Question

A geothermal source provides water at 400°F, 140 psia to a heat exchanger that boils an organic liquid. Condenser cooling water is available at 70°F and there is a 10°F ∆T across the condenser (which is a second counterflow heat exchanger) at the cooling water inlet.

(a) If the minimum allowable ∆T across the geothermal water heat exchanger is 15°F, what is the maximum theoretical thermodynamic efficiency that could be expected of such a cycle?

(b) How many cubic feet of pressurized geothermal source water are required per net megawatthour generated at maximum efficiency?

(c) Re-compute the geothermal water volume requirements if the turbine–generator operates at an efficiency of 80% and the feed pump work is neglected

Answers: a: 0.361. b: 28300ft^3

Answer 1

Refer to the given problem Consider the given data: Source water: Temperature 400°F Pressure140 psi Also, the Condenser cooling water is available at 70°F and there is a 10°F, AT across the condenser (a) Now for the minimum allowable AT of 15°F across the geothermal water heat exchanger Find T TH 400°F -15°F 385°F Find Tr T, 70°F10°F 80°F Consider the expression for the maximum theoretical thermodynamic efficiency that could be expected of such a cycle TL n =1- TH 460°F 80 F (460°F273°C) 460°F 385 F 540°F 1- 845°F 1-0.639 0.361 Thus, the required theoretical thermodynamic efficiency is 0.361

(b) Refer to the steam table Consider the following data At 400°F (Or, 204.44°C) Specific Enthalpy of Saturated Water, h,-872.474 kJ/kg Specific Enthalpy of Saturated Steam, h 2794.6 kJ/kg At 385°F (Or, 196.11°C) Specific Enthalpy of Saturated Water, h 834.898 kJ/kg Specific Enthalpy of Saturated Steam, h 2789.6 kJ/kg From thermodynamics efficiency Wnet Q net W. m(hg-h) net W net m= (ч- ) (0.361) (2789.6-834.898) 1MW-h m= (109 (3600) (0.361) (2789.6-834.898) m= m - 5101.698x10 Kg

From heat balance equation M(h 400F35 m(h,400F-38F (hg 400-Fh385F) g,385°F M m hf400F f385°F 2794.6 kJ/kg-2789.6 kJ/ kg M (5101.698x 10" kg ) \872.474 kJ/kg-834.898 kJ/kg) 2794.6-2789.6 (5101.698x103 872.474-834.898 M 678850.596 kg (convert kg ft M-28134 ft3 M 28300 ft Therefore, around M 28300 ft cubic feet of pressurized geothermal source water are required per net megawatt hour generated at maximum efficiency (c) Now in order to re-compute the geothermal water volume requirements if the turbine generator operates at an efficiency of 80% Let the feed pump work is neglected Calculate the thermal efficiency when turbine-generator operates at an efficiency of 80%

80 n(s0%) 100 80 (0.36 ,361) 100 =0.288 780%) 28.8% Thus, n/80%) 28.8% Consider the expression for re-computation of geothermal water requirement From thermodynamics efficiency. W net Q Wnet Q = n(80%) n(80%) W net m2 7(s0%) net m' 1MW-h m2 (0.288)(2789.6- 834.898) (10*)(3600) m' (0.288) (2789.6-834.898) m' 6394.8x103 Kg

From heat balance equation M'(h, 00F =m'(h,400F385 (h g,385 F M'm400Fhe 385 f,400F hf385 F g,385°F 2794.6 kJ/kg 2789.6 kJ/kg 872.474 kJ/ kg -834.898 kJ/kg ) (6394.8x10' Kg М" - 2794.6-2789.6 M' (6394.8x 103 872.474-834.898 M' 850915.477 kg (convert kg ft M'-35265.02 ft3 M' 35300 ft Therefore, the Re-compute amount of geothermal water volume requirements if the turbine-generator operates at an efficiency of 80% is around M'~ 35300 ft