# 7333585 - = mass flow rate of fluid (lb/hr)*-= h3 h4 600 ....

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We have, Process 1-2: Pump work Process 2-3: Boiler Process 3-4: Turbine generator Process 4-1: Condenser Now, State 3: (exit of boiler) = P3 1450 psi and 900degF From steam table, we get:- = h3 . / 1431 3 Btu lbm = . - s3 1 5617Btulbm R State 2: (exit of generator) = . P4 0 7 psi = = . - s4s s3 1 5617Btulbm R , = - = . - quality x s4s sfsfg 1 5617 . . = . 0 11161 8966 0 765 = . & = + = . / h4s h @0 7 psia s4s hf xhfg 855 68 Btu lbm Efficiency = - - = . h3 h4h3 h4s 0 95 => = . / h4 884 461 Btu lbm

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State 1: (exit of condenser) = . = . h1 hf @ 0 7 psi 58 06 / Btu lbm State 2: (exit of pump) - = - = . - . = . h2 h1 vP3 P4 0 016X1450 0 7 23 33 = . + . = . / h2 23 33 58 06 81 39 Btu lbm Now, Work output of the generator, Wout
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Unformatted text preview: = mass flow rate of fluid (lb/hr)*-= h3 h4 600 . = . MW0 95 631 58MW = . × 2 15504041 10 / 9 btu hr Hence, mass flow rate of motive fluid, lb/hr = . × 2 15504041 10 9 . -. = . / 1431 3 884 461 3940904 745 lb hr Cooling water flow rate = . ( .-. ) . = . / 3940904 745X 884 461 58 06 5X1 007 646825744 2 lb hr Cooling water flow rate = * * = ρ vel Area 646825744 . / 2 lb hr . = . 62 42X10Xπd24 646825744 2 Inside diameter, = . d 33 93 ft Power input for the feed water pump= = . Wpump 3940904 745 . = . / v ∆P 91941307 7 Btu hr = . Wpump 36134 2927 hp...
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## This note was uploaded on 07/22/2011 for the course ME 3 taught by Professor Prof.ramachandran during the Spring '11 term at Indian Institute of Technology, Kharagpur.

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7333585 - = mass flow rate of fluid (lb/hr)*-= h3 h4 600 ....

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