The Carnot Cycle•Idealized thermodynamic cycle consisting of four reversible processes (any substance):! Reversible isothermal expansion (1-2, TH=constant)! Reversible adiabatic expansion (2-3, Q=0, TH"TL)! Reversible isothermal compression (3-4, TL=constant)! Reversible adiabatic compression (4-1, Q=0, TL"TH)1-22-33-4 4-1The Carnot Cycle-2Work done by gas = ∫PdV, area under the process curve 1-2-3.123321Work done on gas = ∫PdV, area under the process curve 3-4-1subtractNet work1234dV>0 from 1-2-3∫PdV>0Since dV<0∫PdV<0The Carnot Principles•The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. ηth, irrev< ηth, rev• The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. (ηth, rev)A= (ηth, rev)B• Both Can be demonstrated using the second law (K-P statement and C-statement). Therefore, the Carnot heat engine defines the maximum efficiency any practical heat engine can reach up to.• Thermal efficiency ηth=Wnet/QH=1-(QL/QH)=f(TL,TH) and it can be shown that ηth=1-(QL/QH)=1-(TL/TH). This is called the Carnot efficiency.• For a typical steam power plant operating between TH=800 K (boiler) and TL=300 K(cooling tower), the maximum achievable efficiency is 62.5%.ExampleLet us analyze an ideal gas undergoing a Carnot cycle between two temperatures THand TL.! 1 to 2, isothermal expansion, ∆U12 = 0QH= Q12 = W12 = ∫PdV = mRTHln(V2/V1)! 2 to 3, adiabatic expansion, Q23 = 0(TL/TH) = (V2/V3)k-1 " (1)! 3 to 4, isothermal compression, ∆U34 = 0QL= Q34 = W34 = - mRTLln(V4/V3)! 4 to 1, adiabatic compression, Q41 = 0(TL/TH) = (V1/V4)k-1 " (2)From (1) & (2), (V2/V3) = (V1/V4) and (V2/V1) = (V3/V4)ηth= 1-(QL/QH )= 1-(TL/TH) since ln(V2/V1) = ln(V4/V3)It has been proven that ηth= 1-(QL/QH )= 1-(TL/TH) for all Carnot engines since the Carnot efficiency is independent of the working substance.Carnot EfficiencyA Carnot heat engine operating between a high-temperature source at 900 K and reject heat to a low-temperature reservoir at 300 K. (a) Determine the thermal efficiency of the engine. (b) If the temperature of the high-temperature source is decreased incrementally, how is the thermal efficiency changes with the temperature.ηηηthLHthHHth HLTTKTTKTT=− =− = ===−==−113009000 667 66 7%30013009001900..()()()()Fixed T and lowering TThe higher the temperature, the higher the "quality"of the energy: More work can be doneFixed T and increasing TLHHL200 400 600 800 100000.20.40.60.81Temperature (TH)EfficiencyTh( )TT200 400 600 800 100000.20.40.60.81Temperature (TL)EfficiencyTH( )TLTLLower THIncrease TLCarnot Efficiency• Similarly, the higher the temperature of the low-temperature sink, the more difficult for a heat engine to transfer heat into it, thus, lower thermal efficiency also. That is why low-temperature reservoirs such as rivers and lakes are popular for this reason.•To increase the thermal efficiency of a gas power turbine, one would like to increase the temperature of the combustion chamber. However, that sometimes conflict with other design requirements. Example: turbine blades can not withstand the high temperature gas, thus leads to early fatigue. Solutions: better material research and/or innovative cooling design.• Work is in general more valuable compared to heat since the work can convert to heat almost 100% but not the other way around. Heat becomes useless when it is transferred to a low-temperature source because the thermal efficiency will be very low according to ηth=1-(TL/TH). This is why there is little incentive to extract the massive thermal energy stored in the oceans and
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