Question 3 of 6: Belledune Heat Balance Diagram — cycle efficiency, HP turbine power and feed pump performance at 430 MW

Question 3: Belledune Heat Balance Diagram — cycle efficiency, HP turbine power and feed pump performance at 430 MW (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. The Belledune heat balance diagram of page 12, at a rated generator output of 430 MW. The streams that bear on the six parts are listed below; G denotes a mass flow in kg/h, H a specific enthalpy in kJ/kg, P a pressure in MPa and T a temperature in °C, following the annotation convention of the diagram itself.

Find. The steam cycle efficiency on boiler heat input, the HP turbine shaft power, the steam power supplied to the boiler feed pump turbine, the shaft and hydraulic powers of the feed pump, and hence the feed pump efficiency.

Approach. The boiler heat input is the sum of two duties, the main steam pass and the reheat pass, each computed from its own mass flow and enthalpy rise. The HP turbine power is then a steady-flow energy balance on a control volume enclosing the HP cylinder, with the leak-off flows removed before the expansion because they never do work. The feed pump is analysed twice, thermodynamically from its enthalpy rise and hydraulically from its pressure rise, and the ratio of the two is the pump efficiency.

1. Part (a) — boiler heat input. The boiler adds heat in two passes. The main pass takes the final feedwater to main steam conditions and carries the whole feedwater flow, and the reheat pass takes cold reheat steam to hot reheat conditions and carries only the reheated flow: $$\dot{Q}_{b}=\frac{G_{ms}}{3600}(h_{ms}-h_{fw})+\frac{G_{rh}}{3600}(h_{hrh}-h_{crh})$$ The main pass gives $(1\,287\,309/3600)\times(3397.5-1231.4)=774\,567\ \text{kW}$ and the reheat pass $(1\,039\,160/3600)\times(3542.5-3002.7)=155\,816\ \text{kW}$.
2. Part (a) continued — cycle efficiency. The total is $$\dot{Q}_{b}=774\,567+155\,816=930\,383\ \text{kW}$$ so with 430 MW at the generator terminals $$\eta_{cycle}=\frac{430\,000}{930\,383}=\boxed{0.4622}$$ that is 46.2 %, equivalent to a station heat rate of $3600/0.4622=7789\ \text{kJ}\cdot(\text{kW}\cdot\text{h})^{-1}$ on boiler heat. The reheat pass supplies 16.7 % of the boiler duty, and note that this efficiency is measured on heat into the steam, so it excludes the boiler loss and is properly compared with the 48 % assumed in Question 1, not with an overall plant figure.
3. Part (b) — HP turbine shaft power, control volume. Take a control volume around the HP cylinder. Steam enters at the throttle, three streams leave: the gland and valve-stem leak-offs, which are tapped upstream of the blading and so leave at throttle enthalpy having done no work; the extraction to the top heater at state $h_{ext}$; and the exhaust to the cold reheat header at $h_{crh}$. Removing the leak-offs first, the flow that actually expands is $1\,287\,309-18\,100=1\,269\,209\ \text{kg}\cdot\text{h}^{-1}$, and the diagram closes exactly on $1\,269\,209=1\,141\,480+127\,729$, which confirms the leak-off total.
4. Part (b) continued — energy balance. The shaft power is the enthalpy the expanding steam gives up: $$\dot{W}_{HP}=\frac{1}{3600}\left[(G_{ms}-G_{leak})(h_{ms}-h_{ext})+G_{exh}(h_{ext}-h_{crh})\right]$$ The first group is the whole working flow dropping 3397.5 to 3175.9, that is 221.6 kJ/kg, and the second is the surviving exhaust flow dropping a further $3175.9-3002.7=173.2\ \text{kJ}\cdot\text{kg}^{-1}$. Substituting, $$\dot{W}_{HP}=\frac{1\,269\,209\times221.6+1\,141\,480\times173.2}{3600}=\boxed{133\,045\ \text{kW}}$$ The HP cylinder therefore accounts for 30.9 % of the 430 MW at the terminals, the IP and LP cylinders supplying the rest.
5. Part (c) — steam power into the feed pump turbine. The boiler feed pump turbine is a small back-pressure machine taking 50 140 kg/h of steam at 0.542 MPa and exhausting to the main condenser: $$\dot{W}_{steam}=\frac{G_{bfpt}}{3600}(h_{in}-h_{out})=\frac{50\,140}{3600}\times(3068.9-2432.6)=\boxed{8862\ \text{kW}}$$ an available enthalpy drop of 636.3 kJ/kg. This is the steam power presented to the turbine, not yet the power delivered to the pump.
6. Part (d) — shaft power into the pump. Measured on the water side, the shaft power is whatever raises the enthalpy of the whole feedwater flow by the stated 24.45 kJ/kg across the pump: $$\dot{W}_{shaft}=\frac{G_{fw}}{3600}\,\Delta h_{pump}=\frac{1\,287\,309}{3600}\times24.45=\boxed{8743\ \text{kW}}$$ The diagram itself prints 8741 kW at the feed pump turbine coupling, so the two agree to 0.02 %. Comparing with part (c), the feed pump turbine converts its 8862 kW of steam power into 8743 kW at the coupling, an internal and mechanical efficiency of 98.7 %; on a real machine that number would be nearer 80 %, and the near unity here indicates that the diagram treats the small turbine as a nearly ideal expander whose exhaust enthalpy has already been credited with its losses.
7. Part (e) — hydraulic power. Hydraulically the pump does work against the pressure rise from the deaerator to the discharge, and because water is nearly incompressible that work is $v\,\Delta p=\Delta p/\rho$ per kilogram: $$\dot{W}_{hyd}=\frac{G_{fw}}{3600}\cdot\frac{(p_{d}-p_{o})}{\rho}=\frac{1\,287\,309}{3600}\times\frac{(19.8-0.554)\times10^{3}}{912}$$ The specific work is $19\,246/912=21.10\ \text{kJ}\cdot\text{kg}^{-1}$ against a volume flow of $0.3921\ \text{m}^{3}\cdot\text{s}^{-1}$, giving $$\dot{W}_{hyd}=\boxed{7546\ \text{kW}}$$
8. Part (f) — feed pump efficiency. The pump efficiency is the useful hydraulic output over the shaft input: $$\eta_{pump}=\frac{\dot{W}_{hyd}}{\dot{W}_{shaft}}=\frac{7546}{8743}=\boxed{0.863}$$ The missing 13.7 % appears as the difference between the actual enthalpy rise of 24.45 kJ/kg and the ideal 21.10 kJ/kg, that is as a 3.35 kJ/kg temperature rise of the feedwater, which is exactly what an 86 % efficient boiler feed pump of this size would be expected to return.