Microsoft Power Point - 3 - Thermodynamics & Heat Capacities

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THERMODYNAMICSAND

HEAT CAPACITIES

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Thermodynamics & Heat Capacities

LEARNING OBJECTIVES

� At the end of the day the trainee shall be able to :

- Perform simple heat balances

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W

∆∆∆∆T ∝∝∝∝ W

Heat is a form of energy

THE JOULE’S EXPERIMENT

W: work done on the water

DT: change in water temperature

Joule found that a definite amount

of work was required per unit

mass of water for every degree of

temperature rise caused by the

stirring and there was a

quantitative relationship between

work and heat and, therefore, that

heat is a form of energy.

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• For a closed (constant-mass) system:

DE = Q – WDE : Total energy change of the system

Q : Heat added to the systemW : Work done by the system

• The total energy change of a closed system is equal to the heat transferred to the system minus the work done by the system.

W

Q

THE FIRST LAW OF THERMODYNAMICS

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THE FIRST LAW OF THERMODYNAMICS

The first law of thermodynamics may be stated in

many other ways -

One of these is as follows :

� Although energy assumes many forms, the total

quantity of energy is constant, and when energy

disappears in one form it appears simultaneously

in other forms.

The first law of thermodynamics is a generalization of the law of the conservation of energy. This is a law to

be obeyed by every process engineers.

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INTERNAL ENERGY

• Internal Energy (∆∆∆∆U) :

Energy of the molecules making up the substance

∆E = ∆U + ∆Ek + ∆Ep ∆E = total energy

∆U = ∆E - ∆Ek + ∆Ep ∆U = Internal Energy

∆Ek = kinetic energy

∆Ep = potential energy

• If no change in kinetic and potential energy:

∆E = ∆U = Q - W

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ENTHALPY

•The sum of the internal energy and the product of pressure and volume appears frequently.

That combination is called enthalpy :

• H = U + pV

• ∆H = ∆U + p∆V

•Steam tables and other property tables usually give energy as enthalpy

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ENTHALPY

∆H Q m C dTP= = ∫

∫∫∫ =+=∆100

0

100

50

50

0dTCmdTCmdTCmH PPP

∫∫∫ =−=∆100

0

200

100

200

0dTCmdTCmdTCmH PPP

Based on the standard state at a given temperature

Can be added :

H(0,100) = H(0,50)+H(50,100)

= H(0,200)-H(200,100)

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HEAT CAPACITY

• m = throughput - [kg or Nm³]

• cp = specific heat - [kJ/kg x T°K or kJ/Nm³ x T°K]

at constant pressure

• ∆t = difference of temperature

between the measured

temperature and the

reference of the heat

balance (in °C)

∆∆∆∆h = m cp ∆∆∆∆t [kJ]

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C1

n

dQ

dT= ⋅ Q nC dT= ∫⇒⇒⇒⇒

Heat Capacity of Oxygen

0.21

0.22

0.23

0.24

0.25

0.26

0.27

0 200 400 600 800

Temperature (°C)

Cp

(k

cal/kg

.°K

)

� Heat capacity: the quantity of heat required

to raise the temperature of a given mass (or

mole) of a material by 1 degree.

HEAT CAPACITY

More practically, the heat required

to change the temperature fromT1

to T2 is the area under the curve

(the integral) between those two

temperatures.

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� Heat capacity is on mole basis (kcal/kgmole.°C)

� Some people will use C’ for heat capacity

� Specific heat is the heat capacity per unit mass

(kcal/kg.°C)

� C will be used for specific heat

� Could be confusing. Always make sure of the units

Q n C d T= ∫ '

Q m C d T= ∫

HEAT CAPACITY VS SPECIFIC HEAT

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CV : heat capacity at constant volume

CP : heat capacity at constant pressure

For an ideal gas :

∆U Q m C dTV= = ∫

∆H Q m C dTP= = ∫

C C RP V' '− =

Cp vs Cv

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HEAT CAPACITY OF A MIXTURE

cp mixture = xAcpA + xBcpB + xCcpC

where

xA, xB, xC : mass fractions of component A, B, C

cpA, cpB, cpC : specific heat (mass) of A, B, C

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REFERENCE TEMPERATURE

•For heat calculation, any reference temperature is valid as long as consistent and specified.

•Recommended : (Convenient)

0°C = 273.15°K = 32°F = 491.67°R

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HEAT CAPACITY CORRELATIONS

Lafarge standard form:

in kcal/kg.°K

with T : temperature in °K

²²

T

1dcTbTacp +++=

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( ) ( ) ( )

( ) ( ) ( )

∆ H Q C d T a b T cTd

TdT

a Tb

Tc

Td

T

a T Tb

T Tc

T T dT T

C

a T Tb

T Tc

T T dT T

T T

P

T

T

T

T

T

T

P m ea n

= = = + + +

= + + −

= − + − + − − −

=

− + − + − − −

−

∫ ∫0

1

0

1

0

1

2

2

2 3

1 0 12

02

13

03

1 0

1 0 12

02

13

03

1 0

1 0

2 3

2 3

1 1

2 3

1 1

( ) ( )∆H Q C T T T TPmean= = × −1 0 1 0,

Cpmean

• CP(T) gives the rate of heat content at a given temperature.

• Absolute temperature is required (ºK)

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Cpmean - Example

Calculate CPmean of nitrogen at 1000°C using the Lafarge standard correlation and the constants given below.

Equation constants :

T°limit a b c d

298 to 800°K 2.32210E-1 2.14938E-5 2.76050E-8 6.53840E+2

800 to 2200°K 2.42450E-1 5.44079E-5 -1.05755E-8 -7.11335E+3

1000°C = 1273°K

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Cpmean – Example (Contd.)

First Integration –

From 0°C to 527°C (273°K to 800°K) :

Q1 = [(2.32210E-1)*(800-273)] +

[((2.14938E-5)/2)*(800^2-273^2)] +

[((2.76050E-8)/3)*(800^3-273^3)] –

[(6.53840E+2)*(1/800-1/273)]

= 122.375 + 6.077 + 4.524 + 1.578

= 134.553 kcal/kg

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Second Integration –

From 527°C to 1000°C (800°K to 1273°K) :

Q2 = [(2.42450E-1)*(1273-800)] +

[((5.44079E-5)/2)*(1273^2-800^2)] +

[((-1.05755E-8)/2)*(1273^3-800^3)] –

[(-7.11335E+3)*(1/1273-1/800)]

= 114.679 + 26.674 - 5.467 - 3.304

= 132.582 kcal/kg


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Cpmean :

= (Q1 + Q2) / (T1 - T0)

= (134.554 + 132.582) / (1273 - 273)

= 0.2671 kcal/kg.°K


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THERMODYNAMIC ADD – INCpmean, Cpgas and Cpmat

� Return Cpmean in kcal/kg.ºC

� Standard within Lafarge group

� For all typical gases in cement industry:� O2, N2, H2, SO2, CO, A, Air, Water vapor

� For most materials in cement industry:� SiO2, Al2O3, Fe2O3, CaCO3, CaO, MgCO3,

MgO, CaSO4� Coal, Steel, Raw mix, clinker

� Usage:=Cpgas("CO2",N17)*N17

where :N17 cell contains temperature in °C

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� Calculates the enthalpy of a material at a given

temperature

� Equivalent to

∆H = Cpmean(T) x T

� Returns ∆H in kcal/kg

� Usage:

= Enthalpy("CO2",N17)

where : N17 cell contains temperature in °C

THERMODYNAMIC ADD – INENTHALPY

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� Returns temperature from enthalpy in ºC

� Usage:

= Temperature("CO2",N17)

where : N17 cell contains enthalpy in kcal/kg

THERMODYNAMIC ADD – INTEMPERATURE

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� Enthalpy change between two temperatures T1 and T2:

CP

T0 T1 T2

Q1 Q2

∆H Q C dT C dT C dTP

T

T

P

T

T

P

T

T

= = = −∫ ∫ ∫1

2

0

2

0

1

THERMODYNAMIC ADD – INCpmean, Cpgas and Cpmat

= Cpgas("CO2",T2)*T2 - Cpgas("CO2",T1)*T1 or

= Enthalpy("CO2",T2) - Enthalpy("CO2",T1)

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� Watson correlation:

� Tc : critical temperature

� Critical temperature of water : 374.2°C

∆

∆

H

H

T T

T T

c

c

2

1

2

1

0 38

=−

−

.

PHASE CHANGE

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What is the heat of vaporization of water at 50ºC

knowing that it is 538.7 kcal/kg at 100ºC?

Tc = 374.2ºC

At which temperature is the heat of vaporization 0.0

kcal/kg?

∆H2

0 38

538 7374 2 50

374 2 100574 1= ×

−

−

=.

.

..

.

kcal / kg

PHASE CHANGE - Example

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Temperature (°C)

He

at

(kca

l/k

g)

0 100 200

liquid

vapor

evaporation

Temperature (°C)H

ea

t (k

ca

l/k

g)

0 100 200

liquid

vapor

evaporation

ENTHALPY OF WATER VAPOR

• The standard state of water at reference temperature (0°C) is liquid.

• Water vapor contains sensible heat (heat to raise vapor from 0°C to the specified temperature) and latent heat (heat to evaporate the liquid).

• Many paths are equivalent -

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What is the heat content of water vapor at 200°C using 0°C as the reference temperature?

Exercise 3.1 – WATER VAPOR

Question :

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Solution :

Path-1: Evaporation at 100°C

heat to raise liquid water from 0°C to 100°C : 100.31 kcal/kg

heat to evaporate water 100°C : 538.70 kcal/kg

heat to raise vapor water from 100°C to 200°C : 45.69 kcal/kg

684.70 kcal/kg

Temperature (°C)H

ea

t (k

ca

l/k

g)

0 100 200

liquid

vapor

evaporation

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Solution : (contd.)

Temperature (°C)

He

at

(kca

l/k

g)

0 100 200

liquid

vapor

evaporation

Path-2 : Evaporation at 200°C

heat to raise liquid water from 0°C to 200°C : 202.75 kcal/kg

heat to evaporate water 200°C : 453.40 kcal/kg

656.15 kcal/kg

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Path-1: Evaporation at 100°C 684.70 kcal/kg

Path-2 : Evaporation at 200°C 656.15 kcal/kg

Difference : 4.2% => The method according to first path was found accurate against Steam Tables

Solution : (contd.)

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HEAT BALANCES

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“Energy cannot be created or destroyed but may be converted from one form to another”

Energy in = Energy out

Σ massin = Σ massout

MASS BALANCE

Σ heatin = Σ heatout

HEAT BALANCE

HEAT BALANCE

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3 - Point Junction : Balance around three points

HEAT BALANCE

OR

1

2

3

m1c1+ m2c2= m3c3m1h1+ m2h2= m3h3m1 + m2 = m3

Component BalanceHeat BalanceMass Balance

In order to solve a 3 – junction problem it is sufficient to know only e.g. mass flow, but knowing one property of

all 3 flows (e.g. heat, O2 content, Blaine)

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• In Fredonia, the air for the coal mill is taken from the kiln hood. To prevent excessive temperature to the mill, some cold air is mixed with that hot air with a bleed-in damper. The only location where the air flow could be measured is after that bleed-in damper.

• On one occasion, the combined flow was 12611 Nm³/h and its temperature was 110°C. The air from the kiln hood was at 450°C and the ambient air 25°C. Calculate the air flow from the kiln hood.

Exercise 4.1 : 3 - POINT JUNCTION

Question :

kiln hood

450°C

coal mill

110°C

ambiant air25°C

cp=1,0339 kJ/kg.K

cp=1,0057 kJ/kg.K

cp=1,0071 kJ/kg.K

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Exercise 4.1 : 3 - POINT JUNCTION

Solution :

1805503

16298

12611

1.0071

110

Coal

Mill

1.00571.0339Cp (kJ/kg K)

101572454V (Nm3/h)

03300371475467h (kJ/h)

0131273171m (kg/h)

25450Temperature (°C)

Balance

Check

Ambient

Air

Kiln

Hood

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• Malogozscz plant in Poland is planning to build a Cl-Bypass, where part of the kiln inlet gas is extracted, cooled by fresh air in a mixing chamber and de-dusted in a bag filter. In parallel the plant is executing a second project which covers the complete re-design of the gas conditioning tower (hereinafter called GCT) plus a new kiln filter. The clean bypass gas is mixed with the preheater exit gas after ID fan before GCT. The mixed gases are conditioned in the GCT and de-dusted in the kiln filter before going to the stack.

• The bypass system is well sized, but the kiln filter and GCT need to be mathematically checked, if the proposed water amount for injection is sufficient (11 [m³/h]) for the proposed bag quality (P84: continuous operation temperature 220 [°C]) and the proposed filtration area (6400 [m²]) is sufficient for an air to cloth ratio of 1 [m³/(min m²)].

Exercise 4.2 : HEAT BALANCEQuestion :

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Exercise 4.2 : HEAT BALANCE

Microsoft Power Point - 3 - Thermodynamics & Heat Capacities

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