1 / 17

Irreversibility

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:

rad
Download Presentation

Irreversibility

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Irreversibility Physics 313 Professor Lee Carkner Lecture 16

  2. 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

  3. 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

  4. 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

  5. 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

  6. Irreversible Processes • How do you compute the entropy change for an irreversible process? • What is the change in entropy for specific irreversible processes?

  7. 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

  8. Adiabatic W to U • Friction or stirring of insulated substance • System will increase in temperature DS =  dQ/T =  CPdT/T = CPln (Tf/Ti)

  9. 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

  10. 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

  11. 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)

  12. 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

  13. 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

  14. 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:

  15. 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

More Related