Gas Turbine: 22-Mec-B3 Energy Conversion and Power Generation, December 2018, Q1

Question 1: Gas Turbine (10 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
Compressor pressure ratio	$r$	12
Air mass flow rate	$\dot{M}_a$	142 kg/s
Fuel mass flow rate	$\dot{M}_f$	2.68 kg/s
Fuel calorific value	$CV$	40 000 kJ/kg
Compressor isentropic efficiency	$\eta_c$	0.90
Turbine isentropic efficiency	$\eta_t$	0.88
Ambient state	$p_1,\ T_1$	100 kPa, 15 °C (288.15 K)
Cold air (compressor) properties	$c_{p,a},\ k_a$	1.005 kJ/kg·K, 1.3997
Hot gas (combustor and turbine) properties	$c_{p,g},\ k_g$	1.148 kJ/kg·K, 1.333

Find. The T–s locus of the cycle, the actual temperatures at compressor exit, turbine inlet and turbine exhaust, and the net shaft power and thermal efficiency of the unit.

Part (a) — the T–s diagram. The four numbered points to be calculated are 1 at the compressor inlet, 2 at the compressor discharge, 3 at the turbine inlet and 4 at the turbine exhaust; the isentropic ideals 2s and 4s are shown as well, because the machine efficiencies are defined against them.

Figure 1.1 — T–s diagram of the open simple-cycle gas turbine. Solid red = the actual path; grey dashed = the isentropic ideals 2s and 4s used to apply the machine efficiencies. The lower isobar is atmospheric (100 kPa), the upper one the combustor pressure (1.20 MPa). The process 4 → 1 is not a heat exchanger; the machine is open and the exhaust is discharged, so the dotted return is a bookkeeping closure only.

Approach. Compress isentropically to find $T_{2s}$ and correct it with $\eta_c$; close a steady-flow energy balance on the combustor using the product mass flow and the hot-gas $c_p$ to obtain the turbine inlet temperature; expand isentropically to $T_{4s}$ and correct with $\eta_t$; then difference the turbine and compressor powers.

Part (b) — ideal compressor discharge temperature. For an isentropic compression of an ideal gas, $$\frac{T_{2s}}{T_1}=r^{(k_a-1)/k_a},\qquad \frac{k_a-1}{k_a}=\frac{1.3997-1}{1.3997}=0.28556 .$$ The compressor works on cold air, so $k_a=c_{p,a}/c_{v,a}=1.005/0.718=1.3997$ from the paper's page-14 constants. Substituting, $T_{2s}=288.15\times 12^{0.28556}=288.15\times 2.0331=585.84\ \text{K}=312.7\,{}^{\circ}\text{C}$.
Actual compressor discharge temperature. The isentropic efficiency of a compressor compares the ideal work with the real work at the same pressure ratio, so $$\eta_c=\frac{T_{2s}-T_1}{T_2-T_1} \quad\Longrightarrow\quad T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=288.15+\frac{585.84-288.15}{0.90}.$$ The bracket is a 297.69 K ideal rise, which the 90 per cent efficiency inflates to 330.77 K, so $$\boxed{T_2=618.92\ \text{K}=345.8\,{}^{\circ}\text{C}}$$ at the compressor discharge.
Combustor energy balance — turbine inlet temperature. The paper's Note requires the changed mass flow and specific heat to be carried through, and the turbine inlet temperature is one of the "conditions of the gas in the turbine". The products leaving the combustor are air plus fuel, $$\dot{M}_g=\dot{M}_a+\dot{M}_f=142+2.68=144.68\ \text{kg/s},$$ and the heat released is $\dot{Q}_{in}=\dot{M}_f\,CV=2.68\times 40\,000=107\,200\ \text{kW}$. Applying the steady-flow energy equation across the combustor with the hot-gas specific heat, $$\Delta T_{comb}=\frac{\dot{Q}_{in}}{\dot{M}_g\,c_{p,g}} =\frac{107\,200}{144.68\times 1.148}=\frac{107\,200}{166.09}=645.4\ \text{K},$$ so that $$\boxed{T_3=618.92+645.4=1264.3\ \text{K}=991.2\,{}^{\circ}\text{C}}$$ at the turbine inlet.
Ideal turbine exhaust temperature. The turbine expands the same 12:1 pressure ratio back to atmosphere, but with the hot-gas index, $$\frac{k_g-1}{k_g}=\frac{0.333}{1.333}=0.24981,\qquad 12^{0.24981}=1.8603 ,$$ so $T_{4s}=T_3/1.8603=1264.34/1.8603=679.66\ \text{K}=406.5\,{}^{\circ}\text{C}$. The exponent is materially smaller than the cold-air one, which is precisely why the note insists on it: using the air value here would put the ideal exhaust temperature some 58 K too low (621.9 K instead of 679.7 K).
Actual turbine exhaust temperature. For a turbine the efficiency is the ratio of real to ideal work, so the real temperature drop is the smaller one: $$T_4=T_3-\eta_t\,(T_3-T_{4s})=1264.34-0.88\times(1264.34-679.66),$$ and with the ideal drop equal to 584.68 K the actual drop is 514.52 K, giving $$\boxed{T_4=749.82\ \text{K}=476.7\,{}^{\circ}\text{C}}$$ at the turbine exhaust flange. This is the temperature the heat recovery steam generator of Question 2 would see. The 560 °C that Question 2 quotes belongs to a different, hotter machine (that question states it can be completed without reference to this one), so no agreement is expected; ducting losses could only lower this 477 °C, never raise it to 560 °C.
Part (c) — turbine and compressor powers. Each machine is a steady-flow device handling its own stream: $$\dot{W}_t=\dot{M}_g\,c_{p,g}\,(T_3-T_4)=144.68\times 1.148\times 514.52 =85\,451\ \text{kW},$$ $$\dot{W}_c=\dot{M}_a\,c_{p,a}\,(T_2-T_1)=142\times 1.005\times 330.77 =47\,204\ \text{kW}.$$ The compressor handles only air, the turbine air plus fuel — that 2.68 kg/s difference is worth about 1.6 MW of turbine work and is the reason the note exists.
Net power output and thermal efficiency. Differencing the two shaft powers, $$\boxed{\dot{P}_{net}=85\,451-47\,204=38\,247\ \text{kW}=38.2\ \text{MW}}$$ and, referring that to the fuel energy admitted, $$\boxed{\eta_{th}=\frac{\dot{P}_{net}}{\dot{Q}_{in}} =\frac{38\,247}{107\,200}=0.357=35.7\ \text{per cent}} .$$ As a plausibility check, the back-work ratio is $\dot{W}_c/\dot{W}_t=47\,204/85\,451=0.552$, comfortably inside the 0.40–0.60 band typical of a simple-cycle industrial machine; a value outside it would mean a machine efficiency had been applied the wrong way round.

Check: the combustor is closed on the product flow. The paper's Note 6 invites a statement of assumptions. Here the combustor is closed on $\dot{M}_g=144.68$ kg/s at $c_{p,g}=1.148$, treating the incoming air as already having become products. Closing it instead on the same 144.68 kg/s but at the cold-air $c_p=1.005$ raises $T_3$ to about 1083 °C and the output to 44.5 MW, which credits the cycle with energy the fuel never supplied, because the air would then be carried at a different specific heat from the products it becomes. The reading used here is the one the Note's wording requires.

Quantity	Symbol	Result
Ideal compressor discharge temperature	$T_{2s}$	312.7 °C
Actual compressor discharge temperature	$T_2$	345.8 °C
Turbine inlet temperature	$T_3$	991.2 °C
Ideal turbine exhaust temperature	$T_{4s}$	406.5 °C
Actual turbine exhaust temperature	$T_4$	476.7 °C
Gross turbine power	$\dot{W}_t$	85.45 MW
Compressor absorbed power	$\dot{W}_c$	47.20 MW
Net power output	$\dot{P}_{net}$	38.2 MW
Thermal efficiency	$\eta_{th}$	35.7 per cent
Back-work ratio	$\dot{W}_c/\dot{W}_t$	0.552

Topic. The open simple-cycle (Brayton) gas turbine with irreversible compression and expansion, and variable gas properties.

Key relations. Isentropic ideal-gas relation $T_2/T_1=r^{(k-1)/k}$; compressor efficiency $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$; turbine efficiency $\eta_t=(T_3-T_4)/(T_3-T_{4s})$; steady-flow energy balance on the combustor $\dot{M}_f\,CV=\dot{M}_g\,c_{p,g}\,(T_3-T_2)$; net work $\dot{W}_{net}=\dot{M}_g c_{p,g}(T_3-T_4)-\dot{M}_a c_{p,a}(T_2-T_1)$.

Why this works. The gas turbine is a chain of steady-flow control volumes, each of which surrenders a temperature difference to a first-law balance. Because the working fluid is very nearly an ideal gas throughout, every enthalpy difference collapses to $c_p\,\Delta T$ and the whole cycle can be solved on temperatures alone, without any property tables. The pressure ratio enters only twice, once for each machine, through the isentropic relation; the machine efficiencies then convert those two ideal temperature changes into real ones, in opposite directions — the real compression is hotter than ideal, the real expansion cooler than ideal, because both irreversibilities generate entropy and therefore heat. The combustor is where the two property sets meet: it accepts air and delivers products, so the mass flow steps up by the fuel flow and the specific heat steps up from 1.005 to 1.148 kJ/kg·K as triatomic combustion products (carbon dioxide and water vapour) replace part of the diatomic air.

Common pitfall. Two mistakes dominate. The first is holding $T_3$ fixed as though it were an input: it is an output of the combustor balance, so any change to the compressor changes it. The second is inverting one of the machine efficiencies — dividing where one should multiply — which usually still yields a plausible-looking answer but drives the back-work ratio outside the 0.40–0.60 band, and that band is the cheapest check available.

Source. Çengel and Boles, Thermodynamics: An Engineering Approach, 9th ed., Chapter 9 (Gas Power Cycles), sections on the Brayton cycle and on deviation of actual gas-turbine cycles from idealised ones; El-Wakil, Powerplant Technology, Chapter 8 (Gas Turbine and Combined Cycles).

Question 1 of 8: Gas Turbine

Question 1: Gas Turbine (10 marks)