1

IDEAL MIXING – IDEAL GASES

( ) )6(13 V,Tpp Law Daltons - Mixing Ideal

V

TRnp V

TRmp Gas Ideal

mixmixi

imix

___

iii

iii

−=

==

∑

+ +…+ =V)(T,

p 1

V)(T,p 2

V)(T,p n

V)(T,p mix

( ) )73(1 p,TVV Law Agamats - Mixing Ideal

p

TRnV p

TRmV Gas Ideal

mixmixi

imix

___

iii

iii

−=

==

∑

+ +…+ =p)(T,

V 1

p)(T,V 2

p)(T,V n

p)(T,V mix

2

MIXTURES

72lbs 100% 2 38.9% lbs 28 28 50% 1 N61.1% lbs 44 44 50% 1 CO

analysis analysis cgravimetri weightMWT molar moles

N mole 1CO mole 1

(13.2) n

nnny fraction, mole

(12.3) MWTnmMWTmolesmass

mm

mmfraction mass

(13.1) mm

2

2

22

ii

ii

ii

ii

ii

+

==

×=×=

==

=

∑

∑

∑

i

i

i

iii

iii

ii

ii

i

iii

i

iii

vv

nn

RTv

vRT

pp

nn

RTvppp

nn

TT

RR

VV

pp

pvRT

RTVp

nn

RTVpn

nRTpV

=

==

=

=

==

=

=

3 WeightMolecular molesMass

O2O2O OxygenProduct MassOxygenReactant Mass

2HH HydrogenProduct MassHydrogenReactant Mass

CC CarbonProduct MassCarbonReactant Mass

Combustion of Products Mass Oxidizer Mass Fuel MassProducts MassReactants Mass

OH2COO2CH

22

24

2224

×=

+==

==

==

=+=

+⇒+

COMBUSTION

Chemical Equation Balance is a Mass Balance

4

.HC of poundper air al theoreticof ratio the theDetermineair. icalin theoret HC of combustion completefor equation theBalance

3617

3617

lbs 3810.4lbs 3810.18182897.844173570240

Products Mass Air Mass Fuel MassBalance Overall

fuel air/lb lb 14.87240

3570Ratio FuelAir

fuel mole lbs/lb 2401361712Mass Fuelfuel mole lblbs/ 35702897.83226eightMolecularWmolesAirMass

O18O1726O BlanceOxygen 18HH BalanceHydrogen 17C1C BalanceCarbon

ncalculatio of basis theas fuel of mole 1 assume

97.8NO18H17CO26N21796O2HC

22

236

17

222223617

=×+×+×=+

=+

==

=×+×==×+×=×=

+===

++=×++

5

100%Air lTheoretica %Air Excess %

Air lTheoreticaAir lTheoreticaAir Combustion ActualAir %Excess

Air Combustion lTheoretica

Air Combustion ActualAir ical%Therotret

supplied is Air) cal(TheroretiOxygen lTheoreticaOnly remainsoxidant no and consumed is fuel All

COMBUSTION TRICSTOICHIOME-CONBUSTION COMPLETE

−=

−=

=

6

Methane is burned with 90% Theoretical Air

2222224

2242224

222224

7.52N22OO2HCO2N217922O2CH

AIR)THEORETCAL(200% AIREXCESS 100%

6.771NO1.8H.1CH.9CO22179.92O.9CH

AIR) EXCESS 10%- ( AIR LTHEROETICA 90%

N 7.52O2HCO2N21792OCH

AIR LTHEORETICA

×+++⇒×=×+

+++⇒×+×+

++⇒++

N

7

C 47.74TkPa) 11.11pg T(@

kPa 11.11pp100%at

kPa 11.1110040.54.5p

pp

nny

40.5 OH moles 4.5moles 36

32NO4.5H4COC? 35 tocooled are

products theif condenses much water HowkPa? 100at combustion of products following

theof eraturepoint temp dew theisWhat

dp

gv

v

vvv

2

222

===

===

=×=

==

=+

++⇒

φ

liquid moles 2.352.1474.5n vapormoles 2.147n

n6291.5366291.5n 100100

5.6291n36

nkPa 100

C 35 @pnn

n

p100%on condensatiat

pp

nnn

pp

nny

35C toCooled

l

v

vv

v

v

g

v

v

v

v

v

v

vvv

=−=

=+×=

=+

=+

==

=+

==

gpφ

8

Determine the volumetric analysis of the gaseous products of combustion of methane burning with 30 % excess airAfter the products of combustion have cooled to 90 F.

vgasdry

v

products

v

products

vOH

gv

2222

224

222224

mmm

pp

vapormolesgas moles vapormoles

ppy

psia .69904pp F 90at

N 9.77.6OO2HCO

2N21791.32O1.3CH

AIR) EXCESS 30% ( AIR LTHEROETICA 130%

N 7.52O2HCO2N21792OCH

AIR LTHEORETICA

2

+=

+==

==

+++⇒

×+×+

++⇒++

81.8%439.776/11.9N %5.02%.6/11.943O %4.75%3.568/11.94OH %8.37%1/11.943CO %

.56811.376m

mm%

moles 1.432.5682m vapormoles .568m

.6990414.711.376.69904

ppmp

m

moles 11.376m

2

2

2

2

i

gas

i

liquid

v

vproducts

gasdry vv

gasdry

====

====

+==

=−==

−×

=−

×=

=

9

kJ/kmole 393,300393,300)(10)10(1h

reaction,CO For the26A Table h

,hnhnh

COMBUSTION OF ENTHALPY

F @77 CO lbmole,BTU/ 169,300h

C @25 CO with,kJ/kmole 393,520h

COOC26EA 26,A Table ,h

elements its from formed is compound a when offgiven or absorbedheat FORMATION OF ENTHALPY

RP

__

2

__of

PRODUCTS

__ofREACTANTS

__ofRP

__

2

__of

2

__of

22

__of

=−×−×+×=

−

×−×=

−=

−=

⇒+−−

∑∑

F 77TC 25T

ref

ref

==

H

RP

__h

T

reactants

products

10

−−

−+=

−−

+−

+=

−=

∑∑

∑∑

ref

__

1

__

reactants

__

ref

__

2

__

products

__

RP

__

______of

reactants

______of

products

__

productsreactants

@Thhn@ThhnhQ

23EA23,A Table ,h (13.19), hΔhnhΔhnQ

HHQBALANCE ENERGY COMBUSTION

H1

H

T

1

2

T

2

F 77TC 25T

ref

ref

==

21 HHQ −=21 HHQ −=

reactants

products

11

combustion of products in the remainsenegy combustion theall

@Thhn@Thhnh0Q

ETEMPERATUR FLAME ADIABATIC

ref

__

1

__

reactants

__

ref

__

2

__

products

__

RP

__

−−

−+== ∑∑

H

T

1 2

F 77TC 25T

ref

ref

== Adiabatic

FlameTemperature

12

Propane is burned in air at 25 C and 100. kPa. Combustion is complete. Products of combustion are at at 87 C. Determine the heat supplied andin kJ/ kmole and the adiabatic flame temperature ?

kJ/kmole 1,994,590.400,492,043,990.Q hΔ h Q

49,400 hΔ 2,043,990. h 25.81n

33,896 8669 10,471 0 0 18.81 N

8352 9904 11,992 967,280.- 241,820- 4 OH7152 9364 11,748 1,180,560. - 393,520- 3 CO

hΔn h h hn h n Prod.

103,850 103,850- 1 HC hn h n React.

26A Table h,hnhnh

N 18.81OH 4CO 3N217955OHC

__

RP

__

__

RP

__2

gas2

2

__

25C

__

87C

____of

__of

83

__of

__of

__ofPRODUCTS

__ofREACTANTS

__ofRP

__

222283

=+−=

+=

=−==

××

−×

−×−×=

++⇒++

∑∑∑

∑∑

oncondensati no

C 55TkPa 94.15p25.81

4100p

nn

pp

kPa 641.4780C @p

dp

v

v

v

atm

v

g

≅=

×=

=

=

13

( )

C 2177.4T25C)949.63(T2,043,990

25C)(T33)18.8141.35454.5(32,043,990C 25Tcnhnh

combustion of products theheating into goes h all 0,Q , adiabaticFor

C kJ/kmole 3320

30,12430.784c

C kJ/kmole 41.3520

35,88236,709c

C kJ/kmole 54.520

42,76943,859c

K 1040 K to 1020 @eTemperatur Flame Adiabatic

adiabatic

ad

ad

_

pii

__

RP

__RP

__

____

pN

____

OpH

_____

pCO

2

2

2

=

−=−××+×+×=

−×=×=

=

=−

=

=−

=

=−

=

∑∑

14

Adiabatic flame temperature (constant pressure) of common gases/MaterialsFuel Oxidizer Tad (°C) Tad (°F)Acetylene (C2H2) air 2500 4532Acetylene (C2H2) Oxygen 3480 6296Butane (C4H10) air 1970 3578Cyanogen (C2N2) Oxygen 4525 8177Dicyanoacetylene(C4N2)

Oxygen 4990 9010

Ethane (C2H6) air 1955 3551Hydrogen (H2) air 2210 4010Hydrogen (H2) Oxygen 3200 5792 [1]

Methane (CH4) air 1950 3542Natural gas air 1960 3562 [2]

Propane (C3H8) air 1980 3596Propane (C3H8) Oxygen 2526 4579MAPP gasMethylacetylene(C3H4)

air 2010 3650

MAPP gasMethylacetylene(C3H4)

Oxygen 2927 5301

Wood air 1980 3596Kerosene air 2093 [3] 3801Light fuel oil air 2104 [3] 3820Medium fuel oil air 2101 [3] 3815Heavy fuel oil air 2102 [3] 3817Bituminous Coal air 2172 [3] 3943Anthracite air 2180 [3] 3957Anthracite Oxygen ≈2900 [see 1] ≈5255

Assuming initial atmospheric conditions (1 bar and 20 °C), the following table list the adiabatic flame temperature for various gases under constant pressure conditions. The temperatures mentioned here are for a stoichiometric fuel-oxidizer mixture (i.e. equivalence ratio ).Note this is a theoretical flame temperature produced by a flame that loses no heat (i.e. closest will be the hottest part of a flame) where the combustion reaction is quickest. And where complete combustion occurs, so the closest flame temperature to this will be a non-smokey, commonly bluish flame

1.^ The temperature equal to ≈3200 K corresponds to 50 % of chemical dissociation for CO2 at pressure 1 atm. The latter one stays invariant for adiabatic flame and the carbon dioxide constitutes 97 % of total gas output in the case of anthracite burning in oxygen. Higher temperatures will occure for reaction output while it going under higher pressure (up to 3800 K and above, see e.g. Jongsup Hong a et

15

REVIEWChapters 10,11,12,13All the review notes are also given in the originalChapter notes

16

Vapor Compression Refrigeration CycleOpen systems, Steady flow

s

1

2is

3

4

2T

h=c

1

23

compressor

expansion

evaporator

condenser

( )( )

( )( )31in

41in

43

32out

12in

shaft

2

hhmQ hhmQ 0, W1,4 Process, Evaporator

hh 0, W0,Q, 43 Process, Expansion hhmQ 0, W, 32 Process,Condenser hhm W0,Q 2,1 Process,n Compressio

Wρgh)2

VpvΔ(umQ

EquationEnergy FlowSteady spacein region -SystemOpen Flow,Steady

−=−==⇒

===⇒−==⇒−==⇒

++++×=

WQCOP

WQCOP

RequiredEffort Effect DesiredCOP

out

pumoheat

inref

=

=

=

17

Vapor Compression Refrigeration CycleOpen systems, Steady flow

s

1

2is

3

4

2T

h=c

1

23

compressor

expansion

evaporator

condenser

( )( )

( )( )31in

41in

43

32out

12in

shaft

2

hhmQ hhmQ 0, W1,4 Process, Evaporator

hh 0, W0,Q, 43 Process, Expansion hhmQ 0, W, 32 Process,Condenser hhm W0,Q 2,1 Process,n Compressio

Wρgh)2

VpvΔ(umQ

EquationEnergy FlowSteady spacein region -SystemOpen Flow,Steady

−=−==⇒

===⇒−==⇒−==⇒

++++×=

WQCOP

WQCOP

RequiredEffort Effect DesiredCOP

out

pumoheat

inref

=

=

=

18

1

2

3

4

1

23

4

2is

4is inQoutW

inW

outQReverse Brayton Refrigeration Cycle

( )( )( )( )41in

43out

32out

12in

shaft

2

hhmQ 0, W1,4 Process, AbsorptionHeat hhm W0,Q , 43 Process, Expansion hhmQ 0, W, 32 Process,Rejection Heat hhm W0,Q 2,1 Process,n Compressio

Wρgh)2

VpvΔ(umQ

EquationEnergy FlowSteady spacein region -SystemOpen Flow,Steady

−==⇒

−==⇒−==⇒−==⇒

++++×=

T

s

19

( )

( ) ( )

( )( )

( )( )

( ) ( )( ) ( )

1.4824.1235.61

wqCOP d)

24.12kJ/kgw 204.57280.1.005270369.551.005w

TTcTTcww wb)35.61kJ/kg204.572401.005Q

TTcQ c)kJ/kg 59.85310369.551.005Q

TTcQ b)K 240280310270T

TTcTTchhhh

BalanceHeat r Regenerato

K 204.5741280

ppTT a)

K 369.554270ppTT

net

ineef

net

net

54p12pcnet

in

56pin

out

32pout

6

61p43p

6143

.2857

4

545

.2857

1

213

t

===

=

−−−=

−−−=−==−×==

−==−×=

−==+−=

−=−

−=−

=

=

=

==

=

1

23

4

5 6

1270 K1 bar

24 bar

3310 K

280 K4

5

6

T

s

q

q10.49100 kPa,270 K air is compressed with a pressure ratio of 4 in aregenerated reversed Brayton Cycle. Air enters the regeneratorat 300 K and leaves at 280 K. Find: a) the low temperature b) work/kg, c) capacity/kg, and d) COP.

20

( )

( ) ( )

( )( )

( )( )

( ) ( )( ) ( )

1.4824.1235.61

wqCOP d)

24.12kJ/kgw 204.57280.1.005270369.551.005w

TTcTTcww wb)35.61kJ/kg204.572401.005Q

TTcQ c)kJ/kg 59.85310369.551.005Q

TTcQ b)K 240280310270T

TTcTTchhhh

BalanceHeat r Regenerato

K 204.5741280

ppTT a)

K 369.554270ppTT

net

ineef

net

net

54p12pcnet

in

56pin

out

32pout

6

61p43p

6143

.2857

4

545

.2857

1

213

t

===

=

−−−=

−−−=−==−×==

−==−×=

−==+−=

−=−

−=−

=

=

=

==

=

1

23

4

5 6

1270 K1 bar

24 bar

3310 K

280 K4

5

6

T

s

q

q10.49100 kPa,270 K air is compressed with a pressure ratio of 4 in aregenerated reversed Brayton Cycle. Air enters the regeneratorat 300 K and leaves at 280 K. Find: a) the low temperature b) work/kg, c) capacity/kg, and d) COP.

21

[ ] [ ]

( )

(11.25)vupand (11.24)

suT

bracket,in quantaties thecomparing

dvvuds

sudu

al.differentiexact an isdu ,vs,uu ,propertiesother any two offunction a and

path oft independen property, a isu SincedvpdsTdu

Combining,Law Second Tdsδq

LawFirst pdvduδq

sv

sv

∂∂

=−

∂∂

=

∂∂

+

∂∂

=

=

−+=

=+=

(11.23)sp

vT

ng,substitutivu

ssu

v

and,ation differenti ofresult affect thenot doesorder

vu

ssp

and su

vvT

again, tingdifferenta

vs

sv

sv

vs

∂∂

−=

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

∂∂

=

∂∂

−

∂∂

∂∂

=

∂∂

Maxwell Relation Derivation

22

12.65

( ) ( )

%1.4302985.0129.

T@pp

0129.pp01325.1

p622.008.

ppp622.

008.21.2478

04.884.27)T(

19.674.25452416005.119.678.25300113.)T(

)T(h)T(h)TT(c)T(h)T(

)T(

0113.01818.01325.1

01818.622.)T(

@Tp1.01325@Tp

.622)ω(T

C 16T C, 24T a)

dbg

v

v

v

v

v

v

db

db

wbfdbg

dbwbpwbfgwbdb

wb

wbg

wbgwb

wbdb

===φ

=−

=

−=ω

=−

=ω

−−+−×

=ω

−−+×ω

=ω

=−

=ω

−=

==

( ) ( )

%3.414302.1772.

T@pp

1772.pp7.14

p622.00759.

ppp622.

00759.17.1066

6.369.11)T(

08.2825.1094756024.6.105901104.)T(

)T(h)T(h)TT(c)T(h)T(

)T(

01104.2563.7.14

2563.622.)T(

@Tp14.7@Tp

.622)ω(T

F 60T F, 75T b)

dbg

v

v

v

v

v

v

db

db

wbfdbg

dbwbpwbfgwbdb

wb

wbg

wbgwb

wbdb

===φ

=−

=

−=ω

=−

=ω

−−+×

=ω

−

−+×ω=ω

=−

=ω

−=

==

23

( )

(12.44) vv

pp

RTpV

RTpV

saturationat or water vapmassor water vapmass actual

saturationat or water vapmassor water vapmass actualhumidity Relative

(12.43) pp29

p 18ω

TMW pTMW p

RTpV

RTpV

airdry massor water vapmassω

airdry massor water vapmassω Fraction, Mass Humidity, Specific

airdry kgm 1 air,dry lb 1 -n calculatio of Basis

w

g

g

w

saturation

wambient

w

airair

waterwater

air

water

m

==φ

==φ

=φ=

−=

=

==

=

Specific and Relative Humidity

Tc ωTch ωTch ωhh

kJ/kg ,BTU/lb HHH

pvpvpva

dryairdryairwair

+=+=+=

+=

24

Adiabatic Saturation

( )

( )( )

( )( )

( )( )

( )wbfdbv

wbdbpfg2wb

f2v1

12pfg221

fg22f2v1112p

fgf2v2

21pa2a1

v222afg212v11a1

outvapor airaddedwater invapor air

hhTTchω

ω

hhTTchω

ω

hω)h(hωTTc

hhh

)T(Tchhng,substituti

hωhhωωhωh

EEEEquationEnergy FlowSteady

dryairunit mass 1n calculatio of basis

−

−−=

−

−−=

=++−

=−

−=−

+=−++

=+ ++

( )( ) ( )

airdry /lb1094.80BTUh 4) accuratemost air dry /lb1093.91BTUh 3)

use easiest toair dry BTU/lb 1093.83h 2)Exactair dry lbBTU/ 1094.07h 1)

wbF db,70 75Ffor

T.441061.8TT.44F 0 @hh 4)

TT*.45T @hh 3)

T @hh 2))pp,T@(Tenthalpy vapor saturationh 1)

v1

v1

v1

v1

dbwbdbgv1

wbdbwbgv1

dbgv1

vdbv1

====

+=−+=

−+=

====

dbT

T wbT

s

2wb

1db

2

1v1h

f2h g2h

g1h

0T =

34

25

Methane is burned with 90% Theoretical Air

2222224

2242224

222224

7.52N22OO2HCO2N217922O2CH

AIR)THEORETCAL(200% AIREXCESS 100%

6.771NO1.8H.1CH.9CO22179.92O.9CH

AIR) EXCESS 10%- ( AIR LTHEROETICA 90%

N 7.52O2HCO2N21792OCH

AIR LTHEORETICA

×+++⇒×=×+

+++⇒×+×+

++⇒++

N

26

OPEN CYCLE ANALYSIS TOOL KIT1. Properties

Ideal GasGas TablesFluid Properties- steam, refrigerant tables

2. Processes3. Energy Balances (First Law)4. Component Performance

Turbine EfficiencyCompressor EfficiencyHeat Exchanger EffectivenessApproach Temperature

5. Cycle PerformanceCycle EfficiencyCarnot Efficiency Cycle COPCarnot COP

27

( )( )

−−=−

+−=−

===

==

−−

−

=−

+

=−

−==

−==

==

∫∫

1

21

02

012

1

21

02

012

2

1

2

1

r2

r1

2

1

r2

r1

rro

1

2

1

2p12

1

2

1

2v12

12pp

12vv

ppln R)(Ts)(Tsss

230 pg vvln R)(Ts)(Tsss

407 pg VV

vv

vv ,

pp

pp

112 pg eTemperatur of functions as v,p,su,h,p

RTmV ,p

RTv

22EA22,A TABLE, GASppln R

TTln css

231 pg vvln R

TTln css

TTcdTch

107 pg TTcdTcu

107 pg p

RTmV ,p

RT vGAS IDEAL

p and T of functions as s, andh,u, v, -Table Liquid Compressed

p Tand of functions as s, andh,u,v, -TableSuperheat mm

xquality,

sor h u, v,becan φ here w

89 pg φ

φφ x,φxφφ

eTemperatur of functions as Ps,h,u, v, -Table eTemperatur

Pressure of functions as Ts,h,u, v, - Table Pressure

134aR steam, -PROPERTIESFLUID

g

fg

ffgf

=

−=×+=

−

1. PROPERTIES

28

2.PROCESSES

142 pg

Wρgh)2

VpvΔ(u mQ

LawFirst of formEquation FlowSteady SystemsOpen

ss

vv

pp

vv

TT

417 258, 257, pg pp

TT

constantpvIReversible 0,ΔS 0,Q Adiabatic,

shaft

2

21

k

2

1

1

2

1k

2

1

1

2

k1k

1

2

1

2

k

++++×=

=

=

=

=

=−

==−

−

−

Gas Real

Gas deal

PROCESSISENTROPIC( )

( ) ( )

( )

( ) ( )

in

netcycle

cyclecycle

332211

321

inhot outhot hotin coldout coldcold

fluidhot fluid cold

21outin

21outin

1221

21outin

QWη WQ

ForCyclehmhmhm

mmm Mixing

hhmhhm ΔQΔQ

ExchangersHeat hh ,HH

158 pg valvesprocess, tling throt0,Q and with W systemsOpen

hhmQ Q,HH 350 pg heaters water feed ,condensersboiler, ,exchangersheat

0, with WsystemsOpen ppvhhm W

352 270, pg pumps hhm W W,HH

349 pg scompressor es, turbin0,Q with systemsOpen

==

=+=+

−×=−×=

==

=−×=+=

=−=−×=

−×=+=

=

∑∑

3. ENERGY BALANCES

29

4. COMPONENT PERFORMANCE

in coldouthot out coldinhot

in coldinhot

in coldout cold

in coldinhot

outhot inhot

12

isentropic 2COMPRESSOR

actual

isentropicCOMPRESSOR

isentropic 21

21TURBINE

isentropic

actualTURBINE

TT andTTETEMPERATUR APPROACH

hhh-h

hhhhε

425,426 pg Q Maximum

Q Actualε

ESSEFFECTIVEN EXCHANGER HEAThh

h-hη

3263,359,43 pg W

Wη

EFFICIENCY PUMPAND COMPRESSOR

hhhhη

8,433261,262,35 pg WWη

EFFICIENCY TURBINE

−−

−=

−−

=

=

−=

=

−−

=

=

T

T

Ts

s

1

22is2

2is

1

hot in

hot outcoldout cold in

30

5. CYCLE PERFORMANCE

LH

H

pumpheat CARNOT

LH

L

ionrefrigeratCARNOT

outin

outoutheatpump

outin

ininionrefrigerat

H

L

H

LH

CYCLECARNOT

out

inCYCLE

in

outin

in

netCYCLE

outin

TTTCOP

TTTCOP

QQQ

WQCOP

206 pg QQ

QWQCOP

EPERFORMANC OF TCOEFFICIEN CYCLE

205 pg TT1

TTT η

QQ1η

350 63, pg Q

QQQWη

QQWEFFICIENCY CYCLE

−=

−=

−==

−==

−=−

=

−=

−==

−=

T

s

inQ

outQnetW

T

sinQ

outQ

netW