ideal mixing ideal gases - university at buffaloideal mixing ideal gases ideal mixing - daltons law...
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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
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
∂∂
=−
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=
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+
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=
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−+=
=+=
(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
∂∂
−=
∂∂
∂∂
∂∂
=
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∂∂
∂∂
∂∂
=
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−
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=
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Maxwell Relation Derivation
22
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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
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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)
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v1
v1
v1
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−+=
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s
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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