1}}$, which is 18.2 % of the main steam flow; the remaining $8.238\ \text{kg}\cdot\text{s}^{-1}$ passes to the condenser.
4. Part (d) — net gas turbine power. The gross turbine work and the compressor work are both ideal-gas enthalpy changes: $$\dot{W}_{gt}=\dot{M}_{g}c_{p}(T_{3}-T_{4})=100\times1.005\times(1000-418)=58\,491\ \text{kW}$$ and $\dot{W}_{c}=100\times1.005\times(422-30)=39\,396\ \text{kW}$. The compressor is on the same shaft, so the net output of the gas set is $$\dot{W}_{G}=58\,491-39\,396=\boxed{19\,095\ \text{kW}}$$ The back work ratio, $39\,396/58\,491=0.674$, is the characteristic weakness of the simple Brayton cycle: two thirds of the turbine output is consumed driving its own compressor.
5. Part (e) — steam turbine power. The turbine must be taken in two sections because the mass flow changes at the extraction belt. Above the belt the full main steam flow expands from 10 to 11, below it only the flow that survives the bleed expands from 11 to 12: $$\dot{W}_{S}=\dot{M}_{s}(h_{10}-h_{11})+(\dot{M}_{s}-\dot{m})(h_{11}-h_{12})$$ The two contributions are $10.07\times(3196-2719)=4801\ \text{kW}$ and $8.238\times(2719-2201)=4267\ \text{kW}$, giving $$\dot{W}_{S}=\boxed{9068\ \text{kW}}$$ Using the full flow all the way to the condenser would overstate this by about 10 %.
6. Parts (f) and (g) — gas machine efficiencies. For an ideal gas with constant specific heat the enthalpy differences are proportional to temperature differences, so the isentropic efficiencies reduce to temperature ratios. The compressor consumes more work than the ideal machine, so the ideal rise goes on top: $$\eta_{c}=\frac{T_{2S}-T_{1}}{T_{2}-T_{1}}=\frac{344-30}{422-30}=\boxed{0.801}$$ and the turbine delivers less work than the ideal machine, so the actual drop goes on top: $$\eta_{gt}=\frac{T_{3}-T_{4}}{T_{3}-T_{4S}}=\frac{1000-418}{1000-353}=\boxed{0.900}$$ A check on the table itself: the pressure ratio is $1.2/0.1=12$, and $12^{(k-1)/k}=2.034$ against $(344+273.15)/(30+273.15)=2.036$, so the quoted isentropic temperatures are consistent with $k=1.4$ to two parts in a thousand.
7. Part (h) — internal efficiency of the steam turbine. The internal (isentropic) efficiency compares the actual work with the work the same mass flows would deliver expanding down the single isentropic condition line 10 to 11S to 12S. The ideal contributions are $10.07\times(3196-2634)=5657\ \text{kW}$ and $8.238\times(2634-2025)=5017\ \text{kW}$, so $$\eta_{S}=\frac{9068}{5657+5017}=\frac{9068}{10\,674}=\boxed{0.850}$$ The two sections are almost equally good, $0.849$ above the belt and $0.851$ below it, which is why the overall figure is insensitive to how the ideal work is apportioned. Note that 12SS in the table (blank in the source) is the isentropic drop from the actual state 11 and would be 2088 kJ/kg; using it would give a stage-by-stage rather than an overall internal efficiency.
8. Part (i) — pump work. The feedwater pump raises the full steam flow from state 8 to state 9, and the tabulated enthalpies give that directly: $\dot{W}_{fp}=10.07\times(610-605)=50.3\ \text{kW}$. The table rounds both condenser and condensate-pump enthalpies to 136 kJ/kg, so the condensate pump must be taken from $w=v\,\Delta p$ with $v_{f}=0.001005\ \text{m}^{3}\cdot\text{kg}^{-1}$ at 0.005 MPa: $w=0.001005\times(400-5)=0.397\ \text{kJ}\cdot\text{kg}^{-1}$ and $\dot{W}_{cp}=8.238\times0.397=3.3\ \text{kW}$. Together $$\dot{W}_{pumps}=50.3+3.3=\boxed{53.6\ \text{kW}}$$ less than 0.2 % of the plant output, which is why pump work is often neglected in a first-pass Rankine calculation but should not be here, because the question asks for it.
9. Part (j) — overall plant efficiency. Charging the pumps against the steam turbine as the question directs, the net plant output is $$\dot{W}_{net}=\dot{W}_{G}+\dot{W}_{S}-\dot{W}_{pumps}=19\,095+9068-53.6=28\,110\ \text{kW}$$ and the only heat input is the combustor duty, so $$\eta_{plant}=\frac{28\,110}{58\,089}=\boxed{0.484}$$ The gas cycle alone would return $19\,095/58\,089=0.329$; adding the bottoming cycle lifts the plant from 33 % to 48 % and the steam side contributes 32 % of the net output while burning no additional fuel. That is the entire case for combined-cycle plant.