MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Exergetic Efficiency

Consider a closed system that receives heat, Qs , from the

combustion of a fuel-air source at a temperature Ts. Part of this
heat flow is delivered for use as Qu at the use temperature of Tu.
The other part of Qs is lost to the surroundings as Ql across part
of the system boundary at Tl.

where the subscripts : s = source

u = use
l = loss

This could be thought of, for example, as a gas-fired space

heater.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

In the steady state case, the Energy and Exergy rate balances
are (with no work):
0
dE 0
= (Qs - Qu - Ql ) - W
dt
0 0
0
dEx T T T dV
= 1 − o Qs − 1 − o Qu − 1 − o Ql - W − Po − Ed
dt Ts Tu Tl dt

which can be rearranged as:

Qs = Qu + Ql [6.1]

To T T
1− Qs = 1 − o Qu + 1 − o Ql + E d [6.2]
Ts Tu Tl

Equation 6.1 clearly shows that the energy carried in by the

source heat transfer Qs is either used as Qu or lost to the
surroundings as Ql . The corresponding efficiency in terms of
product/input is:

Qu
η= [6.3]
Qs

for example for ηmax = 1 (100%), this would correspond to the case
in which Ql = 0 .

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Equation 6.2 shows that exergy is transferred into the system with
Qs and is transferred from the system with Qu or Ql or destroyed
through irreversibilities (as reflected in Ed ). This can be described
by an exergetic efficiency, ε , in terms of a product/input:

To
1− Qu
Tu
ε= [6.4]
To
1− Qs
Ts

Equation 6.3 can be introduced into eq. 6.4 to give another form
of the equation for exergetic efficiency:

To
1−
Tu
ε =η [6.5]
T
1− o
Ts

Note that both η and ε gauge how effectively the input is

converted to the product; η does so on an energy basis and ε
does so on an exergy basis.

Note that ε increases as the use temperature approaches the

source temperature (i.e., as Tu approaches Ts). ε also increases
with η .

Thus, for an effective utilization of exergy:

η should be as close to unity as possible.

There should be a good (i.e., close) match between the use
and source temperatures (Tu and Ts).

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Exergetic Efficiency vs. Use Temperature, Tu (assuming η = 1), for

Ts= 2200 K.

Clearly ε is higher for higher use-temperature processes. This

can be understood by considering a low Tu process (e.g., a
domestic space heater) for which fuel is burned to produce only
slightly warm air, which from the standpoint of exergy has much
less utility.

This is why, to achieve a high ε , there should be a good match

between Ts and Tu.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Exergetic Efficiencies of Common Components

(text pg. 325-327)

Exergetic efficiencies can be written for common components of

thermodynamic systems. They are derived from exergy rate
balances based upon the following assumptions:

- There is no heat flow from the control volume to the

surroundings (i.e., Ql = 0.)
- The control volume is at steady-state.

1. Turbines

With the above assumption, the exergy rate balance is:

0
T
0= 1 − o Q j − Wcv + m ( e f 1 − e f 2 ) − E d
j Tj

Wcv E d
ef1 − ef 2 = +
m m

Thus, the flow exergy is converted into work and is also

destroyed. A gauge of the effectiveness of this conversion of flow
exergy is the exergetic turbine efficiency (or turbine
effectiveness):

Wcv / m
ε=
ef1 − ef 2

Note: This should not be confused with isentropic turbine

efficiency, ηt .

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

2. Compressors and Pumps

The exergy rate balance, with Q j = 0, can be written as:

Wcv E
− = ef 2 − ef1 + d
m m

i.e., the work input is accounted for as an increase in flow exergy

or as exergy destruction. The effectiveness of the conversion of
work into flow exergy, the exergetic compressor (or pump)
efficiency is:

ef 2 − ef1
ε=
( −Wcv / m)

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

3. Closed Feedwater Heaters (i.e., heat exchanger - no mixing)

The exergy rate balance is:

0
0
To
0= 1− Q j − W cv + ( mh e f 1 + mc e f 3 ) − ( mh e f 2 + mc e f 4 ) − E d
j T j

No work done in
heat exchanger

which can be re-arranged to:

mh ( e f 1 − e f 2 ) = mc ( e f 4 − e f 3 ) + E d

Thus, the exergy of the hot stream serves to increase the exergy
of the cold stream and supplies exergy which is destroyed. A
logical efficiency in terms of product input is the exergetic heat
exchanger efficiency (for no mixing):

mc (e f 4 − e f 3 )
ε=
mh (e f 1 − e f 2 )

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

4. Open Feedwater Heaters ((i.e., heat exchanger - direct

contact/mixing)

The exergy rate balance is:

0
0
To
0= 1− Qj − W cv + m1e f 1 + m2e f 2 − m3e f 3 − E d
j Tj
No work in heat
exchanger

which can be rearranged as:

m1 ( e f 1 − e f 3 ) = m2 ( e f 3 − e f 2 ) + E d
Decrease in Increase in Exergy
exergy of hot exergy of destruction
stream cold stream

Regarding the hot stream as the “input” which supplies the

exergy, the exergetic efficiency for a direct contact heat
exchanger can be written as:

m2 (e f 3 − e f 2 )
ε=
m1 (e f 1 − e f 3 )

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Example

A compressor operating at steady state takes in 1 kg/s of air at

1 bar and 25oC and compresses it to 8 bar and 160oC. Q between
the compressor and surroundings occurs at an average surface
temperature of 50oC. WC = −230 kW

a) Perform a full exergy accounting of the power input to the

compressor.
b) Calculate the exergetic efficiency for the compressor.

Qc
Tb = 50oC
m = 1 kg / s

P1= 1 bar P2= 8 bar

T1 = 25oC T2 = 160oC
WC = −230 kW

Solution

Assumptions

- ∆KE , ∆PE = 0
- Air is modeled as an ideal gas
- To = 25oC, Po = 1 bar

We will use an exergy rate balance:

To
0 = 1− Qcv − Wcv + m ( e f 1 − e f 2 ) − E d [6.6]
Tb

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Step 1: Evaluate properties

P1 = 1 bar h1 = 298.18 kJ/kg

T1 = 25oC so(T1) = 1.69528 kJ/kg K

P2 = 8 bar h2 = 434.48 kJ/kg

T2 = 160oC so(T2) = 2.07234 kJ/kg K

Step 2: Calculate Qcv

Applying an energy balance:

Qcv = Wcv + m ( h2 − h1 )

1 kg kJ
Qcv = −230 kW + ( 434.48 − 298.18 )
s kg

Qcv = −93.69 kW

Step 3: Calculate exergy transfer with Q

To
Eq = 1 − (Qcv )
Tb

298 K
Eq = 1 − ( −93.69 kW )
323 K

E q = −7.252 kW

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Step 4: Calculate net flow exergy

e f 2 − e f 1 = (h2 − h1 ) − To ( s2 − s1 )

P2
e f 2 − e f 1 = (h2 − h1 ) − To s o (T2 ) − s o (T1 ) − R ln
P1

e f 2 − e f 1 = (434.48 − 298.18)kJ / kg
(2.07239 − 1.69528)kJ / kg
− 298 K 8.314 kJ 8 bar
− ln
28.97 kg ⋅ K 1 bar

kJ kJ
e f 2 − e f 1 = 201.8 or e f 1 − e f 2 = −201.8
kg kg

Step 5: Calculate exergy destroyed

Re-arrange eq. [6.6] and substitute in values from above steps:

To
0 = 1− Qcv − Wcv + m ( e f 1 − e f 2 ) − Ed
Tb

To
recalling 1 − (Qcv ) ≡ Eq and WC = −230 kW
Tb
then

E d = −7.252 kW − ( −230 kW ) + (1 kg / s )( −201.8 kJ / kg )

E d = 20.95 kW

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 06

Finally: Exergy Accounting

Input:

Exergy carried in with work: 230 kW (100%)

Output + Destruction:

Flow exergy: 201.8 kW (87.74%)

Exergy with heat transfer: 7.252 kW (3.15%)
Exergy destroyed: 20.95 kW (9.11%)
230 kW (100%)

and Exergetic compressor efficiency:

ε = net flow exergy increase of air

power input

ef 2 − ef1
ε=
( −Wcv / m )

201.8 kW
ε=
230 kW

ε = 0.8774 (or 87%)

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