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Thermodinamic, termodinamika, D3 Teknik Kimia Universitas Riau
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THERMODYNAMICTHERMODYNAMIC
CHEMICAL ENGINEERING DIII URCHEMICAL ENGINEERING DIII UR
By 5th Group:
Ryan Tito
Desi Erika Putri
Alim Kurniawan
Rizky Maulana
Didi Siswanto
© UR 2012© UR 2012
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Quest. 1 :Quest. 1 :
A gas contained within a piston-cylinder assembly undergoes three processes in series :
Process 1-2 : Compression with pV = constant from p1 = 1 bar, V1 = 1.0 m3 to V2 = 0.2 m3.
Process 2-3 : Constant-pressure expansion to V3 = 1.0 m3
Process 3-4 : Constant volume.
Sketch the processes in series on a p-V diagram labeled with pressure and volume values at each numbered state.
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Solution :Solution :
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Thermodynamic cycle :
Process 1-2 : pV = constant from
p1 = 1 bar, V1 = 1.0 m3, V2 = 0.2 m3
Process 2-3 : p = Constant, V > V2 (expansion),
V3 = 1.0 m3
Process 3-4 : V = Constant
For processes 1-2, pV = constant. The constant can be evaluated using data at state 1 :
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cont. :cont. :
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pV = constant
= p1 V1
= (1 bar) (1.0 m3)
= 1 bar m3
According on a pressure-volume plot process 1-2 is described by :
p = 1 bar m3 / V
In particular, when V2 = 0.2 m3, p = 5 bar.
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cont. :cont. :
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The thermodynamic cycle takes the form :
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p = constantp = constant
V = constantV = constant
pV = constantpV = constant
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Quest. 2 :Quest. 2 :
A gas in a piston-cylinder assembly undergoes a process for which the relationship between pressure and volume is pV2 = constant. The initial pressure is 1 bar, the initial volume is 0,1 m3, and the final pressure is 9 bar. Determine :
a.The final volume, in m3
b.The work for the process, in kJ
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Solution :Solution :
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p1 = 1 bar = 105 N/m2 pV2 = constant
V1 = 0.1 m3
p2 = 9 bar = 9 x 105 N/m2
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pVpV22 = constant = constant
WW
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a.The final volume :
b.Work :
Assumptions :
1.The gas is a closed system
2.The moving boundary is the only work made
3.The compression is polytropic
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The constant C can be evaluated at either end state : C = p1V1
2 = p2V22, giving
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© UR 2012© UR 2012
Quest. 3 :Quest. 3 :
As shown figure above, a reversible power cycle receives energy QH by heat transfer from a hot reversoir at TH and rejects energy QC by heat transfer to a cold reservoir at TC.
a.If TH = 1,200 K and TC = 300 K, what is the thermal efficiency?
b.If TH = 500 oC, TC = 20 oC, and Wcycle = 1,000 kJ, what are QH and QC ? (each in kJ)
c.If η = 60% and TC = 40 oF, what is TH ? (in oC)
d.If η = 40% and TH = 727 oC, what is TC ? (in oC)
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Solution :Solution :
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a. TH = 1,200 K, TC = 300 K
η = 1 – (TC/TH)
= 1 – (300 K / 1200 K)
= 1 – 0.25
= 0.75
Efficiency is 75%
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b. TH = 500 oC = 773 K, TC = 20 oC = 293 K,
Wcycle = 1,000 kJ
η = 1 – (TC/TH)
= 1 – (293 K / 773 K)
= 0.62 (means 62%)
QH = Wcycle / η
= 1,000 kJ / 0,62
= 1,612.9 kJ
QC = QH – Wcycle = (1,612,9 – 1,000) kJ = 612.9 kJ
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c. η = 60%, TC = 40 oF = 277.6 K
η = 1 – (TC/TH)
0.6 = 1 – (277.6 K / TH)
0.6 TH= TH - 277.6 K
-1.6 TH = -277.6 K
TH = 694 K
TH = 789.53 oF
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cont. :cont. :
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d. η = 40%, TH = 727 oC = 1,000 K
η = 1 – (TC/TH)
0.4 = 1 – (TC / 1,000 K)
400 K = 1,000 K - TC
TC = 600 K
TC = 327 oC
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THERMODYNAMICTHERMODYNAMIC
CHEMICAL ENGINEERING DIII URCHEMICAL ENGINEERING DIII UR
Mission : CompletedMission : Completed
THANK YOUTHANK YOU
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© UR 2012© UR 2012
