l4 energy balance reactive system

8/10/2019 L4 Energy Balance Reactive System

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Energy Balance-in Reactive System

Heat of Reaction Method &Heat of Formation Method


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LEARNING OBJECTIVES

By the end of this topic, you should be able to:

Performed energy balance for reactive system


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ENERGY BALANCES ON REACTIVE PROCESSES

ENERGY BALANCES ON REACTIVE PROCESSES (forBoth Methods)

* Generally it is the same as the procedures used in solving the energy

balances for the non-reactive system:

- FLOWCHART!

- MATERIAL BALANCES TO DETERMINE STREAM COMPONENTS

- REFERENCE STATES

- INLET-OUTLET ENTHALPY TABLE

- CALCULATEH

- SOLVE THE ENERGY BALANCES BY INCORPORATING THE HEATOF REACTION INTO THE EQUATION


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ENERGY BALANCE EQUATION

Generally, the energy balance equation for any process unit isgiven as:

Input + Generation - Output - Consumption = Accumulation

Entersthroughsystem

boundaries

Producedwithin

system onlyfor reactivesystem (+)

Leavesthroughsystem

boundaries

Consumedwithin

system onlyfor reactivesystem (-)

Buildup withinsystem only for

transient operation


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Revision on using the heat capacities table(Table B.2)

Heat capacities are functions of temperature and areexpressed in polynomial form as follow:

Cp

= a + bT + CT 2 + dT 3 (Form 1)

Be sure to use the correct function form ( Form 1 orForm 2 ).

Temperature unit is sometimes K and sometimes oC.

If given in the table heading, a x 10 3 = 123,a =123 x 10 -3


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EXAMPLE 1

100 mol/s of propane at 25o

C is completely burned in afurnace with 20% excess air. The air is fed at 300 oC and aflue gases stream at 1000 oC is discharged from thefurnace. 2256 mol/s of nitrogen gas is fed to the furnaceas inlet gas. Calculate the amount of energy produce fromthe combustion.

Furnace

Propane, 25 oC

Air, N 2300 oC

Flue Gas, 1000o

C


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1. Perform the Mass Balance using the stochiometry equation

C 3H8 (g)+ 5 O 2(g) 3 CO 2(g) + 4 H 2O(l)

n in nout

C 3H8 100 -

O 2 600 100

N2

CO 2 -

H2O -

600500

5002.0

500)100(5

excessO

2

2

2

2

22

2

feed O

feed O

used Oused O

used O feed O

EXAMPLE 1


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600 mol O 2(g)/s

2256 mol N 2(g)/s

300 oCkJ/sQ

100 mol O 2(g)/s

2256 mol N 2(g)/s

300 mol CO 2(g)/s

400 mol H 2O (v)/s

1000 oC

Furnace

100 mol/s, Propane, 25 oC

Air, 300 oC

Flue Gas, 1000 oC

EXAMPLE 1


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ininout

out r o H n H n

Standard Conditionat 25oC and 1 atm Respective reaction

temperature

Method 1: Heat of Reaction Method


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1. Calculate the extent of reaction

A

A,inoutA

oor

n-n where

)P,(THH

,

5mol/s 600mol/s 100n-n

2

22

O

in,Oout,O

EXAMPLE 1


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2. Determine the Heat of Reaction at standard condition:

C3H8 (g)+ 5 O 2(g) 3 CO2(g) + 4 H 2O(l)

reactantsfi

o

i products

fi

o

ifi

o

iir

o

H H H H

r o

r o

8322fi

or

o

H

kJ/mol103.8)](285.84)4(393.5)[3(H

)H()H(4)H(3H H(g)HC

f o

(l)OHf

o

(g)COf

o

ii

EXAMPLE 1


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ininout

out r o H n H n

Method 1: Heat of Reaction Method

Proceed

Completed


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3. Choose a reference state: reactants and products at25 oC and 1 atm in states where or is known: [C 3H8(g), O 2(g), CO 2(g) and H 2O(l) in the example] and the

non-reacting at any convenient temperature.

EXAMPLE 1


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4. Performed energy balances

INPUT PRODUCT

n in T H n out T H

C 3H8 100 25 oC 0 - - -

O2 600 300 oC H2 100 1000 oC H4

N2 2256 300 oC H3 2256 1000 oC H5

CO 2 - - - 300 1000 oC H6

H2O - - - 400 1000 oC H7

Ref state for all component: 25 oC and 1 atm

EXAMPLE 1


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Use table B.8 for obtaining the specific enthalpiesexcept for water at ref. point 25 oC and 1 atm.

INPUT PRODUCT

n in T H n out T H

C3H8 100 25 oC 0 - - -

O2 600 300 oC 8.47 100 1000 oC

N2 2256 300 oC 8.12 2256 1000 oC

CO 2 - - - 300 1000 oC

H2O - - - 400 1000 oC H7=?

EXAMPLE 1


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For water, from 25 oC (l) to 1000 oC(v)

dT c H dT c H oC

o

o

o C vapor pvap

C

C liquid p

1000

100,

100

25,7

Liquid Cp=75.4x10 -3 (kJ/mol.K) 5.655

Vapor Cp=33.46x10 -3 + 0.6880x10 -5T+ 0.7604x10 -8T2 - 3.593x10 -12 T3(kJ/mol.K)

Hvap 40.656 kJ/mol 40.656

Total


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INPUT PRODUCT

n in T H n out T H

C3H8 100 25 oC 0 - - -

O2 600 300 oC H2= 8.47 100 1000 oC

N2 2256 300 oC H3= 8.12 2256 1000 oC

CO 2 - - - 300 1000 oC

H2O - - - 400 1000 oC

EXAMPLE 1


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INPUT PRODUCT

n in H nH n out H nH

C3H8 100 0 0 - - -

O2 600 H 2=8.47 5,082 100 H 4=32.47

N2 2256 H 3=8.12 18,318 2256 H 5=30.56

CO 2 - - - 300 H 6=48.60

H2O - - - 400 H 7=81.46

Total 23,400 Total

EXAMPLE 1


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* ENERGY BALANCES ON REACTIVE PROCESSES

ininout

out r o H n H n

skJ

H n H n ininout out

rj jo

/1026.1

400,23)()220,2(1005

EXAMPLE 1


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CLASS ACTIVITY

Solve the previous question

using the Heat of Formation method .


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Reference State for elementary component

: C, O 2, N 2, H 2 at 25 C, 1 atm.

INPUT PRODUCT

n in T H n out T H

C 3H8 100 25 oC H1 - - -

O 2 600 300 oC H2 100 1000 oC H4

N2 2256 300 oC H3 2256 1000 oC H5

CO 2 300 1000 oC H6

H2O 400 1000 oC H7


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CLASS ACTIVITY

Inlet

H1 = Hf C3H8 = -103.8kJ/mol

H2 = Hf O2 + = kJ/mol

H3 = Hf N2 + = 8.12 kJ/mol

300

25CpdT

300

25

CpdT


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CLASS ACTIVITYOutlet

H4

= Hf O2 + = 32.47 kJ/mol

H5 = Hf N2 + = 30.52 kJ/mol

H6 = Hf CO2 + = -393.5 + 48.6 = -344.9 kJ/mol

H7 = Hf H2O +

1000

25

CpdT

1000

25

CpdT

1000

25

CpdT

1000

100

100

25

CpdT HvCpdT

= -285.84 + 5.655 + 40.66 + 35.06 = -204.5 kJ/mol


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Reference State for elementary component

: C, O 2, N 2, H 2 at 25 C, 1 atm.

INPUT PRODUCT

n in T H n out T H

C 3H8 100 25 oC -103.8 - - -

O 2 600 300 oC 100 1000 oC 32.47

N2 2256 300 oC 8.12 2256 1000 oC 30.52

CO 2 - - - 300 1000 oC -344.9

H2O - - - 400 1000 oC -204.5

EXAMPLE 1


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* ENERGY BALANCES ON REACTIVE PROCESSES

ininou t ou t

H n H n

skJ

skJ

/1026.1

/3.1259815

EXAMPLE 1


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CONCLUSION

You have learn:

Energy balance for reactive system

l4 energy balance reactive system

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