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.

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, η II = 30%/70% = 43%

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

Second Law Efficiencies For heat engines = η th /η th,revFor heat engines = η th /η th,rev For work - producing devices = W u /W revFor work - producing devices = W u /W rev For work – consuming devices = W rev /W uFor work – consuming devices = W rev /W u For refrigerators and heat pumps = COP/COP revFor refrigerators and heat pumps = COP/COP rev W rev should be determined using the same initial and final states as actual.W rev should be determined using the same initial and final states as actual. And for general processes = Exergy recovered/Exergy supplied = 1 – Exergy destroyed/Exergy suppliedAnd for general processes = Exergy recovered/Exergy supplied = 1 – Exergy destroyed/Exergy supplied

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

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

Exergy of a fixed mass or closed system. For a reversible process, the system work: δW = PdV = (P – P 0 )dV + P 0 dV = δW b,useful + P 0 dV For the system heat through a reversible heat engine: δW HE = (1 - T 0 /T) δQ = δQ – T 0 /T δQ = δQ – (-T 0 dS) which gives: δQ = δW HE – T 0 dS Plug the heat and work expressions into: -δQ – δW = dU and integrate to get: W total useful = W HE + W b,useful =(U–U 0 ) + P 0 (V–V 0 ) – T 0 (S–S 0 ) = W rev = X (exergy)

Exergy Change of a Closed System ΔX = (U 2 - U 1 ) + P 0 (V 2 - V 1 ) – T 0 (S 2 - S 1 ) + m(۷ ۷ 1 2 ) +mg(z 2 - z 1 )ΔX = (U 2 - U 1 ) + P 0 (V 2 - V 1 ) – T 0 (S 2 - S 1 ) + m(۷ ۷ 1 2 ) +mg(z 2 - z 1 ) Can also do it on a per-mass basis, Δφ = ΔX/m.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 changeThe 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.The exergy of a closed system is either positive or zero.

Even if T<T 0 and/or P<P 0 the exergy of the system is positive.

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 (x flow = Pv – P 0 v.

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.

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: X 1 = m[(u 1 -u 0 ) + P 0 (v 1 -v 0 ) – T 0 (s 1 -s 0 ) +V 1 2 /2 + gz] Why? Then use ideal gas law relations and T1 = T0 to get X1.

Exergy Change During a Compression. Change in exergy equation for flow systems: Δψ = (h 2 – h 1 ) – T 0 (s 2 – s 1 ) + (V 2 2 – V 1 2 )/2 + g(z 2 – z 1 ) 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.