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,
(
)(
)
3
3
.80 c pollutant,ex V˙ex .80 .06 gmol ft 15 ft min
gmol
c pollutant, ao =
=
= 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
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.9:
The diagram for this problem is
filtration
facility
Sea water
V̇ sw = 2300 L min
cNaCl,sw = 1.4 M
Drinking water
V̇dw = ?
cNaCl,dw = ?
Brine
V̇ br = 560 L min
cNaCl,br = 5.6 M
m˙ NaCl,sw = m˙ NaCl,dw + m˙ NaCl,br
Balance on NaCl:
c NaCl,sw V˙sw = c NaCl ,dw V˙dw + c NaCl ,br V˙br
or
(1)
m˙ sw = m˙ dw + m˙ br
Balance on total mass:
ρ sw V˙sw = ρ dw V˙dw + ρ br V˙br
or
V˙sw = V˙dw + V˙br
For equal densities
Solving equation 2 for the drinking water flow rate
V˙dw = V˙sw − V˙br = 2300 L min − 560 L min = 1740 L min
Solving equation 1 for the NaCl concentration in the drinking water
c NaCl,dw =
c NaCl,sw V˙sw − c NaCl, br V˙br
V˙dw
⎛ gmol ⎞⎛
L ⎞ ⎛ gmol ⎞⎛
L ⎞
⎜1.4
⎟⎜ 2300
⎟ − ⎜ 5.6
⎟⎜560
⎟
⎝
L ⎠⎝
min ⎠ ⎝
L ⎠⎝
min ⎠
=
L
1740
min
= .048
gmol
= .048M
L
(2)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.10:
The diagram for this problem is
Gas from tank
V̇ tank = T
cpoison,tank = cT
Room
cpoison = cR
Air from conditioner
V̇ air = A
cpoison,air = 0
Air out of window
V̇window = W (unknown)
cpoison,window = cR
The mass balance on the poison is
m˙ poison,T = m˙ poison,W
which could be written as
MWpoisonc poison,tank V˙ from tank = MWpoison c poison,window V˙window
where the molecular weight cancels out on both sides of the equation. The remainder of
the equation can be rewritten in terms of the given symbols as
cTT = cRW
(1)
A balance on total mass gives
m˙ from tank + m˙ air conditioner = m˙ window
which can also be written
ρ from tank V˙ from tank + ρ air conditioner V˙air conditioner = ρ windowV˙window
But, the densities of all streams are equal, so they cancel out of the equation. Re-writing
the remainder of the equation in terms of given symbols,
W=T+A
Combining equations 1 and 2
cTT = cR(T + A)
Rearranging,
c R = cT
T
T+ A
(2)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.11:
The diagram for this problem is
pure
. air out
mair-out = ?
xair,air-out = 1.000
dirty
. air
mda = 3,500 kg/hr
xair,da = .98
xacetone,da = .02
Acetone-rich
liquid
.
ma-rich = ?
xacetone,a-rich = .99
xwater,a-rich = .01
water
. in
mw-in = ?
xwater,w-in = 1.000
Water-rich
liquid
.
mw-rich = 650 kg/hr
xacetone,w-rich = .04
xwater,w-rich = .96
In this case, we’ll treat “air” as a species (even though it contains several gases), so that
“dirty air” contains air as one of its components (acetone is the other component).
A mass balance on air gives
m˙ air,in = m˙ air,out
or, in more convenient form,
x air, da m˙ da = x air,air − out m˙ air− out
At this point, we can solve this equation to determine the flow rate of pure air out.
(.98) ( 3500 kg hr )
x
m
m air − out = air,da da =
= 3430 kg hr
xair,air − out
1.0
A mass balance on acetone gives
m˙ acetone,in = m˙ acetone,a− rich + m˙ acetone,w − rich
or, in more convenient form,
˙ a − rich + x acetone,w − rich m˙ w − rich
x acetone,da m˙ da = x acetone,a− rich m
Again, solving for the flow rate of the acetone-rich stream
x
m˙ − x acetone,w − rich m˙ w − rich
m˙ a − rich = acetone,da da
x acetone,a − rich
(.02) ( 3500 kg hr ) − (.04) ( 650 kg hr )
= 44.4 kg hr
.99
Finally, a mass balance on water gives
m˙ water,in = m˙ water,a − rich + m˙ water,w −rich
=
which, in more convenient terms is
x water, w −in m˙ w − in = x water,a− rich m˙ a− rich + xwater,w − rich m˙ w − rich
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.11 (continued):
and the flow rate of water can then be calculated
x water,a −rich m˙ a− rich + xwater,w − rich m˙ w − rich
m˙ w − in =
x water, w −in
=
(.01) ( 44.4 kg hr ) + (.96) ( 650 kg hr )
= 624 kg hr
1.000
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.12:
feed stream
ṁfeed = 10,000 kg hr
x EB,feed = 0.15
x p-x,feed = 0.20
x m-x,feed = 0.35
x o-x,feed = 0.30
Exit Stream 1
ṁExit1 = 1,377 kg hr
x EB,Exit1 = 0.97
x p-x,Exit1 = 0.01
x m-x,Exit1 = 0.02
x o-x,Exit1 = 0.00
Exit Stream 2
ṁExit2 = 5,750 kg hr
ṁEB,Exit2 = ?
ṁo-x,Exit2 = 297 kg hr
Exit Stream 3
x EB,Exit3 = 0
ṁp-x,Exit3 = ?
ṁo-x,Exit3 = ?
x m-x,Exit3 = ?
Species Balances
EB:
m˙ EB, feed = m˙ EB,Exit1 + m˙ EB,Exit2 + m˙ EB,Exit3
Which, in more convenient form, is x EB, feed m˙ feed = x EB,Exit1m˙ Exit1 + m˙ EB,Exit2
o-x:
m˙ o− x, feed = m˙ 0− x,Exit1 + m˙ o− x,Exit2 + m˙ o− x,Exit3
Which, in more convenient form, is
x o− x, feed m˙ feed = x 0− x,Exit1m˙ Exit1 + m˙ o− x,Exit2 + m˙ o− x,Exit3
p-x:
(1)
(2)
m˙ p− x, feed = m˙ p− x,Exit1 + m˙ p− x,Exit2 + m˙ p− x,Exit3
Which, in more convenient form, is
x p− x, feed m˙ feed = x p− x,Exit1m˙ Exit1 + m˙ p− x,Exit2 + m˙ p− x,Exit3
(3)
Total Mass Balance: m˙ feed = m˙ Exit1 + m˙ Exit2 + m˙ Exit3
(4)
Sum of components in Exit 3: m˙ Exit3 = m˙ p− x,Exit3 + m˙ m− x,Exit3 + m˙ o− x,Exit3
(5)
Additional Given Relationship: m˙ p− x,Exit2 = 95%x p− x, feed m˙ feed
(6)
From Equation 1,
⎛
⎛
kg ⎞
kg ⎞
kg
m˙ EB,Exit2 = x EB, feed m˙ feed − x EB,Exit1m˙ Exit1 = (0.15)⎜10,000 ⎟ − (0.97)⎜1,377 ⎟ = 164
⎝
⎝
hr ⎠
hr ⎠
hr
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.12 (continued):
From Equation 2, m˙ o− x,Exit3 = x o− x, feed m˙ feed − x 0− x,Exit1m˙ Exit1 − m˙ o− x,Exit2
⎛
kg ⎞
kg
kg
= (0.30)⎜10,000 ⎟ − 0 − 297 = 2,703
⎝
hr ⎠
hr
hr
⎛
kg ⎞
kg
From Equation 6, m˙ p− x,Exit2 = 95%x p− x, feed m˙ feed = .95(0.20)⎜10,000 ⎟ = 1,900
⎝
hr ⎠
hr
From Equation 3, m˙ p− x,Exit3 = x p− x, feed m˙ feed − x p− x,Exit1m˙ Exit1 − m˙ p− x,Exit2
⎛
⎛
kg ⎞
kg ⎞
kg
kg
= (0.20)⎜10,000 ⎟ − (0.01)⎜1,377 ⎟ −1,900 = 86
⎝
⎝
hr ⎠
hr ⎠
hr
hr
From Equation 4,
kg
kg
kg
kg
m˙ Exit3 = m˙ feed − m˙ Exit1 − m˙ Exit2 = 10,000 −1,377 − 5,750 = 2,873
hr
hr
hr
hr
From Equation 5,
kg
kg
kg
kg
m˙ m− x,Exit3 = m˙ Exit3 − m˙ p− x,Exit3 − m˙ o− x,Exit3 = 2,873 − 2,703 − 86 = 84
hr
hr
hr
hr
So
x m− x,Exit3 =
m˙ m− x,Exit3
84kg /hr
=
= 0.03
m˙ Exit3
2,873kg /hr
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.13:
The diagram for this problem is
Methane input
ṅ meth,in = 30 gmol/s
burner
Oxygen input
ṅ O2,in = 75 gmol/s
CH4 + 2O2 A CO2 + 2H2O
Mole balance on methane:
Flue gas
ṅ flue = ?
yCO2,flue = ?
n˙meth,in = n˙ meth, flue + rconsumption,meth
However, we are told that all of the methane is burned. One way to use that information
is to recognize that the flow rate of methane in the flue gas is zero, so
n˙meth,in = rconsumption,meth
(1)
Mole balance on carbon dioxide: rformation,CO2 = n˙ CO2 , flue
(2)
Mole balance on oxygen:
n˙O 2 ,in = n˙O2 , flue + rconsumption,O2
(3)
Mole balance on water:
rformation,H 2 O = n˙ H 2 O, flue
(4)
The stoichiometric relationships are
rformation,CO2
rconsumption,meth
=
1
1
r formation,H 2 O
rconsumption,meth
=
2
1
rconsumption,O2
rconsumption,meth
=
2
1
From Equation 1: rconsumption,meth = n˙meth,in = 30gmol s
From the stoichiometric relationships,
rformation,CO2 = rconsumption,meth = 30 gmol s
rformation,H 2 O = 2 rconsumption,meth = 60gmol s
rconsumption,O2 = 2 rconsumption,meth = 60gmol s
From Equation 2: n˙CO 2 , flue = rformation,CO2 = 30 gmol s
From Equation 3: n˙O 2 , flue = n˙O2 ,in − rconsumption,O2 = 75gmol s − 60gmol s = 15gmol s
From Equation 4: n˙ H 2 O, flue = rformation, H 2 O = 60 gmol s
The total molar flow rate of the flue gas is the sum of the flow rates of the constituent gases,
so
n˙ flue = n˙CO2 , flue + n˙O 2 , flue + n˙ H 2 O, flue = 30 + 15+ 60 = 105gmol s
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.13 (continued):
Finally, the mole fraction of carbon dioxide in the flue gas is
yCO 2 , flue =
n˙CO 2 , flue
n˙ flue
=
30 gmol s
= 0.286
105gmol s
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.14:
The diagram for this problem is
Species A input
ṅ A,in = 50 kgmol/hr
reactor
Species B input
ṅ B,in = 5 kgmol/hr
2A A B
Reactor outlet
ṅ A,out = 20 kgmol/hr
ṅ B,out = ?
Mole balance on A:
n˙ A,in = n˙ A, out + rconsumption, A
(1)
Mole balance on B:
n˙ B,in + rformation,B = n˙ B,out
(2)
Stoichiometry:
rformation, B
rconsumption,A
=
1
2
(3)
From Equation 1:
rconsumption, A = n˙ A,i n − n˙ A,out = 50 − 20 = 30kgmol hr
From Equation 3:
rformation,B =
From Equation 2:
n˙B,out = n˙B,i n + rformation,B = 5 + 15 = 20kgmol hr
rconsumption, A
2
=
30 kgmol hr
= 15kgmol hr
2
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.15:
The diagram for this problem is
Gas input stream
ṅ H2,in = 200 gmol/s
yH2,in =.95
yCH4,in =.05
Gas output stream
reactor
Liquid outlet stream
Toluene input
ṅ benzene,out = ?
ṅ toluene,in = 40 gmol/s
ṅ
=?
Toluene + H2 A Benzene + CH4 toluene,out
Mole balance on benzene:
rformation,benzene = n˙benzene,out
(1)
Mole balance on toluene:
n˙toluene,in = n˙toluene, out + rconsumption,toluene
(2)
Stoichiometry:
rconsumption,toluene
r formation,benzene
=
1
1
rconsumption,toluene = .75 n˙toluene,in
Conversion:
From Equation 4:
rconsumption,toluene = .75 n˙toluene,in = .75 (40 gmol s) = 30 gmol s
From Equation 2:
n˙toluene,out = n˙toluene,in − rconsumption,toluene = 40 − 30 = 10 gmol s
From Equation 3:
rformation,benzene = rconsumption,toluene = 30gmol s
From Equation 1:
n˙benzene,out = rformation,benzene = 30gmol s
To determine the mass fraction of benzene, we need the mass flow rates of benzene and
toluene, which are
⎛
g ⎞
m˙ benzene,out = MW benzene n˙ benzene,out = ⎜ 78
⎟(30 gmol s) = 2340 g s
⎝ gmol ⎠
⎛
g ⎞
m˙ toluene,out = MW toluene n˙ toluene,out = ⎜92
⎟(10 gmol s) = 920 g s
⎝ gmol ⎠
So the mass fraction of benzene in the liquid outlet stream is
x benzene =
m˙ benzene
2340
=
= 0.72
m˙ benzene + m˙ toluene 2340 + 920
(3)
(4)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.16: The diagram for this problem is
Air in
ṅ ai = ?
yO2,ai = .21, yN2,ai = .79
fuel cell
Hydrogen in
ṅ H2,hi = 27 gmol/min
2H2 + O2 ! 2H2O
Gas output stream
ṅ H2,go = ?
ṅ N2,go = ?
ṅ O2,go = ?
ṅ H2O,go = ?
Mole balance on hydrogen:
n˙ H 2 ,hi = n˙ H 2 ,go + rconsumption,H 2
Mole balance on oxygen:
yO 2 ,ai n˙ ai = n˙O 2 ,go + rconsumption,O2
(2)
Mole balance on water:
rformation,H 2 O = n˙ H 2 O, go
(3)
Mole balance on nitrogen:
y N2 ,ai n˙ai = n˙ N2 ,go
(4)
(1)
Stoichiometry:
rconsumption,O 2
rconsumption,H 2
rformation,H 2 O
rconsumption,H 2
=
1
2
(5)
=
2
=1
2
(6)
rconsumption,H 2 = 0.85 n˙ H 2 ,hi
n˙ H ,hi
n˙ O 2 ,ai = 150% 2
Given:
2
which can be more conveniently written
n˙ H ,hi
yO 2 ,ai n˙ ai = 1.5 2
2
Conversion:
From Equation 7:
rconsumption,H 2 = 0.85 (27 gmol min ) = 23gmol min
From Equation 5:
rconsumption,O 2 =
From Equation 6:
rformation,H 2O = rconsumption,H 2 = 23gmol min
From Equation 8:
n˙ ai = 1.5
From Equation 1:
n˙ H 2 ,go = n˙ H 2 ,hi − rconsumption,H 2 = 27 gmol min − 23gmol min
= 4 gmol min
From Equation 2:
n˙O 2 ,go = yO2 ,ai n˙ ai − rconsumption,O 2
= .21(96.4 gmol min) −11.5 gmol min = 8.7 gmol min
From Equation 3:
n˙ H 2O,go = rformation,H 2O = 23gmol min
From Equation 4:
n˙ N 2 ,go = y N 2 ,ai n˙ ai = .79(96.4 gmol min) = 76.2 gmol min
rconsumption,H 2
n˙ H 2 ,hi 2
yO 2 ,ai
2
= 1.5
=
23gmol min
= 11.5 gmol min
2
(27 gmol min) /2
= 96.4 gmol min
.21
(7)
(8)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.17:
Gas in
V̇in = 8.5 L min
TMG
AsH3
H2
Mole balance on methane:
Mole balance on GaAs:
Gas out
V̇out = 8.5 L min
cCH4 = ?
TMG
GaAs out
ṁGaAs = 7 mg min AsH3
H2
CH4
solid
surface
rform,CH 4 = n˙ CH 4 ,out = cCH 4 ,outV˙out
m˙
rform,GaAs = n˙ GaAs,out = GaAs,out
MWGaAs
(1)
(2)
Stoichiometry:
rform,CH 4
rform,GaAs
=
3
1
Also, MWGaAs = 69.72 + 74.92 = 144.6 g/gmol
From Equation 2, rform,GaAs =
m˙ GaAs,out 7x10−3 g /min
gmol
=
= 4.84 x10−5
MWGaAs 144.6g /gmol
min
⎛
gmol ⎞
−4 gmol
From Equation 3, rform,CH 4 = 3rform,GaAs = 3⎜ 4.84 x10−5
⎟ = 1.45x10
⎝
min ⎠
min
−4 gmol
rform,CH 4 1.45x10 min
gmol
=
= 1.71x10−5
From Equation 1, cCH 4 ,out =
˙
L
Vout
L
8.5
min
(3)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.18:
The diagram for this problem is
Stream 1
ṅ 1 = 400 lbmol/hr
yN2,1 = .95
yH2,1 = .05
Stream 2
V̇ 2 = 31,000 ft3/hr
l2 = .08 lbm/ft3
process
Output stream
yN2,out = 0
ṅ H2,out = ?
N2 + 3H2 A 2NH3
Mole balance on nitrogen:
Or, more conveniently,
n˙ N2 ,1 = rconsumption,N 2
y N2 ,1n˙1 = rconsumption, N 2
Mole balance on hydrogen:
n˙ H 2 ,1 + n˙ H 2 ,2 = n˙ H 2 ,out + rconsumption,H 2
ρ V˙
y H 2 ,1n˙1 + 2 2 = n˙ H 2 ,out + rconsumption,H 2
MWH 2
Or, more conveniently,
Stoichiometry:
rconsumption,H 2
rconsumption,N 2
=
3
1
From Equation 1:
rconsumption,N 2 = y N 2 ,1n˙1 = .95(400lbmol hr) = 380lbmol hr
From Equation 3:
rconsumption,H 2 = 3 rconsumption, N2 = 3(380lbmol hr ) = 1140lbmol hr
From Equation 2:
n˙ H 2 ,out = y H 2 ,1n˙1 +
ρ 2 V˙2
− rconsumption,H 2
MWH 2
(.08lb m ft3 )(31,000 ft3 hr)
= (.05)(400lbmol hr) +
− 1140lbmol hr
2.02 lbm lbmol
= 20 + 1240 – 1140 = 120 lbmol/hr
(1)
(2)
(3)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 5.19:
Coconut Oil
MW = 659
l!= 878 g/L
.
V!= 140 L/hr
Methanol
MW = 32
l!= 792 g/L
4.2 moles/mole C.O.
NaOH
MW = 40
.
m!= 140 g/hr
Output
MWbiodiesel = 221
MWglycerol = 92
lbiodiesel!= 880 g/L
Reactor
100% conversion
Species Balances:
C.Oil:
n˙ C.O.,in = rcons,C.O. + n˙ C .O.,out
(1)
B.Diesel:
rform,B .D. = n˙ B.D.,out
(2)
Methanol:
n˙ meth,in = rcons,meth + n˙ meth,out
(3)
Glycerol:
rform,glyc = n˙ glyc,out
(4)
NaOH:
n˙ NaOH ,in = n˙ NaOH ,out
and
m˙ NaOH ,in = m˙ NaOH ,out
(5)
Stoichiometric Ratios:
rform,B .D. 3
=
rcons,C .O. 1
(6)
rcons,meth 3
=
rcons,C .O. 1
(7)
rform,glyc 1
=
rcons,C .O. 1
(8)
Conversion:
rcons,C.O. = 100% n˙ C.O.,in
(9)
Given:
n˙ meth,in = 4.2 n˙ C.O.,in
(10)
From Eq. 9,
rcons,C .O. = n˙ C .O.,in =
From Eq. 10,
n˙ meth,in = 4.2(186.5 gmol hr) = 783.3gmol hr
From Eq. 6,
rform,B .D. = 3rcons,C.O. = 3(186.5 gmol hr) = 560 gmol hr
From Eq. 2,
n˙ B.D.,out = rform,B.D. = 560 gmol hr
V˙C .O.,in ρC .O. (140 L hr)(.878 g L)
=
= 186.5 gmol hr
MWC .O.
(659 g gmol)
Chapter 5 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
5.19 (continued)
From Eq. 7,
rcons,meth = 3rcons,C .O. = 3(186.5 gmol hr) = 560 gmol hr
From Eq. 8,
rform,glyc = rcons,C .O. = 186.5 gmol hr
From Eq. 1,
n˙ C.O.,out = n˙ C .O.,in − rcons,C .O. = 560 gmol hr − 560 gmol hr = 0
From Eq. 3,
n˙ meth,out = n˙ meth,in − rcons,meth = 783.3gmol hr − 560 gmol hr = 233.3gmol hr
From Eq. 4,
n˙ glyc,out = rform,glyc = 186.5 gmol hr
From Eq. 5,
m˙ NaOH ,out = m˙ NaOH ,in = 540 g hr
Total mass out = m˙ B.D.,out + m˙ glyc,out + m˙ C .O.,out + m˙ meth,out + m˙ NaOH ,out
gmol ⎞ ⎛
g ⎞ ⎛
gmol ⎞ ⎛
g ⎞
gmol ⎞ ⎛
g ⎞ ⎛
g⎞
⎛
⎛
= ⎜ 560
221
+ ⎜ 186.5
92
+ 0 + ⎜ 223.3
32
+ ⎜ 540 ⎟
⎟
⎟
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎝
hr ⎠ ⎝
gmol ⎠ ⎝
hr ⎠ ⎝ gmol ⎠
hr ⎠ ⎝ gmol ⎠ ⎝
hr ⎠
= 123,760 + 17,158 + 7146 + 540 = 148,604 g/hr
a. Biodiesel mass fraction =
123, 760 g hr
= 0.83
148, 604 g hr
b. Volumetric flow of biodiesel =
m B.D.,out 123, 760 g hr
=
= 141L hr
ρ B.D.
880 g L