Second Law Efficiencies and Exergy Change of a System

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Second Law EfficienciesandExergy Change of a
System
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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.
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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, ?II 30/70 43
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The second law efficiency is 100 percent for all
reversible devices.
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Second Law Efficiencies

• For heat engines ?th/?th,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

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Second Law Efficiencies

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

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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?
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Exergy of a fixed mass or closed system. For a
reversible process, the system work
dW PdV (P P0)dV P0dV dWb,useful P0dV
For the system heat through a reversible heat
engine dWHE (1 - T0/T) dQ dQ T0/T dQ
dQ (-T0dS) which gives dQ dWHE T0dS

• Plug the heat and work
• expressions into
• dQ dW dU and integrate
• to get
• Wtotal useful WHE Wb,useful
• (UU0) P0(VV0) T0(SS0)
• Wrev X (exergy)

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Exergy Change of a Closed System

• ?X (U2 - U1) P0(V2 - V1) T0(S2 - S1)
  m(?22 - ?12) mg(z2 - z1)
• Can also do it on a per-mass basis, ?f ?X/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.

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Even if TltT0 and/or PltP0 the exergy of the system
is positive.
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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.
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Just like with energy, with exergy you can
replace the us with hs and get the exergy of a
flowing system.
Just like we use ? for the energy of a flowing
system, we use the Greek letter psi, ?, for the
exergy of a flowing system.
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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.
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Exergy Change During a Compression. Change in
exergy equation for flow systems
?? (h2 h1) T0(s2 s1) (V22 V12)/2
g(z2 z1)
Now, with the two states given, find hs and ss
and calculate ??. 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.