Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 5.1:
The assumptions inherent in Equation 5.3 are:
a) Mass is conserved (i.e., not created or destroyed)
b) The process is at steady state.
Situations where the above assumptions do not hold:
a) In a process which includes nuclear reactions which convert small amounts of mass to
energy
b)
A transient (non-steady state) process where the accumulation term is significant
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 5.2:
A material balance on a particular chemical species must contain generation and
consumption terms since that species can be created (generated) or destroyed (consumed). In
contrast, a total mass balance does not require consumption or generation terms since mass is
conserved (not created or destroyed). A total mole balance is not typically used since the
total number of moles is not conserved.
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 5.3:
Equation (e) of Example 5.7 is an application of the following relationship found in the
Guidelines for Solving Material Balance Problems Involving Multiple Species:
In this case, butene is being consumed. The information in the problem statement that 84%
of the butene is converted to ethylene was used to provide the fractional conversion (X) of
0.84 for the butene. The mass flow rate and molecular weight of butene were also used in the
equation.
rconsumption, butene is the number of moles of butene per time that are converted to ethylene in the
process.
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 5.4:
a. No, it is not correct to write a total mole balance for this process. The total number of
moles does not remain constant since four moles of reactants produce two moles of
products.
b. The maximum number of material balances equations is equal to the number of species
present. In this case, a maximum of three balances can be written.
c. No, your colleague is wrong. As mentioned in part (b), a maximum of three material
balances can be written. The four equations would not be independent. However, one
can use two species balances and a total balance (for a total of three balance equations).
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 5.5:
One key use of material balances would be to determine the amount of reactants needed to
treat the waste stream. Material balances might also be used to determine the concentration
of waste in the stream(s) leaving the process.
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.1:
The diagram for this problem is
Feed water
V̇ feed = 28 m 3 min
! feed = 1000 kg/m3
boiler
Steam
V̇steam = ?
! steam = 3.7 kg/m3
Residual hot water
V̇resid = 6.5 m3 min
! resid = 960 kg/m3
From the principle that mass is conserved,
∑ m˙ = ∑ m˙
inlet
streams
which, for our problem is
outlet
streams
m˙ feed = m˙ steam + m˙ resid
Since all the values are given as volumetric flow rates and densities, the most convenient
equivalent form of mass flow rate to use for all terms is
m˙ = ρ V˙
ρ feedV˙feed = ρsteam V˙steam + ρ resid V˙resid
so
Solving for the flow rate of steam and inserting known values,
ρ feedV˙feed − ρ resid V˙resid
V˙steam =
ρ steam
=
(1000 kg m3 )(28m3 min) − (960kg m3 )(6.5 m3 min)
3.7 kg m3
= 5880 m3/min ≈ 5900 m3/min
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.2:
The diagram for this problem is
Benzene
ṅBenzene = 1140 kgmol hr
Toluene
ṅToluene = 213kgmol hr
solvent
mixer
Solvent
ṁSolvent = 115,000 kg hr
Phenol
ṁPhenol = ?
∑ m˙ = ∑ m˙
Again, we begin with
inlet
streams
or
outlet
streams
m˙ Benzene + m˙ Toluene + m˙ Phenol = m˙ Solvent
But since the flows of Benzene and Toluene are given as molar flow rates, we need to
express the mass flow rates of those species using the relationship
m˙ = n˙ (MW)
where we can determine the following Molecular Weights:
MWBenzene: 6(12.01)+6(1.01)=78.1 kg/kgmol
MWToluene: 7(12.01)+8(1.01)=92.1 kg/kgmol
So
n˙Benzene (MWBenzene ) + n˙Toluene (MWToluene ) + m˙ Phenol = m˙ Solvent
Solving for the mass flow rate of Phenol
m˙ Phenol = m˙ Solvent − n˙Benzene (MWBenzene ) − n˙Toluene (MWToluene )
= 115,000 kg/hr - (1140 kgmol/hr)(78.1 kg/kgmol) - (213 kgmol/hr)(92.1 kg/kgmol)
= 6349 kg/hr ≈ 6350 kg/hr
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.3:
The diagram for this problem is
sugar
ṅsugar = 1.75 lbmol hr
butter
ṁbutter = 60lbm hr
corn syrup
V̇corn syrup = 3.5gal hr
vanilla extract
mixer-boiler
fudge
ṁ fudge = 830lbm hr
cocoa
ṁcocoa = 17lbm hr
milk
V̇milk = ?
Also Given:
ρcorn syrup = ρmilk = 62.4
lbm ⎛⎜ 1 ft3 ⎞⎟
= 8.34lbm gal
ft 3 ⎝ 7.48 gal ⎠
Once again, the important relationship is
∑ m˙ in = ∑ m˙ out
which, for this problem, is
m˙ sugar + m˙ butter + m˙ corn syrup + m˙ van.extract + m˙ cocoa + m˙ milk = m˙ fudge
Writing each term in more convenient terms because of the information given,
MW sugar n˙ sugar + m˙ butter + ρ corn syrupV˙corn syrup + m˙ van.extract + m˙ cocoa + ρ milkV˙milk = m˙ fudge
Also, we are given
1 ˙
1
m˙ van.extract = 30
msugar = 30
MWsugar n˙ sugar
and we also can determine that
MWsugar = 12(12.01) + 22(1.01) + 11(16.00) = 342.3
Solving the balance for the volumetric flow rate of milk,
m˙ fudge − MW sugar n˙ sugar − m˙ butter − ρ corn syrupV˙corn syrup − 301 MW sugar n˙ sugar − m˙ cocoa
˙
Vmilk =
ρ milk
830
=
lbm 31 ⎛
lb ⎞⎛
lbmol ⎞
lb ⎛
lb ⎞⎛
gal ⎞
lb
− 30 ⎜ 342.3 m ⎟⎜1.75
⎟ − 60 m − ⎜ 8.34 m ⎟⎜ 3.5
⎟ −17 m
⎝
hr
lbmol ⎠⎝
hr ⎠
hr ⎝
gal ⎠⎝
hr ⎠
hr
8.34 lbm gal
= 12.5 gal/hr
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.4:
The diagram for this problem is
feed 1
ṁfeed 1 ! 260,000 kg hr
feed 2
V̇feed 2 ! 283 m3 hr
l feed 2 ! 935 kg/m3
product 1
V̇prod 1 ! 157 m3 hr
l prod 1 ! 721 kg/m3
product 2
V̇prod 2 ! 235 m3 hr
distillation
column
product 3
ṁprod 3 ! 208,000 kg hr
The balance on total mass is
m˙ Feed1 + m˙ Feed 2 = m˙ Prod1 + m˙ Prod2 + m˙ Prod 3
In terms of given quantities
m˙ Feed1 + ρ Feed 2 V˙Feed 2 = ρ Prod1V˙Prod1 + ρ Prod 2 V˙Prod 2 + m˙ Prod 3
Solving for the density of product 2,
ρ Prod 2 =
m˙ Feed1 + ρ Feed 2 V˙Feed 2 − ρ Prod1V˙Prod1 − m˙ Prod3
V˙Prod 2
260,000
=
kg ⎛
kg ⎞⎛
m3 ⎞ ⎛
kg ⎞⎛
m3 ⎞
kg
+ ⎜935 3 ⎟⎜ 283 ⎟ − ⎜ 721 3 ⎟⎜157
⎟ − 208,000
hr ⎝
hr ⎠ ⎝
hr ⎠
hr
m ⎠⎝
m ⎠⎝
235
= 866 kg/m3
m3
hr
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.5:
3
Gas In: V˙gas ,in ! 340 ft /min
lgas,in ! .087lbm /ft 3
Liquid In:
3
Gas Out: V˙gas ,out ! 270 ft /min
lgas,out ! .087lbm /ft 3
Absorber
Liquid Out: ṁliq ,out ! 77lbm /min
ṁliq ,in ! ?
The total material balance is:
m˙ gas,in + m˙ liq,in = m˙ gas,out + m˙ liq,out
Because the information for the gas streams is given in terms of densities and volumetric flow
rates, the balance is more conveniently written:
ρ gas,inV˙gas,in + m˙ liq,in = ρ gas,outV˙gas,out + m˙ liq,out
Solving for the liquid mass flow rate in:
m˙ liq,in = ρ gas,outV˙gas,out + m˙ liq,out − ρ gas,inV˙gas,in
(
)
= ρ gas V˙gas,out − V˙gas,in + m˙ liq,out
= .087
lbm ⎛
ft 3
ft 3 ⎞
lbm
270
−
340
+
77
⎜
⎟
min
min ⎠
min
ft 3 ⎝
= −6.1
lbm
lb
lb
lb
+ 77 m = 70.9 m ≈ 71 m
min
min
min
min
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.6:
The diagram for this problem is
Air into canyon
V̇ ai = ?
cpollutant,ai = 0 gmol/ft3
Air out of canyon
V̇ao = ?
cpollutant,ao = ?
canyon
Exhaust gas
V̇ ex = 15 ft3/min
cpollutant,ex = .06 gmol/ft3
It will be most convenient to write a mole balance on the pollutant, which is
n˙ pollutant,ex = n˙ pollutant, air − out + rconsumption,pollutant
which can be written more conveniently as
c pollutant, ex V˙ex = c pollutant,ao V˙ao + rconsumption, pollutant
(1)
Furthermore, the given conversion information can be written
rconsumption, pollutant = .20 n˙ pollutant,ex
which can be more conveniently written
rconsumption, pollutant = .20 c pollutant, exV˙ex
(2)
Combining Equations 1 and 2 gives
c pollutant, ex V˙ex = c pollutant,ao V˙ao + .20 c pollutant,ex V˙ex
or
.80 c pollutant,ex V˙ex = c pollutant, aoV˙ao
(3)
Finally, a balance on total mass gives
m˙ ex + m˙ air in = m˙ air out
which is more conveniently written
ρ exV˙ex + ρ ai V˙ai = ρ ao V˙ao
but, since the densities are all equal, this becomes
V˙ex + V˙ai = V˙ao
(4)
a. For an inlet wind flow rate ( V˙ai ) of 800 ft3/min,
Equation 4 gives V˙ao = 800 ft 3 min + 15 ft3 min = 815 ft 3 min
From Equation 3,
c pollutant, ao
(
)(
)
3
3
.80 c pollutant,ex V˙ex .80 .06 gmol ft 15 ft min
gmol
=
=
= 8.83x10 −4 3
3
˙
Vao
815 ft min
ft
b. For an outlet pollutant concentration (cpollutant,ao) equal to .0025 gmol/ft3,
Equation 3 gives
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.6 (continued):
V˙ao =
.80 c pollutant,ex V˙ex
c pollutant, ao
=
(
)(
.80 .06 gmol ft 3 15 ft 3 min
.0025gmol ft 3
) = 288 ft 3
min
From Equation 4,
V˙ai = V˙ao − V˙ex = 288 ft 3 min −15 ft 3 min = 273 ft3 min
So any incoming wind flow rate less than 273 ft3/min will produce an outgoing pollutant
concentration greater than .0025 gmol/ft3.
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.7:
The diagram for this problem is
Air in
V̇ ai = ?
cO2,ai = .00934 gmol/L
lungs
Blood out
V̇ bo = 5.0 L min
cO2,bo = .00893 gmol/L
Blood in
V̇ bi = 5.0 L min
cO2,bi = .00670 gmol/L
Balance on oxygen:
Or
Air out
V̇ao = ?
cO2,ai = .00705 gmol/L
m˙ O2 ,ai + m˙ O2 ,bi = m˙ O2 ,ao + m˙ O2 ,bo
cO 2 ,ai V˙ai + c O2 ,bi V˙bi = cO 2 ,ao V˙ao + cO 2 ,bo V˙bo
In this case, the inlet air flow rate equals the outlet air flow rate, so we’ll assign one symbol:
V˙ai = V˙ao = V˙a
Similarly, the inlet blood flow rate equals the outlet blood flow rate, so:
V˙bi = V˙bo = V˙b
The oxygen balance now becomes
cO 2 ,ai V˙a + cO 2 ,bi V˙b = cO2 ,ao V˙a + cO 2 ,bo V˙b
Solving for the air flow rate,
cO ,bo − c O2 ,bi ˙
V˙a = 2
V
cO2 ,ai − cO2 ,ao b
gmol
gmol
− .00670
L
L
=
gmol
gmol
.00934
− .00705
L
L
.00893
= 4.87 L/min
⎛
L ⎞
⎜ 5.0
⎟
⎝ min ⎠
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.8:
m˙ dye,in = x dye,out m˙ out
a.
m˙ out =
m˙ dye,in 20 g /min
=
= 5128 g /min
x dye,out
0.0039
b. The mass flow rate of dye in the outlet blood is the same as the input rate, or 20 g/min.
Hence, the mass flow rate of the blood without the dye is 5128 – 20 = 5108 g/min
c.
m˙ = ρV˙
m 5108 g / min
V = =
= 4819 cm 3 / min
3
ρ 1.06 g / cm