Question 4 of 6: Belledune Heat Balance Diagram — cycle efficiency, HP turbine and feed pump duty

Question 4: Belledune Heat Balance Diagram — cycle efficiency, HP turbine and feed pump duty (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 readings taken from the page 11 heat balance diagram.

Find. The steam cycle efficiency on boiler heat input, the HP turbine shaft power, the steam power delivered to the feed pump turbine, the shaft power into the feed pump, the hydraulic power out of it and hence the pump efficiency.

Approach. Every part is an energy balance on one component, and each flow must first be converted from kg/h to kg/s by dividing by 3600. Before using any enthalpy, the mass balances printed on the diagram are checked, because a closure to the last kilogram per hour proves the readings are right.

1. Step 0 — validate the readings against the diagram’s own mass balances. Three closures are available and all three are exact, so the figures above can be trusted:$$\text{leak-offs: } 12\,220+3\,940+1\,600+340=18\,100\ \text{kg/h}$$$$\text{HP turbine: } 1\,287\,309-18\,100=1\,141\,480+127\,729$$$$\text{cold reheat header: } 1\,141\,480=1\,039\,160+101\,910+410$$
2. Part (a) — steam cycle efficiency on boiler heat input. The boiler adds heat twice: once raising the feedwater to main steam, and once through the reheater:$$Q_{boiler}=\dfrac{M_{ms}}{3600}(h_{ms}-h_{fw})+\dfrac{M_{rh}}{3600}(h_{hrh}-h_{crh})$$$$=\dfrac{1\,287\,309}{3600}(3397.5-1231.4)+\dfrac{1\,039\,160}{3600}(3542.5-3002.7)$$$$=774\,567+155\,816=930\,383\ \text{kW}$$$$\eta_{cycle}=\dfrac{430\,000}{930\,383}=\boxed{0.462\ (46.2\ \%)}$$Forgetting the reheater term would inflate the answer to 55.5 %, which no subcritical steam cycle achieves — that alone flags the omission.
3. Part (b) — shaft power of the high pressure turbine. The leak-offs never enter the blading, so they do no work; the steam that does expand is the main steam less the leak-offs, and it leaves in two streams:$$P_{HP}=\dfrac{(M_{ms}-M_{leak})h_{ms}-M_{ext}h_{ext}-M_{exh}h_{crh}}{3600}$$$$=\dfrac{(1\,269\,209)(3397.5)-(127\,729)(3175.9)-(1\,141\,480)(3002.7)}{3600}=\boxed{133\,045\ \text{kW}}$$Grouping the same balance the other way is a useful check: $[1\,269\,209(3397.5-3175.9)+1\,141\,480(3175.9-3002.7)]/3600$ gives the identical 133 045 kW.
4. Part (c) — steam power into the feed pump turbine. The BFPT is a small back-pressure machine taking 50 140 kg/h:$$P_{BFPT}=\dfrac{50\,140}{3600}\times (3068.9-2432.6)=13.928\times 636.3=\boxed{8862\ \text{kW}}$$
5. Part (d) — shaft power into the feed pump from its enthalpy rise. The whole main steam flow passes through the pump as feedwater, and the diagram gives the enthalpy rise directly:$$P_{shaft}=\dfrac{M_{ms}}{3600}\,\Delta h=357.586\times 24.45=\boxed{8743\ \text{kW}}$$The diagram itself prints 8741 kW at the pump coupling, so the reading of $\Delta h=24.45$ kJ/kg is confirmed to 0.02 %. Comparing (c) and (d), the drive train (turbine coupling to pump shaft) is running at 8743/8862 = 98.7 %, which is the expected mechanical loss of a direct-coupled machine.
6. Part (e) — hydraulic power from the pressure rise. PD is the pump discharge pressure, so the rise across the pump is the discharge less the deaerator pressure at the suction:$$\Delta p=19.8-0.554=19.246\ \text{MPa}$$$$P_{hyd}=\dfrac{M_{ms}}{3600}\cdot\dfrac{\Delta p}{\rho}=357.586\times\dfrac{19.246\times 10^{3}}{912}=\boxed{7547\ \text{kW}}$$
7. Part (f) — feedwater pump efficiency. Useful hydraulic power out over shaft power in:$$\eta_{pump}=\dfrac{P_{hyd}}{P_{shaft}}=\dfrac{7547}{8743}=\boxed{0.863\ (86.3\ \%)}$$which sits in the 82 to 88 % band expected of a barrel feed pump. Taking PD as the pressure rise instead of the discharge pressure would give 88.8 % and would quietly credit the pump with work the deaerator had already done.