HW 12 Solution

# HW 12 Solution - Problem9.87 PROBLEMSTATEMENT:...

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ENGRD 2210 – Thermodynamics PROBLEM SET 12 SOLUTIONS Problem 9.87: PROBLEM STATEMENT: Data are known at principal states of a combined gas turbine vapor power plant as shown in figure below. The net power developed by the gas turbine is 147MW. Use air standard analysis for the gas turbine. FIND: Determine (a) the net power, and (b) the overall thermal efficiency. Develop a full accounting of the net rate of exergy transfer to the air passing through the gas turbine combustor. SCHEMATIC & GIVEN DATA: ENGINEERING MODEL: (1) Each component is a control volume at steady state. (2) The compressor, turbines, and pump operate adiabatically. (3) Kinetic and potential energy effects are negligible. (4) The air is modeled as an ideal gas. (5) There are no stray heat losses. ANALYSIS: First, fix all of the principal states. For the air, use Table A 22. 11 1 22 2 33 3 44 4 55 5 300 ; 300.19 , 1.70203 690 ; 702.52 2.55731 1580 ; 1733.17 3.50829 900 ; 932.93 2.84856 400 ; 400.98 1.99194 TK h k J k g s h k J k g s h k J k g s h k J k g s h k J k g s == = = = = = D D D D D For the vapor cycle, State 7: 77 100 520 P bar T C ° 3425.1 / 6.6622 hk J k g s ⇒= =

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State 8: 88 7 8 8 6.6622 0.5926 0.08 , 0.7949; 2084.1 / 8.2287 0.5926 ss s P bar s s x h kJ kg == = = = () 87 78 8 8 2285.3 / ; 0.0786, 7.3019 / ts h h h h kJ kg x s kJ kg K η =− − = = = State 9: 99 9 0.08 , sat. liquid 173.88 / 0.5926 / P bar h kJ kg s kJ kg K =⇒ = = State 6: 35 2 3 69 9 6 9 3 10 1 173.88 1.0084 10 100 0.08 11 0 183.40 s kJ m N m kJ h h P P bar kg kg bar N m kJ kg υ ⎛⎞ ≈+ + × ⎜⎟ ⎝⎠ = with 6 9 185.78 / ps p h h h h kJ kg ηη =+ − = Interpolating in Table A 5: 0.5968 b s = (a) Begin by determining the mass flow rate of air, as follows. 34 21 gas air Wmh hh h = −−− ⎡⎤ ⎣⎦ ± ± or ( ) 147000 / 1733.17 932.93 702.52 300.19 gas air W kJ s m kJ hh hh kg ± ± Energy and mass balances for the inter connecting heat exchanger give 45 67 0 air st mhh m + ±± or ( ) 76 369.43 932.93 400.98 60.67 / 3425.1 185.78 air st mk g s −− = ± ± The power developed by the steam cycle is 3 1 60.67 3425.1 2285.3 185.78 173.88 68.43 10 st vap p st WW h h h kg kJ MW MW sk g k J s = = ± ± The net power for the plant is 147 68.43 215.43 cycle gas vap W M W =−= + = ±
(b) The heat addition is () ( ) ( ) 32 3 1 369.43 1733.17 702.53 380.75 10 in gas Qmh h =− = = ± ± Thus 215.43 0.566 56.6% 380.75 cycle in W Q η == = ± ±

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Problem 10.9: PROBLEM STATEMENT: An ideal vapor compression refrigeration system operates at steady state with Refrigerant 134a as the working fluid. Superheated vapor enters the compressor at 2 10lbf in , 0 F ° , and saturated liquid leaves the condenser at 2 180lbf in . The refrigeration capacity is 8 tons.
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HW 12 Solution - Problem9.87 PROBLEMSTATEMENT:...

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