ocw chapter 4

8/13/2019 Ocw Chapter 4

1/48

!"#$$%&' )*+,-)./ +01-0++2-01 3*+2,4560.,-)!

*+.3 +##+)3!

,789::9; #9;;?= @989>


2/48

.=:7AB 9== C8D:


3/48

!DGA=D *D9B +SDCB

-GI7F:9K7G 7G EF7CDAAP V?FD A?>AB9GCD )=7AD; AMABD: )7GAB9GB EFDAA?FD -G


4/48

+GDFLM >9=9GCD I7F C=7AD; AMABD:J!Q +!W = dU+ dE

K + dE

P !Q +!W = dU !Q "PdV = dU

H =U +PV

dH = dU +PdV +VdP

dH = dU +PdV

dU = dH! PdV

Q = !H

Also,

For constant P

Substitute and integrate So how do we determine !H?


5/48

# W $X%Y% W $ !"# % & %'()*+ dH= !H

!T

"#$

%&'p

dT +!H

!P

"#$

%&'T

dP

dH=CpdT+!H

!P

"

#$

%

&'TdP

dH =CpdT

!H = H2" H

1 = C

pdT

T1

T2

#

For isobaric process

So this equation for !His only valid for isobaricprocess and no change of phase.

If the process is not isobaric,the equation is also valid for ideal gas

Also applicable for liquid and solid as Hisnot highly dependent on pressure


6/48

!


7/48

+T9:E=D

*7B 7


8/48

!mo C

po

l dTTo

T

! + !ma CpaigdT

To

T

! = 0

!moC

po

l (To2! T

o1) + !m

aC

pa

ig(Ta2! T

a1) = 0

!moC

po

l (To2! T

o1) =! !m

aC

pa

ig(Ta2! T

a1)

Solve for Ta2

Assume air is an ideal gas:

Assume constant CP:

!Q + !W = mH( )! 2

" mH( )! 1

+ #EK + #E

P

0 + 0 = mH( )! 2 " mH( )! 1 + 0 + 00 = !m

oH

o2+ !m

aH

a2" !m

oH

o1" !m

aH

a1

!mo(H

o2"H

o1) + !m

a(H

a2"H

a1) =0

Steady state energy balance for open system:


9/48

*D9B )9E9C


10/48

,D9G *D9B )9E9C


11/48

Cp H=

CpdT

To

T

!

T " To

!H = C

pH

T" T0

( )

Hence,

Since


12/48


13/48

+T9:E=D [J$!7=?K7G I7F 2


14/48

+T9:E=D [J&!7=?K7G I7F (1M


15/48

/9BDGB *D9B 7I ]9E7FAB9GCD -G


16/48

+GDFLM `9=9GCD )=7AD; !MABD:!Q +!W = dU+ dE

K + dE

P

!Q +!W = dU

!Q "PdV = dU

!Q = dU+ PdV = dH

Q = #H, this is delta H of vaporization at temperature T1

note,Q = #H

n, this is delta H of vaporization at normal boiling point

Degree of freedom, F=2-2+1=1 Once we specified one thermodynamic properties,

other properties will depends on this.


17/48

#7F DT9:E=DP A9B?F9BD; *$4 9B 9 AEDC


18/48

.B G7F:9= >7


19/48

.=A7 3F7?B7GA 2?=D 9G; )8DGA +_?9K7G NADD#D=;DFA 3DTB>77\R


20/48

-I OD \G7O3%708/9B (/P OD C7?=; DAK:9BD3%70819B (1?A


21/48

*D9B 7I 2D9CK7G

-GI7F:9K7G 7G EF7CDAA #DD;A NDJLJ .%P .$R 9FD FD9CB9GBA 4?B=DBA 9FD EF7;?CBA NDJLJ .&P .[R 7I B8D FD9CK7G *D9B A7F>D; ;?F


22/48

!1A

1+ !

2A

2" !

3A

3+ !

4A

4

Where |vi| is stoichiometric coefficient

e.g. CH4 + 2O2 ! CO2 + 2H2O

Stoichiometric reaction,

Where the value of viis positive for productsand negative for reactants.

!1 CH4 + !2 O

2" 1CO

2+ 2 H

2O


23/48

Energy balance for steady-state open system

with stoichiometric reaction,

Q =!H

= "3 H3 + "4 H4 # "2 H2 # "1 H1

= "i$ Hi

|v1|, |v2| |v3|, |v4|


24/48

!B9G;9F; *D9B 7I 2D9CK7G3%.(

5DZGD; 9A B8D DGB89=EM C89GLD IF7: 9AB7


25/48

There are many references for tabulated data for"Ho298 for commonly found reactions.

|v1|, |v2| |v3|, |v4|

T=25o

C, P=1bar T=25o

C, P=1bar


26/48

+TJ 7I !B9G;9F; *D9B 7I 2D9CK7G 9B 3W$abJ%c"P d*7$ab

1

2N

2(g) + 3

2H

2(g)! NH

3(g)

"H298o

= #46110

J

mol NH3produced

!H298

o= "46110

J

3

2 mol H2 reacted

!H298

o= "46110

J

1

2mol N

2reacted

N2(g) + 3H

2(g)! 2NH

3(g)

"H298o = #92220 J

2mol NH3produced

!H298o

=

"92220

J

3mol H2reacted

!H298

o= "92220

J

mol N2reacted

Note: The value of standard heat of reactiondepends on stoichiometric coefficients as written


27/48

*7ODHDFP B9>?=9K7G 7I 9== E7AA=D FD9CK7GA 9FDG7B EF9CKC9=J

#?FB8DF:7FDP G7B 9== FD9CK7GA C9G B9\D E=9CD 9B$ab"

!7P 87O B7 C9=C?=9BD AB9G;9F; 8D9B 7I FD9CK7Ge


28/48


29/48

!H298

o

="

i

# !

Hf298,io

So we can calculate standardheat of reaction from standard heat offormation of the reactants and products

!H298

o= "

i# Hio


30/48

+T9:E=D 7I I7F:9K7G FD9CK7GPC s( ) + 1

2O

2 g( ) + 2H2 g( )! CH3OH l( )

where

C,O2,H

2are elements

Heat of formations are usually tabulatedat 298.15K, such as in Table C4. !Hf298

o


31/48

CO2(g) + H

2(g)! CO(g)+ H

2O(g)

Example:

Calculation of std heat of rxn using std heat of formation,

!H298

o= "co2!Hf298,co2

o+ "H2!Hf298,H2

o+ "co!Hf298,co

o+ "H2O!Hf298,H2O

o

= #1( ) #393509( )+ #1( ) 0( )+ 1( ) #110525( )+ 1( ) #241818( )

!H298

o= 41166

J

mol CO2 reacted

!H298o

= "i# !Hf298,i

o

!H298

o= "

i# Hio


32/48

07BD'*$4 ;7DA G7B 9CB?9==M DT9FJ

38


33/48

!B9G;9F; *D9B 7I )7:>?AK7G

3MED 7I 8D9B 7I FD9CK7G B89B DBODDG I?D= 9G; 7TMLDG B7 EF7;?CD O9BDF 9G;

C9F>7G ;


34/48

@D C7?=; C9=C?=9BD AB9G;9F; 8D9B 7I FD9CK7G IF7:AB9G;9F; 8D9B 7I C7:>?AK7G 9A I7==7OAP

!H298o

= " #i$ !H298,io

For example, for the following reaction,4C(s) + 5H

2(g) ! C

4H

10(g)

"H298

o=# 1( ) #2877.4( )+ #5( ) #285.83( ) + #4( ) #393.51( )$% &'

"H298

o=#125.79 kJ

mol C4H

10 produced

Heat of

combustion

Note, coincidentally this is also standard heat offormation of n-Butane.


35/48

,D9A?FD:DGB 7I *D9B 7I )7:>?AK7G

`7:> )9=7F


36/48

3D:EDF9B?FD 5DEDG;DGCD 7I?%.(

,7AB FD9CK7GA


37/48

)7GHDG


38/48

dH473

o= n

i

rxts473!298

" CpdT +dH298o

+ ni

prd298!473

" CpdT

dH473

o= # n

irxt298!473

" CpdT +dH298o

+ ni

prd298!473

" CpdT

dH473

o=dH

298

o+ n

iprd298!473

" CpdT # nirxt298!473

" CpdT


39/48

dH473

o=dH

298

o+ v

iprd298!473

" CpdT + virxt298!473

" CpdT

dH473

o=dH

298K

o+ #

i" CpdT

dH473

o=dH

298K

o+ $C

p

odT

For stoichiometric reaction, ni=|vi|

Where,< !C

p

o>H

" Mean Heat Capacity for Reaction

and

H=R !A+ !B

2T

0(" +1) + !C

3T

0

2("2 + "+1) + !D"T

0

2#$% &

'( (4.20)

and !M = )iM

i* , andA,B,C, Dare constants for heat capacity.


40/48


41/48

*D9B +SDCBA 7I -G;?ABF


42/48

#7F DT9:E=DP

@89B M B8D C7:>?AK7G 7I % :7= 7I :DB89GD O


43/48

`9A


44/48

V9B8

Energy balance becomes,!H =!H

298

o+!H

prd =0

where

!H298o

= "i

# !H298,f

o

!H298

o= 1( ) $393509( ) + 2( ) $241818( ) + $1( ) $74520( ) + $2( ) 0( )

!H298

o=$802625

J

Mol

and

!Hprd

= ni# CpT1

T2

% dT = H T

2$ T

1( )

Rxt(25oC) ! Prd(25

oC)

Prd Tout

( )

!H

prd

!H

298

o


45/48

!Hprd

= ni" CpT1

T2

# dT = H T2$ T1( )

!Hprd =R !A+ !B

2T1(% +1) + !

C

3T12(%2 + %+1) + !

D

%T1

2

&

'()

*+ T2$ T1( )

!M = n

iM

i" n i is mol of each component in products.

!A =1(5.457) + 0.4(3.639) + 9.03(3.28) +2(3.47) =43.471

!B =[1(1.045) + 0.4(0.506) + 9.03(0.593) +2(1.45)]x10"3 =9.502x10"3

!D =[1("1.157)+ 0.4("0.227)+ 9.03(0.04) +2(0.121)]x105 = "0.645x105

Outlet

1 mol CO2

0.4 mol O2

9.03 mol N2

2 mol H2O

!Hprd

=R 43.471+9.502x10"3

2(298)(#+1) "

0.645x105

#(298)2

$

%&'

()T2"298( )

For T1=298K,


46/48

!?>AKB?BD 9=9GCD

!H ="802625 J

Mol

+ H T

2"298( ) =0

"802625 J

Mol+ 8.314

J

MolK43.471 +

9.502x10"3

2(298)(# + 1) "

0.645x105

#(298)2

$

%&'

() T

2"298( ) =0

T2 = 802625

8.314 43.471+9.502x10

"3

2(298)(# + 1) "

0.645x105

#(298)2

$

%&'

()

+298

T2

=96539

43.471+ 1.416(# + 1) "0.7263#

$%& '()

+298

!H = !H

298

o+ !H

prd = 0


47/48

Solve by iteration (Use Excel or Casio fx-570)Answer: T2= 2066K = 1793oC What will happen to the outlet temperature (T2) if no

excess air is used?

T2 =

96539

43.471+1.416(!+1) "0.7263

!

#

$%&

'(

+298

T2 =298 +

96539

43.471+1.416(T2

298+1) "

0.7263

T2

298

#

$

%%%

&

'

(((

T2 =298 +

96539

44.887 + 0.00475T2"216.437

T2

#

$%&

'(


48/48

2DIDFDGCD

!:>7i ,J ,J P

-GBF7;?CK7G B7 )8D:

ocw chapter 4

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