22-Mec-B3 Energy Conversion and Power Generation · May 2017
Question 4 of 6: Steam Turbine Operational Conditions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2017 — 16-Mec-B3 Energy Conversion and Power Generation. Closed book, three hours. Section A is calculative (Questions 1–4) and Section B descriptive (Questions 5–6); a candidate answers three from Section A and one from Section B, four questions of 15 marks each constituting a complete 60-mark paper. Reference data are bound in as pages 8–11, reference formulae and constants as pages 12–15, and the Granet & Bluestein steam tables are supplied. All six questions are solved below, because the set as a whole is the study resource.
Reference texts.
Granet, I. and Bluestein, M., Thermodynamics and Heat Power, 6th ed. — the steam and gas tables bound into this examination.
El-Wakil, M. M., Powerplant Technology — heat balance diagrams, gas-cooled reactor plants, condensers, environmental impact of power generation.
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Brayton and Rankine cycle analysis, isentropic efficiencies.
Lamarsh, J. R. and Baratta, A. J., Introduction to Nuclear Engineering, 4th ed. — reactor heat removal and the CANDU heat balance.
Natural Resources Canada, Energy Fact Book — Canadian installed capacity and generation mix, quoted in Questions 5 and 6.
Given. A throttle-governed condensing turbine supplied at 4 MPa and 400 °C, exhausting to 0.005 MPa, passing 24 kg/s at full load, with a condenser whose full-load temperature profile is printed on page 11.
Given data (question page 5 and attachment page 11)
Throttle steam
4 MPa, 400 °C (h = 3213.6 kJ/kg, s = 6.7690 kJ/kg·K)
Exhaust pressure (parts a–c)
0.005 MPa (Tsat = 32.88 °C)
Full-load steam flow
24 kg/s
Internal efficiency, full and part load
80 %
Internal efficiency, zero load
70 %
Part-load throttle pressure
1 MPa
Zero-load throttle pressure
0.1 MPa
Flow law
steam flow proportional to inlet pressure
Full-load condenser profile (page 11)
cooling water 13.0 → 23.0 °C, steam flat at 32.9 °C
Specific heat of water
4.19 kJ/kg·K
Find. The power developed at full load, at the 1 MPa part-load condition and at the zero-load condition; then the part-load condenser temperature profiles, the part-load back pressure, and an estimate of the zero-load back pressure.
Figure 4.1 — the three expansions plotted on the Mollier chart. Each throttling process runs horizontally (constant enthalpy) before the expansion drops to 0.005 MPa.
Figure 4.2 — answer to part (d). Condenser temperature profiles at full, part and zero load with UA and cooling-water flow held constant.
Approach. Parts (a) to (c) are three expansions from the same enthalpy: throttling is isenthalpic, so it moves the state horizontally on the Mollier chart and raises the entropy, and the isentropic drop available afterwards is smaller each time. Parts (d) to (f) fit the condenser's UA and water flow to the printed full-load profile, then hold both constant and let the exhaust temperature float.
Check: a leftover caption on attachment page 11. The worksheet heads its parts "(e) … for part load conditions (when throttled to 4 MPa" and "(f) … at part load conditions (4 MPa steam inlet)", and it letters them (e), (f), (g) where the question page letters them (d), (e), (f). But 4 MPa is the full-load throttle pressure, so "part load at 4 MPa" is self-contradictory; the worksheet caption is a leftover from an earlier printing. The question page governs, and it defines part load as "conditions as defined in (b) above", i.e. 1 MPa. The answer below is worked at 1 MPa and the zero-load estimate at 0.1 MPa, matching the worksheet's own bracketed "(0.1 MPa steam inlet)".
Part (a) — full load, 4 MPa and 400 °C. From the steam tables the throttle state is $h_1 = 3213.6$ kJ/kg, $s_1 = 6.7690$ kJ/kg·K. Expanding isentropically to 0.005 MPa, where $s_f = 0.4764$ and $s_{fg} = 7.9187$:$$x_{2s}=\frac{s_1-s_f}{s_{fg}}=\frac{6.7690-0.4764}{7.9187}=0.7947$$$$h_{2s}=h_f+x_{2s}h_{fg}=137.82+0.7947(2423.7)=2063.9\ \text{kJ/kg}$$The internal efficiency then fixes the real work:$$w=\eta_i\,(h_1-h_{2s})=0.80\,(3213.6-2063.9)=919.8\ \text{kJ/kg}$$$$\boxed{P_a=\dot{m}\,w=24\times 919.8=22\,074\ \text{kW}\ (22.1\ \text{MW})}$$The actual exhaust enthalpy is $3213.6-919.8=2293.8$ kJ/kg, a wetness of 11 % — about the practical limit for a last stage, which is why the machine is designed this way.
Part (b) — throttled to 1 MPa. Throttling is a constant-enthalpy process, so the state after the throttle valve still has $h = 3213.6$ kJ/kg but now at 1 MPa. Interpolating along the 1 MPa isobar between 350 °C ($h = 3157.7$, $s = 7.3011$) and 400 °C ($h = 3263.9$, $s = 7.4651$):$$t=376.3\ ^\circ\text{C},\qquad s=7.3874\ \text{kJ/kg}\cdot\text{K}$$The entropy has risen by 0.618 kJ/kg·K, and that is the whole cost of throttle governing. Expanding from this state to 0.005 MPa:$$x_{2s}=\frac{7.3874-0.4764}{7.9187}=0.8727,\qquad h_{2s}=137.82+0.8727(2423.7)=2253.2\ \text{kJ/kg}$$$$w=0.80\,(3213.6-2253.2)=768.3\ \text{kJ/kg}$$The flow is proportional to inlet pressure, so $\dot{m}=24\times(1/4)=6.0$ kg/s:$$\boxed{P_b=6.0\times 768.3=4610\ \text{kW}\ (4.61\ \text{MW})}$$Power does not scale with flow. Quarter flow gives only 20.9 % of full-load power, because throttling has also destroyed 16.5 % of the available enthalpy drop — the essential economic objection to throttle governing.
Part (c) — zero generator output, 0.1 MPa inlet. Again the enthalpy is preserved through the throttle. Interpolating along the 0.1 MPa isobar between 300 °C ($h = 3074.3$, $s = 8.2158$) and 400 °C ($h = 3278.2$, $s = 8.5435$) gives $t = 368.3$ °C and $s = 8.4397$ kJ/kg·K. This entropy now exceeds $s_g = 8.3951$ at 0.005 MPa, so the isentropic endpoint lies in the superheated region and a quality is meaningless there. Working the small superheat off the saturated-vapour point with $c_{pv}\approx 1.9$ kJ/kg·K:$$T_{2s}=T_g\exp\!\left(\frac{s-s_g}{c_{pv}}\right)=306.03\exp\!\left(\frac{0.0446}{1.9}\right)=313.2\ \text{K}$$$$h_{2s}=h_g+c_{pv}(T_{2s}-T_g)=2561.5+1.9(7.2)=2575.3\ \text{kJ/kg}$$$$w=0.70\,(3213.6-2575.3)=446.8\ \text{kJ/kg},\qquad \dot{m}=24\times\frac{0.1}{4}=0.6\ \text{kg/s}$$$$\boxed{P_c=0.6\times 446.8=268\ \text{kW}}$$This is the bearing friction and generator windage — 1.2 % of the machine rating, exactly the order of magnitude expected for a turbine spinning at speed on no load.
Part (d) — fit the condenser at full load first. Nothing about the part-load profile can be found until the condenser's two constants are extracted from the printed full-load case. The duty is the heat given up by the exhaust steam as it condenses to saturated liquid:$$\dot{Q}=\dot{m}(h_2-h_f)=24\,(2293.8-137.82)=51\,745\ \text{kW}$$The page-11 profile shows the water rising 13.0 to 23.0 °C, so$$\dot{m}_w=\frac{51\,745}{4.19\,(23.0-13.0)}=1235\ \text{kg/s}$$and, using the average temperature difference the question asks for,$$\theta=T_s-\tfrac{1}{2}(T_{w,in}+T_{w,out})=32.88-18.0=14.88\ \text{K},\qquad UA=\frac{51\,745}{14.88}=3477\ \text{kW/K}$$A useful confirmation that the profile has been read correctly: the printed steam line sits at 32.9 °C, and $T_{sat}$ at 0.005 MPa is 32.88 °C.
Part (d) continued — the part-load profile. Now hold $UA = 3477$ kW/K and $\dot{m}_w = 1235$ kg/s fixed, keep the water inlet at 13.0 °C, and let the exhaust temperature float until the three equations agree:$$\dot{Q}=\dot{m}_b\,(h_{2b}-h_f(T_s)),\qquad \Delta T_w=\frac{\dot{Q}}{\dot{m}_wc_p},\qquad T_s=T_{w,in}+\tfrac{1}{2}\Delta T_w+\frac{\dot{Q}}{UA}$$With $\dot{m}_b = 6.0$ kg/s and $h_{2b}=3213.6-768.3=2445.3$ kJ/kg these converge in a few passes to$$\boxed{T_s=18.46\ ^\circ\text{C},\quad \dot{Q}=14\,207\ \text{kW},\quad \text{water }13.00\rightarrow 15.75\ ^\circ\text{C}}$$These are the profiles plotted in Figure 4.2. Both effects shrink together: the water rise falls from 10.0 K to 2.75 K because there is less heat to absorb, and the mean driving difference falls from 14.9 K to 4.1 K because the same surface now has far less duty to pass.
Part (e) — the part-load condenser pressure. The exhaust steam is saturated, so its pressure is simply the saturation pressure at the temperature just found. Interpolating the saturation table between 15 °C (1.705 kPa) and 20 °C (2.339 kPa):$$\boxed{p_{cond}=p_{sat}(18.46\ ^\circ\text{C})=2.14\ \text{kPa}}$$The back pressure has fallen from 5.0 kPa to 2.1 kPa, which is the reason part-load operation recovers a little of the enthalpy drop that throttling destroyed.
Part (f) — the zero-load estimate. The question asks for this "without further calculation", so argue it physically. At 0.6 kg/s the duty collapses to about 1600 kW, the water rise to roughly 0.3 K and the mean temperature difference to under half a degree, so the exhaust steam can be little more than a degree above the 13.0 °C cooling-water inlet:$$\boxed{T_s\approx 13.6\ ^\circ\text{C},\qquad p_{cond}\approx 1.6\ \text{kPa}}$$The important point is the limit rather than the number. As the load tends to zero the back pressure tends to the saturation pressure at the cooling-water inlet temperature, $p_{sat}(13.0\ ^\circ\text{C}) = 1.51$ kPa; the condenser can never go below it, however good the vacuum equipment is, because there is no colder sink available.
