University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Heat pump
Operate between the same to temperature level but direction of heat transfer are
reversible and work is required ( e.g., refrigeration cycle)
∆ U = Q net + W ( first law)
0 = QC – QH + W
W = ½ QH ½ - QC ½
…….
(1)
Entropy analysis
DS H =
QH
TH
, DS C = -
QC
and , ∆S pump = 0
TC
∆S Total = ∆S H + ∆S C + ∆S pump= 0
D S Total = 0 =
Q
QH
- C +0
TH
TC
….. (2)
ö
æTH
- 1 ÷÷
From equations 1 and, 2 : W heatpump = Q C çç
ø
è TC
COP =
QC
TC
outlet
=
=
W pump
inlet
TH -TC
COP :coefficient performance pump
Ex: Thermodynamic device cools small refrigerator and discards heat to the
surroundings at 298.15K .The maximum electric power to which the device is
designed is 100 W , the heat load on the refrigeration is 350 W what is the
maximum temperature that can be maintained in the refrigerator?
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
If Carnot engine is connected to Carnot pump (refrigerator) so that all the work
produced by engine is used by refrigerator
Q1 is he heat rejected by Carnot engine
Q2 is the heat rejected by Carnot refrigerator
WE=WP
Engine
WE
T
=1QH
TH
WP
T
=
-1
QC
TC
and, pump
WE = QH –Q1 and , WP = Q2 -QC
æT
ö
ö
÷÷ = Q C çç
- 1 ÷÷
ø
èTC
ø
æT -TC ö
æT -T ö
÷÷
÷÷ = Q C çç
Q H çç H
T
T
H
C
è
ø
è
ø
æ
T
Q H çç 1 TH
è
æ T -TC
Q H = Q C çç
èT H -T
öæ T H
÷÷çç
øè T C
ö
÷÷
ø
Relation between QH and QC
Q =Q H +QC
æTH -T
Q H çç
è TH
æTH -T
Q H çç
è TH
æT -TC ö
ö
÷÷
÷÷ = Q C çç
T
C
ø
ø
è
æT -TC
ö
÷÷ = (Q - Q H )çç
ø
è TC
æ T -TC
Q H = Q çç
è T H -TC
öTH
÷÷
øT
ö
÷÷
ø
Relation between QH ( hot reservoir)and Q(heat sink)
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics