C H A P T E R 3 Material and Energy Balances
Steps 1 and 3 are both temperature changes at constant pressure, so dH5CP dT applies
to each, as discussed in Section 2.3.2.
DH 1 5
#
T 5 78.4 8C
DH 3 5
#
T 5 100 8C
C P , liq dT
T 5 25 8C
T 5 78.4 8C
117
The integration
of a temperaturedependent expression
for CP was illustrated
in Example 2-5.
C *P dT
Note that CP* is the ideal gas heat capacity of ethanol, which is available in Appendix C-1.
Because this process is carried out at a low pressure (1 atm), it is reasonable to model
ethanol vapor as an ideal gas, as discussed in Section 2.3.3. The value of CP for liquid
ethanol is available in Appendix C-2. Both heat capacities are functions of T. The full
integration of CP dT is shown in the more detailed solution provided in the electronic
appendices.
DH 1 5 6253 Jymol
DH 3 5 1621 Jymol
Step 4 Calculate Q
Returning to Equation 3.108 to solve for the heat addition:
·
Q 5 n· sH 2 H d
out
1
mol
·
Q 5 1000
min
in
2 16253 1 36, 560 1 1621 mol2 11000 J2 5 44,400 min
J
1 kJ
kJ
Notice the molar
enthalpy of
vaporization is much
larger in magnitude
than the changes
in molar enthalpy
resulting from the
temperature changes.
The previous examples illustrate the process of applying and simplifying the generalized energy balance equation to a variety of physical systems. We close this section
by considering a rocket launching, which revisits the motivational example in a more
quantitative way.
LAUNCH OF A ROCKET
A rocket’s payload (everything EXCEPT fuel) has a mass of 10,000 kg and ĈV 5 2.5 kJ/kg · K.
Initially, the rocket is at rest at ground level, is at ambient temperature (T 5 258C) and
pressure (P 5 1 bar), and contains the payload plus 100,000 kg of rocket fuel. The fuel
has a specific enthalpy of formation Ĥ 5 1000 kJ/kg, using the same reference state that
is used for the data in Appendix C.
The fuel burns to completion, and the emitted exhaust consists of 30,000 kg water
vapor and 70,000 kg carbon dioxide. The exhaust leaves at T 5 258C and P 5 1 bar and
has a velocity of 3 km/s. When the last of the fuel is consumed, the rocket is 5 km above
the Earth, and the payload has T 5 2508C. What is its velocity at this point? Exhaust is
released continuously as the rocket climbs, but a large fraction of the fuel burns close
to the ground. Assume the average height at which the exhaust is released is 0.5 km.
Assume further that the rocket is adiabatic and that it neither produces nor uses shaft
work.
SOLUTION:
Step 1 Define a system
The rocket, and everything inside it, will be the system.
Step 2 Apply and simplify the energy balance
This example describes the initial and final states of the rocket, but doesn’t indicate the time
required to climb to this height, so we use the time-independent form of the energy balance.
EXAMPLE 313
Rocket fuel is here
modeled as if it were
a single compound.
Real rocket fuel
would more likely be
a mixture of one or
more fuels plus an
oxidizer to support
their combustion.
FOOD FOR
THOUGHT
318
Is it plausible that
the rocket itself is
at 2508C but the fuel
exhaust is at 258C?
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118 Fundamentals of Chemical Engineering Thermodynamics
This is the only
example in the
first eight chapters
of the book that
involves mixtures of
gases or liquids: the
exhaust is a mixture
of CO2 and H2O. A
complete treatment
of the mathematical
modeling of mixtures
begins in Chapter 9.
FOOD FOR
THOUGHT
319
If the masses of exiting
water vapor and
carbon dioxide were
not given, how could
you estimate them?
No immediate simplification of the accumulation term is possible; the system experiences significant changes in mass, internal energy, kinetic energy and potential energy.
On the right-hand side, there is no Q or WS. We assume there is also no WEC , because the
size of the rocket doesn’t change. There is no matter entering the system, but the exhaust
gases leave. The energy balance simplifies to
5 1
v2
1 gh
2
D M Û 1
Don’t confuse the
different v terms and
h terms. The left-hand
side represents the
system (rocket) and
the right-hand side
represents material
(exhaust gases)
entering or leaving
the system. Thus, the v
on the left-hand side
is the unknown rocket
velocity, and the v
terms on the righthand side are 3 km/s,
the exhaust velocity.
In Chapter 9, the
properties of a mixture
for ideal gases are
found simply by
summing the properties
of the individual pure
gases that make up the
mixture multiplied by
how much of each gas
is present.
o5
1
mk, out Ĥk 1
k51
26
v 2k
1 ghk
2
(3.110)
While the CO2 and H2O leave the system as a single exhaust stream, we haven’t yet
developed how to model mixtures, so we will model them as two separate exiting
streams:
5 1
v2
1 gh
2
D M Û 1
26
1
5 2mCO ,out ĤCO 1
2
v 2CO
2
2
2
1
2 mH O,out ĤH O 1
2
1 ghCO
v 2H O
(3.111)
2
1 ghH O
2
2
2
2
2
2
Writing out the initial and final energy of the system explicitly gives
5 1
M Û 1
v2
1 gh
2
26
1
5 1
2 M Û 1
final
v 2CO
5 2mCO , out ĤCO 1
PITFALL
PREVENTION
26
k5K
52
2
2
2
2
v2
1 gh
2
2
26
(3.112)
initial
1
1 ghCO 2 mH O, out ĤH O 1
2
2
v 2H O
2
2
2
2
1 ghH O
2
The initial height and velocity are both zero. However, initially, the mass of the system
includes payload and fuel; at the end of the process there is only payload:
5M 1Û
payload
payload
1
1
v2
1 gh
2
5 2mCO , out ĤCO 1
2
2
v 2CO
26
2
2
final
5
6
2 Mpayload Ûpayload 1 Mfuel Ûfuel
2
1
1 ghCO 2 mH O, out ĤH O 1
2
2
2
v 2H O
2
2
(3.113)
initial
2
1 ghH O
2
Step 3 Insert known values and identify what is unknown
vfinal is the unknown we seek to calculate. Most of the other information in Equation
3.113 is given. The only unknowns we need to resolve are Ûpayload, final and Ûpayload, initial, and
ĤCO and ĤH O.
2
2
Step 4 Collect data
Appendix C contains enthalpy of formation data for CO2 and H2O. Recall that enthalpy is
only known relative to a reference state. The “enthalpy of formation” is the molar enthalpy
relative to a reference state in which all elements have H 5 0 at P 5 1 bar and T 5 25°C. Can
we relate these to ĤCO and ĤH O in Equation 3.113? Let us consider this more closely. The exiting stream is a mixture of carbon dioxide and water vapor. We haven’t yet learned how to calculate Ĥ for a mixture, so we are imagining that there are two separate exiting streams. What
are the pressures of these two streams? Since the total pressure of the exiting gas is P 5 1 atm,
then CO2 and H2O each must have a partial pressure that is less than 1 atm. The value of
H 5 2393.5 kJ/mol in Appendix C is valid for CO2 at P 5 1 bar and T 5 258C; is it applicable
to CO2 at 258C and a lower pressure? We will say “yes” because again, real gases act like ideal
gases at low pressure, and for an ideal gas, enthalpy is not a function of pressure. Converting the
enthalpy of formation data from Appendix C to mass basis using the molecular weight gives
2
2
Copyright 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C H A P T E R 3 Material and Energy Balances
1 mol2 1 44 g 2 1 kg 2 5 28943 kg
kJ
1 mol 1000 g
kJ
5 12241.8
5 213, 430
2
1
2
1
2
mol
18 g
kg
kg
kJ
ĤCO 5 2393.5
2
ĤH O
2
1 mol
1000 g
kJ
(3.114)
Meanwhile, the internal energy of the rocket fuel, Ûfuel, initial, is unknown, but the enthalpy
of formation of the fuel at the correct temperature (25°C) and pressure (P 5 1 bar) is
kJ
given, Ĥfuel, initial 5 1000 . Using the general rule stated in Section 2.3.5 that for liquids
kg
and solids, Û ! Ĥ. So
kJ
Ûfuel,initial ! 1000
(3.115)
kg
Step 5 Account for change in internal energy
The left-hand side of Equation 3.114 can be rearranged so that the Ûpayload terms are
grouped together:
5M 1
payload
vfinal
2
2
26 1 {M
1 ghfinal
1
5 2mCO ,out ĤCO 1
2
2
2
2
FOOD FOR
THOUGHT
320
Using only the
information in the
appendix of this book,
how could you find ĤCO
2
and ĤH O if the exhaust
were at T 5 1008C,
rather than at 258C?
2
PITFALL
PREVENTION
You can’t use the data
in the steam tables
for ĤH O. The data for
rocket fuel, carbon
dioxide, and water
vapor must all have
the same reference
state. The data in
the steam tables and
Appendix C use
different reference
states.
2
sÛpayload,final 2 Ûpayload,initiald 2 MfuelÛfuel,initial }
payload
v2CO
119
2
1
1 ghCO 2 mH O,out ĤH O 1
2
2
2
v2H O
2
2
(3.116)
2
1 ghH O
2
Assuming the rocket is a constant-volume system, the change in internal energy can be
related to ĈV:
Mpayload sÛpayload,final 2 Ûpayload,initial d 5 Mpayload sĈV,payload dsTpayload,final 2 Tpayload,initial d
(3.117)
Step 6 Calculate velocity
Inserting Equation 3.117 and all of the known values identified in steps 3 and 4 into Equation 3.116 gives an equation in which the final velocity is unknown. The required algebra
is shown in the more detailed solution given in the supplemental material. The solution is:
vfinal 5 11, 600
m
s
In terms of analyzing a real rocket, Example 3-13 is over-simplified in several ways:
it treated “rocket fuel” as a single compound, assumed the rocket fuel was at ambient conditions initially (real rocket fuel would probably be pressurized), and used an
“average” height at which the fuel was released. A rigorous treatment would include
air resistance, model the rocket’s flight continuously with differential equations, and
combine energy balances with momentum balances.
While simplistic, this example illustrates a system where all three forms of
stored energy—internal, kinetic, and potential—are important and demonstrates
the conversion of chemical energy into kinetic and potential energy. Furthermore, it
illustrates mathematically how a rocket can indeed achieve escape velocity without
violating the first law of thermodynamics.
3.6 Energy Balances for Common Chemical Process Equipment
When ĈV is constant,
the equation
dÛ 5 ĈV dT simply
integrates to
Û2 2 Û1 5 ĈV sT2 2 T1d.
This example is
oversimplified in that
modern rockets are
multi stage; meaning
that portions of the
rocket are fuel tanks
that are jettisoned
when empty. Thus,
Mpayload wouldn’t be
constant.
Recall that the
Motivational Example
stated escape velocity
is ~7 miles/s. Here the
final answer, 11,600 m/s,
is approximately 7.2
miles/s.
The focus of this chapter has been on developing a systematic approach to writing
and solving energy balances that is applicable to any physical system. However, some
specific cases that are of particular interest to chemical engineers are explored here.
Copyright 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.