Thermodynamics Notes 22

# Thermodynamics Notes 22 - ME 311 FALL 2007 CHAPTER 8...

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ME 311 THERMODYNAMICS S. Masutani FALL 2007 CHAPTER 8 (conclusion) Example : Basic open Rankine cycle 3 turbine boiler pump t W & 1 2 4 p W & in Q & Working Fluid: H 2 O p 1 = 1 bar, T 1 = 25 ° C = 298K p 2 = 40 bar T 3 = saturation temperature at 40 bar p 4 = 1 atmosphere = 1 bar 80 . 0 = t η ; 70 . 0 = p m & = 8 kg/s Process representation 1 T s 1 4 3 2 4s 2s P = 40 bar P = 1 bar

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ME 311 THERMODYNAMICS S. Masutani FALL 2007 h s P= 1 bar P = 40 bar 3 4 4s 1 2 2s Objective: determine cycle efficiency (SSSF) in p t Q W W & & & η From previous lectures: | ; ) ( 4 3 3 h h m W t = & & ) ( 1 2 1 h h m W p = & & from 1 st Law analysis | 1 3 m m & & = | ; ) ( 2 3 2 h h m Q in = & & 1 3 2 m m m & & & = = Assumptions: 1-D flow; ignore Δ KE and Δ PE for individual components; adiabatic pump and turbine. So ) ( ) ( 1 ) ( ) ( ) ( 2 3 1 4 2 3 1 2 4 3 h h h h h h h h h h = h 1 and h 3 can be determined directly from the steam tables: State 1: p 1 = 1 bar, T 1 = 25 ° C (compressed liquid) h 1 (compressed liquid) h f (25 ° C) - v f (25 ° C) [1 bar – p sat (25 ° C)] = 104.89 kJ/kg – 1.0029 x 10 -3 m 3 /kg [1 x 10 5 Pa – 3.169 x 10 3 Pa] =104.89 kJ/kg – 97.11 J/kg = 104.79 kj/kg 104.8 kJ/kg 2
ME 311 THERMODYNAMICS S. Masutani FALL 2007 State 2: p 2 = 40 bar; saturated vapor T = 250.4 ° C h g = h 3 = 2801.9 kJ/kg; s 3 = 6.0701 kJ/kg K To find h 4 and h 2 we employ the definitions of t η and p : ( ) () 1 1 2 2 1 2 1 2 ) ( ) ( ) ( h h h h h h h h W W p s s actual p s p p + = = = & & From the Gibbs equation: vdp dh vdp dh Tds = = = 0 ∫∫ = = ss s s vdp h h vdp dh 2 1 2 1 1 2 2 1 but v constant v 1 v f (25 ° C) So h 2s – h 1 v f (25 ° C) [p 2s – p 1 ] = 1.0029 x 10 -3 m 3 /kg [40 x 10 5 – 1 x 10 5 N/m 2 ] = 3911.3 J/kg = 3.9113 kJ/kg h 2 = [(0.70) -1 (3.9113 kJ/kg)] + 104.8 kJ/kg = 110.4 kJ/kg 4 3 4 3 4 3 4 3 ) ( ) ( ) ( h h h h h h h h W W s t s s p actual t t = = & & or ) ( 4 3 3 4 s t h h h h = To find h 4s : s 4s = s 3 = 6.0701 kJ/kg K; p 4s = p 4 = 1 bar From the saturation tables for steam at 1 bar: s g = 7.3594 kJ/kg K s f = 1.3026 kJ/kg K h f = 417.46 kJ/kg h g = 2675.5 kJ/kg Since s f < s 4 < s g state 4s is two-phase (i.e., a mixture of liquid and vapor).

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## This note was uploaded on 05/19/2008 for the course ME 311 taught by Professor Masutani during the Fall '07 term at Hawaii.

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Thermodynamics Notes 22 - ME 311 FALL 2007 CHAPTER 8...

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