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Irreversibility. Physics 313 Professor Lee Carkner Lecture 16. Exercise #15 Carnot Engine. Power of engine h = 1 – Q H /Q L h = W/Q H Source temp h = 1 – T L /T H Max refrigerator COP For a Carnot refrigerator operating between the same temperatures:
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Irreversibility Physics 313 Professor Lee Carkner Lecture 16
Exercise #15 Carnot Engine • Power of engine • h = 1 – QH/QL • h = W/QH • Source temp • h = 1 – TL/TH • Max refrigerator COP • For a Carnot refrigerator operating between the same temperatures: • Since K < KC (8.2<9.9), refrigerator is possible
Entropy • Entropy (S) defined by heat and temperature • Total entropy around a closed reversible path is zero • Can write heat in terms of entropy: dQ = T dS
General Irreversibility • Since DS = Sf - Si Sf > Si • This is true only for the sum of all entropies • Since only irreversible processes are possible, • Entropy always increases
Reversible Processes • Consider a heat exchange between a system and reservoir at temperature T • So: dSs = +dQ/T dSr = - dQ/T • For a reversible process the total entropy change of the universe is zero
Irreversible Processes • How do you compute the entropy change for an irreversible process? • What is the change in entropy for specific irreversible processes?
Isothermal W to U • Friction or stirring of a system in contact with a heat reservoir • The only change of entropy is heat Q (=W) absorbed by the reservoir DS = W/T
Adiabatic W to U • Friction or stirring of insulated substance • System will increase in temperature DS = dQ/T = CPdT/T = CPln (Tf/Ti)
Heat Transfer • Transferring heat from high to low T reservoir • For any heat reservoir DS = Q/T • DS for cool reservoir = + Q/TC • Assumes no other changes in any other system
Free Expansion • Gas released into a vacuum • Replace with a reversible isothermal expansion • Thus, (dQ/T) = (nRdV/V) • Note: • Entropy increases even though temperature does not change
Entropy Change of Solids • Solids (and most liquids) are incompressible • We can thus write dQ as CdT and dS as (C/T)dT • If we approximate C as being constant with T • Note: • If C is not constant with T, need to know (and be able to integrate) C(T)
General Entropy Changes • For fluids that under go a change in T, P or V we can find the entropy change of the system by finding dQ • For example ideal gas: • dQ = CPdT – VdP • dQ = CVdT + PdV • These hold true for any continuous process involving an ideal gas with constant C
Notes on Entropy • Processes can only occur such that S increases • Entropy is not conserved • The degree of entropy increase indicates the degree of departure from the reversible state
Use of Entropy • How can the second law be used? • Example: total entropy for a refrigerator • DS (reservoir) = (Q + W) /TH • The sum of all the entropy changes must be greater than zero:
Use of Entropy (cont.) • We can now find an expression for the work: • Thus the smallest value for the work is: • Thus for any substance we can look up S1-S2 for a given Q and find out the minimum amount of work needed to cool it