1 / 14

Second Law Efficiencies and Exergy Change of a System

Second Law Efficiencies and Exergy Change of a System. Both heat engines have the same thermal efficiency. Are they doing equally well?. Because B has a higher T H , it should be able to do better. Hence, it has a higher maximum (reversible) efficiency.

lilli
Télécharger la présentation

Second Law Efficiencies and Exergy Change of a System

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. Second Law EfficienciesandExergy Change of a System

  2. Both heat engines have the same thermal efficiency. Are they doing equally well? Because B has a higher TH, it should be able to do better. Hence, it has a higher maximum (reversible) efficiency.

  3. The second law efficiency is a measure of the performance of a device relative to what its maximum performance could be (under reversible conditions). Second law efficiency for heat engine A For heat engine B, ηII = 30%/70% = 43%

  4. The second law efficiency is 100 percent for all reversible devices.

  5. Second Law Efficiencies • For heat engines = ηth/ηth,rev • For work - producing devices = Wu/Wrev • For work – consuming devices = Wrev/Wu • For refrigerators and heat pumps = COP/COPrev • Wrev should be determined using the same initial and final states as actual. • And for general processes = Exergy recovered/Exergy supplied = 1 – Exergy destroyed/Exergy supplied

  6. Second Law Efficiencies • For explanations of what these terms mean for a particular device, see text page 401

  7. Example 7-6 Second law efficiency of resistance heaters. Thermal efficiency is 100%. However, COP of a resistance heater is 1. What is the COPHP,rev for these conditions? = 1/(1-TL/TH) It works out to be 26.7 so second law eff. is COP/COPrev = 1/26.7 or .037 or 3.7% See now why resistance heating is so expensive?

  8. Exergy of a fixed mass or closed system. For a reversible process, the system work: δW = PdV = (P – P0)dV + P0dV = δWb,useful + P0dV For the system heat through a reversible heat engine: δWHE = (1 - T0/T) δQ = δQ – T0/T δQ = δQ – (-T0dS) which gives: δQ = δWHE – T0dS • Plug the heat and work • expressions into: • δQ – δW = dU and integrate • to get: • Wtotal useful = WHE + Wb,useful • =(U–U0) + P0(V–V0) – T0(S–S0) • = Wrev = X (exergy)

  9. Exergy Change of a Closed System • ΔX = (U2 - U1) + P0(V2 - V1) – T0(S2 - S1) + m(۷22 - ۷12) +mg(z2 - z1) • Can also do it on a per-mass basis, Δφ = ΔX/m. • The exergy change of a system is zero if the state of the system or of the environment does not change • Example – steady-flow system. • The exergy of a closed system is either positive or zero.

  10. Even if T<T0 and/or P<P0 the exergy of the system is positive.

  11. In flowing systems, you also have flow energy. The exergy of flow energy is the useful work that would be delivered by an imaginary piston in the flow (xflow = Pv – P0v.

  12. Just like with energy, with exergy you can replace the u’s with h’s and get the exergy of a flowing system. Just like we use θ for the energy of a flowing system, we use the Greek letter psi, ψ, for the exergy of a flowing system.

  13. Example 7-7 Work Potential of Compressed Air in a Tank. Assume ideal gas and ke and pe negligible. Can calculate mass by ideal gas law. Exergy equation: X1 = m[(u1-u0) + P0(v1-v0) – T0(s1-s0) +V12/2 + gz] Why? Then use ideal gas law relations and T1 = T0 to get X1.

  14. Exergy Change During a Compression. Change in exergy equation for flow systems: Δψ = (h2 – h1) – T0(s2 – s1) + (V22 – V12)/2 + g(z2 – z1) Now, with the two states given, find h’s and s’s and calculate Δψ. This represents the minimum work required to compress the refrigerant between these two states. This also represents the maximum amount of work you can get from expanding this gas again between the same two states.

More Related