Chapter 4. Heat Effects

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(1)

Chapter 4. Heat Effects

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Introduction

Heat Transfer – Common operation in chemical industry

Sensible Heat effect

Latent Heat

Heat of Reaction

Heat of Mixing, Heat of Solution Covered later

(Thermodynamics II)

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4.1 Sensible Heat Effects

Heat transfer to system with no phase transition, no reaction, no change in composition  T changes

V dV dT U

C V dV

dT U T

dU U

T V

T V

 

 

 

 

 

 

 

 

 

• Constant volume process

• Ideal gases

• Approximately zero for low-pressure gases

dT C

dU

V

U

TT12

C

V

dT

(4)

4.1 Sensible Heat Effects

P dP dT H

C P dP

dT H T

dH H

T P

T P

 

 

 

 

 

 

 

 

 

• Constant pressure process

• Ideal gases

• Approximately zero for low-pressure gases

dT C

dH

P

H

TT12

C

P

dT

(5)

4.1 Sensible Heat Effect

Expressions for constant pressure heat capacities

Heat capacity for a substance is a function of T and P

Calculation of Fluid properties

- Based on ideal gas : Ideal gas properties

- Departure from ideal gas : residual properties

Ideal Gas Heat Capacity

- Heat capacity at P  0

- Valid for low pressure gases

2

2

A BT CT DT R

C

Pig

R C

C

V

P

Values A, B, C, and D for many chemical substances  Table C.1

(6)

Mean Heat Capacity

Calculation of enthalpy value

evaluation of integrals are required

Mean Heat Capacity

2

1 2

1 2

1

1 3

2

2 2

3 2

) (

T

T T

T T

T P

CT DT AT BT

dT DT

CT BT

A dT

C H

 

 

   

1 2 1

2

1 2

2 1

2

)

1

( T T

dT C T

T

H T H

T C

T

T P

p

 

 

 

) ( T

2

T

1

C

H

p

Always two temperature values are required !

(7)

Mean Heat Capacity

Tabulation

Based on a reference temperature T

ref

(8)

Mean Heat Capacity

Calculation of Enthalpy

) ( T

2

T

1

C

H

p

) (

) ( ) (

)

(

2 1

1

2 ref p T ref

T

p

T T C T T

C

H    

) (

) (

)

( T

1

T

2

H T T

2

H T T

1

H   

ref

  

ref

(9)

Enthalpy Calculation

1. Heat Capacity

2. Table

3. Mean heat capacity

Heat capacity for mixtures

 ( )

)

( T y C T C

pm i pi

2

1

)

T

(

T

C

pm

T dT H

y

i

 mole fraction of i

2

1 2

1

T T

2 T 2

T

C

P

dT ( A BT CT DT ) dT

H

) T T

( C

H 

p 2

1

 H n

out

H ˆ

out

n

in

H ˆ

in

H ˆ

: Specific enthalpy (from table)

(10)
(11)
(12)

Example 4.2

Calculate the heat required to raise the temperature of 1 mol of

methane from 260

o

C to 600

o

C in a steady-flow process at a pressure sufficiently low that methane may be considered as an ideal gas.

) T .

T .

. ( R

C

p,CH4

 1 702  9 081  10

3

 2 164  10

6 2

From Table C.1

J , dT

) T .

T .

. ( . H

Q

.

.

778 19 10

164 2 10

081 9 702 1 314 8

15 873

15 533

2 6

3

Energy balance for steady-state flow

Q

H 

s fs

W Q m

zg u

H     

 

 

 

  2  2

1

(13)

4.2 Latent Heat of Pure Substances

Heat Effects of Phase Change

(14)

Latent Heat

Phase changes are usually accompanied by large changes in U and H

fusion

vaporization

sublimation

Latent Heat

the enthalpy changes associated with phase transition of a unit amount of a substances applied heat goes to phase changes

 at constant T and P

no temperature change

- Latent heat of vaporization - Latent heat of fusion - Latent heat of sublimation

v

m

s

(15)

Estimation and Correlation of Latent Heat

Clapeyron Equation

Derivation will be given in Chapter 6

slope of vapor pressure curve, volume change  heat of vaporization

dT V dP T H

sat

(16)

Estimation and Correlation of Latent Heat

Standard Heat of Vaporization

Trouton’s Rule : Rough estimates (30 % accuracy for hydrocarbons)

Chen’s equation (2 % accuracy for hydrocarbons)

Standard Heat of Fusion

(K) 109

0

(K) 088

0 (kJ/mol)

b b

v

. T

T H ˆ .

 

) / ( 07 . 1

) log

0297 .

0 0327 .

0 ) / ( 0331 .

0 (kJ/mol) (

ˆ

10

c b

c c

b b

v

T T

P T

T H T

 

(K) 050

0

(K) 025 0

(K) 0092 0

(kJ/mol)

m m m m

T .

T .

T .

H ˆ

(Non-polar liquids) (water, alcohols)

(metallic elements) (inorganic compounds)

(organic compounds)

(17)

Heat of Vaporization at other temperature

Watson Correlation

38 . 0

1 2 1

2

) ( )

(  

 

 

T T

T T T

H T

H

c c v

v

38 . 0

1 2 1

2

1

) 1 ( )

(  

 

 

r r v

v

T

T T H T

H

(18)

4.3 Standard Heat of Reaction

Heat effects accompanying when reaction occur

Reaction may be carried out in many different ways

Tabulation of all possible reactions is impossible

Convenient way

- “Standard way”  Q = H, and H is the state function - The required data to be a minimum

- Use Hess’s Law

(19)

Heat of Reaction

Heat of Reaction (Enthalpy of Reaction)

Reactants : stoichiometric quantities

Complete Reaction

Reactants are fed at T, P

Products are emerging at T, P

ex) CaC

2

(s) + 2 H

2

O (l)  Ca(OH)

2

(s) + C

2

H

2

(g)

reactants products

r

( T , P ) H H

H ˆ  

mol /

kJ 4 . 125 )

atm 1

, C 25 ( H ˆ

o

r

 

(20)

Standard Condition

A standard state is a particular state of a species at temperature T

and at specified conditions of pressure, composition and physical condition as, e.g., gas, liquid or solid.

• Standard state pressure : 1 atm (101,325 Pa)

• Standard state temperature : 25

o

C (298.15 K)

• Gas : The pure substance in ideal gas state at 1 bar

• Liquid and solid : The real pure liquid or solid at 1 bar

(21)

Heat of Reaction : Per mole of what ?

Example

2A + B  3C

Extent of Reaction, x

Amount of reactants with stoichiometric coefficient of “1” that have been reacted.

produced C

mol 3

50 reacted

B mol 1

50 reacted

A mol 2

ˆ 50 kJ kJ kJ

H

r

     

A A

r

n H H

   n

A

: moles of A consumed or produced

A

: the stoichiometric coefficient of A kJ

H ˆ

r

  50

x

 

 

A r

A

r

n H

H H

(22)

Properties of Heat of Reaction

“Standard” heat of reaction

at reference T and P (25

o

C, 1 atm)

Exothermic : Endothermic :

Value depends on stoichiometric eqn.

CH

4

(g) + 2 O

2

(g)  CO

2

(g) + 2H

2

O (l) H

r

(25

o

C) = -890.3 kJ/mol 2CH

4

(g) + 4 O

2

(g)  2CO

2

(g) + 4H

2

O (l) H

r

(25

o

C) = -1780.6 kJ/mol

Value depends on the state (gas, liquid, solid)

CH

4

(g) + 2 O

2

(g)  CO

2

(g) + 2H

2

O (l) H

r

(25

o

C) = -890.3 kJ/mol CH

4

(g) + 2 O

2

(g)  CO

2

(g) + 2H

2

O (g) H

r

(25

o

C) = -802.6 kJ/mol

 0

H

r

 0

H

r

) C (

H or

H

ro

r

25

o

(23)

Measurement of Heats of Reaction

Measurements of heats of reaction : Calorimeter

Temperature rise or fall of the fluid can be measured and heat of reactions are determined.

A: Small funnel B: Styrofoam lid

C: Two styrofoam cups glued together D: Magnetic stir bar (inside the cups) E: Magnetic stir plate

F: Temperature probe G: Thermometer

(24)

Difficulties for measuring heat of reaction

 Some reactions cannot be accomplished.

 Ex) 2C + O

2

(g)  2 CO (incomplete combustion)

It is impossible to determine the heat of the incomplete combustion only.

The reaction cannot proceed at such low temperature.

 Alternative method

 C + O2  CO2Hr = -393.51 kJ/mol (Easily measured)

 CO + ½ O2  CO2Hr = -282.99 kJ/mol (Easily measured)

C + ½ O

2

(+ ½ O

2

) CO (+ ½ O

2

)

CO

2

3

ˆ

0r

H

0 1

ˆ

r

H H  

  H    H ˆ

0r2

kJ mol

H H

H

r r r

/ 52 . 110

99 . 282 51

. 393

ˆ ) ˆ (

ˆ

0 3 0 1 0 2

(25)

Hess’s Law

If the stoichiometric equation for a reaction can be obtained by algebraic operations (+, –, ×, / ) on the other stoichometric equations (2,3,…), then the heat of reaction can be obtained by performing the same operations on the heats of reactions (2,3,…).

H is state property

 the enthalpy change for a reaction, H

r

, is the same whether it occurs by one step or by a series of steps.

 Only depends on the initial and final state.

(26)

4.4 Heats of Formation

Formation reaction :

Reaction which the compound is formed from its atomic constituents.

Normally occur in nature (O

2

instead of O)

Heat of formation

H associated with the formation of 1 mole of the compound.

Reference state: 25

o

C and 1 atm

Examples

Ammonium Nitrate :

- N2(g) + 2H2(g) + 3/2 O2(g)  NH4NO3 (c )

Benzene :

- 6C (s) + 3H2 (g)  C6H6 (l)

mol / kJ .

H ˆ

of

  365 14

mol / kJ .

H ˆ

of

 48 66

H ˆ

f

(27)
(28)

Determination of Heats of Reaction using Heats of Formation

Heats of reactions can be determined from heats of formation using Hess’s Law

Example

C

5

H

12

(l) + 8O

2

(g)  5CO

2

(g) + 6H

2

O (l)

reactants

0 products

0

0

( ˆ ) ( ˆ )

ˆ

r i

H

f i i

H

f i

H  

) l ( H f C o )

l ( O f H

o )

g ( f CO r o

o

( C H ) ( H ˆ ) ( H ˆ ) ( H ˆ )

H ˆ

12 5 2

2

6

12

5

5

  

   

(29)

4.5 Standard Heats of Combustion

Standard Heat of Combustion

Heat of reaction of the substance with oxygen to yield specified products

Condition : 25

o

C, 1 atm

Products : CO

2

(g), H

2

O (l), SO

2

(g), NO

2

(g)

Using Hess’s Law heat of reaction can be calculated from heats of combustion

Example

- C2H6 C2H4+ H2

products reactants

c i o i

c i o r i

o

( H ˆ ) ( H ˆ )

H ˆ    

2 4

2 6

2 oc H

H c C o H

c C r o

o

( H ˆ ) ( H ˆ ) ( H ˆ )

H ˆ      

(30)

4.6 Temperature Dependence of H o

General Procedure

Choice of reference conditions 1. Heat of Reactionmethod

– Usually for single reaction, Hr is known

– Reactants and products : T0 where Hr is known – Non-reactive species (e.g., N2) : any convenient T

2. Heat of Formationmethod

– Usually for multiple reactions, Hris unknown – Reactants and Products :elements at 25 oC

» Use sum of heats of formation

– Non-reactive species (e.g., N2) : any convenient T

reactant product

i i i

i A

or

AR

H ˆ n H n H

H n

 

reactant product

i i i

i

H n H

n

H

nA: moles of A consumed or produced

A: the stoichiometric coefficient of A

(31)

Method 1 vs. method 2

Heat of reaction method

Heat of formation method

Reactants T

in

Products T

out

H

Reactants 25

o

C

Products 25

o

C

H

ro

reactant product

i i i

i A

r o

AR

H ˆ n H n H

H n

 

Reactants T

in

Products T

out

H

Elements 25

o

C

 H H

of

C

p

dT

reactant product

i i i

i

H n H

n

H

(32)

H Calculations

In the textbook of Thermodynamics

Table C.1 – C3 : Heat capacity of gases, solids, and liquids (constants – A, B, C, D)

Table C4 : Heat of formation

Table F.1 – F.4 : Steam table

In the textbook of Introduction to Chemical & Biological Engineering

Table B.1 : Latent heat, heat of formation, heat of combustion

Table B.2 : Heat capacity (constants – A, B, C, D)

Table B.5 – B.7 : Steam table

Table B.8 & B.9 : Specific enthalpy for simple gases (Air, O2, N2, H2, CO, CO2, H2O)

Table B.10 : Heat capacity for Kopp’s rule

 It is convenient to prepare the enthalpy table.

substance nreactant Hreactant nproduct Hproduct

NH3 n1 H1

O2 n2 H2 n3 H3

NO n4 H4

H2O n5 H5

(33)

Example 4.7

What is the maximum temperature that can be reached by the combustion of methane with 20 % excess air? Both the methane and the air enter the burner at 25

o

C.

Processes with unknown outlet conditions : adiabatic reactors

Adiabatic reactor

- No heat exchange with surrounding  maximum temperature

Q = H = 0, then solve for unknown T

(34)

Example 4.7 - solution

The standard heat of reaction is

CH4(g) + 2 O2(g)  CO2 (g) + 2H2O (g) Hr(25 oC) = -802,256 J/mol Since Hr(25 oC) is known, use the heat of reaction method

Reactants Tin = 25 oC

Products Tout = ?

H

Reactants 25 oC

Products 25 oC

H

ro

Reactants 1 mol CH4 2.4 mol O2 9.03 mol N2

Products 1 mol CO2 2 mol H2O 0.4 mol O2 9.03 mol N2

20 % excess air?

reactant product

i i i

i A

or

AR nH nH

H n

 

(35)

Example 4.7 - solution

Using the mean heat capacity,

Enthalpy table

substance nin Hin nout Hout

CH4 1 0 - -

O2 2.4 0 0.4 34.42(T-298.15)

N2 9.03 0 9.03 32.58(T-298.15)

H2O - - 2 41.42(T-298.15)

CO2 - - 1 52.77(T-298.15)

K mol / J . C

K mol / J . C

K mol / J . C

K mol / J . C

CO , p

O H , p

N , p

O , p

77 52

42 41

58 32

42 34

2 2 2 2

) ( T

2

T

1

C

H

p

K . T

T . .

H n H

H ˆ n

H n

i i i i

A or AR

8 2106 0

6 443 6

934585

0

reactant product

 

* Repeat this problem yourself using the heat capacity (use the value A, B, C, and D from Table C.1)

2

2

A BT CT DT R

CPig

(36)

Example 4.8

One method for the manufacture of “synthesis gas” (a mixture of CO and H2) is the catalytic reforming of CH4 with steam at high temperature and atmospheric pressure:

CH4(g) + H2O(g)  CO (g) + 3H2 (g)

The only other reaction to be considered is the water-gas-shift reaction:

CO(g) + H2O(g)  CO2 (g) + H2 (g)

If the reactants are supplied in the ratio, 2 mol steam (H2O) to 1 mol CH4, and if heat is supplied to the reactor so that the products reach a temperature of 1300K, the CH4 is completely converted and the product stream contains 17.4 mol % CO. Assuming the reactants to be preheated to 600K, calculate the heat requirement for the reactor.

(37)

Example 4.8 - solution

Since there are two reactions, it is convenient to use the heat of formation method.

Reactants T

in

= 600K

Products T

out

= 1300 K

H

Elements 25

o

C

H C dT H

of p

reactant product

i i i

i

H n H

n

H

Reactants 1 mol CH4 2 mol H2O

Products

CH4 = 1 – x1 = 0

H2O = 2 – x1 – x2 = 1 – x2 CO = x1 – x2 = 1 – x2 CO2 = x2

H2 = 3 x1 + x2 = 3 + x2 Total = 5 mol

CH4(g) + H2O(g)  CO (g) + 3H2 (g) x1

CO(g) + H2O(g)  CO2 (g) + H2 (g) x2

To find the amount of product, use the extent of reaction that you have learned from “Intro. to Chem & Bio Eng”

CH4 is completely converted  x1 = 1 mol

(38)

Example 4.8 - solution

Products

CH4 = 1 – x1 = 0

H2O = 2 – x1 – x2 = 1 – x2 = 0.87 mol CO = x1 – x2 = 1 – x2 = 0.87 mol CO2 = x2 = 0.13 mol

H2 = 3 x1 + x2 = 3 + x2 = 3.13 mol Since the product stream contains 17.4 mol % CO,

(1 – x2)/5 = 0.174 x2 = 0.13 mol

substance nin Hin nout Hout

CH4 1 H1 0 0

H2O 2 H2 0.87 H3

CO - - 0.87 H4

CO2 - - 0.13 H5

H2 - - 3.13 H6

H C dT H

of p

K mol / J . C

K mol / J . C

K mol / J . C

K mol / J . C

H , p

CO , p

CO , p

O H , p

78 29

85 49

70 31

66 38

2 2 2

) T .

T .

. ( R

Cp,CH4 17029081103 2164106 2

(39)

Example 4.8 - Quiz

Perform the rest of calculation to obtain Q = H.

0

509 393

525 110

830 285

520 74

2 2 2

4

o H , f o

CO , f o

CO , f o

O H , f o

CH , f

H

mol / J , H

mol / J , H

mol / J , H

mol / J , H

Heat of formation at 298 K is

(from Table C.4)

(naturally occurring substance)

(40)

Example 4.8 - solution

0

509 393

525 110

830 285

520 74

2 2 2

4

o H , f o

CO , f o

CO , f o

O H , f o

CH , f

H

mol / J , H

mol / J , H

mol / J , H

mol / J , H

Heat of formation at 298 K is

(naturally occurring substance)

mol / J , ) (

. H

H

mol / J , )

( . H

H

mol / J , )

( . H

H

mol / J , )

( . H

H

mol / J , )

( . H

H

mol / J , dT

) T .

T .

. ( . H

H

o H , f o

CO , f o

CO , f o

O H , f o

O H , f o

CH , f

588 47 298

1300 78 29

559 343 298

1300 85 49

850 115 298

1300 70 31

092 247 298

1300 66 38

154 274 298

600 66 38

145 61 10

164 2 10

081 9 702 1 314 8

2 6

2 5

4

2 3

2 2

600

298

2 6 3

4 1

J ,

H n H

n Q

H

i i i i

980 397

reactant product

  

(41)

Homework

Problems

4.2, 4.5, 4.21 [(a),(b),(c)], 4.32, 4.38, 4.41

Due:

Other Recommend Problems

4.35, 4.39, 4.46 [(a), (b), (c), (d)]

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