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For the given question and solution, for part (b), finding the work input  of the saturated vapour at the inlet state, why couldnt i  use the same formula as part (a) Win = v*(change in pressure) where the v would be the saturated vapour specific volume instead of saturated fluid volume

For the given question and solution, for part (b), finding the work input  of the saturated vapour at the inlet state, why couldnt i  use the same formula as part (a) Win = v*(change in pressure) where the v would be the saturated vapour specific volume instead of saturated fluid volume

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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For the given question and solution, for part (b), finding the work input  of the saturated vapour at the inlet state,

why couldnt i  use the same formula as part (a) Win = v*(change in pressure) where the v would be the saturated vapour specific volume instead of saturated fluid volume??

(b) Compressor (for gases): Tds = dh – vdP, ds = 0 (isentropic) || From Tds relations vdP = dh -[ vdP -S dh=-(h, – h,) W, = - rev State 1 (Table A-5): P =100 kPa, Sat. vapour s, =7.3589 kJ/kgK S1 h = 2675.0 kJ/kg State 2 (Table A-6): P, =1 MPa, s, = s, = 7.3589 kJ/kgK h, = 3194.5 kJ/kg Wroy = -(3194.5- 2675.0)kJ/kg = -519.5 kJ/kg rev Less energy to pump liquids than gas between same pressure range
Transcribed Image Text:(b) Compressor (for gases): Tds = dh – vdP, ds = 0 (isentropic) || From Tds relations vdP = dh -[ vdP -S dh=-(h, – h,) W, = - rev State 1 (Table A-5): P =100 kPa, Sat. vapour s, =7.3589 kJ/kgK S1 h = 2675.0 kJ/kg State 2 (Table A-6): P, =1 MPa, s, = s, = 7.3589 kJ/kgK h, = 3194.5 kJ/kg Wroy = -(3194.5- 2675.0)kJ/kg = -519.5 kJ/kg rev Less energy to pump liquids than gas between same pressure range
Determine the pump (compressor) work input required to compress steam isentropically from 100 kPa to 1 MPa, assuming that the steam exists as (a) saturated liquid and (b) saturated vapour at the inlet state. T P2 = 1 MPa Process is isentropic: P, = 1 MPa 1 MPa (a) Pump (for liquid, v = const.): PUMP COMPRESSOR (b) V1 = Vf@100kPa = 0.001043 m'/kg (a) 100 kPa P = 100 kPa P = 100 kPa W,ey = -[ vdP = -v, (P, -R) %3| (b) Compressing (a) Compressing a liquid rev a vapor 1 kJ =-(0.001043 m/kg)[(1000-100) kPa] =-0.94 kJ/kg W, 3 1kPa · m rev
Transcribed Image Text:Determine the pump (compressor) work input required to compress steam isentropically from 100 kPa to 1 MPa, assuming that the steam exists as (a) saturated liquid and (b) saturated vapour at the inlet state. T P2 = 1 MPa Process is isentropic: P, = 1 MPa 1 MPa (a) Pump (for liquid, v = const.): PUMP COMPRESSOR (b) V1 = Vf@100kPa = 0.001043 m'/kg (a) 100 kPa P = 100 kPa P = 100 kPa W,ey = -[ vdP = -v, (P, -R) %3| (b) Compressing (a) Compressing a liquid rev a vapor 1 kJ =-(0.001043 m/kg)[(1000-100) kPa] =-0.94 kJ/kg W, 3 1kPa · m rev
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