8 nonreactive process balances
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7/31/2019 8 Nonreactive Process Balances
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Balances on
Nonreactive ProcessesCh E 201Material and Energy Balances
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State Property CalculationsCal
cul
atio
n o
f
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Hypothetical Process PathsSin
ce
stat
e p
r
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Hypothetical Process Paths
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Energy Balance CalculationProcedure
. Per
f or
m al
l r
e
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Energy Balance CalculationProcedure
. Con
str
uct
a t
a
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Energy Balance CalculationProcedure
. Cal
cul
ate:
– open
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Energy Balance CalculationProcedure
. Cal
cul
ate
any
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Condenser Energy BalancePar
tia
l co
nde
n
0 0 0
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Condenser Energy Balance. C
ho
ose
ref
ere
n
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Condenser Energy Balance. C
on
str
uct
inl
e
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Condenser Energy Balance. C
al
cul
ate
unk
n
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Condenser Energy Balanceyp
oth
etic
al
p
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Condenser Energy Balanceyp
oth
etic
al
p
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Condenser Energy Balanceyp
oth
etic
al
p
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Condenser Energy Balanceyp
oth
etic
al
p
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Condenser Energy Balance
h
ang
e in
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Condenser Energy Balanceyp
oth
etic
al
p
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Condenser Energy Balanceyp
oth
etic
al
p
C
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Condenser Energy Balance. C
al
cul
ate
unk
n
35.7
1.16
32.0
-0.10
C
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Condenser Energy Balance. C
al
cul
ate
35.7
1.16
32.0
-0.10
C
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Condenser Energy Balance. C
al
cul
ate
non
z
I
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Changes in P at Constant TIt has been
obs
e
S
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Changes in T at constant PSen
sib
le h
eat
T
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Changes in T at constant P The
sl
ope
of
t
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Changes in T at constant P
• exact for an ideal gas• good approximation for a solid or
liquid• valid for nonideal gas only if V is
constant
C
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Changes in T at constant PCal
cul
ate
Q r
e
C
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Changes in T at constant PCon
sider t
he
e
i
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Hypothetical Pathini
tial st
ate
H
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Heat Capacity FormulasHea
t capac
ity
A
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Cooling of an Ideal GasAss
uming i
dea
l
N
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Cooling of an Ideal Gas. N
2
f lowing
at
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Cooling of an Ideal Gas2. N2 in 5 L flask initially 3 bar, cooled
from 90 to 30°C – get CP for N2 from Table B.2, subtract R
to get CV
– calculate internal energy change
K
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Estimate Heat Capacity of a PureSubstance
K op
p’s Rul
e i
s
E ti t H t C iti fR
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Estimate Heat Capacities of Mixtures
Rul
e 1 – F
or
a
Energ Balance SingleF
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Energy Balance: Single-Phase System
For
heatin
g/c
o
Energy Balance Gas
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Energy Balance: GasPreheater
Ass
ume
ide
a
0 0 0
Energy Balance Gas
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Energy Balance: GasPreheater
Energy Balance: Gas
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Energy Balance: GasPreheater
• ideal gas neglect effect of P on H→
• neglect heat of mixing of gas phasecomponents
Energy Balance: Gas
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Energy Balance: GasPreheater
• ideal gas neglect effect of P on H→
• neglect heat of mixing of gas phasecomponents
•
from Table B.8:
12.09
Energy Balance: Gas
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Energy Balance: GasPreheater
12.09
8.17-0.15
Energy Balance: Waste Heat
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Energy Balance: Waste HeatBoiler
Energy Balance: Waste Heat
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Energy Balance: Waste HeatBoiler
T
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Latent Heats The
specif
ic
e
A
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Heat of VaporizationAt
what ra
te
(
O
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Heat of VaporizationOf t
en a ph
ase
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Heating and Vaporization
0 0 0
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Heating and Vaporization
true path
but don’t knowvap(146°ΔĤ
C) at step E
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Heating and Vaporization
shortest path
don’t knowvap(25°C)ΔĤ
at step C
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Heating and Vaporization
hypothetical path
permits use of knownvap(69°C)ΔĤ
at step D
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Heating and Vaporization
TableB.1
TableB.2
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Heating and Vaporization
TableB.1
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Heating and Vaporization
TableB.2
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Heating and Vaporization
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Heating and Vaporization
note contribution of term to total of ΔĤpath
Estimation/Correlation of LatentH
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Estimation/Correlation of LatentHeats
Hea
t of va
por
i
Estimation/Correlation of LatentW
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Estimation/Correlation of LatentHeats
Wat
son’s C
orr
e
Energy Balance: System w/ PhaseA
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Energy Balance: System w/ PhaseChange
An
equimol
ar
m
Energy Balance: System w/ Phase
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Energy Balance: System w/ PhaseChange
total molebalance:benzenebalance:
Energy Balance: System w/ Phase
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Energy Balance: System w/ PhaseChange
Energy Balance: System w/ Phase
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Energy Balance: System w/ PhaseChange
Energy Balance: System w/ Phase
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Energy Balance: System w/ PhaseChange
5.332
6.34
037.52
42.93
P h i ChA
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Psychrometric Charts p
s y chr om
et r
i
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Psychrometric chart – SI Units.Reference states:H2O (L, 0°C, 1 atm), dry air (0°C,1 atm).
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Dry Bulb Temperature:air temperature asmeasuredby a thermometer.
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absolute humidity, ha:moisture content,measuredin kg H2O/kg DA
easily converted to mass fraction:e.g., if ha = 0.0150 kg H2O/kg DA,then mH2O = 0.0150 kg H2O/(1.015 kghumid air)
= 0.0148 kg H2O/kg
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relative humidity, hr hr =[pH2O/p*H2O(T)]×100%
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Dew Point (saturation temp),Tdp:found by moving horizontallyfroma given air hr and temperature
e.g., air at 29°C, 20% relativehumidity
has a dew point of 4°C
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e.g., volume of 1 kg humid air (HA) 30°C,30% hr:
humid volume
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wet bulb temperature,Twbis the lowest temperature an object may becooled to by the process of evaporation. Waterevaporating from a moistened wick on the wet-
bulb thermometer bulb cools the thermometerbulb and lowers the temperature reading. Thecooling effect of the evaporation from the bulb isinversely proportional to the amount of watervapor present in the air: the more water vaporpresent, the less moisture will evaporate from themoistened wick, and the less cooling of thethermometer bulb will occur.
From the dry- and wet-bulb readings, the dew-point temperature and humidity values may becalculated.
specific enthalpy of saturated air, Ĥ
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Represents enthalpy change of dry airand accompanying water vapor.
Can be obtained by adding the sensibleheat of DA from reference state (0°C)to 25°C,with H2O = H2O(v, 25°C) - H2OΔĤ Ĥ Ĥ(l, 0°C)
enthalpy devation
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used to determine enthalpyof air that is not saturated
e.g., 35°C 10% rel. hum air=ΔĤ 45.0 – 0.52 = 44.45kJ/kg DA
Using the Psychrometric
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Using the PsychrometricChart
Estimate the absolute
humidity, wet-bulbtemperature, humidvolume, dew point, andspecific enthalpy of humidair at 41°C, 10% relative
humidity.
3°C
19°C 0.895
m3/kg
0 . 0
0 4 8
k g
H 2 O / k g
D A
54.2 – 0.7 = 53.5 kJ/kgDA
41°C
1 0 % r e l
.
h u m.
Air Conditioner
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Air Conditioner•
DOF analysis7 unknowns (m1, m2, m3, 1, 2, 3, Q)Ĥ Ĥ Ĥ
- 1 material balance (DA already balancedon flowchart)
- 2 absolute humidities from psychrometriccharts (inlet/outlet)
- 2 enthalpies from psychrometric charts
(inlet/outlet)- 1 enthalpy of condenstate- 1 energy balance- 0 degrees of freedom
Air Conditioner
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Air Conditioner
0.018 lbmH2O/lbm DA
0.0079 lbmH2O/lbm DA
waterbalance:
fractioncondensed:
Air Conditioner
condensate enthalpy( f t t i li id t t
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Air Conditioner
1 = 38.8ĤBTU/lbm DA
2 = 20.9ĤBTU/lbm DA
(reference state is liquid water at32°F):
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1 = 38.8ĤBTU/lbm DA
2 = 20.9ĤBTU/lbm DA
3 = 19.0ĤBTU/lbm H2O EnergyBalance
cooling requirement for
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1 = 38.8ĤBTU/lbm DA
2 = 20.9ĤBTU/lbm DA
3 = 19.0ĤBTU/lbm H2O
cooling requirement for1000 ft3/min of delivered air
Adiabatic CoolingAd
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Adiabatic Coolingdi
abatic
coo
l
Adiabatic CoolingIf
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Adiabatic Coolingf
certain
, w
e
Adiabatic HumidificationA
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Adiabatic Humidification s
tream o
f ai
Adiabatic HumidificationAs
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Adiabatic Humidificationss
ume the
hea
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0.0026 kg H2O/kgDA
13.2°C
0.0063 kg H2O/kg
DA21.2°C
ExitingAir
EnteringAir
evaporationrate: