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

Question 3 of 6: Steam Turbine Operational Conditions

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

Question 3: Steam Turbine Operational Conditions (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.

Find. The shaft power at three throttle pressures, and the condenser temperature profile and pressure that each of the two reduced loads produces.

Approach. Each case is the same three-step calculation: throttle at constant enthalpy to the new inlet pressure to find the new entropy, expand isentropically to 0.005 MPa and multiply the drop by $\eta_i$, then multiply by the flow that the pressure-proportional law allows. For the condenser, fit $UA$ and the water flow to the full-load profile and hold both constant while the duty falls.

1. Part (a) — full-load power. At 4 MPa and 400 °C the tables give $h_1 = 3213.6$ kJ/kg and $s_1 = 6.7690\ \text{kJ/kg}\cdot\text{K}$. Expanding isentropically to 0.005 MPa ($s_f = 0.4764$, $s_{fg} = 7.9187$, $h_f = 137.82$, $h_{fg} = 2423.7$): $$x_{2s} = \frac{6.7690 - 0.4764}{7.9187} = 0.7947 \;\Rightarrow\; h_{2s} = 137.82 + 0.7947 \times 2423.7 = 2063.9\ \text{kJ/kg}$$ The internal efficiency converts that into the real drop: $$\Delta h = \eta_i (h_1 - h_{2s}) = 0.80 \times 1149.7 = 919.8\ \text{kJ/kg}$$ $$P = \dot m \Delta h = 24 \times 919.8 = \boxed{22\,074\ \text{kW} = 22.07\ \text{MW}}$$ The exhaust enthalpy is $3213.6 - 919.8 = 2293.9$ kJ/kg, a dryness fraction of 0.890. Eleven per cent moisture at the last stage is about the practical limit before erosion of the moving blades becomes a maintenance problem, which is one reason real plant reheats.
2. Part (b) — power with the throttle closed to 1 MPa. A throttle valve is adiabatic and does no work, so the enthalpy is unchanged at 3213.6 kJ/kg while the pressure falls to 1 MPa. Interpolating the 1 MPa superheat table between 350 °C ($h = 3157.7$, $s = 7.3011$) and 400 °C ($h = 3263.9$, $s = 7.4651$): $$f = \frac{3213.6 - 3157.7}{3263.9 - 3157.7} = 0.526 \;\Rightarrow\; T = 376.3\ ^{\circ}\text{C},\quad s = 7.3874\ \text{kJ/kg}\cdot\text{K}$$ Expanding from that entropy to 0.005 MPa, $$x_{2s} = \frac{7.3874 - 0.4764}{7.9187} = 0.8727 \;\Rightarrow\; h_{2s} = 2253.3\ \text{kJ/kg},\quad \Delta h = 0.80 \times 960.3 = 768.3\ \text{kJ/kg}$$ The flow follows the inlet pressure, so $$\dot m_b = 24 \times \frac{1}{4} = 6\ \text{kg/s} \qquad P_b = 6 \times 768.3 = \boxed{4\,610\ \text{kW} = 4.61\ \text{MW}}$$ Quarter flow has given not a quarter of the power but 20.9 % of it, because throttling also destroyed 16 % of the available enthalpy drop. That loss is the reason throttle governing was abandoned for nozzle-group governing on large machines.
3. Part (c) — power at zero generator output. Throttling to 0.1 MPa at constant enthalpy, and interpolating the 0.1 MPa table between 300 °C ($h = 3074.3$, $s = 8.2158$) and 400 °C ($h = 3278.2$, $s = 8.5435$): $$f = \frac{3213.6 - 3074.3}{3278.2 - 3074.3} = 0.683 \;\Rightarrow\; T = 368.3\ ^{\circ}\text{C},\quad s = 8.4397\ \text{kJ/kg}\cdot\text{K}$$ This entropy exceeds $s_g = 8.3951$ at 0.005 MPa, so the isentropic end state is superheated rather than wet. Treating low-pressure steam as an ideal gas with $c_{pv} \approx 1.9\ \text{kJ/kg} \cdot\text{K}$ from the saturated-vapour point, $$T_{2s} = 306.03 \exp\!\left(\frac{8.4397 - 8.3951}{1.9}\right) = 313.3\ \text{K}\ (40.1\ ^{\circ} \text{C})$$ $$h_{2s} = 2561.5 + 1.9 \times (40.1 - 32.88) = 2575.3\ \text{kJ/kg}$$ With the reduced internal efficiency and the pressure-proportional flow, $$\Delta h = 0.70 \times (3213.6 - 2575.3) = 446.8\ \text{kJ/kg} \qquad \dot m_c = 24 \times \frac{0.1}{4} = 0.6\ \text{kg/s}$$ $$P_c = 0.6 \times 446.8 = \boxed{268\ \text{kW}}$$ So 268 kW of bearing friction and generator windage — 1.2 % of the full-load rating — is what the machine consumes to spin at synchronous speed with the breaker open. That is the right order for a 22 MW set, and it is the quantity that sets the minimum steam admission needed to keep a unit on turning gear or in spinning reserve.
4. Part (d) — condenser temperature profiles at part load: fitting the full-load condition first. The duty at full load is the exhaust steam condensed to saturated liquid: $$\dot Q_{\text{full}} = \dot m (h_2 - h_f) = 24 \times (2293.9 - 137.8) = 51\,745\ \text{kW}$$ The circulating-water flow that produces the observed 10 K rise is $$\dot m_w = \frac{\dot Q_{\text{full}}}{c_p \Delta T_w} = \frac{51\,745}{4.190 \times 10} = 1\,235\ \text{kg/s}$$ Using the average temperature difference the question stipulates rather than the log mean, $$\theta_{\text{full}} = T_s - \tfrac{1}{2}(T_{w,\text{in}} + T_{w,\text{out}}) = 32.88 - \tfrac{1}{2}(13.0 + 23.0) = 14.88\ \text{K}$$ $$UA = \frac{\dot Q_{\text{full}}}{\theta_{\text{full}}} = \frac{51\,745}{14.88} = \boxed{3\,477\ \text{kW/K}}$$ Both $UA$ and $\dot m_w$ are fixed hardware — the tube surface does not change and the circulating pumps run at constant speed — so they carry over unchanged to every other load.
5. Part (d) continued — the part-load profile. At 6 kg/s the exhaust enthalpy is $3213.6 - 768.3 = 2445.3$ kJ/kg. Three equations close the problem: the duty $\dot Q = \dot m (h - h_f(T_s))$, the water rise $\Delta T_w = \dot Q / (\dot m_w c_p)$, and the transfer equation $\dot Q = UA\,[\,T_s - (T_{w,\text{in}} + \Delta T_w/2)\,]$. Solving them together (the saturation enthalpy $h_f$ makes it mildly implicit, so it converges in two passes): $$\dot Q_b = 14\,207\ \text{kW},\qquad \Delta T_{w,b} = 2.75\ \text{K},\qquad \theta_b = 4.09\ \text{K}$$ $$T_{s,b} = 13.0 + \tfrac{2.75}{2} + 4.09 = \boxed{18.5\ ^{\circ}\text{C}}$$ so the profile to plot is exhaust steam flat at 18.5 °C with cooling water rising from 13.0 °C to 15.8 °C. Both curves collapse towards the water inlet temperature, and the gap between them narrows from 14.9 K to 4.1 K, because the same surface now has only 27 % of the duty to pass.
6. Part (e) — condenser pressure at part load. The condenser pressure is simply the saturation pressure at the exhaust-steam temperature just found. Interpolating the saturation table between 15 °C (1.705 kPa) and 20 °C (2.339 kPa): $$p_b = p_{\text{sat}}(18.5\ ^{\circ}\text{C}) = \boxed{2.14\ \text{kPa} = 0.00214\ \text{MPa}}$$ The vacuum has deepened from 5 kPa to 2.1 kPa purely because the load fell. This is the mechanism behind the familiar observation that a condensing turbine's heat rate improves slightly at part load even as its throttle loss worsens: the back pressure follows the duty down, and every kilopascal of recovered vacuum is worth roughly 1 % on the low-pressure enthalpy drop.
7. Part (f) — condenser pressure at zero load, estimated without further calculation. The reasoning is the limit of step 5. At 0.6 kg/s the duty is only about 1 600 kW — 3 % of full load — so both the water temperature rise ($\approx 0.3$ K) and the mean temperature difference ($\approx 0.5$ K) become negligible, and the exhaust steam temperature must fall to within about a degree of the cooling-water inlet temperature: $$T_{s,c} \approx T_{w,\text{in}} + 0.6 \approx 13.6\ ^{\circ}\text{C} \;\Rightarrow\; p_c \approx p_{\text{sat}}(13.6\ ^{\circ}\text{C}) = \boxed{1.6\ \text{kPa} = 0.0016\ \text{MPa}}$$ In other words, at no load the condenser pressure is pinned by the cooling-water inlet temperature alone, and 13 °C water sets a floor of about 1.5 kPa. In service that floor is never reached, because air in-leakage and the finite capacity of the air-removal equipment hold the back pressure somewhat higher; a real machine on no load would sit nearer 2.5–3 kPa.