1 IDEAL MIXING – IDEAL GASES ( ) ) 6 (13 V , T p p Law Daltons - Mixing Ideal V T R n p V T R m p Gas Ideal mix mix i i mix ___ i i i i i i - = = = + + &+ = V) (T, p 1 V) (T, p 2 V) (T, p n V) (T, p mix ( ) ) 7 3 (1 p , T V V Law Agamats - Mixing Ideal p T R n V p T R m V Gas Ideal mix mix i i mix ___ i i i i i i - = = = + + &+ = p) (T, V 1 p) (T, V 2 p) (T, V n p) (T, V mix
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IDEAL MIXING IDEAL GASES - University at BuffaloIDEAL MIXING IDEAL GASES Ideal Mixing - Daltons Law p p (T ,V) (13 6) V n R T p V m R T ... % Excess Air % Theoretical Air 100% Theoretical
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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
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222
===
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liquid moles 2.352.1474.5n vapormoles 2.147n
n6291.5366291.5n 100100
5.6291n36
nkPa 100
C 35 @pnn
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35C toCooled
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v
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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.
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
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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 ?
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
_____
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2
2
2
=
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=
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∑∑
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
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
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6
61p43p
6143
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4
545
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1
213
t
===
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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
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BalanceHeat r Regenerato
K 204.5741280
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K 369.554270ppTT
net
ineef
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net
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545
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1
213
t
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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
[ ] [ ]
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Maxwell Relation Derivation
22
12.65
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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
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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
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airair
waterwater
air
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m
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Specific and Relative Humidity
Tc ωTch ωTch ωhh
kJ/kg ,BTU/lb HHH
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dryairdryairwair
+=+=+=
+=
24
Adiabatic Saturation
( )
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( )( )
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EEEEquationEnergy FlowSteady
dryairunit mass 1n calculatio of basis
−
−−=
−
−−=
=++−
=−
−=−
+=−++
=+ ++
( )( ) ( )
airdry /lb1094.80BTUh 4) accuratemost air dry /lb1093.91BTUh 3)