Chapter 4. Heat Effects

(1)

Chapter 4. Heat Effects

(2)

(3)

T V

T V

V

TT12

V

(4)

T P

T P

P

TT12

P

(5)

 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

Pig

V

P



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

(6)

2

1 2

1 2

1

1 3

2

2 2

T

T T

T T

T P

1 2 1

2

1 2

2 1

2

1

T

T P

p

2

1

p



Always two temperature values are required !

(7)

ref

(8)

2

1

p

2 1

1

2 ref p T ref

T

p

1

2

2

1

ref

ref

(9)

Enthalpy Calculation

1. Heat Capacity

2. Table

3. Mean heat capacity

Heat capacity for mixtures

pm i pi

2

1

T

T

pm

i

2

1 2

1

T T

2 T 2

T

P

p 2

1

out

out

in

in

Hˆ

: Specific enthalpy (from table)

(10)
(11)
(12)

o

o

p,CH4

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

s fs

W Q m

zg u

H     

 

 

 

  2  2

1

(13)

(14)

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)

 Derivation will be given in Chapter 6

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

sat

(16)

b b

v

10

c b

c c

b b

v

m m m m



(Non-polar liquids) (water, alcohols)

(metallic elements) (inorganic compounds)

(organic compounds)

(17)

38 . 0

1 2 1

2

c c v

v

38 . 0

1 2 1

2

r r v

v

(18)

 Convenient way

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

- Use Hess’s Law

(19)

2

2

2

2

2

(g)

reactants products

r

o

r

(20)

o

(21)

r

A A

r

A

A

r

A r

A

r

(22)

o

4

2

2

2

r

o

4

2

2

2

r

o

4

2

2

2

r

o

4

2

2

2

r

o

r

r

ro

r

o

(23)

 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)

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)

2

2

2

2

3

0r

0 1

r

0r2

r r r

0 3 0 1 0 2

(25)

r

(26)

2

o

 Ammonium Nitrate :

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

 Benzene :

- 6C (s) + 3H2 (g)  C6H6 (l)

of

of

f

(27)
(28)

5

12

2

2

2

reactants

0 products

0

0

r i

f i i

f i

) l ( H f C o )

l ( O f H

o )

g ( f CO r o

o

12 5 2

2

12

5

(29)

o

2

2

2

2

- C2H6 C2H4+ H2



products reactants

c i o i

c i o r i

o

2 4

2 6

2 oc H

H c C o H

c C r o

o

(30)

 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

reactant product

i i i

i

 H

nA: moles of A consumed or produced

A: the stoichiometric coefficient of A

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in

out

o

o

ro

reactant product

i i i

i A

r o

AR

in

out

o

of

p

reactant product

i i i

i

(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

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o

 Processes with unknown outlet conditions : adiabatic reactors

- No heat exchange with surrounding  maximum temperature

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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 = ?

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

2

1

p

i i i i

A or AR

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.

in

out

o

of p

reactant product

i i i

i

 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

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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

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)

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

i i i i

reactant product

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