Irreversibility Physics 313 Professor Lee Carkner Lecture 16
Exercise #15 Carnot Engine Power of engine = 1 – Q H /Q L = W/Q H Source temp = 1 – T L /T H Max refrigerator COP For a Carnot refrigerator operating between the same temperatures: Since K < K C (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 S = S f - S i S f > S i 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: dS s = +dQ/T dS r = - 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 S = W/T
Adiabatic W to U Friction or stirring of insulated substance System will increase in temperature S = dQ/T = C P dT/T = C P ln (T f /T i )
Heat Transfer Transferring heat from high to low T reservoir For any heat reservoir S = Q/T S for cool reservoir = + Q/T C 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 = C P dT – VdP dQ = C V dT + 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 S (reservoir) = (Q + W) /T H 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 S 1 -S 2 for a given Q and find out the minimum amount of work needed to cool it