Microsoft Power Point - 3 - Thermodynamics & Heat Capacities
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Transcript of Microsoft Power Point - 3 - Thermodynamics & Heat Capacities
1
03/07/2008 - 1
THERMODYNAMICSAND
HEAT CAPACITIES
03/07/2008 - 2
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
• 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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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
03/07/2008 - 8
Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
� 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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
HEAT CAPACITY CORRELATIONS
Lafarge standard form:
in kcal/kg.°K
with T : temperature in °K
²²
T
1dcTbTacp +++=
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Thermodynamics & Heat Capacities
( ) ( ) ( )
( ) ( ) ( )
∆ 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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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
Cpmean – Example (Contd.)
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Thermodynamics & Heat Capacities
Cpmean :
= (Q1 + Q2) / (T1 - T0)
= (134.554 + 132.582) / (1273 - 273)
= 0.2671 kcal/kg.°K
Cpmean – Example (Contd.)
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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
� 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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Thermodynamics & Heat Capacities
� 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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Thermodynamics & Heat Capacities
� 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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Thermodynamics & Heat Capacities
� 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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
Exercise 3.1 – WATER VAPOR
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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Thermodynamics & Heat Capacities
Exercise 3.1 – WATER VAPOR
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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Thermodynamics & Heat Capacities
Exercise 3.1 – WATER VAPOR
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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Thermodynamics & Heat Capacities
HEAT BALANCES
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03/07/2008 - 33
Thermodynamics & Heat Capacities
“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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
• 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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Thermodynamics & Heat Capacities
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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Thermodynamics & Heat Capacities
• 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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Thermodynamics & Heat Capacities
Exercise 4.2 : HEAT BALANCE