Exergy Analysis - Åbo Akademi | Startsidausers.abo.fi/rzevenho/APT17-1-XRG.pdf · Exergy Analysis...

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

Advanced Process Thermodynamicscourse # 424520.0 (5 sp) v. 2017

Commented / edited byProf. Özer Arnas,visiting ÅA VST May/June 2014from the US Military Academy

ÅA 424520

8 mar 17Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 2/108

1.1 Exergy vs. Energy

8 mar 17 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 3/108

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


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


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 depends on the deviation from thermodynamic equilibriumwith 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

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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 surroundingenvironment: 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.

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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)


Surroundings - Environment

40.6 Pa would be better – see co2now.org


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 424520


1.2 Reversible Work-Irreversibility


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. (Thanatia: see VV15)

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.


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 ΔV(for example: combustion furnaces, engines)

Note: here W > 0 is definedas work done by the system


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: İ = Wrev - Wu

which is equal to the amount of exergy that is destroyed.

The irreversibility represents ”lostopportunity” to do work.

If İ = 0, no entropy is generated.

For work consuming devices, Wrev ≤ Wu

Reversible Work

Pic: ÇB98

.


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 İ, 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 İ = Wrev - Wu = 195 kW

Note that the 500 kW - 375 kW = 125 kW rejected to the sink was not available for work

Pic: ÇB98

!

8 mar 17 Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku

15/108


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

!



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 İ, 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

!



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 equalsİ = Wrev = 8191 kJ (21% of Qin)

The remaining 38925 - 8191 = 30734 kJ (79% of Qin) is rejected also to the surroundings.

Pic: ÇB98

!


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


19/108

Second-Law Efficiency, η2, Revisited

Second-law efficiency is usually 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, İ, or, expressed as exergies:

carnotη

ηη efficiency thermal possible maximum

efficiency thermal actual

revrev

u

WW

W I

workuseful possible maximum

workuseful 12

supplied exergy

destroyed exergy

supplied exergy

of use made i.e. recovered, exergyη


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 424520


1.3 Physical Exergy


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 completelyand reversibly converted into work Ex(Ep) = Ep

Electric energy Eelec = i·V·t (current × voltage × time)

can** be completely and reversibly converted intowork Ex(Eelec) = Eelec

Similar for energy streams Ė (J/s, W) or specificenergy (streams) e = E/m = Ė/ṁ, ė = Ė/m, with massstream ṁ (kg/s) or mass m (kg)

* if frictional losses (giving heat) can be avoided** if Ohmic losses (giving heat) can be avoided

velocityv

massm

heightz

gravityg


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


Exergy of Internal Energy U 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°)!!!


Exergy of Enthalpy H

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°)

= Ex (U) + (p-p°)V 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

!!


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


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); with M = 29 g/mol for air m = Mp1V1/RT1 ≈ 2323 kg.

Pic:ÇB98


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


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


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

Exergy – entropy: air , air flow

air (~ U) air flow (~H)

8 mar 17Åbo Akademi Univ - Thermal and Flow Engineering, Piispankatu 8, 20500 Turku 31

Pics: GRMRRG90cp = 1 kJ/kgKreference state 25°C 1 atm

Exergy – enhalpy: air flow


Pic: GRMRRG90

cp = 1 kJ/kgKreference state 25°C 1 atm

Exergy - entropy / enthalpy diagramsexample Brayton cycle: process data 1/2


Pic: GRMRRG90

cp = 1 kJ/kgKreference state 25°C 1 atm

Exergy - entropy / enthalpy diagramsexample Brayton cycle: ex – s & ex - h 2/2


Pic: GRMRRG90cp = 1 kJ/kgKreference state 25°C 1 atm

ÅA 424520


1.4 Exergy Destruction; Isolated-Closed-Open Systems


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)


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


Pic: B01



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 :

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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:

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))((1

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genk

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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.

(symbols ψ, Φ also used on slide 26)

Pic:ÇB98


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 424520


1.5 Physical Exergy: Applicationsheat exchangers, tube flow, combustion, endoreversibility


43/108

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


Heat Transfer Efficiency (PTG course 424101)

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

QQ

balanceEnergy

gen

gen

.

See alsovS91


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


46/108

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

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T

Tlncm(T)xEΔxEΔ(xEΔ

Cin

Cout

pCC

Hin

Hout

pHHHC

> 0 < 0

< 0 > 0

< 0 > 0

cancel out



! 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 used on the hot side:

%)(xEΔQ

Q

SΔTQ

Qη

HX

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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: Z13)


Example: Heat Transfer 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


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


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


52/108

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


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, round tube diameter D

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


54/108

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


55/108

Exergy Losses and Tube Flow /4

For a given optimal value for the diameter, Doptimal , the exergy losses are minimal:

Can be applied to district heat networks

)TT(ρ

Tmη.D

.,

,

optimal U


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)


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


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


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


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 424520


Energy and Exergy Analysis-PFBC (Pressurised Fluidised Bed Combustion) : Sub-systems

Pic: ACA95

ÅA 424520


Energy and Exergy Analysis PFBC

Pics: ACA95


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


64/108


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


Thermal efficiency 10 existing power plants


CARNOT

ENDO-REVERSIBLE

B97 p 378

ÅA 424520


1.6 Mixing Exergy

See SAKS04 §7.2, §6.3


67/108

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)


68/108

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 %!


69/108

Exergy of Mixing /3

More general, for mixing of two molar streams n1, n2with molar fractions xi in the original streams and molar fractions xj in the final stream:

iii

iii

jjjmix

xlnxnxlnxnxlnx)nn(RTExΔ

Source: S05


70/108

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 (to) per mole of product gas (pg), 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 at T° = 298K:

J/mol

.

).ln..ln.(.

).ln..ln.(.

.ln..ln.

.exΔmix

Source: S05

ÅA 424520


1.7 Heat Radiation Exergy

See S05 §1.4.6; §2.7

and P10


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

.

.



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



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



More recently (see e.g. W07) it was shown that for the entropyof 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 424520


1.8 Chemical Exergy


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)



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

SMS88, S05



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 exergyfor the reference species are16.3 kJ/mol for CaCO3, 19.87 kJ/mol for CO2 and 3.97 kJ/mol for O2.

This gives for elemental calcium, Ca:ex°chem = b°chem =-738.6 + 16.3 - 19.87 - 1½×3.97 = 729.1 kJ/mol

See also table



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



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.


Standard Chemical Exergy of the Elements 1/5

c = fraction of element in reference species, γ = activity coefficient in seawater, m = molarity in seawater, p = %-vol, x = molar fraction

Source: S05



Source: S05

ÅA 424520



Source: S05

O

ÅA 424520



Source: S05

ÅA 424520



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


Standard exergies

for solids based on referencespecies dissolved in seawater

Source: S05


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


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


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 ,


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 424520



Calculatedresultsusingdata from SMS88 and K95

taken from ZK04

ÅA 424520



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 424520


1.9 Simplified Exergy Analysis With Chemical Reactions


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


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


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


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 424520


1.10 Exergy Analysis vs. PinchAnalysis


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

.


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


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

ÅA 424520


1.11 Final Remarks


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.


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)


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) Ch. 3 & 8 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 DelftUniv. of Technol., Delft (1984)


Sources (cont’d)

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) GRMRRG90: Gómez Ribelles. J.L., Monleón Pradas, M., Ribes Greus, A., Termodínamica –

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