Question

(b) An ocean thermal gradient electric generating system (OTEC) is proposed for siting in Oman. This system would operate a heat engine driven by heat flow from warm ocean water (cooled from 27°C to 25 °C in the heating heat exchanger of the system), and cooled by heat flow to cold ocean water (which is heated from 3 °C to 5 C in the cooling heat exchanger of the system). The ocean water used in the different parts of the heat engine is obtained from different depths of the surrounding sea. Using this heat engine electricity can be produced at 90%ofthe corresponding camo efficiency of the heat engine. 

(i) What is the value of the Camot efficiency of the heat engine? Explain the basis for selecting the parameter values used in calculating this value. 

(ii) Should the cold ocean water flow through the engine at a rate of 100 million kg/min 

 what is the electric power output of the engine?

(iii) What is the corresponding flow rate of warm ocean water? 

(iv)Friction losses arising in pumping seawater through the engine account for 0.33 of the engine's irreversibilities (maximum work-electrical power). What would be the resulting value of the heat engine's efficiency? 

Answer 1

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Answer 2

Solution: 

a. (10 points) What is the value of the Carnot efficiency of the heat engine? Explain the basis for selecting the parameter values used in calculating this value. 

The Carnot efficiency of the heat engine is 1-TC/TH, where Tc and Th are the maximum temperature of the cold reservoir, and the minimum temperature of the hot reservoir, respectively. The minimum absolute temperature of the hot reservoir is 298 ⁰C, and the maximum absolute temperature of the cold reservoir is 276 ⁰C, making the Carnot efficiency 1-(276/298)=0.074. 

b. (15 points) Should the cold ocean water flow through the engine at a rate of 100 million kg/min what is the electric power output of the engine 

The heat carried away from the system is Qout=Qin-W. The work done by the system is W=Qin*efficiency. This means that Qout=W(1/efficiency – 1). 

Since the heat transferred to the working fluid is defined by its flow rate, temperature change, and heat capacity as: Qout=mcp∆T, We find W=mcp∆T/(1/efficiency-1). W=(100 x 106 kg/min)(min/60 s)(4.184 kJ/kg K)(2 ⁰C)/(1/0.074/0.9 -1 ) = 0.995 GWe. 

c. (15 points) What is the corresponding flow rate of warm ocean water? 

In order to find the flow rate of the warm ocean water, we need to balance the heat rejected to the cold water, the heat rejected by the cold water, and the work done by the system. Qin=Qout+W=moutcp∆T+0.995 GWe Qin=mincp∆T min=(moutcp∆T+0.995 GWe)/cp∆T =[(100 x 106 kg/min)(min/60 s)(4.184 kJ/kg)(2 ⁰C) +0.995 GWe)]/[(4.184 kJ/kg K)(2 ⁰C)] =107.134 million kg/min 

d. (10 points) Friction losses arising in pumping seawater through the engine account for 0.33 of the engine’s irreversibilities. Over time we can expect such friction losses to double, due to befouling in the flow circuits. What would be the resulting value of the heat engine’s efficiency? 

Irreversibility is the difference between the maximum theoretical work which can be done by the thermodynamic cycle and the actual work done. The Carnot efficiency calculated for this cycle leads us to a maximum theoretical work of Wmax=Qin*0.074=mincp∆T*0.074=(107.134 x 106 kg)(min/60s)(4.184 kJ/kg)(2 ⁰C) *0.074=1.105 GWe. 

The irreversibility is then 1.105 GWe-0.995 GWe=0.110 GWe. If over time, the 33% of this value representing frictional loss doubles, we add an extra 33% to the irreversibility, for a total of 0.147 GWe lost work. This makes the efficiency of the system (1.105 GWe-0.147 GWe)/Qin =0.064