Solutions_10_1

Solutions_10_1 - Eng 2210 Thermodynamics Homework Set #10...

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Eng 2210 – Thermodynamics Homework Set #10 Solutions Problem 8.49: Problem Statement: Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater heater. The condensate leaving the feed water heater is a saturated liquid, and the feedwater leaving the heater leaves at the same temperature as the condensate that exits the feedwater heater. Determine for the cycle: a. The rate of heat transfer to the working fluid passing through the steam generator, in kJ per kg of steam entering the first-stage turbine. b. The thermal efficiency. c. The rate of heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per kg of steam entering the first-stage turbine. T 1 = 480 ˚C Given Information: P 1 = 10 MPa P 2 = 0.7 MPa P 3 = 6 kPa Engineering Model : Each component is a Control Volume at steady state. There is no loss in heat or pressure between components. All of the processes are internally reversible except for the throttling process between 7 and 8 where h 7 = h 8 . Kinetic and potential energy effects are negligible. Solution:
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The first thing to do is to solve for the enthalpy of every state. This will allow us to compute all the requested values. State 1 kJ kg : From Table A – 4, and using P 1 = 10 MPa and T 1 = 480 ˚C, h 1 = 3321.4 and s 1 = 6.5282 kJ kg K State 2 kJ kg K : Because it is an ideal Rankine cycle the turbine has a 100% isentropic efficiency, so that s 1 = s 2 = 6.5282 . Using Table A-3 we find that at P 2 and s 2 , the fluid is in the two phase region. We need to first solve for x 2 , and use our answer to find h 2 . 2 2 22 2 () f gf s s xs s = +− 2 0.9619 f ss x = 2 2 2 ( ) 2684.8 f h h xh h = + −= kJ kg State 3 kJ kg K : Like State 2 the second stage turbine is ideal so that s 1 = s 2 = s 3 = 6.5282 . We then apply the same procedure to State 3 as we did to State 2 using P 3 as our pressure. 3 3 33 3 f = 3 0.7692 f x = 3 3 3 ( ) 2009.8 f = + kJ kg State 4 kJ kg : We know that at the exit of the condenser the fluid is a saturated liquid, and we know that there is no pressure drop in an ideal condenser so that P 3 = P 4 = 6kPa. Using Table A – 3 we can then find that h 4 = 151.53 . State 5 3 3 5 4 45 4 ( ) 151.53 1.0064*10 (10,000 6) 161.588 kJ m h h v P P kPa kg kg = + + = : To find h 5 we can assume that the water is incompressible, so that the only change in enthalpy occurs due to compression work of the pump. We can then write the equation below that will allow us to solve for h 5 using this assumption. kJ kg State 7 kJ kg : Like State 4 we know that the fluid is in a saturated liquid state at the end of the feedwater heater, so we can use P 2 and Table A – 3 to find that h 7 = 697.22 . State 6 kJ kg : We know that the T 6 = T 7 from the problem statement, so we can assume that the water is incompressible and apply equation 3.20b and say that h 6 = h 7 = 697.22 . State Enthalpy (kJ/kg) h 1 3321.4 h 2 2684.8 h 3 2009.8 h 4 151.53 h 5 161.588 h 6 697.22 h 7 697.22 h 8 697.22
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State 8 kJ kg : Because we are assuming an ideal throttling process between States 7 and 8, we therefore know that h 8 = h 7 = 697.22 .
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This note was uploaded on 03/15/2010 for the course ENGRD 2210 at Cornell.

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Solutions_10_1 - Eng 2210 Thermodynamics Homework Set #10...

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