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! A common practice is to arbitrarily designate a reference state for a
substance at which U or H is declared to equal zero, and then tabulateU and/or H for the substance relative to the reference state. *Note: ReferFelder pp. 339 and 359
! In Chapter 7 (Felder), U and H are state properties of a species; theirvalues depend only on the state of the species – primarily on itstemperature and state of aggregation (solid, liquid or gas) and, to a
lesser extent, on its pressure (and for mixtures of some species, on itsmole fraction in the mixture).
! When a species passes from one state to another, both ΔU and ΔH for
the process are independent of the path taken from the first state tosecond one – Hypothetical Process Path *Note: Refer Felder pp. 360
CHAPTER 4- ENERGY balance for non
reactive system
Reference State
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CHANGES IN P AT CONSTANT T
CHAPTER 4- ENERGY balance for non
reactive system
CHANGES IN T AT CONSTANT P
PHASE CHANGE OPERATIONS
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CHAPTER 4- ENERGY balance for non
reactive system
! Internal energy (U ) is nearly independent of pressure for solids andliquids at a fixed temperature, as is specific volume (V ) . *Note: ReferFelder pp. 365 – 366
! If pressure of a solid and liquid changes at constant temperature
ΔU = 0
ΔH = [ΔU + Δ(PV)] = [ΔU + PΔV + V ΔP] = [V ΔP]
! Both U and H independent of pressure for ideal gases – may assumeΔU = 0 and ΔH = 0 for a gas undergoing an isothermal pressure change
unless gas temperature below 0 0C or well above 1 atm are involved.
CHANGES IN P AT CONSTANT T
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CHAPTER 4- ENERGY balance for non
reactive system
SENSIBLE HEAT AND HEAT CAPACITIES
HEAT CAPACITY FORMULAS
ESTIMATION OF HEAT CAPACITIES
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CHAPTER 4- ENERGY balance for non
reactive system
! The term sensible heat signifies that heat must be transferred to raiseor lower the temperature of a substance or mixture of substances.
*Note: Refer Felder pp. 366
! The quantity of heat required to produce a temperature change:
Q = ΔU (closed system)
Q = ΔH (open system)
! Heat capacity at constant volume – Cv. At constant volume:
SENSIBLE HEAT AND HEAT CAPACITIES
( )dT T C U
T
T
v
^
∫ =Δ
2
1
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CHAPTER 4- ENERGY balance for non
reactive system
! Suppose both temperature and the volume of a substance change. Tocalculate ΔU – break the process into 2 steps ( a change in V at
constant T followed by a changes in T and constant V):
SENSIBLE HEAT AND HEAT CAPACITIES
( ) ( ) ( )
21
222111
21
^ ^ ^
U U
U U U
V ,T AV ,T AV ,T A^ ^
Δ+Δ=Δ
⎯ ⎯→ ⎯ ⎯ ⎯→ ⎯ ΔΔ
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CHAPTER 4- ENERGY balance for non
reactive system
! For ideal gas and (to a good approximation) liquid and solids, U depends only on T. In step 1, T is constant, ΔU 1 = 0.
Step 2 – V is constant:
SENSIBLE HEAT AND HEAT CAPACITIES
( )dT T C U
T
T
v
^
∫ =Δ2
1
Ideal gas: ExactSolid or liquid: good approximationNon ideal gas: valid only if V is constant
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CHAPTER 4- ENERGY balance for non
reactive system
• Heat capacity at constant pressure – C p. At constant pressure:
SENSIBLE HEAT AND HEAT CAPACITIES
( )dT T C H T
T
p
^ ∫ =Δ2
1
! For first step – refer section 8.2 (Felder), as T is constant, ΔH 1 = 0 (for
ideal gas),Δ
H 1 = V Δ
P (for solid or liquid). Step 2 – P is constant:
( )dT T C H
T
T
p
^
∫ =Δ2
1
Ideal gas: ExactNon ideal gas: valid only if P
is constant
( )dT T C P V H
T
T
p
^ ^
∫ +Δ=Δ2
1
Solid or liquid
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CHAPTER 4- ENERGY balance for non
reactive system
! Example
Calculate the heat required to raise 200 kg nitrous oxide from 200C to1500C in a constant – volume vessel. The constant – volume heatcapacity in this temperature range is given by the equation:
CinT whereT 109.420.855CkJ/kgC
0
40v
−
×+=⋅
! Solution
For vessel – closed system. Q = ΔU
For constant volume: ( )dT T C U
T
T
v
^
∫ =Δ2
1
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CHAPTER 4- ENERGY balance for non
reactive system
! Solution
( )
( ) ( )[ ]
( )
kJ 31224kg200kgkJ 121.5591U Δ
kg
kJ 10.4091111.15U Δ
20150104.7120-1500.855T 2109.42 0.855T U Δ
dT T 109.420.855U Δ
^
^
224-
150
20
24-^
150
20
4-^
=×=
+=
−×+=⎥⎦
⎤⎢⎣
⎡ ×+=
×+= ∫
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CHAPTER 4- ENERGY balance for non
reactive system
! Heat capacities are functions of temperature and frequently expressedin polynomial form (C p = a + bT + cT 2 + dT 3). *Note: Refer Felder pp. 369
constantGas:R
RCC: GasesIdeal
CC:SolidsandLiquid
v p
v p
+=
=
HEAT CAPACITY FORMULAS
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CHAPTER 4- ENERGY balance for non
reactive system
! Example (Felder pp369)
Assuming ideal gas behavior, calculate the heat must be transferred ineach of the following cases.
1. A stream of nitrogen flowing at a rate of 100 mol/min is heatedfrom 200C to 1000C
2. Nitrogen contained in 5 – liter flask at an initial pressure of 3 baris cooled from 900C to 300C
Neglect the changes in kinetic energy
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CHAPTER 4- ENERGY balance for non
reactive system
! Solution
Case 1
System: open system → Q = ΔH
From Table B.2, Appendix B, the heat capacity of N2 at constantpressure of 1 atm:
3-122-8-50 p T 102.871T 100.5723T 100.21990.02900CmolkJ/C ×−×+×+=⋅
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CHAPTER 4- ENERGY balance for non
reactive system
! Solution
( )
( ) ( ) ( ) ( )[ ]
minkJ 233.3
minmol100
molkJ 2.333H ΔQ
Finally,
kJ/mol2.333H Δ
0.000070.0020.0112.32H Δ
20100100.717820100100.190820100100.110020-1000.02900H Δ
T 4
102.871T
3
100.5723 T
2
100.2199 0.02900T H Δ
dT T 102.871T 100.5723T 100.21990.02900H Δ
^
^
^
4412-338-225-^
100
20
412-
38-
25-^
100
20
312-28-5-^
=×==
=
−++=
−×−−×+−×+=
⎥⎦
⎤⎢⎣
⎡ ×−
×+
×+=
×−×+×+= ∫
Case 2 and Example 8.3 – 3 (pp 371): DIY
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CHAPTER 4- ENERGY balance for non
reactive system
! Kopp’s rule – simple empirical method for estimating the heatcapacity of a solid or liquid near 200C. *Note: Refer Felder pp. 372
! Use Data in Table B.10 for C p of atom compound
! Example: heat capacity of solid Ca(OH)2
( )
( )( )
( ) ( )
Cmol
J
89.5isvalueTrue
Cmol
J 79
Cmol
J 9.6217226C
C2C2CC
0
00OH Ca p
H paO paCa paOH Ca p
2
2
⋅
⋅
=
⋅
++=
++=
ESTIMATION OF HEAT CAPACITIES
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CHAPTER 4- ENERGY balance for non
reactive system
! For heat capacities of certain mixture – may use these rules:
Rules 1 : For a mixture of gases or liquids, calculate the total enthalpychange as the sum of the enthalpy changes for the pure mixturecomponent
Rules 2 : For a highly dilute solutions of solids or gases in liquids,
neglect the enthalpy change of solute.
ESTIMATION OF HEAT CAPACITIES
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CHAPTER 4- ENERGY balance for non
reactive system
! For heat capacities of certainmixture: (C p)mix (T)
ESTIMATION OF HEAT CAPACITIES
( ) ( )
( )
componentiof capacityheatC
componentiof fractionmoleormass y
mixtureof capacityheatCwhere
T C yT C
pi
i
mix p
componentsmixture
all piimix p
=
=
=
= ∑
! For enthalpycalculation:
( ) ( ) ∫ =Δ2
1
T
T
^
dT H T Cmix p
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CHAPTER 4- ENERGY balance for non
reactive system
! Example (Felder pp373)
Calculate the heat required to bring 150 mol/h of a stream containing60% C2H6 and 40% C3H8 by volume from 00C to 4000C.
Solution
System: open system → Q = ΔH
From Table B.2, Appendix B, the heat capacity of component:
C pi yi C pi (C p)mix
kJ/mol.0CC2H6 C3H8 C2H6 C3H8
a 0.04937 0.06803 0.02962 0.0272 0.05682
b 13.92 x 10-5 22.59 x 10-5 8.352 x 10-5 9.036 x 10-5 17.388 x 10-5
c -5.618 x 10-8 -13.11 x 10-8 -3.3708 x 10-8 -5.244 x 10-8 -8.6148 x 10-8
d 7.280 x 10-12
31.71 x 10-12
4.368 x 10-12
12.684 x 10-12
17.052 x 10-12
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CHAPTER 4- ENERGY balance for non
reactive system
!Solution
( )
( ) ( ) ( )
( )
h
kJ 5237
h
mol150
mol
kJ 34.91H Δ
mol
kJ 34.910.10911.837813.910422.728H Δ
mol
kJ
0-400
4
1017.052
0-4003
108.6148 0-400
2
1017.388 0-4000.05682
H Δ
T 4
1017.052
T 3
108.6148
T 2
1017.388
0.05682T H Δ
dT T 1017.052T 108.6148T 1017.3880.05682H Δ
^
^
4412-
338-
225-
^
400
0
412-
38-
25-^
400
0
312-28-5-^
=×=
=+−+=
⎥
⎥⎥⎥
⎦
⎤
⎢
⎢⎢⎢
⎣
⎡
×+
×−
×+
=
⎥⎦
⎤
⎢⎣
⎡ ×+
×
−
×+=
×+×−×+= ∫
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CHAPTER 4- ENERGY balance for non
reactive system
! Latent heat – the specific enthalpy change associated with thetransition of a substance from one phase to another at constanttemperature and pressure. *Note: Refer Felder pp. 378
! Latent heats for the two most commonly encountered phase changesare defined as follows:
1. Heat of fusion (or heating of melting) ΔH m (T,P) – specific enthalpy difference between the solid and liquid forms of aspecies at T and P
2. Heat of vaporization ΔH v (T,P) – specific enthalpy
difference between the liquid and vapor forms of a species at T andP
PHASE CHANGE OPERATIONS
1) Latent Heats
C G f
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CHAPTER 4- ENERGY balance for non
reactive system
! Trouton’s rule – a simple formula for estimating a standard heat of
vaporization (ΔH v at normal boiling point); provide an estimate of ΔH v accurate to within 30%. *Note: Refer Felder pp. 381
PHASE CHANGE OPERATIONS
2) Estimation and Correlation of Latent Heats
liquidtheof pointboilingnormal:T
alcoholsweightmolecularlowwater, (K)0.109T
liquidsnonpolar(K)0.088T (kJ/mol)
b
b
b
→≈
→≈Δ
^
v H
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Chen’s equation – provides roughly 2% accuracy. *Note: Refer Felder
pp. 381
PHASE CHANGE OPERATIONS
2) Estimation and Correlation of Latent Heats
( )( )
(atm) pressurecritical
(K)etemperaturcritical
(K) pointboilingnormal
T T 1.07
Plog0.02970.0327T T 0.0331T (kJ/mol)
cb
c10cbb
: P
:T
:T
H
c
c
b
^
v
−
+−=Δ
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! A formula for approximating a standard heat of fusion. *Note: ReferFelder pp. 381
PHASE CHANGE OPERATIONS
2) Estimation and Correlation of Latent Heats
solidtheof pointmeltingnormal:T compoundsorganic (K)0.050T
compoundsinorganic (K)0.0025T
elementsmetallic (K)0.0092T (kJ/mol)
m
m
m
m
→≈
→≈
→≈Δ
^
m H
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Watson’s colleration – a useful approximation for estimating ΔH v at
T 2 for known value at T 1. *Note: Refer Felder pp. 382
PHASE CHANGE OPERATIONS
2) Estimation and Correlation of Latent Heats
etemperaturcritical
T T
T T )(T )(T
1c
2c12
:T
H H
c
.^
v
^
v
380
⎟⎟
⎠
⎞⎜⎜
⎝
⎛
−
−Δ=Δ
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Example (Felder pp379)
100 mole/h of liquid n-hexane (C6H14) at 250C and 7 bar is vaporizedand heated to 3000C at constant pressure. Neglecting the effect ofpressure on enthalpy, estimate the rate at which heat must be supplied.Given the boiling temperature of n-hexane at 7 bar is 1460C. Use data
provided in Table B.1 to solve the problem.
HO W ? ? ? Analyze the information….
! P1 , T 1 = 7 bar, 250C → Liquid Phase
! P2 , T 2 = 7 bar, 3000C → Gas Phase
! T b at 7 bar = 1460C
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Example (Felder pp379)
! There is phase change operation, however the ΔH v at 1460C and7 bar is not provided in the question.
! From Table B.1, T b = 68.740C, ΔH v = 28.85 kJ/mol at P = 1 atm or1.013 bar
! As stated in question, the effect of pressure on
enthalpy is neglected
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Example (Felder pp379)
! Path method
C6H14 (l)
250
C, 7 bar
C6H14 (l)
1460
C, 7 bar
C6H14 (v)
1460
C, 7 bar
1
DH1=òCpdt (liq) C6H14 (v)
3000
C, 7 bar
2
DH2=DHv
3
DH3=òCpdt (vap)
C6H14 (l)
250C, 1 atm
C6H14 (l)
68.740C, 1 atm
C6H14 (v)
68.740C, 1 atm
C6H14 (v)
3000C, 1 atm
4
DH4= VDP (liq)
5
DH5=òCpdt (liq)
6
DH6=DH v
7
DH7=òCpdt (vap)
8
DH8= 0 (vap)
! True process path – 1 → 2 → 3 : ΔH v at 1460C and 7 bar is notprovided
! Alternative process path – 4→ 5→ 6→ 7→ 8
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Example (Felder pp379)
! Calculate the enthalpy for each of stream number.
! Stream 4
! Since the effect of pressure on enthalpy is neglected, ΔH = 0
04
4
=Δ
Δ=Δ
H
P V H
CHAPTER 4 ENERGY b l f
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CHAPTER 4- ENERGY balance for non
reactive system
! Stream 5
! Find C p for liquid at Table B.2
( )
mol
kJ 9.461ΔH
2568.7410216.3dT 10216.3ΔH
5
3
68.74
25
3
5
=
−×=×=
−−
∫
! Stream 6
mol
kJ 28.85ΔH ΔH v6 ==
CHAPTER 4 ENERGY balance for non
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CHAPTER 4- ENERGY balance for non
reactive system
! Stream 7
! Find C p for gas at Table B.2
molkJ 47.191ΔH
dT T 1057.66T 1023.92T 1040.8510137.44ΔH
7
300
68.74
3122853
7
=
×+×−×+×= ∫ −−−−
! Stream 8
0=8
ΔH
CHAPTER 4 ENERGY balance for non
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CHAPTER 4- ENERGY balance for non
reactive system
! Total enthalpy
mol
kJ 85.502ΔH
047.19128.859.4610ΔH
ΔH ΔH ΔH ΔH ΔH ΔH 87654
=
++++=
++++=
! For overall process
kW 2.375Qs3600
h1
mol
kJ
85.502h
mol
100H nQ=
××=Δ=