Tut3 Q8.pdf
2
h f164.97degC 697.22 kJ kg ⋅ := h g164.97degC 2763.5 kJ kg ⋅ := h 1 .9 h g164.97degC h f164.97degC - ( ) ⋅ h f164.97degC + := h 1 2.557 10 3 × kJ kg = Point 2: h f100degC 419.04 kJ kg ⋅ := h g100degC 2676.1 kJ kg ⋅ := h f81.33degC 340.49 kJ kg ⋅ := h g81.33degC 2645.9 kJ kg ⋅ := Solution: c p 4.19 kJ kg degC ⋅ ⋅ := Q m w c p ⋅ T wo T wi - ( ) ⋅ := Q 100.56 kW = m c1 Q h 1 h f164.97degC - ( ) := m c1 0.054 kg s = p 2 1bar = x h f164.97degC h f100degC - h g100degC h f100degC - := x 0.123 = m g2 xm c1 ⋅ := m g2 6.665 10 3 - × kg s = Tutorial 3 - Question 8 How much "flash" steam at amospheric pressure was released from each kg of condensate leaving te heat exchanger steam traps in (11) above? If this steam had been taken off at 0.5 bar, how much would be available? T [oC] s [kJ/kg.K] 1 2 Definitions: MW 1000000 W ⋅ ≡ MPa 1000000 Pa ⋅ ≡ kPa 1000 Pa ⋅ ≡ bar 100 kPa ⋅ ≡ degC K 273.16 K ⋅ - ≡ kJ 1000 J ⋅ ≡ Given: To simplify the problem, assume that atmospheric pressure is 1 bar p 1 7 0 + ( ) bar ⋅ := p 1 7 bar = T 1 164.97 degC ⋅ := m w 2.4 kg s ⋅ := T wi 70 degC ⋅ := T wo 80 degC ⋅ := Thermophysical Properties: Used Steam Tables in Cengel & Boles: Point 1:
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Transcript of Tut3 Q8.pdf
hf164.97degC 697.22kJ
kg⋅:= hg164.97degC 2763.5
kJ
kg⋅:=
h1 .9 hg164.97degC hf164.97degC−( )⋅ hf164.97degC+:= h1 2.557 103×
kJ
kg=
Point 2: hf100degC 419.04kJ
kg⋅:= hg100degC 2676.1
kJ
kg⋅:=
hf81.33degC 340.49kJ
kg⋅:= hg81.33degC 2645.9
kJ
kg⋅:=
Solution:
cp 4.19kJ
kg degC⋅⋅:= Q mw cp⋅ Two Twi−( )⋅:= Q 100.56 kW=
mc1Q
h1 hf164.97degC−( ):= mc1 0.054kg
s=
p2 1bar= xhf164.97degC hf100degC−
hg100degC hf100degC−:= x 0.123= mg2 x mc1⋅:= mg2 6.665 10
3−×kg
s=
Tutorial 3 - Question 8
How much "flash" steam at amospheric pressure was released from each kg of condensate leaving te heat exchanger steam traps in (11) above? If this steam had been taken off at 0.5 bar, how much would be available?
T [oC]
s [kJ/kg.K]
1
2
Definitions:
MW 1000000 W⋅≡ MPa 1000000 Pa⋅≡ kPa 1000 Pa⋅≡ bar 100 kPa⋅≡ degC K 273.16 K⋅−≡ kJ 1000 J⋅≡
Given:
To simplify the problem, assume that atmospheric pressure is 1 bar
p1 7 0+( ) bar⋅:= p1 7 bar= T1 164.97 degC⋅:=
mw 2.4kg
s⋅:= Twi 70 degC⋅:= Two 80 degC⋅:=
Thermophysical Properties:
Used Steam Tables in Cengel & Boles:
Point 1:
2 hg100degC hf100degC− g2 c1 g2 s
p2 0.5bar= xhf164.97degC hf81.33degC−
hg81.33degC hf81.33degC−:= x 0.155= mg2 x mc1⋅:= mg2 8.367 10
3−×kg
s=
