Introduction to the concept of available work and exergy 2.83/2.813 T. G. Gutowski Feb 24, 2006 Readings and Handouts

1. Ch.2 “Thermodynamics Revised” Jakob de Swaan Arons

2. Ch.10 “Exergy” Norio Sato

3. Appendix, tables of standard chemical exergy, Szargut

4. Derivation of Mixing Entropy, Gutowski Why available work?

1. rigorous framework 2. based on 2 nd law (lost Vs transformed) 3. can account for materials energy potential 4. allows combined accounting of fuel and non-fuel materials 5. other advantages when dealing with waste heat (quality of waste heat) Outline • 1st and 2 nd Laws • Exergy • Examples – hot & cold water, gas expansion & extraction • Chemical Exergy – chemical reactions; burning carbon, and oxidizing Aluminium – Iron Ore Processing • Homeworks Energy from 1 st Law

• U = internal energy of molecules and atoms

• E = U + K.E. + P.E. = U + ½mv² + mgz

• “isolated” systems ∆E=0, or ∆U=0 (v=const, z=const)

• “closed” system ∆U = Qin –Wout • “open system” (mass flow) … The 1st Law

∆U = Qin −Wout dU = δQin −δWout ∫ dQ − ∫ dW = 0 Availability

“The First Law states that in every cyclic process either work is converted into heat or heat is converted into work. In this sense it makes no distinction between work and heat except to indicate a means of measuring each in terms of equivalent units. Once this technique of measurement is established, work and heat become entirely equivalent for all applications of the First Law.”

Keenan, 1941 Availability

• “The Second Law, on the other hand, marks the distinction between these two quantities by stating that heat from a single source whose temperature is uniform cannot be completely converted into work in any cyclic process, whereas work from a single source can always be completely converted into heat.”

Keenan, 1941 Availability

“The efforts of the engineer are very largely directed toward the control of processes so as to produce the maximum amount of work, or so as to consume the minimum amount of it. The success of these efforts can be measured by comparing the amount of work produced in a given process with the maximum possible amount of work that could be produced in the course of a change of state identical with that which occurs in the process.” Keenan, 1941 State Variables

2 dU = 0 dφ = φ −φ ∫ ∫ 2 1 1

dQ = 0 d(U + PV ) = 0 ∫ T ∫ State Variables

• T = temperature intensive variables • P = pressure • V = volume extensive • U = internal energy and intensive • H = enthalpy (H = U + PV) variables • S = entropy Enthalpy H=U+PV

Here the Work done is W = P(V 2 –V1)

The First Law can be written as Q = (U+PV) 2 – (U + PV) 1

The quantity in parenthesis is Enthalpy H = U + PV

The First Law can be written as Qin = ∆H

Constant Pressure Equilibrium Process Open flow system First Law for a Flow System

 dU   u 2  &  i    = ∑ mi hi + + gzi   dt cv in  2   u 2  &  i  − ∑ m jhj + + gz j  out  2 

+ Q& − Q& + W& − W& ∑ in ∑ out ∑ sh,in ∑ sh,out

 ∆u 2  &   & & m∆H + + g∆z = Qin −Wout EQ 1  2 

one stream steady state The Second Law

• ∆W and ∆Q are not equivalent • something gets lost Heat Engine

TH

Qin Wout = Qin − Qout

W Q − Q η = out = in out Qin Qin Qout

TL The Second Law

W Q − Q T −T η = out = in out η = H L Qin Qin TH

Q Q dQ Q Q in = out = in − out = 0 T T ∫ H L T T H T L Entropy: State Variable

dQ dS = T

∫ dS rev = 0 ∫ dS ≥ 0 Second Law for a Flow System

 dS  & &   = ∑ mi Si − ∑ m j S j  dt cv in out δQ& δQ& + k − 1 + S& ∑∫ ∑∫ generated in T out T

Q& m& ∆S + surr + S& = 0 EQ 2 T generated

one stream steady state Reversible Work

TH Wout = Qin − Qout Qin  T   L  Wrev = Qin 1−   TH 

Qout Qin Wrev = Qin − TL TH TL Exergy “B”(also “E” and Ex”)

Qin Wmax = Qin −To TH

Wmax = ∆H −To∆S

B = (H −ToS) − (H −ToS)o

o Standard ref.valuesT0 = 298.2 K, Po =101.3kPa Alternatively, From Flow EQ 1 & 2

 ∆u 2  &   & & m∆H + + g∆z = Q −Wout  2  Q& m& ∆S + + S& = 0 T generated & eliminate Q and set & Sgenerated=0

 ∆u 2  & &   & Wrev = m∆H + + g∆z − mTo∆S  2  Exergy

W& For the flow system: B = max m&  ∆u 2  then:   B = ∆H + + g∆z −To∆S  2 

Wmax Alternatively: B = B = ∆H −T ∆S m o Exergy

System State

Maximum work obtainable between System and Reference States.

Reference State Exergy

System State

Maximum work obtainable between System and Reference States; or minimum work needed to raise System from the reference state to the System State

Reference State Definition of Exergy “B”

“Exergy is the amount of work obtainable when some matter is brought to a state of thermodynamic equilibrium with the common components of the natural surroundings by means of reversible processes, involving interaction only with the above mentioned components of nature” [Szargut et al 1988]. Exergy Accounting

Bout Bin Blost

Bin − Bout = Blost Example: mixing hot and cold water

1 kg 100°C 2 kg 50°C

1 kg 0°C

st 1 Law: H 1 + H 2 = H 3

recall; ∆H = ∫m C p ∆T Example: mixing hot and cold water Το calculate the 2 nd law property;

∆B = ∆H – To∆S 1 kg 100°C Tref = 25°C 2 kg 50°C Pressure = constant 1 kg 0°C We need to know the temperature dependence of enthalpy and entropy temperature and pressure dependence of H and S

  ∂V   dH = cpdT + V −T  dP   ∂T  p 

c p  ∂V  dS = dT −   dP T  ∂T  p for constant pressure

  ∂V   dH = cpdT + V −T  dP   ∂T  p 

c p  ∂V  dS = dT −   dP T  ∂T  p calculating ∆H-T0∆S

373 2. H −T S = C ()75 − 298 2. C ln = 1.8 C 1 o 1 p p 298 2. p

273 2. H −T S = C ()− 25 − 298 2. C ln = 1.1 C 2 o 2 p p 298 2. p

323 2. H −T S = 2C ()25 − 2x298 2. C ln = 2C 3 o 3 p p 298.2 p Bin –Bout = Blost

Blost = (H 2 −To S2 )+ (H1 −ToS1 )− (H3 −To S3 )

Blost = ()1.8 + 1.1 − 2 C p = 2.7 C p Ref Sussmann ‘80 and de Swaan Arons ‘04 Example : Gas Expansion

Isothermally expansion of n moles of a gas

V1 V2 V1 V2

∆B = ∆H −T0 ∆S temperature and pressure dependence of H and S

T = constant ideal gas

  ∂V   dH = cpdT + V −T  dP   ∂T  p 

c p  ∂V  dS = dT −   dP T  ∂T  p

T = constant substitute ideal gas law yields…. Gas Expansion

V +V ∆S = n R ln 1 2 V1

∆B = ∆H - To ∆S

V1 ∆B = n RT0 ln V1 +V2

Note Blost = −∆B Gas Expansion/Extraction

Note the volume concentrations are :

n1 n1 State :1 c1 = ; State 2 =c2 = V1 V1 +V2 ∆S c c 1 = R ln 1 = R ln 1 n1 c2 cref Extraction from cref

c1 ∆Sextraction = - R ln cref c ∆Bextraction = +To R ln cref Chemical Exergy

Material Transformations – chemical reaction • oxidation, reduction – physical transformations • mixing • separation • extraction • disposal Chemical Reactions stoichiometric mass balance va Ra + vb Rb +.... → v j Pj + vk Pk +... exergy "balance" v B + v B +....− v B − v B = B a Ra b Rb j Pj k Pk lost where exergy B is given in kJ/mole Example: Burning Carbon C + O
CO 2 2

B + B -B = B C O2 CO 2 ∆ 410.3 kJ + 3.97 kJ – 19.9 kJ = 394.4kJ mole mole mole

The heating value of carbon is 394.4 kJ = 32.9 MJ mole of carbon kg Example: Oxidation of Aluminum

3 2Al + O → Al O 2 2 2 3

kJ 3 kJ kJ 2×888 4. + × .3 97 − 200 4. = B mole 2 mole mole lost

kJ Blost = ()1776 8. + 0.6 − 200 4. =1582 4. mole(Al2O3 )

See Appendix of Szargut for exegy values Materials can do work?

System State (T H)

QH

Insert reversible heat engine between Wrev high and low temperatures

QL

Reference State (T L) Chemical Properties referenced to the “environment” Crust

Oceans

Atmosphere

T0 = 298.2 K, P 0 = 101.3 kPA Exergy Reference System

pure metal, element

chemical reactions oxides, sulfides…

crustal component extraction earth’s crust (ground state) Exergy Reference System

Aluminum (c=1) 888.4 kJ/mole

Al 2O3 (c=1) 200.4 kJ/mole

Al 2SiO5 (c=1) 15.4kJ/mole -3 Al 2SiO5 (c = 2 x 10 ) 0 kJ/mole (ground) Example; making pure iron from the crust

Fe (c = 1) 376.4 kJ/mole

reduction

Fe 2O3 (c=1) 16.5 kJ/mole extraction -3 Fe 2O3 (c = 1.3 x 10 ) 0 kJ/mole (ground)

Extraction from the crust

−3 Extracting Fe2O3 from c =1.3x10 (crust) to c =1

1 B = T R ln o 3.1 x10−3

J 1 kJ B = 298 2. o K × .8 314 ×ln =16 5. moleo K 1.3×10−3 mole Reduction of Fe 2O3 (Hematite)

2Fe 2O3 + 3C
4Fe + 3CO 2

2 x 16.5 + 3 x 410.3 – 4 x 376.4 – 3 x 19.9 =

Blost = - 301.4 kJ this is an endothermic reaction i.e. minimum energy required to reduce 4 mole of hematite Iron Ore Reduction

Recall C + O 2
CO 2 produces 394.8 kJ/mole C

301 4. We need = .0 76 moles of carbon 394 8.

2Fe2O 3 + 3.76C + 0.76O 2
4Fe + 3.76 CO 2 Iron Ore Reduction

but the efficiency is only 33% therefore the actual reaction is

2Fe 2O3 + 12.42C + 9.42O 2
4Fe + 12.42CO 2 33kJ + 5095.9 + 37.7 - 1505.6 - 247.2kJ = 3,413.8 kJ for 4 mole of Fe this is 15.2 MJ/kg (Fe) Iron Ore Reduction

12.42×394 8. .4 903MJ MJ Fuel (C) Intensity = = = 22 4×55.85g .2234kg kg

12.42× 44 kg CO CO Intensity = = 5.2 2 2 4×55.85 kg Fe

Exergy value of pure Fe is 376.4 kJ/mole = 6.7 MJ/kg Summary for Iron Ore

fuel used = 22 MJ/kg ≈ 15.2(B lost ) + 6.7(B Fe )

“Credit” for producing pure iron from the crust

Lost exergy from making iron

from Fe 2O3 Energy data from Smil Exergy in Mixing Substances

∆ B mix= ∆ H mix−To∆ S mix for a perfect mixture()athermal

∆ H mix = 0 l ∆ B mix≅ −To∆Smix = RTo ∑ xi n xi

Ωmixed ∆Smix = R ln Ωseparated

see hand out Summary

• Exergy = Available work referred to the “dead state” • Both fuel and non-fuel materials have exergy equivalents • Exergy is not conserved

Bin − Bout = ∆Blost Homeworks

1. How much energy is required for photosynthesis? 2. How much energy is produced from auto-respiration? 3. How much energy is produced from burning octane? methane? methanol?

hydrogen? How much CO 2 is generated for each? Applications

• Exergy Analysis of Industry (Ayres et al) • Exergy Analysis of Natural Systems (Bakshi) • Exergy of Mfg Processes (YOU!) • Exergy Analysis of Recycling Flows