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Heat Engines Efficiency of a Heat Engine Carnot Cycle

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Heat Engines Efficiency of a Heat Engine Carnot Cycle Second Law of Thermodynamics Three Statements of the second law Heat engines (Kelvin-Planck Law) Second Law of Thermodynamics. 9.1 Refrigerators (Claussius Statement) Second Law of Thermodynamics Entropy Principle Carnot cycle efficiency -Heat engines - Refrigerators Kelvin-Planck Statement of Heat Engines the second law of thermodynamics Perfect heat engine A heat engine takes in heat at a high temperature and exhausts heat at a low temperature. In the process of heat flow some of the input heat is converted to work Tc It is impossible to construct a heat engine operating in a cycle that extracts heat from a reservoir and delivers an equal amount of work. First Law (for a cycle) No perfect heat engine Q = Qh + Qc = W Second Law ( puts limits on Qh and QC) Efficiency of a heat engine Carnot Cycle The efficiency is the Sadi Carnot, a French engineer (1796-1832) proposed a fraction of the heat input cycle set the limits to the efficiency of a heat engine operating at high temperature converted between two temperatures. to work. Th W QQ− The cycle consists of 4 reversible steps. Tc e hc 1 == Qh QQhh P 1. Isothermal expansion at Th 4 Q 2. Adiabatic expansion from T to T 2 e1=− c h c 3. Isothermal compression at Tc Qc Qh 4. Adiabatic compression from Tc to Th. 3 In calculating efficiencies Qh and Qc are taken as positive quantities. V (i.e. the magnitude of the heat) The second law says that a heat engine cannot be 100% efficient. 1 Heat engine undergoing the Carnot Heat engine. cycle. Work is done by the gas in expansion Work must be done on the gas to compress it to the initial state Efficiency of the Carnot cycle Heat transfer in the Carnot cycle Th In the Carnot cycle the magnitude of the heat Th Q T transferred at T is proportional to the absolute e1=− c c A 1 temperature T. T Qh c A 1 Q P Q h B P h becomes 4 QThh(here Q is B 2 = 4 D always positive) T QT 2 c Qc C cc D e1= − 3 Qc C T From the relations 3 h V Isothermal expansion and compression V The efficiency only depends on the ratio of the absolute VB VC temperatures. QRTlnhh= QRTlncc= VA VD Adiabatic expansion and compression The efficiency would be 100% if Qc =0. T V γ-1 = T V γ-1 This is only possible if Tc = 0 K (i.e absolute Zero) h B c c V V Q Q B = C h = c T V γ-1 = T V γ-1 A temperature of absolute zero cannot be attained. h A c D VVAD TThc (Third law of thermodynamics) Carnot’s Theorem Stirling Engine Maximum efficiency is less All Carot engines operating between temperatures Th and Tc than the Carnot efficiency. have the same efficiency. Isothermal Th Q4 T w2 =w4 =0 Q2 = Q4 P Q1 e1=−c T h then Q − Q Q − Q e = in out = 1 3 Q Q + Q in 1 4 Q2 No other heat engine operating between these temperatures Q3 can have a greater efficiency For isothermal Q Q Isothermal Tc 1 = 3 processes same T T volume change h c V ⎛ ⎞ ⎜ ⎟ ⎜ 1 ⎟ thene = ecarnot Efficiency lower due to extra heat added. ⎜ Q4 ⎟ ⎜1+ ⎟ ⎝ Q1 ⎠ 2 Clausius Statement of the Second Law Refrigerator and heat pump of Thermodynamics. Th T c A 1 Q It is impossible to construct a refrigerator P h B operating in a cycle whose sole effect is to 4 2 transfer heat from a cooler object to a hotter D one. Qc C 3 V Heat always flows from high temperature to low temperature. A heat engine run in reverse is a refrigerator and heat pump. Work is done to move heat from a cold temperature source to a hot sink. This device and be used for cooling or heating. Equivalence of the Kelvin-Planck and Proof of the Carnot Principle Clausius statements. higher efficiency Carnot engine engine reversed If a perfect refrigerator were possible (Clausius) then a perfect If a heat engine with a higher efficiency than a Carnot engine heat engine could be constructed (Kelvin-Planck). Thus, the could exist. Then it could convert heat to work with 100% efficiency. impossibility one implies the impossibility of the other. The Carnot engine has the highest efficiency for any heat engine acting between two temperatures. Real Heat Engines Example work done in expansion In a wood burning power plant the steam in the turbine operates between the high temperature of 810 K and a low temperature of 366 K. What is the Carnot efficiency for this plant. T 366 e1= − c =−1 =05.5 Th 810 Compare this to the efficiency calculated from the electrical power output of 59 MW and heat power input of 165 MW (see prob. 60) Q h Qc WP 59MW e === =0.35 heat must be removed Qh H 165MW to condense the gas The actual efficiency is less than the Carnot efficiency. Power Plant 3 How to improve the efficiency Cogeneration of a heat engine. Use of waste heat Increase Th This requires high temperature materials Decrease Tc This requires efficient heat transfer. This power plant in Denmark uses the waste heat to heat green houses nearby. Refrigerators Refrigerator The refrigerator uses a liquid with a low Coefficient of performance boiling point. The evaporation takes up heat COP- the heat removed from the 4 from the refrigerator and the condensation of cold source divided by the work the gas releases the heat outside the refrigerator. done. 1 Compressor – gas is compresses and heated QQ COP ==cc wQQhc− 2 Heat exchange coils - the heat is released from the gas and exhausted outside of the 1 The maximum Carnot COP refrigerator. T COP = c 2 TT− 3 Expansion valve – The pressure is decreased hc after going through the valve. The gas is cooled 3 since Q Q h = c 4 Cooling coils- Heat is absorbed from within the TThc refrigerator. The gas is heated. Refrigerator Summary A freezer is kept at a temperature of 0o F. What is the maximum COP for a Carnot refrigerator with output temperature of 85o F. If the • The second law of thermodynamics limits the efficiency electrical energy use is 500kWh/year how much heat is removed in of heat engines to less than 100% one year, assume 90% conversion of electrical energy to work. • The Carnot cycle is a reversible cycle taking in heat at COP high T and exhausting heat at low T. T = ( 0 -32)(5/9) +273=255 K T 255 c COP = c ==54. • The maximum efficiency of a heat engine working TThc− 302 − 255 Th =(85-32)(5/9)+273=302 K between two temperatures is the Carnot efficiency that depends only on the ratio of the absolute temperatures. Work • Refrigerators and heat pumps are heat engines run in 3 ⎛⎞⎛⎞60min 60s 9 WxWhr=⋅0.9(500 10 )⎜⎟⎜⎟= 1.62xJ 10 reverse. ⎝⎠⎝⎠hr min • The maximum coefficient of performance is determined Heat Q by a Carnot cycle. COP = c w 9 9 Heat removed QCOPwc ==( ) 5.4 x 1.62 x 10 =8.7x 10 J 4.
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