22-Mec-B3 Energy Conversion and Power Generation · Undated paper
Question 3 of 8: Steam Turbine Operational Conditions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2019 — 16-Mec-B3 Energy Conversion and Power Generation. Three hours, closed book. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four questions from Section A and two from Section B, six questions of ten marks each for a total of sixty. Reference data for particular questions are bound in as pages 10–17, reference formulae and constants as pages 18–21, and steam tables from Thermodynamics and Heat Power are supplied. All eight questions are solved here.
Reference texts.
Granet & Bluestein, Thermodynamics and Heat Power, 6th ed. — steam tables, vapour and gas power cycles (the tables bound into this paper).
El-Wakil, Powerplant Technology — steam generators, gas-turbine and combined-cycle plant, wind and hydro conversion, plant siting and environmental impact.
Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed. — regenerative Rankine cycles, Brayton-cycle modifications, isentropic and internal efficiencies.
Çengel & Ghajar, Heat and Mass Transfer — heat-recovery steam generator temperature profiles and pinch analysis.
Manwell, McGowan & Rogers, Wind Energy Explained, 2nd ed. — actuator-disc theory, the Betz limit and power coefficients.
Natural Resources Canada, Canadian Renewable Energy Atlas, and CNSC REGDOC series — Canadian generation mix, siting and licensing context for Question 8.
The combustion balance of Question 1 returns a gas mass flow of 125.4 kg/s, which matches the 125 kg/s that Question 2 states. Readings taken from printed charts are identified explicitly wherever they occur.
Check: water and steam properties used below are IAPWS values, the formulation the bound Granet & Bluestein tables tabulate; every reading agrees with those tables to better than 0.1 %, comfortably inside the paper’s own rounding. Where a value had to be read off a printed chart (the Page 13 power curve and the Page 14 efficiency curves) the reading is stated explicitly and carries roughly ±1 % of graph-reading uncertainty. Every boxed result is recomputed from the question’s own data for this paper.
Given. A condensing turbine throttle-governed down to no load, with the exhaust pressure held constant throughout.
Question 3 — given data
Quantity
Symbol
Value
Throttle-valve inlet state
$p_0,\ T_0$
4 MPa, 400 °C
Exhaust pressure
$p_c$
0.005 MPa
Internal efficiency, parts (a) and (b)
$\eta_i$
0.80
Internal efficiency, part (c)
$\eta_i$
0.70
Full-load steam flow
$M$
24 kg/s
Part-load flow law
$M \propto p_{in}$
given
Find. The power developed at full load, at a throttled inlet pressure of 1 MPa, and at the no-load condition with a 0.1 MPa inlet pressure and a reduced internal efficiency.
Question 3 — the three cases plotted on h-s coordinates. Throttling moves the state horizontally to the right at constant enthalpy from state 1; each expansion then drops vertically to 0.005 MPa for the isentropic end point 2s, and the actual end state lies 20 % (or 30 % in case c) of that drop above it.
Approach. Throttling is isenthalpic, so each part-load case starts from the same enthalpy at a lower pressure and a higher entropy; expand isentropically to the condenser pressure, apply the internal efficiency, and multiply by the flow the pressure law allows.
Fix the throttle inlet state. From the superheated tables at 4 MPa and 400 °C, $$h_1 = 3214\ \text{kJ/kg},\qquad s_1 = 6.771\ \text{kJ/kg}\cdot\text{K}.$$ At the exhaust pressure of 0.005 MPa the saturation properties are $T_{sat} = 32.9$ °C, $h_f = 137.8$, $h_g = 2560.7$ kJ/kg, $s_f = 0.476$ and $s_g = 8.394$ kJ/kg·K.
Part (a) — isentropic expansion at full load. Expanding at constant entropy to 0.005 MPa lands inside the dome, so the quality follows from the entropy balance: $$x_{2s} = \frac{s_1-s_f}{s_g-s_f} = \frac{6.771-0.476}{8.394-0.476} = 0.795.$$ $$h_{2s} = h_f + x_{2s}(h_g-h_f) = 137.8 + 0.795 \times 2422.9 = 2064\ \text{kJ/kg}.$$ The isentropic drop is $3214 - 2064 = 1150$ kJ/kg.
Full-load power. The internal efficiency converts the isentropic drop into the actual one, $\Delta h = 0.80 \times 1150 = 920$ kJ/kg, and the full flow passes: $$\boxed{P_a = M\,\eta_i\,\Delta h_{is} = 24 \times 920 = 22\,085\ \text{kW} = 22.1\ \text{MW}}$$
Part (b) — throttled to 1 MPa. The throttle valve does no work and passes no heat, so the enthalpy is unchanged: at 1 MPa and $h = 3214$ kJ/kg the steam is still superheated, at 376.5 °C, and its entropy has risen to $s = 7.391$ kJ/kg·K. Expanding from there, $$x_{2s} = \frac{7.391-0.476}{7.918} = 0.873,\qquad h_{2s} = 137.8 + 0.873 \times 2422.9 = 2254\ \text{kJ/kg},$$ so the isentropic drop has fallen to $3214-2254 = 960$ kJ/kg and the actual drop to $0.80 \times 960 = 768$ kJ/kg.
Part-load flow and power. The flow follows the inlet pressure, $M_b = 24 \times (1/4) = 6.0$ kg/s, hence $$\boxed{P_b = 6.0 \times 768 = 4611\ \text{kW} = 4.61\ \text{MW}}$$ Quartering the pressure has cut the output to 21 % of full load rather than 25 %, because throttling destroys availability: the same enthalpy now sits at a higher entropy, so less of it can be converted to work.
Part (c) — the no-load condition. Throttling to 0.1 MPa at constant enthalpy gives 368.9 °C and $s = 8.448$ kJ/kg·K. That entropy exceeds $s_g$ at the condenser pressure (8.394 kJ/kg·K), so the isentropic end point now lies just outside the dome, in the superheat region at 41.6 °C with $h_{2s} = 2577$ kJ/kg. The isentropic drop is $3214-2577 = 637$ kJ/kg and, at the reduced internal efficiency, the actual drop is $0.70 \times 637 = 446$ kJ/kg.
Friction and windage power. The flow allowed by a 0.1 MPa inlet pressure is $M_c = 24 \times (0.1/4) = 0.60$ kg/s, so $$\boxed{P_c = 0.60 \times 446 = 268\ \text{kW}}$$ This is the power the machine absorbs in bearing friction and generator windage while spinning at synchronous speed with no electrical output. At 1.2 % of the rated 22.1 MW it is exactly the order of magnitude expected of a machine of this size, which is the physical check on the answer.
