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Exergy Analysis
Ron Zevenhoven
Åbo Akademi UniversityThermal and Flow Engineering Laboratory / Värme- och strömningsteknik
tel. 3223 ; [email protected] material: http://users.abo.fi/rzevenho/kursRZ.html
Process EngineeringThermodynamicscourse # 424304.0 v. 2015
Commented / edited byProf. Özer Arnas,visiting ÅA VST May/June 2014from the US Military Academy
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 2/102
1.1 Exergy vs. Energy
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 3/102
Energy Conversion /1
In a hydraulic power plant, the potential energy difference canbe completelyconverted into work
For a thermal plant Carnot (C1824) discovered that onlya portion of heat Q at temperature T1can be convertedinto work, with coldsink temperature T2 :
Pic: S05
T
TQ
T
TTQW
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 4/102
Energy Conversion /2
The preferable wayto convert heat Q at T1 into work is to use a natural, unlimited cold sink, such as the naturalenvironment at temperature T0.
With T2 > T0 no full use of Q is made; for T2 < T0 a coolingeffort is needed.
Pic: S05
T
TQ
T
TTQW
max
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 5/102
Exergy /1
EXERGY quantifies how much of a certainamount of energy contained in an energysource can be extracted as useful work.
The potential of (natural) energy resources for driving a mechanical, thermal, chemical or biological process dependson the deviation from thermodynamic equilibrium with the surrounding environment.
This allows evaluation of the ability to perform maximum work under certainenvironmental conditions.
This involves quality assessments, for example for heat Q its temperatureT, relating it to temperature T0 of the
surrounding environment. Pic
ture
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://cp
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11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 6/102
Exergy /2
Note that work is a path function: W = ƒ(initial state, path, final state)
For maximising the work output:– The process should be reversible– The end of the process should be at a dead state which
is in equilbrium with the surrounding environment: temperature T0 and pressure p0 as the environment; no kinetic or potential energy relative to the environment; no chemical reactivity with the environment. (And no electrical, magnetic etc. effects.)
This requires a description of the state and, ifchemistry is involved, the composition of the localenvironment for the prevailing chemical species.
Pic
ture
: http
://up
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11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 7/102
Exergy /3
While the quantities of energy allow for ”bookkeeping” with the help of the First Law of Thermodynamics (”energycannot be produced or destroyed”); the exergy conceptfollows from the Second Law of Thermodynamics, whichinvolves the quality of energy and its degradation duringenergy conversion.
A few useful definitions: Exergy expresses the maximum work output attainable in the natural environment, or the minimum work input necessary to realise the reverse process (Rant, 1956);Exergy expresses the amount of mechanical work necessary to produce a material in its specified state from components in the natural environment, in a reversible way, heat being exchanged only with the environment (Rickert, 1974)
During the 1940s, the term ”availability” had been introduced for a similar purpose (Keenan, 1941, K41) – see p. 9
Some care should be considered with respect to ”the surroundings”, which is everything outside the system’sboundaries. Part of the surroundings, referred to as the ”immediate surroundings” are affected by the process, the ”environment” refers to the region beyond that.
”Normal conditions” for the reference environment:T° = 298.15 K, p°= 101.325 kPa, relative humidity 70%, atmospheric CO2 concentration 330 - 380 ppm, sea water salinity 3.5 %-wt, etc. etc. etc.
Some data for air: N2 = 75.78 kPa; O2 = 20.39 kPa, Ar = 906 Pa, CO2 = 33.5 Pa , H2O = 2200 Pa, He = 1.77 Pa, and Ne, D2O, Kr, Xe < 1 Pa (SAKS04)
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 8/102
Surroundings - Environment
39.7 Pa would be better – see co2now.org
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 9/102
Notation For Exergy In the literature, symbols like Ex, or X for exergy, or ex
or x for specific exergy (i.e. exergy per mass, Ex/m = X/m = ex = x) are most common; but
Also the symbols B and b are used, especially in (and by followers of) the very important work by Szargut et al.
And, in some (U.S.) literature, availability, B, is defined as BAV = H - T°·S, which can be related to exergy by Ex = BAV (T, p) – BAV (T°, p°), where BAV (T°,p°) = (H - T°·S)T°, p°
= GT°,p° which ≠ 0 necessarily (depends on UT°,p° in HT°,p° ) Ex (T,p) = (H - T°·S)T,p - (H - T°·S)T°,p° (SAKS04)
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 10/102
1.2 Reversible Work-Irreversibility
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11/102
Exergy and Work Potential Thus: a system will deliver the maximum possible work in a
reversible process from a specified initial state to the stateof its environment, i.e. the dead state.
This useful work potential is called exergy. It is also the upper limit of work from a device without violating the Lawsof Thermodynamics. The part of the energy that cannot be converted into work is called unavailable energy or anergy: energy = exergy + anergy
Since exergy depends on the surrounding environment, it is not a state variable like U, H, S, V, p, G.
In some special cases a changing environment can be considered, for example when analysing air flight, or largeseasonal or daily changes (day/night).
Unlike energy, exergy is consumed or destroyed.
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 12/102
Surroundings Work During the expansion of a work W producing closed system,
with volume change ΔV = Vfinal – Vstart > 0, work Ws is doneon the surroundings at pressure p°:
Ws = p°·ΔV
Similarly, ΔV < 0 is possible. In many cases this surroundingswork cannot be recovered, reducing the useful work Wu of the process:
Wu = W - Ws = W - p°·ΔV
This has no significance for 1) cyclic processes and 2) systems with fixed boundaries, but is important for constant pressure processes with large volume changes(for example: combustion furnaces, engines)
Note: here W > 0 definedas work done by the system
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 13/102
Reversible work Wrev is the maximum amount of work that can be obtained from a process; it is equal to exergy if the final state is the dead state (which usually is not so!)
The difference between Wrev and Wu
is referred to as the irreversibibility: I = Wrev - Wu
which is equal to the amount of exergy that is destroyed.
The irreversibility represents ”lostopportunity” to do work.
If I = 0, no entropy is generated. For work consuming devices, Wrev ≤ Wu
Reversible Work
Pic: ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 14/102
Example: Irreversibility 1/1
A heat engine (HE) receives Qin = 500 kW heat at TH = 1200 K and rejectswaste heat at TL = 300 K. The power output of the engine is 180 kW. Calculate the reversible power Wrevand irreversibility rate I, in kW.
Wrev follows from the Carnot efficiencyηcarnot = 1-TL/TH = 0.75 Wrev = ηcarnot · Qin = 375 kW.
The actual power output Wu = 180 kW, thus I = Wrev - Wu = 195 kW
Note that the 500 kW - 375 kW = 125 kW rejected to the sink was not available for work
Pic: ÇB98
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
15/102
Example: Irreversibility 1/2
The 125 kW rejected to the sink follows from Carnot efficiency; this is the anergy part of the input energy:energy = anergy + exergy
500 kW = 125 kW + 375 kW
The entropy part of the energy cannot be converted into usefulwork.
s)J/(K K
kW
4171200
500
inT
Q
s)J/(K K
kW
417300
125
outT
Q
Pic: ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 16/102
Example: Irreversibility 2/1
A block of iron (m = 500 kg) is cooledfrom TH = T1 = 200°C to TL = 27°C by transferring heat to the surrounding air (T0 = 27°C). Calculate reversible work Wrev and irreversibility I, in kJ
The maximum work equals (assuming a series of reversible heat engines):
kJ 8191
gives which where
TT
lnT)TT(cmWδW
dTcmQδ
Qδ)T
T(Qδ)
T
T(QδηWδ
avg
T
Trevrev
avgin
inin
H
L
increv
Pic: ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 17/102
Example: Irreversibility 2/2
Here, cavg is an average value for the specific heat of iron, ~ 450 J/(kg·K)
In the result, the first part (m·cavg·ΔT) is the heat transferred from the ironblock which equals 38925 kJ.
The 8191 kJ calculated as Wrev could havebeen converted into useful work
Since no work was obtained, Wu = 0 and the irreversibility therefore equalsI = Wrev = 8191 kJ (21% of Qin) .
The remaining 38925 - 8191 = 30734 kJ (79% of Qin) is rejected also to the surroundings.
Pic: ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 18/102
Irreversiblities and Exergy Loss Typical examples of irreversibilities that result in exergy
destruction, or exergy losses:– Friction, including friction in fluid flows
which results in pressure drop– Mixing (which increases entropy)– Chemical reactions– Heat transfer, especially through
large temperature differences– Unrestrained expansion; throttling– Fast compression– Electric resistance– Adiabatic processes are often more reversible
Note that fast processes can be more reversible than slow processes; for example discharging a battery
Slower does not automatically mean more reversible !
Rule of thumb:irreversible processes are processes that look absurd when run backwards
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
19/102
Second-Law Efficiency, η2, Revisited
Second-law efficiency is defined as:
Equally, it can be defined as (for work producing devices):
for work done (for work -power- consumed η2 = Wrev / Wu, input )with irreversibility, I, or, expressed as exergies:
carnotη
ηη efficiency thermal possible maximum
efficiency thermal actual
revrev
u
WWW
ηI
work useful possible maximum work useful
supplied exergy
destroyed exergy
supplied exergy
of use made i.e. recovered, exergyη
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 20/102
Example: Efficiency A house, like many in Finland, is electrically
heated, using an electric resistance heatingelement that converts electricity for 100% into heat at room temperature. For indoor T = 21°C and outdoor T = 10°C, determine the Second-law efficiency.
The performance can be described by the so-called coefficient of performance, COP, defined as
which here is equal to COP = 1.
For the most efficient, Carnot, process:with result COPrev = 294/(294-283) = 26.7.
Thus, the Second-law efficiency η2 = COP/COPrev = 0.037 = 3.7%
input work
output heatheaterCOP
lowhigh
highheater TT
TCOP
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 21/102
1.3 Physical Exergy
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 22/102
Exergy of ”Non-Heat” Energy Kinetic energy Ek = ½m(v-vref)2 can* be completely and
reversibly converted into work Ex(Ek) = Ek
Potential energy Ep = mg(z-zref) can* be completely and reversibly converted into work Ex(Ep) = Ep
Electric energy Eelec = i·V·t (current × voltage × time) can** be completely and reversibly converted into work Ex(Eelec) = Eelec
Similar for energy streams Ė (J/s, W) or specific energy(streams) e = E/m = Ė/ṁ, ė = Ė/m, with mass stream ṁ
(kg/s) or mass m (kg)
* if frictional losses (giving heat) can be avoided** if Ohmic losses (giving heat) can be avoided
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 23/102
Exergy of Heat The exergy of heat follows directly from Carnot’s
analysis (C1824): the maximum amount of reversible work that can be obtained from conversion of heat Q at temperature T, with temperature T° of the surroundings, equals
with of course T’s in K!!
As noted above: the entropy part of energy (anergy), for heat Q equalto Q/T, cannot be converted into work.
QQT
TQEx factor Carnot)()(
0
1
Sadi Carnot1796-1832
see C1824
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 24/102
Exergy of Internal Energy Internal energy U is composed of sensible heat (∫cvdT), latent
heat (ΔUphase transitions), chemical energy and nuclear energy. Chemical (reaction) exergy will be addressed separately (sections 1.8 and 1.9), nuclear (reaction) exergy will not be addressed here.
Remember that for internal energy, ΔU = Q + W or dU = δQ + δW, where, as shown above,- Ex(δW) = δWu = -(p-p°)dV = -pdV + p°dV = δW + p°dV,
with useful work Wu> 0, gained by the system (!!) and- with Ex(δQ) = (1 - T°/T)· δQ and δQ = dS/T it follows that
Ex(δQ) = (1 - T°/T)·TdS = δQ -T°dS- Ex(dU) = Ex(δWu) + Ex(δQ) = dU + p°dV - T°dS
This gives finally: Ex(U)=T°,V°,p°∫T,p,VdU+p°dV-T°dS=(U-U°)+p°(V-V°)-T°(S-S°)!!!
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 25/102
Exergy of Enthalpy
For enthalpy, H = U + pVThus, Ex(H) = Ex(U) + Ex(pV)
As shown above, for internal energy: Ex(U) = (U-U°) + p°(V-V°) - T°(S-S°)and since Ex(pV) = (p-p°)V (see above: surroundings work p°ΔV)
for work done by the system the result is: Ex(H) = U+pV - (U°+p°V°) - T°(S-S°) = (H-H°) - T°(S-S°)
Combined with Ek + Ep for kinetic and potential energy this is also known as ”flow exergy”, with symbol ψ typically used for Ex(H) + Ek + Ep,
Similarly, symbol Φ is often used for Ex(U) + Ek + Ep
!!
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 26/102
Exergy of a Flow / Non-flow System The exergy of a moving or non-moving (or flow or non-
flow) system is a combination of the kinetic, potential and thermo-mechanical exergy
see Figures below for non-flowing and flowing systems:
Pics:ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 27/102
Example: Internal Energy Exergy /1
Calculate the maximum amount of work to be obtained from expanding200 m3 compressed air at p1=10 bar, T1 = 300 K to ambient conditions(T° = 300K, p° = 1 bar).
The maximum work is equal to the exergy of the closedsystem of compressed gas: Ex = Ex(U) = m·ex(u), where m can be found using ideal gas law (temperature and pressure are far from Tcrit, pcrit); m = Mp1V1/RT1 ≈ 2323 kg.
Pic:ÇB98
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 28/102
Example: Internal Energy Exergy /2
No kinetic energy, no potential energyto be considered.
Ex(U) = (U-U°) + p°(V-V°) - T°(S-S°) T1=T° ΔT=0 U1 - U0 = 0
Note: s1 = s° + cp·ln(T1/T°) - R·ln(p1/p°)
using δq = du - δw Tds = du + pdV = dh - Vdp = cpdT - (RT/p)dp
p°(v1-v°) = p°·R(T1/p1-T°/p°) = RT°(p°/p1-1); (T1=T° !) T°(s1-s°) = T°·[cp·ln(T1/T°)-R·ln(p1/p°)] = - R·T°·ln(p1/p°)
Ex(u1) = 120.76 kJ/kg; Ex(U1) = m· ex(u1) = 280.53 MJPic: ÇB98
= 0
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 29/102
Open System Exergy, h-s diagram /1 As for an alternative derivation: Consider an open
system that produces work w, which is maximal, wmax, for a reversible process. Neglect kinetic and potential energy effects.
For reversible heat exchange q, this must occur at a temperature T° q = T°·Δs = T°·(s2 - s1)
Also, the outflow 2 must be at equilibrium with the surroundings: h2 = h°, T2 = T°, p2 = p°, s2 = s°
Pic: B01
For thosewho useh,s diagrams
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 30/102
Open System Exergy, h-s diagram /2 The energy balance for this system gives:
wmax = h1 – h2 + q = h1 – h2 - T°·(s2 – s1) , and thusfor the maximum work, wmax = ex, in general: ex = h – h° - T°·(s – s°)
This can be conviniently plotted in h-s diagrams, with reference line dh/ds = T°
Pics: B01
T0
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 31/102
1.4 Exergy Destruction; Isolated-Closed-Open Systems
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 32/102
Exergy Destruction /1
Consider an isolated system: no energy or masstransfer across the system boundaries is possible
The energy and entropy balance equations give, for time t1 t2: ΔU = U2 - U1 = 0, Sgen = S2 - S1.
Combined, and ×T° T°·Sgen= T°·(S2 - S1) For the total exergy of this system:
Ex2 - Ex1 = (U2 -U1) + p°·(V2 - V1) - T°·(S2 - S1) = - T°·(S2-S1) = -T°·Sgen ΔEx = -T°·Sgen ≤ 0with exergy destruction –ΔEx = T°·Sgen ≥ 0
For systems in general, this is known as the Gouy-Stodola relation referring to work reported by Gouy(1889) and by Stodola (1910)
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 33/102
Considering again an opensystem as in the figure ,for a reversible processwrev = h1 - h2 + q°, where q°= T°(s2-s1), while for an irreversible process wirrev = h1 - h2 + q with q < q°, indicating a net entropy production Δs = s2-s1 – q°/T° >0 Thus, Δex = wrev – wirrev = T°·Δs (Gouy – Stodola)
Similar to the principle of increasing entropy, this is known as the principle of decreasing exergy
For an entropy production rate the exergy destructionrate is equal to -ΔĖx = T°·Ṡgen ≥ 0
Exergy Destruction /2
Pic: B01
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 34/102
Exergy Destruction /3
For a closed system, which does not involve massflows a general balance equation reads:
Exheat + Exwork – Exdestroyed = ΔExsystem,
where Exdestroyed = T°·Sgen ≥ 0
which gives, for several heat (streams) Qi :
120120
2
1
0
ii
00
0
120120
0
))((1
:Q for T constantnon withor
)(1 :rate as or
))((1
ExExSTVVpWQT
T
dt
dExST
dt
dVpWQ
T
T
ExExSTVVpWQT
T
genii i
genii i
genii i
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://w
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11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 35/102
Exergy Destruction /4
For an open system, which does involve mass flows a general balance equation reads:
Exheat + Exwork + Exin - Exout - Exdestroyed = ΔExsystem,
where Exdestroyed = T°·Sgen ≥ 0which gives, for several heat (streams) Qi and incoming and outgoing enthalpy streams with exergy ψj, ψk:
120120
2
1
0
ii
00
0
120120
0
))((1
:Q for T constantnon withor
)(1 :rate as or
))((1
ExExSTmmVVpWQT
T
dt
dExSTmm
dt
dVpWQ
T
T
ExExSTmmVVpWQT
T
genk
kkj
jjii i
genk
kkj
jjii i
genk
kkj
jjii i
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 36/102
Example: Exergy Analysis-Open System /1
A 200 m3 storage tank for air, at p1=100 kPa, T1= 300K is to be used for air storage at p2 =1 MPa, T2 =300 K. Air is supplied by a compressor that takes in air at p° =100 kPa, T°= 300 K. Calculate the minimum work requirement for this process.
Ideal gas behaviour can be assumed; no kinetic or potential energy effects.
Note that for the incoming air, ψ1=ψ°=Φ°= 0 Final mass follows from ideal gas law m2 ≈ 2323kg.
Pic:ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 37/102
Example: Exergy Analysis-Open System /2
For the final state, Ex(U2) = m2·Φ2 is to be calculated, where u2 - u° = 0 since ΔT = 0
p°(v2-v0) = RT°(p°/p2 -1) and T°(s2-s1) = -RT°(ln(p°/p2))
Thus, Φ2 = ex(u) = (u2-u°) + p°(v2-v°) - T°(s2-s°)= 0 + RT°(p°/p2 -1) + RT°(ln(p°/p2))= 120.76 kJ/kg
Wrev = m2·Φ2 = 280.53 MJ
Pic:ÇB98
see alsoslides 27 – 28...
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 38/102
1.5 Physical Exergy: Applicationsheat exchangers, tube flow, combustion, endoreversibility
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
39/102
Heat Transfer and Exergy Losses During the transfer of heat Q
from medium 1 to medium 2, with temperatures T1 > T2, the entropy of the heat Q increasesfrom to .
The entropy increase ΔS = Sgen
is equal to
And the exergy losses are equalto –ΔEx = T°·Sgen
T
Q
Pic:ÇB98
T
Q
T
Q
TTQ
T
Q
T
QS
gen
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 40/102
Heat Transfer Efficiency (PTG course)
A simple steady-state heat transfer process; heat is transported from medium 1 to medium 2 by conduction through a material that separates them.
Temperature T1 > T2
Thermodynamic analysis
This shows that Sgen is large for large temperaturedifferences (T1-T2) and lowtemperatures T1 and T2
Q1
.Q2
.
T = T1 T = T2
TT
TTQ
TTQS
T
QS
T
Q
balanceEntropy
balanceEnergy
gen
gen
.
See alsovS91
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 41/102
Quality Diagram : 1/T versus Q
Heat exchange: hot side T1 T2, cold side T3 T4
Exergy loss = T0 the surface between lines in the quality diagram
4
13
2
Q
1/T
1/T0
1/T = 0
1
2
3
4
Source: TUD92
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
42/102
Heat Exchanger Exergy Losses /1
A heat exchanger involves two media at temperatures TCinand TCout on the cold side, THin and THout on the hot side. (Note: TCin = TCout or THin = THout is possible!)
For a heat exchanger transferring a heat stream Q, to a cold mass stream ṁC from hot mass stream ṁH:
and the exergy losses are then equal to
with specific heats cpH, cpC for the hot and cold streams.
Hin
HoutHinHoutpHH
THout
THin
pHc
THout
THin
H
Cin
CoutCinCoutpCC
TCout
TCin
pCc
TCout
TCin
C
T
TTTTcmdTcm
T
TQ
T
TxE
T
TTTTcmdTcm
T
TQ
T
TxE
ln)()1()1(
ln)()1()1(
000
000
)T
Tlncm
T
Tlncm(T)xEΔxEΔ(xEΔ
Cin
Cout
pCC
Hin
Hout
pHHHC
> 0 < 0
< 0 > 0
< 0 > 0
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 43/102
Heat Exchanger Exergy Losses /2
! Also, -ΔEx = T°∙ [ṁH· (sHout-sHin) + ṁC· (sCout-sCin)] The efficiency of a heat exchanger, ηHX, can then be
defined as
In cases where electric power Pelec is (for example) used on the hot side:
%)(xEΔQ
Q
SΔTQ
Qη
HX
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://ol
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)TT(T
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CinCout
Cin
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11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 44/102
Heat Exchanger Exergy Losses /3
Heat exhanger exergy analysis shows that the temperaturedifference between the flows (or with the flow, for only on medium flow) should be as small as possible (but too small ΔT requires much surface A!).
This shows that counter-current heat exchangers performmuch better than co-current heat exchangers.
Ideally, the flows aquire each others temperature: the exergy losses will then be zero. For this, the heat capacitiesṁ·cp for the streams should be equal: ṁC·cpC = ṁH·cpH.
This is anyhow a requirement for a high ”effectiveness” of the heat exchanger, which depends on the ratio(ṁC·cpC) / (ṁH·cpH) (see course PTG 424101 # 4.3)
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 45/102
Example: Heat Exchanger Consider heat flow Q through a 5m×6m
wall with thickness dwall = 0.3 m and heat conductivity λ = 0.69 W/(m·K), from an indoor temperature of 27°C to an outside temperature of 0°C. The insideand outside wall temperatures are 20°C and 5°C, respectively.Calculate the heat flow Q and the exergydestruction rate –ΔEx in W.
Q = -λ(ΔTwall/dwall) = 1035 W - ΔEx = Ex(Qinside) – Ex(Qoutside)
= Q·[(1-T°/Twall,in) – (1-T°/Twall,out)] = 52.0 W in the wall, and = Q·[(1-T°/Tinside) – (1-T°/Toutside)] = 93.2 W in total
Pic:ÇB98
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 46/102
A Process Example Bilge water (waste
oil/water mixtures) processing for recycled fuel oil (in Turku)
Significant savings could be obtained by– Replacing the heating
centre and, more importantly the electric heating (!) by heat from the local district heat network
– Making use of hot raw material deliveries
Coldstorage
Heatedstorage
Separator
Sludge
Productstorage
Waterdisposal
Heatingcenter
Hot water
Hot water
Hot water
Surfactant
Fueloil
Electricity
Rawmaterial
Product
hea
t
heat
heat
heat
heat
Coldstorage
Heatedstorage
Separator
Sludge
Productstorage
Waterdisposal
Heatingcenter
Hot water
Hot water
Hot water
Surfactant
Fueloil
Electricity
Rawmaterial
Product
hea
the
at
heatheat
heatheat
heatheat
heatheat
Source:ZHPSS07
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 47/102
Heat Exchange and Driving Force Heat transfer theory states that heat transfer is driven by a
temperature difference ΔT = TH-TL : Q ~ A·(TH-TL), for heat flow Q through a surface A from higher temperature TH to lower temperature TL.
Exergy analysis shows that which suggests a thermodynamicdriving force equal to Δ(1/T) instead, and
the exergy losses are the product of three factors, being1) surroundings temperature T0, 2) the heat flow Q, and 3) the driving force Δ(1/T).
As irreversible, or non-equilibrium thermodynamicsshows: Ṡgen = ∑ Ji·Xi for flows Ji and driving forces Xi
HL
loss TTTQExΔ
See later partof thiscourse
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
48/102
Exergy Losses and Tube Flow /1 For fluid flow (m³/s) through a tube, viscous friction results
in pressure drop Δp (Pa), which must be compensated for by pumping power . This energy is converted to heat at fluid flow temperature T (so not all is lost!)
Thus the exergy losses can be calculated as: ΔĖx = Δp· - Δp· ·(1-T°/T) = Δp· ·T°/T = mechanical
dissipation × loss factor T°/T; or Δėx = (Δp/ρ)·T°/T
Note that ΔĖx > Δp· if T < T°. (No cooling for free!)
1 2
-τw
x
rp1
p2
L
R
A
S
1 2
-τw
x
rp1
p2
L
R
1 2
-τw
x
rp1
p2
1 2
-τw
x
rp1
p2
L
R
A
S
For pressure drop Δp:(p1 - p2 ) = -Δp = τw∙L·(4 /Dh) = 4ƒ∙½∙ρ∙<v>2 ∙L/Dh
with hydraulic diameter Dh = 4·A/SA = ”wet” area, S = ”wet” perimeter
Note: 4ƒ = ζ = Blasius, Darcy friction factorƒ = Fanning friction factor
V
VpW
V V V
V
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 49/102
Exergy Losses and Tube Flow /2 The pressure drop can be related to the fluid flow with cross
sectional averaged velocity <v> = /A, with friction factor ƒ, where 4ƒ = 64/Re for laminar circular tube flow, or for turbulent flow in smooth tubes with hydraulic diameter Dh : 4ƒ= 0.316·Re-0.25 (2500 < Re < 105) (Blasius).
This gives dissipation losses for turbulent flow along round tubelength x: δ /dx = (δ /ṁ)/dx = (-dp/dx)/ρ = cp·dT/dx, which is equal to
for fluid density ρ, dynamic viscosity η, mass flow ṁ.
1 2
-τw
x
rp1
p2
L
R
A
S
1 2
-τw
x
rp1
p2
L
R
1 2
-τw
x
rp1
p2
1 2
-τw
x
rp1
p2
L
R
A
S
75,4275,1
75,125,079.1
D
m
dx
w
V
Ww
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
50/102
Exergy Losses and Tube Flow /3 For a round tube with diameter D, with insulation material, giving
overall heat transfer coefficient U (W/m2·K), carrying turbulent flow at temperature T (for example a heat exchangertube) the heat exchange per meter with the environment is:
which gives for the total exergy losses per length section dx:
which shows that losses due to pressure drop decrease with diameter, while those for heat losses increase with diameter !
m
DTT
dxm
Q
dx
q
)(U 0
Tm
DTT
T
T
D
m
dx
dex
200
75,4275,1
75,125,0 )(U79.1
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
51/102
Exergy Losses and Tube Flow /4
For a given optimal value for the diameter, Doptimal , the exergy losses are minimal:
)TT(ρ
Tmη.D
.,
,
optimal U
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 52/102
Energy Conversion Stages in “conventional” Electricity Production, i.e. condensing power plant
Energy in chemical bonds of fuel
Furnace
Thermal energy of hot flue gas
Heat transfer to steam cycle
Thermal energy of water / steam
Steam cycle generator
Electrical energy
Sources of exergy loss: Friction in flow lines,
pressurising and de-pressurising
Heat transfer at low temperatures and/or across large temperature differences
Conversion of chemical (potential) energy into thermal energy (heat)
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 53/102
Exergy Loss in Expansion Turbine p1, T1, S1 p2, T2, S2 is non-isotropic
due to friction frictional heat Q = surface 1-2’-S2’-S1-1
in T,S diagram frictional heat to fluid = surface 1-2’-4-3-1
= (1-T0/Tav) Q
exergy loss Ex = surface 3-4-S2’-S1-3 = (T0/Tav)Q
loss factor T0/Tav i.e. less serious at increasing average temperature Tav
Transformationof Work into Heat
1
2
Pic:vL77
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 54/102
Exergy Losses in Combustion /1 Combustion = Oxidation, i.e. transfer of electrons
from fuel to oxygen Two “extreme” options :1. “Conventional electric power generation” Mix fuel and oxygen at (adiabatic) temperature
T, and produce heat Q at temperature T. Fuel exergy Ex
exergy of hot gas + exergy loss Exergy of hot gas = (1-T0/T) Q
2. “Electric power generation” using fuel cells Separate the release of electrons from the fuel from the take-up of
electrons by oxygen, connect this, allow for the positive ions from the fuel to combine with the negative oxygen ions.
Fuel exergy Ex Electrical energy We + exergy losses Exergy loss due to heat production Qp : Ex = We + (1-T0/T) Qp
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 55/102
Fuel: area = exergy / T0 Product gases: calculate exergy from Tfurnace to T0
Exergy Losses in Combustion /2Quality Diagram for Furnace
Q
1/T
0
1/T0
1/Tfurnace
1/TstackFuel
Ex flue gasesEx fuelFlue gases
Pic:TUD92
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 56/102
Exergy input (fuel) = Exergy loss in combustion + Stack losses + Heat transfer loss to steam cycle + Exergy taken up by steam cycle
Exergy Losses in Combustion /3
Quality Diagram for Total System
Q
1/T
0
1/T0
1/Tfurnace
1/Tstack
Fuel
Ex flue gasesEx fuel
Combustionlosses
Exergy tosteam cycle
Stack losses
Pic:TUD92
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 57/102
Energy and Exergy Analysis-PFBC (Pressurised Fluidised Bed Combustion) : Sub-systems
Pic: ACA95
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 58/102
Energy and Exergy Analysis PFBC
Pics: ACA95
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 59/102
Endoreversibility /1
The maximum work output from a Carnot engineoperating on heat input Qin, between T=TH and T=T° is equal to Wmax = Qin(1-TH/TL), with isothermal transfer of heat Qin·T°/TH to the environment at T=T°.
However, isothermal heat transfer cannot exist; a driving force ΔT or, morecorrectly, Δ(1/T) is needed, which then gives losses
Qin·T°·Δ(1/T)
Qin
Qout
W
TH
T°
Carnotengineoperatingbetween TH and T°
Qin
Qout
W
THc
TLc
Endo-reversibleCarnotengineoperatingbetween THc and TLc
TH
T°Pics: SAKS04
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
60/102
Endoreversibility /2
If these are the only losses, the process is referred to as endoreversible.
If Qin→0, W→0, the endore-versible process → reversible with THC≈TH , TLC≈T°, W = 0
If Qin→Qin,max, temperatures THC≈ TLC and W → 0.
Maximum power Wmax is achieved for
at efficiency
H
o
optimum
HC
optimum
LC
T
T
T
Tη
Qin
Qout
W
THc
TLc
Endo-reversibleCarnotengineoperatingbetween THc and TLc
TH
T°
W
Qin
ηcycle
ηcarnot
Qin,max
Wmax
See also Curzon & Ahlhorn Am. J. Phys. 43 (1975) 22
H
o
optimum
HC
optimum
LC
TT
TT
Pics: SAKS04
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 61/102
1.6 Mixing Exergy
See SAKS04 §7.2, §6.3
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
62/102
Exergy of Mixing /1
Assuming two gases, each at temperature and pressure T0and p0, separated by a barrier;
Removing the barrier results in (diffusive) mixing, giving a homogeneous mixture in which the chemical potential(i.e. Gibbs energy) of each of the components decreases.
The exergy change is given by
For ideal mixing (no changes in p, T, total V, and activity coefficients γi = 1), which means also ΔHmix=0:
This is important for separation processes (e.g. distillation!)
)T,p(GΔSΔTHΔExΔmixmixmixmix
i
iimixi
iimixxlnxRTExΔxlnxRSΔ
µ(x) = µpure + RT ln (γ∙x)
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
63/102
Exergy of Mixing /2
Consider adiabaticmixing of cold and hot water, at T°=25°C, p=p° =1 atm
Ėxi = ṁi·((hi-h°) - T°(si-s°)), with enthalpy h°, entropys° for liquid water at T°, p°
p=constant; si-s°=cp· ln(Ti/T°) Water/steam table data
give the results as in the so-called Grassmannexergy flow diagram
Ėxlost=T°· (ṁ3s3 - ṁ2s2 - ṁ1s1)Pics: SAKS04
1
2 3
η = Ėxout / Ėxin = 0.216Ėxlost = 78.4 %!
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
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Exergy of Mixing /3
More general, for mixing of two molar streams n1, n2 with molar fractions xi in the original streams and molar fractions xjin the final stream:
iii
iii
jjjmix
xlnxnxlnxnxlnx)nn(RTExΔ
Source: S05
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku
65/102
Exergy of Mixing /4
For example: oxygen-rich (x=30%-vol O2) gas for a blast furnace is produced by mixing technical oxygen (x=95%-vol O2) and atmospheric air (x=21%-vol O2).
The consumption of technical oxygen per mole of product gas, nto/npg, follows from: 0.3 = 0.95·nto/npg + 0.21·(1- nto/npg) nto/npg= 0.122 mol/mol (= m3/m3)
for the exergy loss per mole of product gas
J/mol
.
).ln..ln.(.
).ln..ln.(.
.ln..ln.
.exΔmix
Source: S05
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 66/102
1.7 Heat Radiation Exergy
See S05 §1.4.6; §2.7
and P10
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 67/102
Thermal Radiation /1
The radiation QR (W) from a surface with emissivity ε (-) surface A (m2) and temperature T(K) equals
QR = ε· σ·A·T4
with Stefan-Boltzmann coefficientσ = 5.67×10-8 W/m2K4
For a blackbody surface– ε =1 in the Stefan-Boltzmann Law– all incident radiation is absorbed– radiation is maximum for its temperature at
each wavelength, λ (m)– the intensity of the emitted radiation is
independent of direction: it is a diffuse emitter
Pictures: T06
.
.
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 68/102
Thermal Radiation /2
For a blackbody, the entropy flux associated with thermalradiation is not, as might be expected, ṠR,BB= QR/T = σ·A·T3, but
ṠR,BB= (4/3)·σ·A·T3 = (4/3)· QR/T
For emitting and absorbing blackbody surfaces with equal surfaceAe = Aa that can ”see” each other completely (i.e. view factors Fa→e = Fe→a = 1) the entropy generation during heat radiation equals
a
e
ea
a
ae
eaBB,gen
T
TTTσ
T
TTTTσS
.
.
See also Petela, R., J. of Heat & Mass. Transf. Ser. C, 86 (1964) 187-192; and S05 section 2.7 Pic: WRSH02
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 69/102
Thermal Radiation /3
For gray surfaces, emissivity ε = ε(λ,T,θ), absorptivity α = α(λ,T,θ), reflectivity ρ = ρ(λ,T,θ), transmissivity τ = τ(λ,T,θ), are all 0 < .. < 1.
Energy balance α + ρ + τ = 1, and Kirchhoff’s law (from Second law)gives ε = α for gray bodies and not-too-large temperature differences.
The exergy of (blackbody) radiation, exBB, with respect to itsenergy content, eBB:
which does not depend on ε°= εenvironment, and can be used for gray bodies if ε ≠ ε(λ)
Pic: after KJ05
Incident angle θ with respect to normal
λ = wavelength; here of interest is the range 0.1 < λ < 100 µm
)T
T(
T
T
e
ex
BB
BB
See also Petela, R., J. of Heat & Mass. Transf. Ser. C, 86 (1964) 187-192and S05 section 2.7
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 70/102
Thermal Radiation /4
Recently (see e.g. W07) it was shown that for the entropy of gray radiation
Pics, source: W07To be continued in Part 2 Heat Radiation, Solar Energy
Net radiation from a black-body (BB) with incident blackbody radiation (BR)
Net radiation from a BB with incident BR, with gray-body radiation (GR)
Net radiation for incident GR on a blackbody
)εln()ε..(π
)ε(n,T
Qε)ε(nS
;)εln()ε..(π
Tσε)ε(Iπ
S)ε(IπT
QS
BB,R
GB,R
BR,R
BB,R
GB,R
where general in or
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 71/102
1.8 Chemical Exergy
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 72/102
Exergy of Chemical Substances /1
Chemical exergy gives the exergy of a substancewith respect to a reference environment. Note, again, that the environment is not a dead state at thermodynamicequilibrium, due to a constant influx of solar energy.
For each chemical element a reference species that contains, it is chosen, which gives the lowest exergylevel in nature (yet indeed appearing!), i.e. the mostcommon component of the environment (which canbe the air, seawater, the earth’s crust)
”Normal conditions” for the reference environment (see also slide 8):T° = 298.15 K, p°= 101.325 kPa, relative humidity 70%, atmosphericCO2 concentration 330 - 380 ppm, sea water salinity 3.5 %-wt, etc. Some data for air: N2 = 75.78 kPa; O2 = 20.39 kPa, Ar = 906 Pa, CO2 = 33.5 Pa, H2O = 2.2 kPa, He = 1.77 Pa, Ne, D2O, Kr, Xe < 1 Pa (SAKS04)
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 73/102
Exergy of Chemical Substances /2
Examples for reference species are atmospheric O2 and CO2for oxygen and carbon, CaCO3 for calcium, Fe2O3 for iron, etc. at 298.15 K, 101.325 kPa
The standard chemical exergy ex°chem = b°chemof a compound or element follows from an exergy balanceof a reversible standard reference reaction, where :
ttanreac
o
ttanreac,chemproduct
o
product,chem
o
r
o
chembbGΔb
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 74/102
Exergy of Chemical Substances /3
For example, for Ca+½O2+CO2 CaCO3, with -ΔrG° = 738.6 kJ/mol; for the elements (see Tables on following pages)the normal standard values for the chemical exergy for the referencesspecies are16.3 kJ/mol for CaCO3, 19.87 kJ/mol for CO2 and 3.97 kJ/mol for O2.
This gives for calcium, Ca:ex°chem = b°chem =-738.6 + 16.3 - 19.87 - 1½×3.97 = 729.1 kJ/mol
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 75/102
Exergy of Chemical Substances /4
For chemical compounds, the standard chemicalexergy can be more easily calculated from the reversible standard formation reaction, with ΔG = ΔfG°
where nelement is the number of moles of the element per mole of compound, and ex°chem, element = b°ch from tables.
elements
o
element,chemelement
o
f
o
chem
o
chemexnGΔexb
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 76/102
Exergy of Chemical Substances /5
For the constituents of air (N2, O2, Ar, CO2, ...), at T° = 298.15 K and the average pressure p°avg = 99.31 kPathe exergy value = 0.
Note that ideal gas law usually applies.
Thus, for a pure component at T°,p°, since ΔExseparate = ΔHseparate -T°·ΔSseparate = -T°·ΔSseparate
= T°·ΔSmix, and ΔS1,2 = -Rln(p2/p1)
gives E°chem,i = RT°ln(p°/pi) for these.
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 77/102
Standard Chemical Exergy of the Elements 1/5
γ = activity coefficient in seawater, m = molarity in seawater, p = %-vol, x = molar fraction
Source: S05
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 78/102
Standard Chemical Exergy of the Elements 2/5
Source: S05
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 79/102
Standard Chemical Exergy of the Elements 3/5
Source: S05
O
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 80/102
Standard Chemical Exergy of the Elements 4/5
Source: S05
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 81/102
Standard Chemical Exergy of the Elements 5/5
Remember the meaning of chemical exergy: it gives the maximum amount of work that can be made availablefrom a mole of the species in a given environmental surroundings
Source: S05
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 82/102
Standard exergies
for solids based on referencespecies dissolved in seawater
Source: S05
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 83/102
Chemical Exergies of Fuels
Ratio of the standard chemical exergy to the lower and higher heating value for several hydrocarbon fuels (HL = LHV; HH = HHV)
Source: S05
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 84/102
Example: Standard Chemical Exergy
Calculate the standard chemical exergy of larnite, Ca2SiO4, for which the energy of formation is -ΔfG° = 2191.6 kJ/mol.
b°chem = -2191.6 + 2×b°chem,elCa + 2×b°chem,elO2 + b°chem,elSi
= - 2191.6 + 2×729.1 + 2×3.97 + 854.9 = 129.4 kJ/mol
Tabelised data in S05 give b°chem = 95.7 kJ/mol .......
(mixing exergy cannot explain this)
Larnite
Pic
ture
: http
://w
ebm
iner
al.c
om/d
ata/
Larn
ite.s
htm
l
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 85/102
Example: Mineral Carbonation /1
(1) xMgO.ySiO2.zH2O(s) + (x-z)H2O …. xMg(OH)2(s) + ySiO2(s)
(2) Mg(OH)2 (s) + CO2 MgCO3 (s) + H2O
Chemical exergies of the chemical species are calculatedusing exchem(T, p) = exchem + exchem (T,pT,p) where
exchem(T,pT,p) = h(T,pT,p) - Ts(T,p T,p)
For liquids and solids: exchem(T,pT,p) = exchem(TT) = h(TT) - Ts(TT); and exchem(T, p) = exchem(T)
Standard exergies of formation are calculated using:
where nelement is the number of moles of the element in a mole of a certain compound, and e°chem, element from Tables
elements
oelementchemelement
of
ochem exnGex ,
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 86/102
Example: Mineral Carbonation /2 The exergy of CO2 as a function of temperature and pressure can
be calculated, with h = h(T) and s = s(T) (for ideal gas) as
exChem,CO2 (p,T) = ex°Chem,CO2+(h(T)-h°)- T°·(s(T)-s°)+RT ·ln(p/p°)
with reference ex°Chem,CO2 = RT° ln (p°/p°°) = 19.587 kJ/mol
using a reference concentration of 0.0375 %-vol of CO2 in the dry
atmosphere (p°° = 0.000375·p°)*.
R = 8.314 J/(mol·K), T = 298.15 K, p = 101.325 kPa. For all compounds “reaction exergies” can be calculated as
exergy differences between reactant and product exergies With entropy difference ΔS(T) for the carbonation reaction at
temperature T, the irreversibility or exergy loss is defined as
ΔExloss(T) = -ΔEx(T) = T°·ΔS(T)* In 2014, atmospheric CO2 concentration has risen to 0.0397 %-vol
ÅA 424304
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Example: Mineral Carbonation /3
Calculatedresultsusingdata from SMS88 and K95
taken from ZK04
ÅA 424304
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 88/102
Example: Mineral Carbonation /4
taken from ZK04
Irreversibility of the reaction of CO2 with several MgO-
containing species, per mol CO2. (1 bar)
-60
-50
-40
-30
-20
0 100 200 300 400 500Temperature °C
Irre
vers
ibili
ty k
J/m
ol C
O2
MgO + CO2(g) <=> MgCO3
Mg(OH)2 + CO2(g) <=> MgCO3 + H2O(g)
½ Mg2SiO4 + CO2(g) <=> MgCO3 + ½ SiO2
1/3 Mg3Si2O5(OH)4 + CO2(g) <=> MgCO3 + 2/3 SiO2 + 2/3 H2O
-20
0
20
40
60
80
100
0 100 200 300 400 500
Temperature °C
Ch
emic
al e
xerg
y o
f su
bst
ance
kJ/
mo
l M
g
MgOMg(OH)2Mg2SiO4Mg3Si2O5(OH)4MgCO3
Chemical exergies of MgO, Mg(OH)2, Mg2SiO4,
Mg3Si2O5(OH)4 and MgCO3, per mol Mg.
ÅA 424304
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1.9 Simplified Exergy Analysis With Chemical Reactions
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 90/102
Simplified Process Analysis /1 Exergy optimisation of processes with many heat streams
(in / out), giving an overall excess output:
From this it can be assessed if
indeed gives an exergy output (or if <0 : minimal losses...)
For processes with chemical conversions (if exotherm, ΔH < 0 Qout > 0, if endotherm ΔH > 0 Qin > 0), these typically willgive the largest exergy losses due to entropy changes ΔS ΔExloss = T°·ΔS, or = T°·∑ΔSi
The optimisation becomes, for a process with k chemicalconversion steps:and calculate the values for ΔSk and from that ΔExloss
Q)T
T()Q(Ex)Q(Ex)Q(Ex
ii,in
jj,out
with , max
0i
iinj
jout QQ ,,
kk
k
max)HΔ(T
T
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Simplified Process Analysis /2
This is very effective for finding the optimum temperature combinations for a process with severalreactions/reactors at different temperatures.
Example: the ÅA route for serpentinite carbonation
1. Magnesium extraction solid/solid ~ 400-450°C
– Mg3Si2O5(OH)4 + 3(NH4)2SO4 + heat ↔3MgSO4 + 2SiO2 + 5H2O(g) + 6NH3(g)
2. Mg(OH)2 production aqueous solution
– MgSO4 + 2NH4OH(aq) ↔(NH4)2SO4(aq) + Mg(OH)2
3. Mg(OH)2 carbonation gas/solid, 20-30 bar, ~500°C
– Mg(OH)2 (s) + CO2 (g) ↔MgCO3 (s) + H2O (g) + heat
ZFNRBH12
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Simplified Process Analysis /3
Example: the ÅA route for serpentinite carbonationGrassman diagram
B10
Exergy consumption5.54 GJ (~1.5 MWh)/ton CO2
ÅA VSTDI thesisT Björklöf2010
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 93/102
Simplified Process Analysis /4 Another approach: various temperatures and electrolysis.
BZ12
CO2
Serpentine “Serpentine” MgCl2CO2
Mg5(OH)2(CO3)4∙4H2O
Side prods.
NaCl
NaHCO3 CO2 ⑤
capture
Pre‐ ①
treatment
Electrolysis
②
HCl ③
production
Extraction
④
Carbonation
⑥
NaOH
Cl2, H2
HCl
T (°C)
ΔH (kJ/mol)
ΔG (kJ/mol)
ΔEx (kJ/mol)
R1 2 3 2 5 4 → 3 2 4 2 4 2 650 279.411 ‐248.034 267.3
R2 2 2 2 → 2 2 2 85 426.608 412.834 423.2
R3 2 2 → 2 25 ‐351.935 ‐254.468 ‐263.4
R4 4 2 4 → 2 2 2 2 2
6 3 2 5 4 → 3 2 2 2 5 2 or
70
‐236.137
‐280.632
‐195.95
‐252.296
‐15.0
‐12.0
R5 2 → 3 25 ‐68.865 ‐36.586 0.4
R6 5 2 10 3
→ 5 2 3 4 ∙ 4 2 10 6 2 55 488.753 ‐104.527 ‐6.7
ÅA 424304
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1.10 Exergy Analysis vs. PinchAnalysis
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Process Integration, Pinch /1
Process integration deals primarily with process energyand efficient design of heat exchanger networks(HENs)
Important for the system is the so-called pinch point; heat should not be ”transferred across the pinch”
Combining the hot streams and cold streams of a system into two composite curves (temperature versus enthalpy) shows the minimum temperature: the pinch
The pinch divides the system in two ”regions”
Picture: CR93
Example process with two hot streams to be cooled and two cold streams to be heated up.CP = stream heat capacity = m·cp (W/°C)with cp averaged between Tin and Tout
.
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Process Integration, Pinch /2
↑ Grid for the four-stream network Proposed heat exchanger network for ΔTmin = 10 K ↑
↑ Hot stream temperatures versus enthalphy Hot and cold stream composite curves ↑
Pictures: CR93
For the example
on previousslide
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Exergy Analysis vs. Pinch Analysis
It has been argued by Linnhoff (1987) that pinch analysis is superior to exergy analysis.
A study (1991) on the process integration of a new nitric acid plant, however, showed that exergy analysis allowed for energy savings 3× higher than what pinch analysis suggested.
Note, however, that pinch analysis is limited to heat exchanger network (HEN) optimisation, and then with certain restrictions, such as– so-called threshold problems (no pinch!)– integrating with heat pumps (where heat is transported
to higher temperatures) Source: S08
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1.11 Final Remarks
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Common-sense 2nd Law Guidelines
Taken from: Sama, 2008 (S08)
1 Do not use excessively large or excessively small thermodynamic driving forces in process operations.2 Minimize the mixing of streams with differences in temperature, pressure or chemical composition. 3 Do not discard heat at high temperatures to the ambient, or to cooling water. 4 Do not heat refrigerated streams with hot streams or with cooling water. 5 When choosing streams for heat exchange, try to match streams where the final temperature of one is close to the initial temperature of the other. 6 When exchanging heat between two streams, the exchange is more efficient if the flow heat capacities of the streams are similar. If there is a big difference between the two, consider splitting the stream with the larger flow heat capacity. 7 Minimize the use of intermediate heat transfer fluids when exchanging heat between two streams. 8 Heat (or refrigeration) is more valuable, the further its temperature is from the ambient 9 The economic optimal ΔT at a heat exchanger decreases as the temperature decreases, and vice versa, 10 Minimize the throttling of steam, or other gases. 11 The larger the mass flow, the larger the opportunity to save (or to waste) energy. 12 Use simplified exergy (or availability) consumption calculations as a guide to process modifications. 13 Some second law inefficiencies cannot be avoided; others can. Concentrate on those which can.
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Closing Remarks The First and Second laws of thermodynamics as
formulated by Baehr (B88):– 1. The sum of anergy and exergy is always
constant– 2. Anergy can never be converted into exergy
”Accepting exergy losses should always have someeconomic justification. If such a justification does not exist, it indicates that the exergy loss results only from an error in the art of engineering” (S05)
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 101/102
Sources ACA95: Pressurised Fluidised Bed Combustion, M. Alvarez Cuenca, E.J. Anthony, Black Acad. & Profess.,
Glasgow (1995) B88: Baehr, H.D. Thermodynamik Springer Verlag, Berlin (1988) B01: Bart, G.C.J. Advanced Thermodynamics (in Dutch) course compendium Delft Univ. of Technol., Delft
(2001) B97: Bejan, A. Advanced Engineering Thermodynamics John Wiley & Sons (1997) Chapter 3 B10: Björklöf, T. An energy efficiency study of carbon dioxide mineralization . MSc (Eng) thesis, ÅA (2010) BZ12: Björklöf, T., Zevenhoven, R. “Energy efficiency analysis of CO2 mineral sequestration in magnesium
silicate rock using electrochemical steps” Chem. Eng. Res. and Design 90 (2012) 1467-1472 C1824: Carnot, S., Reflections on the motive power of fire. Paris, 1824; Dover Publ. Mineola (NY) 1980 CB98: Çengel, Y.A., Boles, M.A. Thermodynamics. An Engineering Approach, McGraw-Hill (1998) CR93: Sinnott, R.K. ”Coulson & Richardson’s Chemical Engineering”, vol. 6, 2nd ed., Pergamon Press
(1993) dWetal08: DeWulf, J. et al., ”Exergy: its potential and limitations in environmental science and technology”
Env. Sci. & Technol. 42(7) (2008) 2221-2232 FÖ97: Finnveden, G., Östlund, P. ”Exergies of natural resources in life-cycle assessment and other
applications” Energy - The International Journal 22(9) (1997) 923-931 H84: Hoogendoorn, C.J. Advanced Thermodynamics (in Dutch) course compendium Delft Univ. of
Technol., Delft (1984) K41: Keenan, J.H. Thermodynamics, MIT press (1941) Chapter 17 KJ05: D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids Engineering”, Wiley (2005) K95: Kotas, T.J. The Exergy Method of Thermal Plant Analysis. Krieger Publ. Co., Malabar (FL) (1995) P03: Petela, R. ”Exergy of undiluted thermal radiation” Solar energy 74 (2003) 469-488 P10: Petela, R. Engineering Thermodynamics of Thermal Radiation” McGraw-Hill (2010) RG06: Rivero, R., Garfias, M. ”Standard chemical exergy of elements updated” Energy - The International
Journal 31 (2006) 3310-3326
11 jan 15Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 102/102
Sources (cont’d) S08: Sama, D. ”Thermodynamic insights; an unusually stubborn attempt to think clearly”, Proc. of
ECOS2008, Kraków-Gliwice, Poland, June 2008, 19-40 SAKS04: de Swaan Arons, J., van der Kooi, H., Sankarana-rayanan, K. Efficiency and
Sustainability in the Efficiency and Chemical Industries. Marcel Dekker, New York 2004 S05: Szargut, J. Exergy Method. WIT Press (2005) SMS88: Szargut, J., Morris, D., Steward, F.R. Exergy Analysis of Thermal, Chemical and
Metallurgical Processes. Hemisphere Publishing Co, New York, (1988) SW07: Sciubba, E., Wall G. ”A brief commented history of exergy from the beginnings to 2004”. Int. J.
Thermodyn., 10(1) (2007), 1-26. T06: S.R. Turns ”Thermal – Fluid Sciences”, Cambridge Univ. Press (2006) vL77: Van Lier, J.J.C. Thermodynamic processes in the power plant and possiblities to improve these
processes, (in Dutch) course compendium Delft Univ. of Technol., Delft (1977) vS91: von Schalien R. Teknisk termodynamik och modellering, 6. ed, Åbo Akademi Univ. (1991) TUD92: Energy and exergy, Symposium Delft Univ. of Technol., Delft, Nov. 3,1992 (in Dutch) WRSH02: Wright, S.E., Rosen, M.A., Scott, D.C., Haddow, J.B. ”The exergy flux of radiative heat transfer
for the special case of heat radiation”, Exergy 2 (2002) 24-33 W07: Wright, S. ”Comparative analysis of entropy of radiative heat transfer” Int. J. Thermodyn., 10(1)
(2007), 27-35 Z99: Zevenhoven, R. ”Advanced combustion and gasification technology” course material ENY-47.210
Helsinki Univ. of Technol., Espoo (1999-2004) ZK04: Zevenhoven, R., Kavaliauskaite, I. ”Mineral carbonation for long-term CO2 storage: an exergy
analysis” Int. J. Thermodyn., 7(1) (2004), 23-31. ZHPSS07: Zevenhoven, R., Helle, H., Pettersson, F., Söderman, J., Saxén H., Heat optimisation of recycled
oil production from waste oils and oily waste waters. Proc. of ECOS'2007, Padova, Italy, June 2007, 839-846
ZFNRBH12: “Carbon storage by mineralisation (CSM): serpentinite rock carbonation via Mg(OH)2reaction intermediate without CO2 pre-separation” Zevenhoven, R., Fagerlund, J., Nduagu, E, Romão, I, Bu, J, Highfield, J, presented at GHGT-11, Kyoto Japan, Nov. 18-22, 2012