Exergy Analysis
Process EngineeringThermodynamics course # 424304.0 v. 2015
Exergy Analysis
Commented / edited by Ron Zevenhoven Prof. Özer Arnas, visiting ÅA VST May/June 2014 from the US Military Academy Åbo Akademi University Thermal and Flow Engineering Laboratory / Värme- och strömningsteknik tel. 3223 ; [email protected] course material: http://users.abo.fi/rzevenho/kursRZ.html
ÅA 424304
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 2/102 Energy Conversion /1
. In a hydraulic power plant, the potential energy difference can be completely converted into work . For a thermal plant Carnot (C1824) discovered that only a portion of heat Q
at temperature T1 can be converted into work, with cold T T T Pic: S05 W Q Q sink temperature T : 2 T T
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 3/102
Energy Conversion /2
. The preferable way to convert heat Q at
T1 into work is to use a natural, unlimited cold sink, such as the natural environment at
temperature T0.
. With T2 > T0 no full use of Q is made; T T T Pic: S05 for T2 < T0 a cooling W Q Q max effort is needed. T T
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 4/102 Exergy /1 . EXERGY quantifies how much of a certain amount of energy contained in an energy
source can be extracted as useful work. jpg . The potential of (natural) energy resources for driving a mechanical, thermal, chemical or biological process depends on the deviation from thermodynamic equilibrium with the surrounding environment. . This allows evaluation of the ability to perform maximum work under certain environmental conditions. . This involves quality assessments, for example for heat Q its temperature
T, relating it to temperature T0 of the
surrounding environment. Picture:http://cpcclibraries.files.wordpress.com/2008/04/earth.
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 5/102
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 b1/Dead_possum.jpg/800px-Dead_possum.jpg – 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. org/wikipedia/commons/thumb/b/ (And no electrical, magnetic etc. effects.) . This requires a description of the state and, if chemistry is involved, the composition of the local environment for the prevailing chemical species. Picture: http://upload.wikimedia. Picture:
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 6/102 Exergy /3 . While the quantities of energy allow for ”bookkeeping” with the help of the First Law of Thermodynamics (”energy cannot be produced or destroyed”); the exergy concept follows from the Second Law of Thermodynamics, which involves the quality of energy and its degradation during energy 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) –seep. 9
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 7/102
Surroundings - Environment
. Some care should be considered with respect to ”the surroundings”, which is everything outside the system’s boundaries. 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) 39.7 Pa would be better – see co2now.org
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 8/102 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)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 9/102
ÅA 424304
1.2 Reversible Work-Irreversibility
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 10/102 Exergy and Work Potential . Thus: a system will deliver the maximum possible work in a reversible process from a specified initial state to the state of 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 Laws of 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 large seasonal or daily changes (day/night). . Unlike energy, exergy is consumed or destroyed.
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 11/10 2
Surroundings Work
. During the expansion of a work W producing closed system,
with volume change ΔV = Vfinal –Vstart > 0, work Ws is done on the surroundings at pressure p°: W = p°·ΔV Note: here W > 0 defined s as work done by the system . Similarly, ΔV < 0 is possible. In many cases this surroundings
work 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)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 12/102 Reversible Work
. 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 ”lost opportunity” to do work. . If I = 0, no entropy is generated. . For work consuming devices, W ≤ W rev u Pic: ÇB98
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 13/102
Example: Irreversibility 1/1
. A heat engine (HE) receives Qin = 500 kW heat at TH = 1200 K and rejects waste heat at TL = 300 K. The power output of the engine is 180 kW.
Calculate the reversible power Wrev and irreversibility rate I, in kW.
. Wrev follows from the Carnot efficiency ηcarnot = 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 Pic: ÇB98 rejected to the sink was not available for work
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 14/102 Example: Irreversibility 1/2
. The 125 kW rejected to the sink follows from Q 500 kW Carnot efficiency; this is T 1200 K the anergy part of the in 417 J/(K s) input energy: energy = anergy + exergy 500 kW = 125 kW + 375 kW Q 125 kW . The entropy part of the T out 300 K energy cannot be 417 J/(K s) converted into useful work. Pic: ÇB98
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 15/102 Turku
Example: Irreversibility 2/1
. A block of iron (m = 500 kg) is cooled
from 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):
TL T δWrev ηc δQin ( ) δQin ( ) δQin TH T
where δQin m c avg dT which gives
T T W δW m c (T T ) T ln 8191 kJ rev rev avg T T Pic: ÇB98
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 16/102 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 iron block which equals 38925 kJ.
. The 8191 kJ calculated as Wrev could have been converted into useful work
. Since no work was obtained, Wu = 0 and the irreversibility therefore equals
I = Wrev = 8191 kJ (21% of Qin) . . The remaining 38925 - 8191 = 30734 kJ
(79% of Qin) is rejected also to the surroundings. Pic: ÇB98
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 17/102
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 Rule of thumb: large temperature differences irreversible processes are – Unrestrained expansion; throttling processes that look absurd when run backwards – 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 !
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 18/102 Second-Law Efficiency, η2, Revisited
. Second-law efficiency is defined as: actual thermal efficiency η η maximum possible thermal efficiency ηcarnot
. Equally, it can be defined as (for work producing devices):
useful work Wu I η maximum possible useful work Wrev Wrev for work done (for work -power- consumed η2 = Wrev / Wu, input ) with irreversibility, I, or, expressed as exergies: exergy recovered, i.e. made use of exergy destroyed η exergy supplied exergy supplied
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 19/102 Turku
Example: Efficiency
. A house, like many in Finland, is electrically heated, using an electric resistance heating element 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 heat output COP which here is equal to COP = 1. heater work input Thigh . For the most efficient, Carnot, process: COPheater Thigh Tlow with result COPrev = 294/(294-283) = 26.7.
. Thus, the Second-law efficiency η2 = COP/COPrev = 0.037 = 3.7%
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 20/102 ÅA 424304
1.3 Physical Exergy
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 21/102
Exergy of ”Non-Heat” Energy
2 . Kinetic energy Ek = ½m(v-vref) 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 22/102 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 T 0 Ex(Q ) (1 ) Q Carnot factor Q T with of course T’s in K!!
. As noted above: the entropy part Sadi Carnot of energy (anergy), for heat Q equal 1796-1832 to Q/T, cannot be converted into work. see C1824
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 23/102
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: T,p,V !!! Ex(U)=T°,V°,p°∫ dU+p°dV-T°dS=(U-U°)+p°(V-V°)-T°(S-S°)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 24/102 Exergy of Enthalpy
. For enthalpy, H = U + pV Thus, 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 25/102
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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 26/102 Example: Internal Energy Exergy /1
. Calculate the maximum amount of work to be obtained from expanding 3 200 m 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 closed system 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 = Mp V /RT ≈ 2323 kg. 1 1 1 Pic: ÇB98
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 27/102
Example: Internal Energy Exergy /2
. No kinetic energy, no potential energy to 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°) = 0
Ex(u1) = 120.76 kJ/kg; Ex(U1) = m· ex(u1) = 280.53 MJ Pic: ÇB98
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 28/102 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°
For those who use h,s diagrams
Pic: B01
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 29/102
Open System Exergy, h-s diagram /2 . The energy balance for this system gives:
wmax = h1 –h2 + q = h1 –h2 - T°·(s2 –s1) , and thus for 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°
T0
Pics: B01
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 30/102 ÅA 424304
1.4 Exergy Destruction; Isolated-Closed-Open Systems
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 31/102
Exergy Destruction /1 . Consider an isolated system: no energy or mass transfer 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 ≤ 0
with 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)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 32/102 Exergy Destruction /2 . Considering again an open system as in the figure , for a reversible process
wrev = h1 -h2 + q°, where q°= T°(s2-s1), while for an irreversible process Pic: B01
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 destruction
rate is equal to -ΔĖx = T°·Ṡgen ≥ 0
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 33/102
Exergy Destruction /3 . For a closed system, which does not involve mass flows a general balance equation reads:
Exheat + Exwork –Exdestroyed = ΔExsystem,
where Exdestroyed = T°·Sgen ≥ 0
which gives, for several heat (streams) Qi : 0 T org.uk/images/radiation.jpg 1 Q (W p (V V )) T S Ex Ex i 0 2 1 0 gen 2 1 i Ti T 0 dV dEx or as rate : 1 Q (W p ) T S i 0 0 gen i Ti dt dt
or with non constant Ti for Qi : 2 T 0 1 Q (W p (V V )) T S Ex Ex i 0 2 1 0 gen 2 1 T http://www.mikecurtis. Picture: i 1 i
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 34/102 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 ≥ 0
which gives, for several heat (streams) Qi and incoming and outgoing enthalpy streams with exergy ψj, ψk: T 0 1 Q (W p (V V )) m m T S Ex Ex i 0 2 1 j j k k 0 gen 2 1 i Ti j k T 0 dV dEx or as rate : 1 Q (W p ) m m T S i 0 j j k k 0 gen i Ti dt j k dt
or with non constant Ti for Qi : 2 T 0 1 Q (W p (V V )) m m T S Ex Ex T i 0 2 1 j j k k 0 gen 2 1 i 1 i j k
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 35/102
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 Pic: requirement for this process. ÇB98
. 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.
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 36/102 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 Pic: . T°(s2-s1) = -RT°(ln(p°/p2)) ÇB98
. 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 see also slides 27 – 28...
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 37/102
ÅA 424304
1.5 Physical Exergy: Applications heat exchangers, tube flow, combustion, endoreversibility
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 38/102 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 increases Q
from Q to Q . T
T T
. The entropy increase ΔS = Sgen is equal to
Q Q Pic: S Q gen ÇB98 T T T T
. And the exergy losses are equal
to –ΔEx = T°·Sgen
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 39/102 Turku
Heat Transfer Efficiency (PTG course)
. A simple steady-state heat . Thermodynamic analysis transfer process; heat is Energy balance transported from medium 1 Q Q to medium 2 by conduction Entropy balance through a material that Q Q separates them. S gen T T . Temperature T1 > T2 T T S Q Q gen T T T T . . This shows that Sgen is large for large temperature . . differences (T -T ) and low Q1 Q2 1 2 See also temperatures T1 and T2 T = T1 T = T2 vS91
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 40/102 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
Q Source: 1/T0 TUD92
3 4
4 2 2 1 1/T 1 3
1/T = 0
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 41/102
Heat Exchanger Exergy Losses /1
. A heat exchanger involves two media at temperatures TCin and 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: > 0 < 0 TCout T TCout T T Ex (1 0 )Q (1 0 )m c dT m c (T T ) T ln Cout C c pC C pC Cout Cin 0 TCin T TCin T TCin < 0 > 0 THout T THout T T Ex (1 0 )Q (1 0 )m c dT m c (T T ) T ln Hout H c pH H pH Hout Hin 0 THin T THin T THin . and the exergy losses are then equal to < 0 > 0
THout TCout ΔE x (ΔE x C ΔE xH ) T (m Hc pH ln m Cc pC ln ) THin TCin with specific heats cpH, cpC for the hot and cold streams.
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 42/102 Turku 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 e_heat_exchanger_1-pass.png Q Q ηHX (%) Q T ΔS Q ΔE x . In cases where electric
power Pelec is (for example) used on the hot side:
ΔE x Pelec T m c T ln Cout (T T ) C pC Cout Cin TCin http://oldtacomamarine.files.wordpress.com/2008/01/straight-tub
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 43/102
Heat Exchanger Exergy Losses /3
. Heat exhanger exergy analysis shows that the temperature difference 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 perform much better than co-current heat exchangers. . Ideally, the flows aquire each others temperature: the exergy losses will then be zero. For this, the heat capacities
ṁ·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)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 44/102 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 inside and outside wall temperatures are 20°C and 5°C, respectively. Calculate the heat flow Q and the exergy destruction rate –ΔEx in W. . Q = -λ(ΔT /d ) = 1035 W Pic: wall wall ÇB98 . - 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 45/102
A Process Example . Bilge water (waste
oil/water mixtures) Cold Heated storage heat Hot water processing for recycled storage
fuel oil (in Turku) Raw material . Significant savings could heat Hot water Fuel oil be obtained by Water disposal – Replacing the heating heat Heating centre and, more center importantly the electric heat Electricity heating (!) by heat from Product the local district heat Surfactant network Product storage Separator – Making use of hot raw Source: heat Sludge material deliveries Hot water ZHPSS07
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 46/102 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 Δ Ex Q T loss which suggests a thermodynamic TL TH driving force equal to Δ(1/T) instead, and . the exergy losses are the product of three factors, being
1) surroundings temperature T0, 2) the heat flow Q, and 3) the driving force Δ(1/T).
See later. As irreversible, or non-equilibrium thermodynamics part of this shows: Ṡgen = ∑ Ji·Xi for flows Ji and driving forces Xi course
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 47/102
Exergy Losses and Tube Flow /1
. For fluid flow V (m³/s) through a tube, viscous friction results in pressure drop Δp (Pa), which must be compensated for by pumping power W p V . 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·V - Δp·V ·(1-T°/T) = Δp·V ·T°/T = mechanical dissipation × loss factor T°/T; or Δėx= (Δp/ρ)·T°/T . Note that ΔĖx> Δp·V if T < T°. (No cooling for free!) Note: 4ƒ = ζ = Blasius, Darcy friction factor -τ w ƒ = Fanning friction factor p p1 R 2 r For pressure drop Δp: A (p1 -p2 ) = -Δp = τw∙L·(4 /Dh) x 2 = 4ƒ∙½∙ρ∙
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 48/102 Turku Exergy Losses and Tube Flow /2
. The pressure drop can be related to the fluid flow with cross sectional averaged velocity
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 tube length x: δw/dx = (δW /ṁ)/dx = (-dp/dx)/ρ = cp·dT/dx, which is equal to 0,25 1,75 w 1.79 m for fluid density ρ, dynamic viscosity η, dx 1,75 2 D4,75 mass flow ṁ.
-τw p p1 R 2 r A x S L 1 2 11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 49/102
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 exchanger tube) the heat exchange per meter with the environment is: q Q U(T T 0)D dx m dx m which gives for the total exergy losses per length section dx: dex 1.790,25m1,75 T 0 U(T T 0)2D dx 1,75 2D4,75 T m T which shows that losses due to pressure drop decrease with diameter, while those for heat losses increase with diameter !
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 50/102 Turku Exergy Losses and Tube Flow /4
For a given optimal value for the diameter,
Doptimal , the exergy losses are minimal:
η, m. T D, . optimal U ρ (T T )
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 51/102 Turku
Energy Conversion Stages in “conventional” Electricity Production, i.e. condensing power plant
. Energy in chemical bonds of fuel . Sources of exergy loss: Furnace . Friction in flow lines, pressurising and de- . Thermal energy of hot flue gas pressurising Heat transfer to steam cycle . Heat transfer at low . Thermal energy of water / steam temperatures and/or across large temperature Steam cycle generator differences . Electrical energy . Conversion of chemical (potential) energy into thermal energy (heat)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 52/102 Exergy Loss in Expansion Turbine
Pic:vL77 . 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 1
Transformation of Work into Heat 2
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 53/102
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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 54/102 Exergy Losses in Combustion /2 Quality Diagram for Furnace
. Fuel: area = exergy / T0 . Product gases: calculate exergy from Tfurnace to T0 Q 0
1/Tfurnace
1/T Pic: TUD92
Fuel 1/Tstack 1/T 0 Ex flue gases Flue gases Ex fuel
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 55/102
Exergy Losses in Combustion /3 Quality Diagram for Total System
. Exergy input (fuel) = Exergy loss in combustion + Stack losses + Heat transfer loss to steam cycle + Exergy taken up by steam cycle Fuel Q 0 1/T Combustion furnace losses Pic: 1/T TUD92
Exergy to 1/Tstack steam cycle 1/T 0 Ex flue gases Stack losses Ex fuel
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 56/102 ÅA 424304 Energy and Exergy Analysis-PFBC (Pressurised Fluidised Bed Combustion) : Sub-systems
Pic: ACA95
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 57/102
ÅA 424304 Energy and Exergy Analysis PFBC
Pics: ACA95
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 58/102 Endoreversibility /1 Qin . The maximum work output Carnot from a Carnot engine TH engine operating on heat input Qin, operating W between T=TH and T=T° is between T and T° equal to W = Q (1-T /T ), H max in H L T° with isothermal transfer of Qout heat Qin·T°/TH to the environment at T=T°. Endo- T H Qin . However, isothermal heat reversible transfer cannot exist; a Carnot THc driving force ΔT or, more engine W operating correctly, Δ(1/T) is needed, between TLc which then gives losses THc and TLc Qout Qin·T°·Δ(1/T) Pics: SAKS04 T°
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 59/102
Endoreversibility /2 See also Curzon & Ahlhorn Am. J. Phys. 43 (1975) 22 Endo- T . If these are the only losses, H Qin the process is referred to as reversible Carnot THc endoreversible. engine W operating . If Qin→0, W→0, the endore- between versible process → reversible TLc T and T with T ≈T , T ≈T°, W = 0 Hc Lc HC H LC Qout T° . If Qin→Qin,max, temperatures THC ≈ TLC and W → 0. . Maximum power W is max ηcarnot achieved for Toptimum T o LC W ηcycle optimum THC TH W at efficiency T optimum T o max η LC Pics: SAKS04 optimum Q THC TH in,max Qin 11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 60/102 Turku ÅA 424304
1.6 Mixing Exergy
See SAKS04 §7.2, §6.3
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 61/102
Exergy of Mixing /1
. Assuming two gases, each at temperature and pressure T0 and 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 µ(x) = µpure + RT ln (γ∙x)
ΔExmix ΔHmix T ΔSmix ΔGmix (p,T )
. For ideal mixing (no changes in p, T, total V, and activity
coefficients γi = 1), which means also ΔHmix=0:
ΔSmix R xi lnxi ΔExmix RT xi lnxi i i . This is important for separation processes (e.g. distillation!)
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 62/102 Turku Exergy of Mixing /2
. Consider adiabatic mixing of cold and hot 1 water, at T°=25°C, p=p° =1 atm 2 3
. Ėxi = ṁi·((hi-h°) - T°(si-s°)), with enthalpy h°, entropy s° 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 Grassmann exergy flow diagram η= Ėxout / Ėxin = 0.216 . Ėxlost=T°· (ṁ3s3 - ṁ2s2 - ṁ1s1) Ėx = 78.4 %! lost Pics: SAKS04
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 63/102 Turku
/3 Exergy of Mixing Source: S05
. More general, for mixing of two molar streams n1, n2 with molar fractions xi in the original streams and molar fractions xj in the final stream:
ΔExmix RT (n n ) x j lnx j n xi lnxi n xi lnxi j i i
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 64/102 Turku /4 Exergy of Mixing Source: S05
. 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:
3 3 0.3 = 0.95·nto/npg + 0.21·(1- nto/npg) nto/npg= 0.122 mol/mol (= m /m )
. for the exergy loss per mole of product gas
.ln. .ln. Δex mix . . (. ln. . ln.) . J/mol . (.ln. . ln.)
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 65/102 Turku
ÅA 424304
1.7 Heat Radiation Exergy
See S05 §1.4.6; §2.7
and P10
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 66/102 Thermal Radiation /1
. . The radiation QR (W) from a surface with emissivity ε (-) surface A (m2) and temperature T(K) equals . 4 QR = ε· σ·A·T with Stefan-Boltzmann coefficient σ = 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 67/102
Thermal Radiation /2
. For a blackbody, the entropy flux associated with. thermal 3 radiation is not, as might be expected, ṠR,BB= QR/T = σ·A·T , but . 3 ṠR,BB= (4/3)·σ·A·T = (4/3)· QR/T
. For emitting and absorbing blackbody surfaces with equal surface Ae = Aa that can ”see” each other completely (i.e. view factors Fa→e = Fe→a = 1) the entropy generation during heat radiation equals
T T e a Sgen,BB σ Ta Te Ta
Te σ Ta Te Ta
See also Petela, R., J. of Heat & Mass. Transf. Ser. C, 86 (1964) 187-192; and S05 section 2.7 Pic: WRSH02
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 68/102 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 Incident angle θ with respect 0 < .. < 1. to normal . Energy balance α + ρ + τ = 1, and Pic: after Kirchhoff’s law (from Second law) KJ05 gives ε = α for gray bodies and not- λ = wavelength; here of interest too-large temperature differences. is the range 0.1 < λ < 100 µm
. The exergy of (blackbody) radiation, exBB, with respect to its energy content, eBB: ex T T BB ( ) e T T See also Petela, R., J. of BB Heat & Mass. Transf. Ser. which does not depend on ε°= εenvironment, C, 86 (1964) 187-192 and can be used for gray bodies if ε ≠ ε(λ) and S05 section 2.7
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 69/102
Thermal Radiation /4
. Recently (see e.g. W07) it was shown that for the entropy of gray radiation
Q R,BB S I(ε) S I(ε) σεT (. .ε)ln(ε); R,GB T π R,BR π π
Q R,BB or in general S n(ε) ε , where n(ε) (. .ε)ln(ε) R,GB T π
Net radiation from a black- Net radiation from a BB Net radiation for incident body (BB) with incident with incident BR, with GR on a blackbody blackbody radiation (BR) gray-body radiation (GR) To be continued in Part 2 Heat Radiation, Solar Energy Pics, source: W07
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 70/102 ÅA 424304
1.8 Chemical Exergy
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 71/102
Exergy of Chemical Substances /1 . Chemical exergy gives the exergy of a substance with respect to a reference environment. Note, again, that the environment is not a dead state at thermodynamic equilibrium, due to a constant influx of solar energy. . For each chemical element a reference species that contains, it is chosen, which gives the lowest exergy level in nature (yet indeed appearing!), i.e. the most common component of the environment (which can be 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%, atmospheric
CO2 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)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 72/102 Exergy of Chemical Substances /2
. Examples for reference species are atmospheric O2 and CO2 for oxygen and carbon, CaCO3 for calcium, Fe2O3 for iron, etc. at 298.15 K, 101.325 kPa
. The standard chemical exergy ex°chem = b°chem of a compound or element follows from an exergy balance of a reversible standard reference reaction, where :
o o o o bchem ΔrG bchem,product bchem,reac tan t product reac tan t
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 73/102
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 references species are
16.3 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 74/102 Exergy of Chemical Substances /4
. For chemical compounds, the standard chemical exergy can be more easily calculated from the reversible
standard formation reaction, with ΔG = ΔfG°
o o o o bchem ex chem Δ f G nelement ex chem,element elements
where nelement is the number of moles of the element per mole of compound, and ex°chem, element = b°ch from tables.
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 75/102
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 kPa the 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 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 76/102 Standard Chemical Exergy of the Elements 1/5 Source: S05
γ = activity coefficient in seawater, m = molarity in seawater, p = %-vol, x = molar fraction
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 77/102
Standard Chemical Exergy of the Elements 2/5 Source: S05
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 78/102 ÅA 424304 Standard Chemical Exergy of the Elements 3/5 Source: S05
O
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 79/102
ÅA 424304 Standard Chemical Exergy of the Elements 4/5 Source: S05
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 80/102 ÅA 424304 Standard Chemical Exergy of the Elements 5/5 Source: S05
Remember the meaning of chemical exergy: it gives the maximum amount of work that can be made available from a mole of the species in a given environmental surroundings
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 81/102
Standard exergies for solids based on reference species dissolved in seawater
Source: S05
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 82/102 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 83/102
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 ...... om/data/Larnite.shtml om/data/Larnite.shtml (mixing exergy cannot explain this)
Larnite Picture: http://webmineral.c
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 84/102 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 calculated
using exchem(T, p) = exchem + exchem (T,pT,p) where
exchem(T,pT,p) = h(T,pT,p) - Ts(T,pT,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: o o o exchem fG nelement exchem,element elements where nelement is the number of moles of the element in a mole of a certain compound, and e°chem, element from Tables
11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 85/102
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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 86/102 ÅA 424304 Example: Mineral Carbonation /3
Calculated results using data from SMS88 and K95
taken from ZK04
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 87/102
ÅA 424304 Example: Mineral Carbonation /4
-20 MgO + CO2(g) <=> MgCO3 Mg(OH)2 + CO2(g) <=> MgCO3 + H2O(g) ½ Mg2SiO4 + CO2(g) <=> MgCO3 + ½ SiO2
2 -30 1/3 Mg3Si2O5(OH)4 + CO2(g) <=> MgCO3 + 2/3 SiO2 + 2/3 H2O
-40
Chemical exergies of MgO,
Irreversibility kJ/mol CO -50 Mg(OH)2, Mg2SiO4, Mg Si O (OH) and MgCO , -60 3 2 5 4 3 0 100 200 300 400 500 per mol Mg. Temperature °C 100 MgO Irreversibility of the reaction Mg(OH)2 80 Mg2SiO4 of CO2 with several MgO- Mg3Si2O5(OH)4 MgCO3 containing species, per mol 60
CO2. (1 bar) 40 kJ/mol Mg kJ/mol 20
Chemical exergy of substance 0
-20 0 100 200 300 400 500 taken from ZK04 Temperature °C
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 88/102 ÅA 424304
1.9 Simplified Exergy Analysis With Chemical Reactions
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 89/102
Simplified Process Analysis /1 . Exergy optimisation of processes with many heat streams (in / out), giving an overall excess output:
T Ex(Qout,j ) Ex(Qin,i ) max , with Ex(Q) ( )Q j i T . From this it can be assessed if Qout ,j Qin,i 0 j i 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 will give the largest exergy losses due to entropy changes ΔS
ΔExloss = T°·ΔS, or = T°·∑ΔSi . The optimisation becomes, for a process with k chemical conversion steps: T (ΔHk ) max and calculate the values for k Tk ΔSk and from that ΔExloss 11 jan 15 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 90/102 Simplified Process Analysis /2 . This is very effective for finding the optimum temperature combinations for a process with several reactions/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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 91/102
Simplified Process Analysis /3 B10 . Example: the ÅA route for serpentinite carbonation Grassman diagram
Exergy consumption
5.54 GJ (~1.5 MWh)/ton CO2
ÅA VST DI thesis T Björklöf 2010
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 92/102 Simplified Process Analysis /4 . Another approach: various temperatures and electrolysis.
NaOH
HCl ③ Cl2, H2 Electrolysis production ②
CO2 CO2 ⑤ NaHCO3 capture NaCl HCl
CO2 Serpentine Pre‐ ① “Serpentine” Extraction MgCl2 Carbonation treatment ④ ⑥
Mg5(OH)2(CO3)4∙4H2O Side prods. BZ12
T ΔH ΔG ΔEx (°C) (kJ/mol) (kJ/mol) (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 ‐236.137 ‐195.95 ‐15.0 4 2 4 →2 2 2 2 2 R4 or 70 6 3 2 5 4 → 3 2 2 2 5 2 ‐280.632 ‐252.296 ‐12.0
R5 2 → 3 25 ‐68.865 ‐36.586 0.4
5 10 R6 2 3 55 488.753 ‐104.527 ‐6.7 → 5 2 3 4 ∙4 2 10 6 2
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 93/102
ÅA 424304
1.10 Exergy Analysis vs. Pinch Analysis
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 94/102 Process Integration, Pinch /1
. Process integration deals primarily with process energy and 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 Example process with two hot streams to minimum temperature: the be cooled and two cold streams to be pinch heated up. . . The pinch divides the system CP = stream heat capacity = m·cp (W/°C) in two ”regions” with cp averaged between Tin and Tout Picture: CR93
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 95/102
Process Integration, Pinch /2
For the example on previous slide
↑ Hot stream temperatures versus enthalphy Hot and cold stream composite curves ↑
↑ Grid for the four-stream network Proposed heat exchanger network for ΔTmin = 10 K ↑ Pictures: CR93
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 96/102 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 97/102
ÅA 424304
1.11 Final Remarks
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 98/102 Common-sense 2nd Law Guidelines 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. Taken from: Sama, 2008 (S08)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 99/102
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 some economic 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)
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 100/102 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 101/102
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)2 reaction 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
Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 11 jan 15 102/102