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Professor: Euiwon Bae
Lecture Hours : MWF 1:30-2:20 in ME1130
Office Hours : MWF 2:30-3:30 in ME1091
Email : [email protected]
Phone : 765-494-6849
Course website : http://engineering.purdue.edu/ME200
ME200 : Thermodynamics I
Lecture 23 : Temperature scales &
maximum performance measures
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Outline
The Kelvin and temperature scales
Maximum performance measures for cycles operating
Between Two thermal reservoirs
Examples
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Kelvin scales
From the Carnot corollary:
Hot reservoir
System A
Cold reservoir
Q
Q
W
ππ΄
Hot reservoir
System B
Cold reservoir
Q
Q
W
ππ΅
TH TH
TC TC
πΌπ¨ πΌπ© =
=
=
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Kelvin scales
Ξ· = 1 βπππ’π‘πππ
Can argue that efficiency for the reversible cycle is the function of (TH, TC)
From the Carnot corollary:
All reversible power cycles with the same two thermal reservoirs
have the same πΌ
(Regardless of substance making up the system (air, steam etc))
-> Temperature is the important factor
ππ
πππππ£ = π(TH, TC)
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Kelvin scales
In Kevlin scale, the ratio of temperature is equal to the amount of heat transfer for an reversible cycle.
Define a Kelvin scales as :
ππ
πππππ£ =
ππΆππ»
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Maximum performance measures
For power cycles
Ξ·πππ₯ =πππ¦πππ
πππ= 1 β
πππ’π‘
πππ = 1 β
ππΆ
ππ»
Maximum achievable efficiency between two thermal reservoir : Carnot efficiency
Wcycle
= Qin- Qout
TH
(Hot)
Tc
(Cold)
W=Qin-Qout
Qin System
ππ
πππππ£ =
ππΆππ»
Qout
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Maximum performance measures
For power cycles with Tc=298 K and
TH
Ξ·
Ξ·πππ₯= 1 βππΆ
ππ»
Rate of efficiency increase Is high at tower TH
Decreasing Tc below certain
level is impractical
Real power plant 40% compared to theoretical max.
Ξ·πππ₯= 1 βππΆ
ππ»
= 1 β298
745= 0.6 (60%)
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Maximum performance measures
For power cycles with TH=298 K and
TC
Ξ·
TC=TH
Ξ·πππ₯= 1 βππΆ
ππ»
Rate of efficiency decreases as TC increases
Tc=Th, efficiency is zero
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Maximum performance measures
For Refrigeration and heat pump cycles
Maximum achievable efficiency between two thermal reservoir
ππ
πππππ£ =
ππΆππ»
Wcycle =
Qout- Qin>0
TH Tc
W=Qout-Qin
Qout Qin System
Ξ²πππ₯ =πππ
(πππ’π‘;πππ)=
ππ
ππ»;ππ; Refrigeration cycle
Ξ³πππ₯ =πππ’π‘
(πππ’π‘;πππ)=
ππ»ππ»;ππΆ
; Heat pump cycle
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Maximum performance measures
For Refrigeration cycles with TH=298 K and
TC
π½ π½πππ₯ =
1
ππ»ππ
β 1
COP will gradually increase until Tc gets closer to the Th
TH
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Maximum performance measures
For Refrigeration cycles with TL=298 K and
TH
π½ π½πππ₯ =
1
ππ»ππ
β 1
COP will gradually increase when TH is closer to the TL
TL
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Example 1 Power cycle
5.30 An inventor claims to have developed a power cycle operating between TH=1000 K and Tc=250 K that develops net work equal to a multiple of the amount of energy Qc rejected to the cold reservoir β that is Wcycle=NQc, where all quantities are positive. What is the theoretical maximum N?
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Example 1 Power cycle
Given : TH, TC, W=N Qc
Find : maximum N
System :
TH
(Hot)
Tc
(Cold)
W=Qin-Qout
Qin System
Qout
Sol: π = πππ¦πππ
ππ»= 1 β
ππ
ππ»= 1 β
ππ
ππ»
πππ¦πππ= πππ
πππ
ππ»β€ 1 β
ππΆ
ππ»β π 0.25 β€ 1 β 0.25 β π β€ 3
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Example 1 Power cycle
5.38 At steady state, a 750 MV power plant receives energy by heat transfer from the combustion of fuel at an average T of 317 C. The plant discharges energy by heat transfer to a river whose mass flow rate is 1.65x105 kg/s. Upstream of the power plant the T is 17 C. Determine the increase in the T of the river if the thermal efficiency of the power plant is 40%
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Example 1 Power cycle
Given : 750 MW power plant, TH=317 m=1.65x10^5 kg/s
Tup=17, efficiency 0.4
Find : ΞT
System :
TH
(Hot)
Tc
(Cold)
W=750 MW
Qin System
Qout
T1=17 C
T2=?
M=1.65e5 kg/s
Assumption : steady state, incompressible flow with constant specific heat
Sol: π =
πππ¦πππ
ππ»= 1 β
ππππ»
; ππ» =750
0.4ππ = 1875 ππ
π = 1 βππππ»
; ππΆ = 0.6 ππ» = 1125 ππ
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Example 1 Power cycle
ππ = ππΆ π2 β π1 ; π€ππππ πΆ = 4.2ππ½
ππ π ππππ πππππ π΄ β 19
Qout
T1=17 C
T2=?
m=1.65e5 kg/s Tc
(Cold)
Heat exchanger; Qcv=m (h2-h1)
π2 =ππππΆ
+ π1 =1125 MW
1.65x105kgs 4.2
ππ½ππ π
+ 290 πΎ = 291.62 πΎ
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Example 2 Refrigeration cycle
5.60 The refrigerator operates at steady state with a COP of 4.5 and a power input of 0.8 kW. Energy is rejected from the refrigerator to the surroundings at 20C by heat transfer from metal coils whose average surface temperature is 28 C. Determine (a) π in KW (b) Lowest theoretical T inside the refrigerator in K (c) Maximum theoretical power in kW that could be developed by a power cycle operating between coils and the surroundings. Would you recommend making use of this opportunity for developing power?
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Example 2
Given : COP 4.5 Win=0.8 kW TH=20 C Tcoil=28 C
Find : Q, Tc, max theoretical power
System :
TH=20C Tc
W=0.8 kW
Qout Qin System
COP=4.5
Tcoil=28C
Sol Ξ²πππ₯ =
ππππππ
=πππ
(πππ’π‘βπππ); πππ = 3.6 ππ
πππ’π‘ =Ξ² β 1
Ξ²πππ = 4.4 ππ
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Example 2
TH=20C Tc
W=0.8 kW
Qout Qin System
COP=4.5
Tcoil=28C
Ξ²πππ₯ =πππ
(πππ’π‘βπππ)=
ππ
ππ» β ππ;
π πππ£πππ πππ ππ =ππ»1
Ξ²πππ₯
:1= 239.72 πΎ = β33 πΆ
π = 1 βππΆππ»
= 1 βππ»πππππ
= 1 β293
301= 0.0266 = 2.6%
π =πππ¦πππ
ππ»β€ 1 β
ππΆππ»
β π β€ 0.12 ππ
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Example 3 Heat pump cycle
5.66 A heat pump with a COP of 3.8 provides energy at an average rate of 75,000 kJ/h to maintain a building at 21 C on a day when the outside temperature is 0 C. If electricity cost is 8 cents per kWh (a) Actual operating cost and the minimum theoretical operating
cost in $/day (b) Compare this with the cost of electrical resistance heating
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Example 3 Heat pump cycle
Given : heat pump to maintain temperature Find : theoretical and actual cost/ cost for using electrical resistance System : W
Qin Qout
Ξ³=3.8
Qout=75000 kJ/h
T=0 C
Tin=21 C
Assumption : electricity cost 8 cent KWH Sol :
Since COP is given π =π ππ’π‘
πΎ=
75000ππ½
β
3.8= 19737
ππ½
β
Since 1 KWH=3600 kJ => $0.08 per 3600 kJ Therefore for 24 h work input 19737 kJ/h*24h=473688 kJ => 473688 kJ/3600 kJ*0.08=$10.52
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Example 3 Heat pump cycle
If we need the minimum cost -> when the COP is maximum Using two temperature only
πΎπππ₯ =ππ»
ππ» β ππΆ=
1
1 βππΆππ»
= 14
Since COP is given π =π ππ’π‘
πΎ=
75000ππ½
β
14= 5357
ππ½
β
Since 1KWH=3600 kJ => $0.08 per 3600 kJ Therefore for 24 h work input 5357 kJ/h*24h=128568kJ => 128568 kJ/3600 kJ*0.08=$2.85
For resistance heating 75000kJ/h/3600*24*0.08=$40/day