lecture23_temperature scales maximum performance measures
TRANSCRIPT
1
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
2
Outline
The Kelvin and temperature scales
Maximum performance measures for cycles operating
Between Two thermal reservoirs
Examples
3
Kelvin scales
From the Carnot corollary:
Hot reservoir
System A
Cold reservoir
Q
Q
W
๐๐ด
Hot reservoir
System B
Cold reservoir
Q
Q
W
๐๐ต
TH TH
TC TC
๐ผ๐จ ๐ผ๐ฉ =
=
=
4
Kelvin scales
ฮท = 1 โ๐๐๐ข๐ก๐๐๐
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 ๐ผ
(Regardless of substance making up the system (air, steam etc))
-> Temperature is the important factor
๐๐
๐๐๐๐๐ฃ = ๐(TH, TC)
5
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 :
๐๐
๐๐๐๐๐ฃ =
๐๐ถ๐๐ป
6
Maximum performance measures
For power cycles
ฮท๐๐๐ฅ =๐๐๐ฆ๐๐๐
๐๐๐= 1 โ
๐๐๐ข๐ก
๐๐๐ = 1 โ
๐๐ถ
๐๐ป
Maximum achievable efficiency between two thermal reservoir : Carnot efficiency
Wcycle
= Qin- Qout
TH
(Hot)
Tc
(Cold)
W=Qin-Qout
Qin System
๐๐
๐๐๐๐๐ฃ =
๐๐ถ๐๐ป
Qout
7
Maximum performance measures
For power cycles with Tc=298 K and
TH
ฮท
ฮท๐๐๐ฅ= 1 โ๐๐ถ
๐๐ป
Rate of efficiency increase Is high at tower TH
Decreasing Tc below certain
level is impractical
Real power plant 40% compared to theoretical max.
ฮท๐๐๐ฅ= 1 โ๐๐ถ
๐๐ป
= 1 โ298
745= 0.6 (60%)
8
Maximum performance measures
For power cycles with TH=298 K and
TC
ฮท
TC=TH
ฮท๐๐๐ฅ= 1 โ๐๐ถ
๐๐ป
Rate of efficiency decreases as TC increases
Tc=Th, efficiency is zero
9
Maximum performance measures
For Refrigeration and heat pump cycles
Maximum achievable efficiency between two thermal reservoir
๐๐
๐๐๐๐๐ฃ =
๐๐ถ๐๐ป
Wcycle =
Qout- Qin>0
TH Tc
W=Qout-Qin
Qout Qin System
ฮฒ๐๐๐ฅ =๐๐๐
(๐๐๐ข๐ก;๐๐๐)=
๐๐
๐๐ป;๐๐; Refrigeration cycle
ฮณ๐๐๐ฅ =๐๐๐ข๐ก
(๐๐๐ข๐ก;๐๐๐)=
๐๐ป๐๐ป;๐๐ถ
; Heat pump cycle
10
Maximum performance measures
For Refrigeration cycles with TH=298 K and
TC
๐ฝ ๐ฝ๐๐๐ฅ =
1
๐๐ป๐๐
โ 1
COP will gradually increase until Tc gets closer to the Th
TH
11
Maximum performance measures
For Refrigeration cycles with TL=298 K and
TH
๐ฝ ๐ฝ๐๐๐ฅ =
1
๐๐ป๐๐
โ 1
COP will gradually increase when TH is closer to the TL
TL
12
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 โ that is Wcycle=NQc, where all quantities are positive. What is the theoretical maximum N?
13
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: ๐ = ๐๐๐ฆ๐๐๐
๐๐ป= 1 โ
๐๐
๐๐ป= 1 โ
๐๐
๐๐ป
๐๐๐ฆ๐๐๐= ๐๐๐
๐๐๐
๐๐ปโค 1 โ
๐๐ถ
๐๐ปโ ๐ 0.25 โค 1 โ 0.25 โ ๐ โค 3
14
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%
15
Example 1 Power cycle
Given : 750 MW power plant, TH=317 m=1.65x10^5 kg/s
Tup=17, efficiency 0.4
Find : ฮT
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: ๐ =
๐๐๐ฆ๐๐๐
๐๐ป= 1 โ
๐๐๐๐ป
; ๐๐ป =750
0.4๐๐ = 1875 ๐๐
๐ = 1 โ๐๐๐๐ป
; ๐๐ถ = 0.6 ๐๐ป = 1125 ๐๐
16
Example 1 Power cycle
๐๐ = ๐๐ถ ๐2 โ ๐1 ; ๐ค๐๐๐๐ ๐ถ = 4.2๐๐ฝ
๐๐ ๐ ๐๐๐๐ ๐๐๐๐๐ ๐ด โ 19
Qout
T1=17 C
T2=?
m=1.65e5 kg/s Tc
(Cold)
Heat exchanger; Qcv=m (h2-h1)
๐2 =๐๐๐๐ถ
+ ๐1 =1125 MW
1.65x105kgs 4.2
๐๐ฝ๐๐ ๐
+ 290 ๐พ = 291.62 ๐พ
17
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) ๐ 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?
18
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 ฮฒ๐๐๐ฅ =
๐๐๐๐๐๐
=๐๐๐
(๐๐๐ข๐กโ๐๐๐); ๐๐๐ = 3.6 ๐๐
๐๐๐ข๐ก =ฮฒ โ 1
ฮฒ๐๐๐ = 4.4 ๐๐
19
Example 2
TH=20C Tc
W=0.8 kW
Qout Qin System
COP=4.5
Tcoil=28C
ฮฒ๐๐๐ฅ =๐๐๐
(๐๐๐ข๐กโ๐๐๐)=
๐๐
๐๐ป โ ๐๐;
๐ ๐๐๐ฃ๐๐๐ ๐๐๐ ๐๐ =๐๐ป1
ฮฒ๐๐๐ฅ
:1= 239.72 ๐พ = โ33 ๐ถ
๐ = 1 โ๐๐ถ๐๐ป
= 1 โ๐๐ป๐๐๐๐๐
= 1 โ293
301= 0.0266 = 2.6%
๐ =๐๐๐ฆ๐๐๐
๐๐ปโค 1 โ
๐๐ถ๐๐ป
โ ๐ โค 0.12 ๐๐
20
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
21
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
ฮณ=3.8
Qout=75000 kJ/h
T=0 C
Tin=21 C
Assumption : electricity cost 8 cent KWH Sol :
Since COP is given ๐ =๐ ๐๐ข๐ก
๐พ=
75000๐๐ฝ
โ
3.8= 19737
๐๐ฝ
โ
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
22
Example 3 Heat pump cycle
If we need the minimum cost -> when the COP is maximum Using two temperature only
๐พ๐๐๐ฅ =๐๐ป
๐๐ป โ ๐๐ถ=
1
1 โ๐๐ถ๐๐ป
= 14
Since COP is given ๐ =๐ ๐๐ข๐ก
๐พ=
75000๐๐ฝ
โ
14= 5357
๐๐ฝ
โ
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