3.1 Heat Engines, Heat Pumps - The Carnot cyclelira/liranotes/chap03.pdf · 3.1 Heat Engines, Heat...
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3.1 Heat Engines, Heat Pumps - The Carnot cycle
V
P
a
b
cd
isotherm
isotherm
adiabatadiabatremovableinsulation
Hot Reservoir
Cold ReservoirW· S
2
Table 3.1 Illustration of Carnot cycle calculations for an ideal gas.
Step Type W
a → b isotherm (ig) (ig)
b → c adiabat 0 ΔU = nCV (TC − TH) < 0 (*ig)
c → d isotherm (ig) (ig)
d → a adiabat 0 ΔU = nCV(TH − TC) > 0 (*ig)
Q
QH nRTH
VbVa------ln 0>= Q
H– n– RTHVbVa------ln 0<=
QC nRTC
VdVc------ln 0<= Q
C– n– RTCVdVc------ln 0>=
3
Carnot Heat Pump
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3.2 Distillation Columns
VR
VR LR
VS LS
F
LS
VS
B
Stripping Section
Reboiler
Feed Section
Rectifying Section
Total
LR D
Q· cond VR HvapΔ=
Q· reboiler VS HvapΔ=
LR = (1 – qR)VR
D = qRVR
VR
CondenserPartial
Q· cond 1 qR–( )VR HvapΔ=
Tray 11
Tray 12
Tray 13
L11
L12
L13
V13
V11
V14
V12
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Constant Molar Overflow
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3.3 Introduction to Mixture Propertiesthe energy of mixing is
3.20
The enthalpy of mixing is
3.21
The volume of mixing is the volume of the mixture relative to the volumes of thecomponents before mixing.
3.22
UmixΔ U xiUii–=
HmixΔ H xiHii–=
VmixΔ V xiVii–≡
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3.23
3.24
3.25
3.4 Ideal Gas Mixtures
or (ig) 3.26
(ig) 3.27
U xiUii UmixΔ+=
H xiHii HmixΔ+=
V xiVii VmixΔ+=
Uig yiUiig
i= Uig niUi
ig
i=
Vig RT nii
P⁄=
8
(ig) 3.28
(ig) 3.29
3.30
3.5 Ideal Solutions
Hig niUiig
i RT ni
i+ ni
i Ui
ig RT+ ni
i Hi= = =
Hig yii Hi
ig=
UmixigΔ 0= Vmix
igΔ 0= HmixigΔ 0=
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Example 3.3 Condensation of mixed vapor stream
F
E
B
Q· C
20°C Tbliquid
HvapΔHΔ L CPL Td
20
Tb
=
HΔ V CPV Td
Tb
90
=
10
Reacting Systems
νi -1 - 1 1 3
νi is stoichiometric number is stoichiometric coefficient
νi
dn dn1
1
2
2ν ν=
11
Extent of reaction can be negative
12
Multiple ReactionsCH3OH CO 2H2+↔
2CH3OH CH3OCH3 H2O+↔
13
Heat of reaction + = +
+ +
for
CH4 H2O CO 3H2
C 3H212---O2
H2 O2 N2 C s( ), , ,
3ΔHf H2,oΔHf CO,
oΔHf CH4,
o– ΔHf H2O,o–
Enthalpyof formation
ΔHfo 0=
14
Specified Temperature (usually 298.15K) becomes reference state,
3.45HRoΔ viHR i,
o≡ vi HfR i,
oΔ
νi HfR i,oΔ
products νi HfR i,
oΔreac tstan–
= =
HTo 0≥Δ
HTo 0≤Δ
Endothermic
Exothermic
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HToΔ HR
oΔ CPΔ TdTR
T
+=
ΔCP υiai υibi( )T υici( )T2 υidi( )T3+ + += }
Δ Δ Δ Δ Δ Δ
Δ Δ Δ Δ
H H a T T b T T c T T d T T
J aT b T c T d T
To
Ro
R R R R= + − + − + − + −
= + + + +
( ) ( ) ( ) ( )2 3 4
2 3 4
2 2 3 3 4 4
2 3 4
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Energy Balances - Heat of Reaction Method0 Hinn· in Houtn·out– Q· W· S+ +=
Hn· H·= =
n· icomponents
Hi R,o H∂
T∂-------
Po i,TdTR
T
H∂P∂
-------
T i,Pd
PoP+ +
n· HmixΔ+=
n· icomponents
≈ Hi R,o CPTR
T dT+
smallsmall
n· iin n· i
out–( )Hi R,o
components ξ· νiHi R,
o
components– ξ· HR
oΔ–= =
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for a given ξ, Tin, find
a)
b)
0 n· iin
components CP i,TR
Tin
dT n· iout
components CP i,TR
Tout
dT–=
Q· W· S ξ· HRoΔ–+ + 3.51
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Energy Balances - Heat of Formation Method
where all enthalpies based on elements at 298.15 K
and
Better if stoichiometry not known or many reactions.
0 Hinn· in Houtn·out– Q· W· S+ +=
Hinn· in n· iin
components HΔ fR i,
o CPTR
Tin
dT+ =
Houtn·out n· iout
components HΔ fR i,
o CPTR
Tout
dT+ =
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Adiabatic Reactors (Heat of Reaction Method)
0 n· iin
components CP i,TR
Tin
dT n· iout
components CP i,TR
Tout
dT– ξ· HRoΔ–=
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