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Lecture 19
Carnot Cycle The Carnot cycle was first proposed in 1824, so it is hardly a new development. The interest in the cycle is largely theoretical, as no practical Carnot cycle engine has yet been built. Nevertheless, it can be shown to be the most efficient cycle possible, so that considerable attention has been given at discovering ways of making the more practical cycles look, as much as possible, like the Carnot. Process 1-2 Process 2-3 Process 3- Process 4- (Process 1-2) A constant T heat addition. (Process 2-3) An adiabatic expansion (Process 3-4) A constant T heat rejection. (Process 4-1) An adiabatic compression Thermo I EGN 3343 Fall 2000 Energy Source @ TH Energy Sink @ TL TH = Const Insulation TH TL TL = Const Insulation TH TL 1 2 P 4 3 V
Carnot Cycle Efficiency Like other heat engines, the Carnot cycle efficiency can be attained from the relationship: = Wout, net/Qin = (Wout - Win)/ Qin = (Qin – Qout)/Qin = 1 – Qout/Qin Since the heat comes into the system from a high temperature reservoir and is discharged to a low temperature reservoir: = 1 – QL/QH Carnot cycle principles: The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. Thermo I EGN 3343 The Carnot Cycle
f (TA, TC) = f (TA, TB) f (TB, TC) A careful examination of this equation reveals that the left hand side is a function of TA and TC, and therefore the right hand side must also be a function of TA and TC only and not, TB. This condition will be satisfied only if the function has the following form: (TA)/ (TC) = [(TA)/ (TB)][(TB)/ (TC)] From this relationship Kelvin proposed a temperature scale in which (T) = T, such that: = 1 – TL/TH Example : Problem 5. An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air. Assume: Steady State Operating Conditions. Analysis: (a). Determine the COP: COPRef = QL/Win = COP = (750 kJ/min)/(60 s/min)(6 kW) COP = 2. (b). Determine the rate of heat transfer to the outlet: From the 1 st Law of Thermodynamics: Thermo I EGN 3343 6 kW House AC Note that the temperatures must be in absolute Units 750 kJ/min Environment
E= Qin - Wout 0 = [(750 kJ/min)/(60 sec/min) – Qout] – [-6 kW] Qout = 18.5 kW Example: Problem 5. A Carnot heat engine receives 500 kJ of heat from a source of unknown temperature and rejects 200 kJ of it to a sink at 17 o C. Determine (a) the temperature of the source and (b) the thermal efficiency of the heat engine. Assume: Steady State Process E = Qin - Wout Wout, net = Qin - Qout = (500 – 200) kJ Wout, net = 300 kJ = Wout,net/Qin = 300 kJ/500 kJ = 60% = 60% For the Carnot Cycle, = 1 – TL/TH 0.6 = 1 – (290 K)/TH TH = 725 K
Three Applications of the Carnot Cycle
1. Carnot Cycle Engine
Thermo I EGN 3343 Carnot Cycle Engine High Temperature Heat Source, TH Low Temperature Heat Sink, TL = 17oC 500 kJ Wout 200 kJ High Temperature Region
2. Carnot Cycle Refrigerator
Using the thermodynamic temperature scale:
Thermo I EGN 3343 Low Temperature Region
Q
L
W
COP
Q
W
Q
Q Q Q
Q
L L H L H L
From Conservation
of Energy, energy
flows must balance:
Q^ ^ Q^ W
H ^ L o
Carnot
Cycle
High Temperature Region
Q
H
COP
T
T
H L
From the definition of Coefficient of Performance we find
for the Carnot refrigeration cycle:
2. Carnot Cycle Heat Pump
Using the thermodynamic temperature scale:
Thermo I EGN 3343 Low Temperature Region
Q
L
W
COP
Q
W
Q
Q Q Q
Q
H H H L L H
From Conservation
of Energy, energy
flows must balance:
Q^ ^ Q^ W
H ^ L o
Carnot
Cycle
High Temperature Region
Q
H
COP
T
T
L H
From the definition of Coefficient of Performance we find
for the Carnot refrigeration cycle:
Now, let us put the 2000 of power into the heat pump. The COP of the heat pump is 1/[1 – TL/TH] = 1/[1 – 500/1500] = 1.5. We see then that the heat pump returns exactly 1.5· kW = 3000 kW to the high temperature source. Conclusions The Carnot cycle is truly reversible. We can recreate exactly the same conditions that we started with. We can see that this is the general case in that: · COPHP = {1 – TL/TH} · {1/[1 – TL/TH]} = 1 If an engine could be devised which could produce a greater efficiency than a Carnot engine, then the excess power could be used the power, say a car, while the Carnot cycle power could be used to power a Heat Pump, returning the initial energy to the gas tank. If that were possible, we could build a car in which the gas tank was initially filled and sealed – it would never require gas again! If a heat pump could be built which had a greater COP than the Carnot Heat pump, we could create high temperature energy (analogous to gasoline) driving it with a motor powered from the same source. That is, we could all have our own free gas pump outside our own houses which would never need to be powered from the outside. Homework: 4 th Edition: Chapter 5: 80, 84 E, 85, 87, 101E, 120, 136 Answers: 6.7%; 25%, 31.3%; 0.79 hp; 0.937 hr; 17, kJ/hr Thermo I EGN 3343
Thermo I EGN 3343
