22-Mec-B3 Energy Conversion and Power Generation · May 2014

Question 2 of 6: Steam Injected Gas Turbine

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Closed book, three hours. Two sections: Section A (Calculative) Questions 1–4 and Section B (Descriptive) Questions 5–6. Candidates answer three questions from Section A and one from Section B; four questions of 15 marks each constitute a complete 60-mark paper. Reference data for particular questions are bound in as pages 10–17, reference formulae and constants as pages 18–21, and the steam tables from Granet & Bluestein are supplied. All six printed questions are solved below.

Reference texts for 22-Mec-B3 Energy Conversion and Power Generation

Granet, I. and Bluestein, M., Thermodynamics and Heat Power, 6th ed. — the source of the steam tables (Tables A.1–A.4 SI) bound into this paper.
El-Wakil, M. M., Powerplant Technology — steam-station cycles, heat balance diagrams, gas turbines, wind and solar conversion (Figure 5.6 of this text is paper page 14).
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Rankine, Brayton, Otto, Diesel and combined cycles.
Fox, R. W., Pritchard, P. J. and McDonald, A. T., Introduction to Fluid Mechanics, 10th ed. — the energy equation and head-loss accounting used for the hydro station.
Rayaprolu, K., Boilers for Power and Process — boiler efficiency and heat-rate conventions.

Units and conventions used throughout. The paper's own constant sheet (page 19) is used without substitution: $g = 9.81\ \text{m/s}^2$, $c_p$ air $= 1.005\ \text{kJ/kg}\cdot \text{K}$, $c_v$ air $= 0.718\ \text{kJ/kg}\cdot\text{K}$, $c_p$ water $= 4.190\ \text{kJ/kg} \cdot\text{K}$, $\rho_{\text{water}} = 1000\ \text{kg/m}^3$, $p_{\text{atm}} = 100\ \text{kPa}$. Monetary amounts are in Canadian dollars, as the paper is an Engineers Canada national examination.

Question 2: Steam Injected Gas Turbine (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.

Quantity	Symbol	Value
Gas-cycle pressure ratio	$r_p$	12
Compressor inlet air	$T_1$, $p_1$	30 °C, 100 kPa
Turbine gas inlet temperature	$T_3$	1000 °C
Compressor / turbine isentropic efficiency	$\eta_c$, $\eta_t$	0.80, 0.90
Gas mass flow	$\dot m_g$	100 kg/s
Fuel calorific value	$CV$	42 000 kJ/kg
Steam delivered by the generator (state 8)	—	1.2 MPa, 300 °C
Feedwater into the generator (state 6)	$T_6$	30 °C
Gas leaving the generator (state 5)	$T_5$	160 °C
Steam exhaust pressure (state 9)	$p_9$	0.1 MPa
Cold-air-standard properties	$k$, $c_p$	1.4, 1.005 kJ/kg·K

Find. The fuel and steam flows the plant needs, the net electrical power with injection, and the resulting fuel-to-busbar efficiency.

Figure 2.1 — the plant with the paper's own point numbering. Air is compressed 1→2, fired to 3, and expands to 4. Steam raised in the generator leaves at 8 at 1.2 MPa — exactly the compressor delivery pressure — and is injected at the turbine inlet, expanding as a separate stream to 9 at 0.1 MPa. Both streams then traverse the generator, gas leaving at 5 and steam at 10.

Approach. Work the air stream first: isentropic compression corrected by $\eta_c$ fixes $T_2$, the combustor energy balance gives the fuel flow, and isentropic expansion corrected by $\eta_t$ fixes $T_4$. The gas then gives up heat between $T_4$ and 160 °C in the generator, and that duty divided by the enthalpy rise of the water fixes the steam flow. Finally add the work of the two expanding streams and subtract the compressor and pump work.

Compressor delivery temperature. For an ideal gas with constant specific heats, $$\frac{T_{2s}}{T_1} = r_p^{(k-1)/k} = 12^{0.2857} = 2.0340$$ $$T_{2s} = 303.15 \times 2.0340 = 616.6\ \text{K}$$ and correcting for the 80 % isentropic efficiency, $$T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_c} = 303.15 + \frac{616.6 - 303.15}{0.80} = 694.96\ \text{K}\ (421.8\ ^{\circ}\text{C})$$ Note that $r_p = 12$ on a 100 kPa inlet puts the combustor at 1.2 MPa, which is precisely the steam generator delivery pressure the question specifies. That is not a coincidence: the steam has to be raised to combustor pressure before it can be injected, and the agreement is a free check that the data set is self-consistent.
Part (a) — fuel mass flow. An energy balance on the combustion chamber, with the fuel's own mass neglected in the gas stream as the note permits, gives $$\dot m_f CV = \dot m_g c_p (T_3 - T_2)$$ $$\dot m_f = \frac{100 \times 1.005 \times (1273.15 - 694.96)}{42\,000} = \frac{58\,108}{42\,000} = \boxed{1.384\ \text{kg/s}}$$ Carrying the fuel mass through the turbine instead — that is, solving $\dot m_f CV = (\dot m_g + \dot m_f) c_p (T_3 - T_2)$ — gives 1.403 kg/s, a 1.4 % difference. The paper's note that iteration is unnecessary sanctions the simpler figure. The air–fuel ratio of 72:1 is typical of a gas turbine, where most of the air is there to keep the blade metal temperature down rather than to burn fuel.
Turbine exhaust temperature of the gas stream. Expanding from 1.2 MPa back to atmosphere, $$T_{4s} = \frac{T_3}{r_p^{(k-1)/k}} = \frac{1273.15}{2.0340} = 625.9\ \text{K}$$ $$T_4 = T_3 - \eta_t (T_3 - T_{4s}) = 1273.15 - 0.90 \times 647.23 = 690.65\ \text{K}\ (417.5\ ^{\circ}\text{C})$$ A 417 °C exhaust is exactly why a heat-recovery steam generator is worth fitting: there is more than a third of the fuel energy still in the gas after the turbine.
Part (b) — steam mass flow. The generator's hot side is the gas cooling from 417.5 °C to the stated 160 °C stack temperature: $$\dot Q_{HRSG} = \dot m_g c_p (T_4 - T_5) = 100 \times 1.005 \times (690.65 - 433.15) = 25\,878\ \text{kW}$$ On the water side the pump raises 30 °C saturated water to 1.2 MPa, which costs almost nothing: $$w_p = v_f (p_7 - p_6) = 0.001004 \times (1200 - 4.25) = 1.20\ \text{kJ/kg} \;\Rightarrow\; h_7 = 125.79 + 1.20 = 126.99\ \text{kJ/kg}$$ and the generator delivers 1.2 MPa, 300 °C steam at $h_8 = 3045.8$ kJ/kg, so $$\dot m_s = \frac{\dot Q_{HRSG}}{h_8 - h_7} = \frac{25\,878}{3045.8 - 126.99} = \boxed{8.87\ \text{kg/s}}$$ Only the gas stream is credited with raising steam. The exhaust steam at state 9 also passes through the generator, but as step 5 shows it leaves the turbine at 99.6 °C — well below the 160 °C gas outlet — so it has no heat to give and simply leaves at 10. The injected steam is 8.9 % of the air flow, which is the usual order for a steam-injected (Cheng-cycle) machine.
Steam expansion work through the same turbine. The steam enters at state 8 and expands to 0.1 MPa. From the tables, $s_8 = 7.0317\ \text{kJ/kg}\cdot\text{K}$, and at 0.1 MPa $s_f = 1.3026$, $s_{fg} = 6.0568$, $h_f = 417.46$, $h_{fg} = 2258.0$, so the isentropic end state is wet: $$x_{9s} = \frac{7.0317 - 1.3026}{6.0568} = 0.9459 \;\Rightarrow\; h_{9s} = 417.46 + 0.9459 \times 2258.0 = 2553.3\ \text{kJ/kg}$$ Applying the 90 % expansion efficiency, $$\Delta h_s = \eta_t (h_8 - h_{9s}) = 0.90 \times (3045.8 - 2553.3) = \boxed{443.3\ \text{kJ/kg}}$$ $$h_9 = 3045.8 - 443.3 = 2602.5\ \text{kJ/kg} \quad (x_9 = 0.968,\ T_9 = 99.6\ ^{\circ}\text{C})$$
Part (c) — net power output. Collecting the four work terms, $$\dot W_c = \dot m_g c_p (T_2 - T_1) = 100 \times 1.005 \times 391.81 = 39\,377\ \text{kW}$$ $$\dot W_{t,\text{gas}} = \dot m_g c_p (T_3 - T_4) = 100 \times 1.005 \times 582.50 = 58\,542\ \text{kW}$$ $$\dot W_{t,\text{steam}} = \dot m_s \Delta h_s = 8.866 \times 443.3 = 3\,930\ \text{kW} \qquad \dot W_p = 8.866 \times 1.20 = 11\ \text{kW}$$ With unity generator efficiency the net electrical output is $$P_{\text{net}} = 58\,542 + 3\,930 - 39\,377 - 11 = 23\,085\ \text{kW} = \boxed{23.1\ \text{MW}}$$ The compressor absorbs 67 % of the gas-turbine work, which is why gas-turbine output is so sensitive to compressor condition and to inlet air temperature.
Part (d) — cycle efficiency with injection. The only energy purchased is the fuel: $$\eta = \frac{P_{\text{net}}}{\dot m_f CV} = \frac{23\,085}{1.384 \times 42\,000} = \frac{23\,085}{58\,108} = \boxed{0.397\ (39.7\ \%)}$$ The comparison the question invites is with the same machine running dry. Without injection the net output is $58\,542 - 39\,377 = 19\,165$ kW at the same fuel flow, an efficiency of 33.0 %. Injecting 8.87 kg/s of steam therefore buys 3.9 MW and 6.7 efficiency points for no extra fuel, because the steam is raised entirely from exhaust heat that would otherwise go up the stack and then does work in a turbine that already exists. The price is 8.87 kg/s of demineralised water thrown away every second — roughly 280 000 m³ a year — which is why steam injection suits peaking and cogeneration duty rather than continuous base load.

Part	Quantity	Result
—	Compressor delivery temperature $T_2$	694.96 K (421.8 °C)
(a)	Fuel mass flow	1.384 kg/s
—	Gas turbine exhaust temperature $T_4$	690.65 K (417.5 °C)
—	Heat recovered in the steam generator	25 878 kW
(b)	Injected steam mass flow	8.87 kg/s
—	Steam specific work (state 8 to 9)	443.3 kJ/kg, $x_9 = 0.968$
(c)	Net power output with injection	23.1 MW
(d)	Cycle efficiency with injection	39.7 %
—	Same machine dry, for comparison	19.2 MW at 33.0 %

Topic. Steam-injected (Cheng-cycle) gas turbine analysed on the cold-air-standard assumption with two separate expanding streams.

Key relations. $T_{2s}/T_1 = r_p^{(k-1)/k}$; $\eta_c = (T_{2s}-T_1)/(T_2-T_1)$; $\eta_t = (T_3-T_4)/(T_3-T_{4s})$; $\dot m_f CV = \dot m_g c_p (T_3-T_2)$; $\dot m_s = \dot m_g c_p (T_4-T_5)/(h_8-h_7)$; $P_{\text{net}} = \dot W_{t,\text{gas}} + \dot m_s \Delta h_s - \dot W_c - \dot W_p$; $\eta = P_{\text{net}}/(\dot m_f CV)$.

Why this works. The paper's note to treat gas and steam as non-interacting streams turns a genuinely two-phase mixing problem into two independent one-dimensional ones: the air stream is handled entirely with $c_p$ and the isentropic pressure-temperature relation, while the steam stream is handled entirely with the steam tables. Each stream expands through the same nozzles from the same 1.2 MPa to the same 0.1 MPa, so each is entitled to its own share of the shaft work. The injected steam adds work without adding compressor duty, which is the whole point of the cycle: the compression work for the water was paid by a feed pump at about 1.2 kJ/kg instead of by a compressor at about 394 kJ/kg.

Common pitfall. Applying the gas-turbine pressure ratio to the steam, or expanding the steam only to the generator gas-outlet temperature, both misplace the steam end state. The second frequent error is crediting the heat-recovery steam generator with heat from the exhaust steam as well as the gas — check the steam exhaust temperature first, because at 99.6 °C it is colder than the 160 °C gas leaving the unit and can supply nothing.

Source. El-Wakil, Powerplant Technology, Ch. 8 (gas-turbine and combined cycles, steam injection); Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed., §9-8 to §9-11 and §10-9; steam properties from Granet & Bluestein Table A.3 (SI).

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