Thermodynamics A/C Techniques Dep. 1st Year Class First Term 2018-2019
Lecture 16 : The Carnot Cycle (Examples) by: Asst. lect. Karrar Al-Mansoori
( Analysis of a Carnot Heat Engine ) Example 16 – 1 / A Carnot heat engine, shown in ( Fig. 16 -1 ), receives 500 kJ of heat per cycle from a high-temperature source at 652°C and rejects heat to a low- temperature sink at 30°C. Determine (a) the thermal efficiency of this Carnot engine and (b) the amount of heat rejected to the sink per cycle.
Solution: The heat supplied to a Carnot heat engine is given. The thermal efficiency and the heat rejected are to be determined.
Analysis: (a) The Carnot heat engine is a reversible heat engine, and so its efficiency can be determined as :
푇퐿 (30+ 273) ηth, rev = 1 − = 1 − = 0.672 푇퐻 (652+ 273) That is, this Carnot heat engine converts 67.2 percent of the heat it receives to work.
(b) The amount of heat rejected QL by this reversible heat engine is easily determined from :
푇퐿 푄퐿 ηth, rev = 1 − and , ηth = 1 − 푇퐻 푄퐻
푇퐿 (30+ 273) QL = QH = (500 kJ) = 164 kJ 푇퐻 (652+ 273)
Figure (16-1): Schematic for Example 16-1.
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Thermodynamics A/C Techniques Dep. 1st Year Class First Term 2018-2019
Lecture 16 : The Carnot Cycle (Examples) by: Asst. lect. Karrar Al-Mansoori
(A Carnot Refrigeration Cycle Operating in the Saturation Dome) Example 16 - 2 / A Carnot refrigeration cycle is executed in a closed system in the saturated liquid–vapor mixture region using 0.8 kg of refrigerant-134a as the working fluid (Fig. 15-2). The maximum and the minimum temperatures in the cycle are 20 and 288C, respectively. It is known that the refrigerant is saturated liquid at the end of the heat rejection process, and the net work input to the cycle is 15 kJ. Determine the fraction of the mass of the refrigerant that vaporizes during the heat addition process, and the pressure at the end ofthe heat rejection process.
Solution: A Carnot refrigeration cycle is executed in a closed system. The mass fraction of the refrigerant that vaporizes during the heat addition process and the pressure at the end of the heat rejection process are to be determined.
Assumption: The refrigerator operates on the ideal Carnot cycle.
Analysis: Knowing the high and low temperatures, the coefficient of performance of the cycle is :
푻푳 ퟏ ퟏ COPR = = = 푻푯−푻푳 (푻푯⁄푻푳)−ퟏ ( ퟐퟎ+ퟐퟕퟑ⁄−ퟖ+ퟐퟕퟑ) −ퟏ
The amount of cooling is determined from the definition of the coefficient of performance to be:
QL = COPR * Win = (9.464) * (15 kJ) = 142 kJ
The enthalpy of vaporization R-134a at - 8°C is hfg = 204.59 kJ/kg (Table A-11). Then
the amount of refrigerant that vaporizes during heat absorption becomes : 2
Thermodynamics A/C Techniques Dep. 1st Year Class First Term 2018-2019
Lecture 16 : The Carnot Cycle (Examples) by: Asst. lect. Karrar Al-Mansoori
QL = mevap hfg@-8°C mevap = 142 kJ / 204.59 kJ/kg = 0.694 kg
Therefore, the fraction of mass that vaporized during heat addition process to the refrigerant is:
Mass fraction = mevap / mtotal = 0.694 kg / 0.8 kg = 0.868 or 86.8%
The pressure at the end of heat rejection process is simply the saturation pressure at heat rejection temperature,
P4 = Psat@20°C = 572.1 kPa Figure (16-2): Schematic for Example 16-2.
Example 16 - 3 / A Carnot heat engine receives 650 kJ of heat from a source of unknown temperature and rejects 250 kJ of it to a sink at 24°C. Determine (a) the temperature of the source and (b) the thermal efficiency of the heat engine.
Solution: The sink temperature of a Carnot heat engine and the rates of heat supply and heat rejection are given. The source temperature and the thermal efficiency of the engine are to be determined. Assumptions : The Carnot heat engine operates steadily.
Q T Analysi : (a) For reversible cyclic devices we have H = H Q T L rev L
Thus the temperature of the source TH must be :
Q 650 kJ T = H T = (290 K) = 942.5 K H Q L 200 kJ L rev
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Thermodynamics A/C Techniques Dep. 1st Year Class First Term 2018-2019
Lecture 16 : The Carnot Cycle (Examples) by: Asst. lect. Karrar Al-Mansoori
(b) The thermal efficiency of a Carnot heat engine depends on the source and the sink
temperatures only, and is determined from source 650 kJ T 290 K =1− L =1− = 0.69 or 69%. HE th,C T 942.5 K H 200 kJ 17°C
Example 16 - 4 / An inventor claims to have developed a refrigeration system that removes heat from the closed region at -12°C and transfers it to the surrounding air at 25°C while maintaining a COP of 6.5. Is this claim reasonable? Why?
Analysis The highest coefficient of performance a refrigerator can have when removing heat from a cool medium at -5°C to a warmer medium at 25°C is
1 1 25C COPR,max = COPR,rev = = = 8.9 (TH /TL )−1 (25+ 273K)/(− 5 + 273K)−1 COP=6.5 The COP claimed by the inventor is below this maximum R value, thus the claim is reasonable. -5C
Example 16 - 5 / The performance of a heat pump degrades (i.e., its COP decreases) as the temperature of the heat source decreases. This makes using heat pumps at locations with severe weather conditions unattractive. Consider a house that is heated and maintained at 20°C by a heat pump during the winter. What is the maximum COP for this heat pump if heat is extracted from the outdoor air at (a) 10°C, (b) -5°C, and (c) -30°C?
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Thermodynamics A/C Techniques Dep. 1st Year Class First Term 2018-2019
Lecture 16 : The Carnot Cycle (Examples) by: Asst. lect. Karrar Al-Mansoori
Analysis: The coefficient of performance of a heat pump will be a maximum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined for all three cases above to be :
20C 1 1 COPHP,rev = = = 29.3 1− (TL / TH ) 1− (10 + 273K)/(20 + 273K) HP 1 1 COPHP,rev = = = 11.7 1− (TL / TH ) 1− (−5+ 273K)/(20 + 273K) TL 1 1 COPHP,rev = = = 5.86 1− (TL / TH ) 1− (−30 + 273K)/(20 + 273K)
Example 16 - 6/ A Carnot engine operates with air, using the cycle shown. Determine the thermal efficiency and the work output for each cycle of operation.
Solution: The thermal efficiency is found to be :
푇퐿 300 퐾 휂푡ℎ, 퐶 = 1 − = 1 − = 0.4 표푟, 40 % 푇퐻 500 퐾 To find the work output we can determine the heat added during the constant 푊 temperature expansion and determine the work W from : ηth = , We find QH 푄퐻 from the first law using :
푉3 푅 푇 푉3 QH = W2–3 = ∫ 푃 푑푉 = ∫ dV = R TH ln 푉2 푉 푉2
To find V2 first we must find V1: 5
Thermodynamics A/C Techniques Dep. 1st Year Class First Term 2018-2019
Lecture 16 : The Carnot Cycle (Examples) by: Asst. lect. Karrar Al-Mansoori
3 R T1 (278) (300) m 휈1 = = = 1.076 P1 (80000) kg
T V 2 = ( 1 ) γ−1 we have : T1 V2
1 T 300 1 m3 1 γ−1 휈2 = V1( ) = (1.076) ( ) 1.4− 1 = 0.3 T2 500 kg
1 T 300 1 m3 4 γ−1 Like wise : 휈3 = V4( ) = (10) ( ) 1.4− 1 = 2.789 T3 500 kg
V3 2.789 QH = R TH ln = (287) (500) ln = 320 kJ/kg V2 0.3
W ∵ ηth = QH
∴ W= ηth * QH = (0.4) ( 320) = 128 kJ/kg
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