1

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

4

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

5

Constant Molar Overflow

6

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–≡

7

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=

9

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

15

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

16

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–= =

17

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

18

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+ =

19

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–=

20