Energy: Quantity & Quality
Exergy
Energy has both quantity and quality.
Quality of energy is its potential to produce useful work.
Dr. Md. Zahurul Haq, Ph.D., CEA, FBSME, FIEB
First Law of Thermodynamics:
energy is conserved in all (non-nuclear) processes.
Professor
Department of Mechanical Engineering Second Law of Thermodynamics:
Bangladesh University of Engineering & Technology (BUET)
Dhaka-1000, Bangladesh
the quality of energy is reduced in all real processes.
http://zahurul.buet.ac.bd/ ⇒ During transformation and transfer, energy is both conserved and
degraded.
ME 203: Engineering Thermodynamics Exergy provides a direct relationship between the thermodynamic
http://zahurul.buet.ac.bd/ME203/ state of a system and its capability to do useful work.
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 1 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 2 / 24
Datum Condition & Useful Work
012
P > P0
Standard atmosphere:
4567
P0 = 101.325 kPa, T0 = 298.15 K T1097
Species RH = 60% RH=100%
80 012 012 30 012345
N2 0.7662 0.7564
80 30 T1093
O2 0.2055 0.2029 089 91 5
CO2 0.0003 0.0003
H2 O 0.0188 0.0313 1 5
55 55 P < P0
Other 0.0092 0.0091 657 657 0678
T1095 803
When the pressure, temperature, composition, velocity, or elevation of a T1096 75 T1094
system is different from the environment, there is an opportunity to
Useful work could be produced by Energy contains exergy when – and
develop work.
utilizing temperature deviation from only when – that energy is not in
the environment. equilibrium with its environment
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 3 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 4 / 24
Useful Work (Wu ) & Datum State (T0 , P0 ) Dead State & Exergy (eX )
A system in a dead state is in thermal & mechanical equilibrium
with environment at T0 & P0 (T0 = 298.15 K, P0 = 101.325 kPa).
P0 Datum state Exergy of a system in a closed system in a given state is the
P F P0 = 101.325 kPa maximum useful work output that may be obtained from a
T0 = 298.15 K system-environment combination as the system proceeds from a
T345
specified equilibrium state to the dead state, while exchanging
R2 heat solely with the environment.
If P ≈ P0 ⇒ W = PdV = P0 ∆V 6= 0;
1
Exergy, eX ≡ Ex m is the sum of thermo-mechanical, KE, PE,
−→But useful work, Wu = 0. chemical exergies:
−
→ →
δWu = F .d −
x = (P − P0 )As dx = (P − P0 )dV = δW − P0 dV
eX = eXTM + eXKE + eXPE + eXCH + · · ·
⇒ δWu = δW − P0 dV = δW − δWsurr
Exergy is a function of both the state of the system & the local
As a closed system expands, some work needs to be done to push
environment. Once the environmental conditions are standardized,
the atmospheric air out of the way (and, vice versa) Wsurr .
exergy is treated as a property of the system alone.
⇒ Surrounding work (Wsurr ) is not recoverable & can’t be utilized.
At dead state, exergy of the system is zero.
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 5 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 6 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
Exergy of Heat (eXQ )
T341 T487
T0
ΦQ ≡ Wrev ,HE = Q 1 − T R
T1104 T1105
Pn
ΦQ = j =0 Qj 1 − T 0
Tj
h i
IQ = ΦQ,1 − ΦQ,2 = T0 Q T12 − T11 = T0 σQ
T1099
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 7 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 8 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
Heat Conduction Through Wall
Moran, Ex. 7-3 ⊲
T2 −T1
q̇ = −k L = 0.2 kW m2
T384 T385
h i
T0
ηcarnot = 1 − 300/1000 = 0.7 ηcarnot = 1 − 300/800 = 0.625 φ̇Q,in = q 1 − T 1
= 0.1 kW
m2
| W |= 1000(0.7) = 700 J | W |= 1000(0.625) = 625 J
h i
φ̇Q,out = q 1 − T
T2 = 0.01 m 2
0 kW
| QL |= 1000 − 700 = 300 J | QL |= 1000 − 625 = 375 J
kW
IQ = φ̇Q,in − φ̇Q,out = 0.09 m2
⇒ Exergy destruction for heat transfer from 1000 K to 800 K, ExQ
⇒ ExQ = 1000(300)(1/800 − 1/1000) = 75 J
T1112
⇒ Reversible work loss = 700 J - 625 J = 75 J.
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 9 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 10 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
Equations: CM & CV Systems Example: Exergy of Air
CM System: φ = (u − u0 ) + P0 (v − v0 ) − T0 (s − s0 )
Q − W = ∆U u − u0 = cv (T − T0 )
Wu = W − P0 ∆V = W − W0 v = RT
P
φ = (u − u0 ) + P0 (v − v0 ) − T0 (s − s0 ) s − s0 = cV ln TT0 + R ln vv0 = cP ln T 2
− R ln P2
P T1 P1
ΦQ = nj=1 Qj 1 − T 0
Tj Environment: T0 = 298.15 K, P0 = 101.325 kPa.
∆Φ = ΦQ − Wu − Icm ⊲ Air at 298.15 K & 101.325 kPa: φ = 0 kJ/kg
SSSF CV System: ⊲ Air at 298.15 K & 50 kPa: φ = 27.4 kJ/kg
Q − Wsf = m(∆h + ∆pe + ∆he) = m∆h ⊲ Air at 298.15 K & 200 kPa: φ = 16.0 kJ/kg
✯
✟ 0 ⊲ Air at 200 K & 101.325 kPa: φ = 20.8 kJ/kg
Wu = Wsf − P0✟✟
∆V = Wsf
2
⊲ Air at 400 K & 101.325 kPa: φ = 14.4 kJ/kg
ψ = (h − h0 ) − T0 (s − s0 ) + V2 + gz
P When the pressure, temperature, composition, velocity, or elevation of a
ΦQ = nj=1 Qj 1 − T 0
Tj system is different from the environment, there is an opportunity to
P P develop work.
∆(mψ) = e ṁe ψe − i ṁi ψi = Φ̇Q − Ẇu − İcv
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 11 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 12 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
CM System: Expansion of Steam inside Cylinder Evaluating the Exergy of Exhaust Gas
Moran Ex. 7.1 ⊲ A cylinder of an internal combustion engine contains
2450 cm3 of gaseous combustion products at a pressure of 7 bar and a
Cengel, Ex. 8-11 ⊲ Piston-cylinder assembly contains 0.05 kg steam. temperature of 867oC just before the exhaust valve opens. Determine the
specific exergy of the gas, in kJ/kg. Assume, the combustion products as air
Q − W = m(∆u + ∆KE + ∆PE ) ≃ m∆u
as ideal gas.
φ ≡ (u − u0 ) + P0 (v − v0 ) − T0 (s − s0 )
Wu = W −W0 = 8.826 −3.509 = 5.317 kJ T0 = 300 K, P0 = 1.0 bar
φQ = 0: Heat loss to T0 φ ≡ (u − u0 ) + P0 (v − v0 ) − T0 (s − s0 )
∆φ = −9.648 kJ u − u0 = cv (T − T0 ) = 600 kJ/kg
ICM = φQ − Wu − ∆φ = 4.331 kJ P0 (v − v0 ) = R( PP0 T − T0 ) = 39.36 kJ/kg
T1113
Wu
ǫ= −∆φ = 0.551 s −s0 = cp ln(T /T0 )−R ln(P/P0 ) = 0.7870 kJ/kg
φ = 324.54 kJ/kg
T1435
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 13 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 14 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
Second Law Efficiency (ηII ) or Effectiveness (ǫ) Adiabatic Compression & Pumping
A performance parameter based on the exergy concept is know as
Second Law Efficiency (ηII ) or as Second Law Effectiveness (ǫ).
A first-law efficiency gages how well the energy is used when CM process: CV process:
compared against an ideal process, whereas an effectiveness
∆Φ = ΦQ − Wu − Icm ∆Ψ = ΦQ − Wu − Icv
indicates how well exergy is utilized.
useful exergy out exergy destruction Wu = Wact + P0 ∆V Wu = Wact
ηII ≡ ǫ ≡ exergy in =1− exergy in ∆Φ ∆Ψ
ǫ≡ Wact ǫ≡ Wact
Effectiveness (ǫ) is defined as the increase in the specific
Wout 12 availability of the fluid per unit of actual work input.
ηI = QHT +QMT = 25+50 = 16%
Ws
293 First law efficiency, η ≡ Wact .
ExHT = 25 1 − 1098 = 18.33
293
ExMT = 50 1 − 513 = 21.44
Wout
ηII = ExHT +ExMT = 30.2%
T349
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Exergy Concepts: Examples Exergy Concepts: Examples
Steam/Gas Turbine, Throttling & Nozzle Heat Exchange without Mixing
Turbines:
wa
η= ∆h
∆ψ = φQ − wu − icv
wa
ǫ= ∆ψ
T476
T475
w = 0, q = 0, ∆ke = 0, ∆pe = 0
Throttling: Nozzle: P P
0 SSSF Energy: 0 = 0 − 0 + i (mh)i − e (mh)e
0
w✟
✼− ✟
q✓ ✯+ hi − he − ✘
0 ✘✘
✿ 0
w✟
✓+ ✟
q✼ ✯=0
he − hi + ∆ke
0 =✓ ∆ke ✓ ⇒ m1 h1 + m3 h3 = m1 h2 + m4 h4 → mc (h4 − h3 ) = −mh (h2 − h1 )
0 ∆kea
η=
φQ✚
❃− w✟
✟✯
Exergy balance:
0
∆ψ = ✚ u − Icv ∆kes
0
W✟
✯− I cv → m (ψ − ψ ) + m (ψ − ψ ) = −I
0
❃+ w✟
0 ✚❃− ✟
✚ ∆(mΨ) = ✚
✟✯
0 ΦQ c 4 3 h 2 1 cv
u − Icv
ψe ∆ψ = ✚
ΦQ
ǫ= ψi
mc (ψ4 −ψ3 ) mc ψ4 +mh ψ2
ǫ= Ψe ⇒ ǫ≡ −mh (ψ2 −ψ1 ) or ǫ≡ mc ψ3 +mh ψ1
Ψi
The first form of ǫ for heat exchanger is usually preferred.
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 17 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 18 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
SSSF Compressor Throttling Process
Holman, Ex. 5.10: ⊲ A steady-flow compressor is used to compress air from
1 bar, 25o C to 8 bar in an adiabatic process. The first-law efficiency, ηI , of
the process is 87%. Calculate the irreversibility and ηII of the process if
Cengel, P. 8-130 ⊲ Argon gas expands from 3.5 MPa and 100oC to 500 kPa in
T0 = 293 K.
an adiabatic expansion valve. For environment conditions of 100 kPa and
Ψ ≡ (h − h0 ) − T0 (s − s0 ) + V2 ∼ (h − h0 ) − T0 (s − s0 )
+ gz = 25o C, determine (a) the exergy of argon at the inlet, (b) the exergy
2
wa = h1 − h2 = −cP (T2 − T1 ) : ws = h1 − h2s = −cP (T2s − T1 ) destruction during the process, and (c) the second-law efficiency.
ws
ηI = wa = 0.87 ψ1 = 223.8 kJ
(k −1)/k ) ψ2 = 103.3 kJ
P2 Wa
T2s = T1 P1 = 540 K → T2a = Cp + T1 = 571 K
Icv = −∆ψ = 120.5 kJ
wa = −279.0 kJ/kg T1122
ψ2
ǫ= ψ1 = 0.461
wmin = −∆Ψ = −[(h2a − h1 ) − T0 (s2a − s1 )] = −259.8 kJ/kg
isf = T0 (s2a − s1 ) = 19.2 kJ/kg
∆Ψ
ηII = wa = 0.931 ◭
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 19 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 20 / 24
Exergy Concepts: Examples Exergy Concepts: Examples
Nozzle Air Cooled Condenser
Cengel, P. 8-71 ⊲ Hot combustion gases enter the nozzle of a turbojet engine. Cengel, P. 8-63 ⊲ Determine (a) the rate of heat rejected in the condenser, (b)
Assuming the nozzle to be adiabatic and the surroundings to be at 20o C, the COP of this refrigeration cycle if the cooling load at these conditions is 6
determine (a) the exit velocity and (b) the decrease in the exergy of the gases. kW, and (c) the rate of exergy destruction in the condenser.
Take air properties for the combustion gases.
p QL = 6 kW
V2 = V12 − ∆KE = 627 m/s
QH = ṁ(h2 − h1 ) = 9.98 kW
ψ1 = 368.9 kJ
QL QL
COP = Win = QH −QL = 1.5
ψ2 = 339.4 kJ
Icv = −∆ψ = 0.0998 kW
Icv − ∆ψ = 29.5 kJ
ψe
ψ2 ǫ= ψi = 0.955
T1124 ǫ= ψ1 = 0.92 T1116
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Exergy Concepts: Examples Exergy Concepts: Examples
SSSF Turbine Boiler
Borgnakke, Ex. 8.5: ⊲ Borgnakke, Ex. 8.6: ⊲ Determine the second-law efficiency for this process
and the irreversibility per kilogram of water evaporated. Assume, cp of the
products of combustion is 1.155 kJ/kg K.
T351
Ψ ≡ (h − h0 ) − T0 (s − s0 ) + V2 ∼ (h − h0 ) − T0 (s − s0 )
+ gz =
2
ẇs = ṁ1 h1 − ṁ2 h2s − ṁ3 h3s = 25.27 MW
T1111 T1436
ẇa = ṁ1 h1 − ṁ2 h2 − ṁ3 h3 = 20.18 MW h i
mgas h2 −h1
ηI = ẇa
= 0.799 ◭ mwater = h3 −h4 = 3.685 kg/kg
ẇs
∆(ṁ Ψ) = ṁ1 Ψ1 − ṁ2 Ψ2 − ṁ3 Ψ3 = 24.65 MW mgas (ψ2 −ψ1 )
ǫ= mwater (ψ3 −ψ4 ) = 0.458
ẇa
ηII = ∆(ṁΨ) = 0.819 ◭ Icv = −∆ψ = 9.09 kW
© Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 23 / 24 © Dr. Md. Zahurul Haq (BUET) Exergy ME 203 (2022-23) 24 / 24
