At a particular location in the ocean, the temperature near the surface is 80oF,

Question

At a particular location in the ocean, the temperature near the surface is 80oF, and the temperature at a depth of 1500 ft is 46oF. A power plant based on the Rankine cycle, with ammonia as the working fluid, has been proposed to utilize this naturally occurring temperature gradient to produce electrical power. The power to be developed by the turbine is 8.2 X 108 Btu/h. A schematic diagram is shown. For simplicity, the properties of seawater can be taken as those of pure water.Assume the isentropic efficiency of the turbine is 80% and the isentropic efficiency of the pump is 70%. Assume that the ammonia leaving the boiler is saturated vapor and that the ammonia leaving the condenser is saturated liquid.
a.) Estimate the net power output of the plant, in Btu/h if the pumps used to circulate seawater through the evaporator and condenser heat exchangers require a total power input of 2.55 X 108 Btu/h.





b.) Determine the seawater flow rates through the boiler and condenser, in lb/h.



c.) Estimate the thermal efficiency of the proposed cycle and compare it to the thermal efficiency of a reversible power cycle operating between thermal reservoirs at 80 and 46oF.

Explanation / Answer

The thermal efficiency of a power cycle can be determined using : Eqn 1 We need to evaluate Q12 and Q34 so we can use Eqn 1 to evaluate the thermal efficiency of the cycle. Apply the 1st Law to the boiler, assuming it oreates at steady-state, changes in kinetic and potential energy are negligible and no shaft work crosses the boundary of the boiler. Eqn 2 We can get a value for H1 from the steam tables or NIST Webbook because we know that the boiler effluent in a Rankine Cycle is a saturated vapor at 75oF. P2 140.59 psia H2 699.48 Btu/lbm In order to fix state 1 and evaluate H1, we must use the isentropic efficiency of the pump. The feed to the pump is a saturated liquid at 50oF. So, we can lookup S4 in the steam tables or NIST Webbook. P4 89.205 psia S4 0.39148 Btu/lbm-oR H4 167.66 Btu/lbm The definition of isentropic efficiency for a pump is : Eqn 3 We can solve Eqn 2 for H1 : Eqn 4 Next, we need to determin H1S. For an isentropic pump : S1S 0.39148 Btu/lbm-oR Now, we know the value of two intensive properties at state 1S: S1S and P1 (because the boiler is isobaric in a Rankine Cycle, P1 = P2. At P = 140.59 psia : T (oF) H Btu/lbm S Btu/lbm- 50 167.75 0.39117 T1S H1S 0.39148 T1S 50.14 oF 55 173.34 0.40209 H1S 167.91 Btu/lbm Now ,we can plug values into Eqn 4 to determine H1 : H1 168.02 Btu/lbm Next, we can plug values into Eqn 2 to evaluate Q12 : Q12 531.46 Btu/lbm We can determine Q34 by applying the 1st Law, with all the same assumptions made about the boiler. Eqn 5 We already know H4, so we need to evaluate H3. To do this, we use the isentropic efficiency of the turbine. In order to fix state 3 and evaluate H3, we must use the isentropic efficiency of the turbine. The feed to the turbine is a saturated vapor at 75oF. So, we can lookup S2 in the steam tables or NIST Webbook. P4 89.205 psia S2 1.3869 Btu/lbm-oR H2 699.48 Btu/lbm The definition of isentropic efficiency for a turbine is : Eqn 6 We can solve Eqn 6 for H3 : Eqn 7 Next, we need to determin H3S. For an isentropic turbine : S3S 1.3869 Btu/lbm-oR Now, we know the value of two intensive properties at state 3S: S3S and P3 (because the condenser is isobaric in a Rankine Cycle, P3 = P4. At P = 89.205 psia : Ssat liq 0.39148 Btu/lbm-oR Since Ssat liq

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