Parciales

CHAPTER FOUR
a.

Continuous, Transient

b.

4.1

Input – Output = Accumulation
No reactions ⇒ Generation = 0, Consumption = 0

6.00
c.

t=

dn
kg
kg dn
kg
− 3.00
=
⇒
= 3.00
dt
dt
s
s
s

100 m3 1000 kg 1 s
.
= 333 s
1 m3 3.00 kg

a.

Continuous, Steady State

b.

k = 0 ⇒ C A = C A0

c.

4.2

Input – Output – Consumption = 0
Steady state ⇒Accumulation = 0
A is a reactant ⇒ Generation = 0

FG m IJ C FG mol IJ = V FG m IJ C FG mol IJ + kVC FG mol IJ ⇒ C
HsK
H sK Hm K H sK Hm K
3

V

4.3

k = ∞ ⇒ CA = 0

3

A0

A

3

b

a.

mv kg / h
100 kg / h
0.550 kg B / kg
0.450 kg T / kg

b

A

=

CA0
kV
1+
V

g

0.850 kg B / kg
0.150 kg T / kg
ml kg / h

A

3

g

Input – Output = 0
Steady state ⇒Accumulation = 0
No reaction ⇒ Generation = 0, Consumption = 0

0.106 kg B / kg
0.894 kg T / kg

(1) Total Mass Balance: 100.0 kg / h = mv + ml

(2) Benzene Balance: 0.550 × 100.0 kg B / h = 0.850mv + 0106ml
.
Solve (1) & (2) simultaneously ⇒ mv = 59.7 kg h, ml = 40.3 kg h
b.

The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by
masses (kg). Thebalance equations are also identical (initial input = final output).

c.

Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state,
the feed composition is incorrect, the flow rates are not what they are supposed to be, other
species are in the feed stream, measurement errors.

4- 1

4.4

b.

b

n (mol)
0.500 mol N 2 mol
0.500 mol CH 4 molc.

100.0 g / s

nE =

b
g
bg C H gg
bg C H gg

x E g C2 H 6 g
xP
xB

d.

3

b

g

g

100 x E g C 2 H 6 1 lb m lb - mole C2 H 6 3600 s
s
h
453593 g 30 lb m C2 H 6
.

b

g

10

b
g
Rn blb - mole DA sg
| 0.21 lb - moleO lb - mole DA U
|
S
| 0.79 lb - mole N lb - mole DAV
|
T
W

nO2 = 0.21n2 ( lb-mole O 2 / s )

n1 lb - mole H 2 O s
2

xH2O =2

2

e.

b

= 26.45x E lb - mole C 2 H 6 / h

8

4

g

0.500 n mol N 2 28 g N 2 1 kg
= 0.014 n kg N 2
mol N 2 1000 g

xO2 =

n ( mol )

n1 ⎛ lb-mole H 2 O ⎞
n1 + n2 ⎜ lb-mole ⎟
⎝
⎠

0.21n2 ⎛ lb-mole O 2 ⎞
n1 + n2 ⎜ lb-mole ⎟
⎝
⎠

nN2O4 = n ⎡0.600 − yNO2 ⎤ ( mol N 2 O4 )
⎣
⎦

0.400 mol NO mol
yNO2 ( mol NO 2 mol )
0.600 − yNO2 ( mol N 2 O 4 mol )
4.5

a.Basis: 1000 lbm C3H8 / h fresh feed
(Could also take 1 h operation as basis flow chart would be as below except
that all / h would be deleted.)
1000 lb m C3H 8 / h

b

b

n6 lb m / h

n7 lb m / h

g

0.02 lb m C3H8 / lb m
0.98 lb m C3H 6 / lb m

g

0.97 lb m C3H8 / lb m
0.03 lb m C3H 6 / lb m

Still

Compressor

b
n blb
n blb
n blb

g
C H / hg
CH / h g
H / hgb
b

n1 lb m C3H 8 / h

n1 lb m C3H 8 / h

Reactor

2

m

3

m

4

m

3

4

2

Note: the compressor and the off gas from
the absorber are not mentioned explicitly
in the process description, but their presence
should be inferred.

b
b

n3 lb m CH 4 / h
n4 lb m H 2 / h

g

g

b

n5 lb m / h

g

Stripper

Absorber

b
b
n blb

n1 lb m C3H 8/ h

g
g

n2 lb m C3H 6 / h
5

4- 2

g
g

n2 lb m C3H 6 / h

6

m

oil / h

g

4.5 (cont’d)
b. Overall objective: To produce C3H6 from C3H8.
Preheater function: Raise temperature of the reactants to raise the reaction rate.
Reactor function: Convert C3H8 to C3H6.
Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other
components.Stripping tower function: Recover the C3H8 and C3H6 from the solvent.
Distillation column function: Separate the C3H5 from the C3H8.
4.6

a.

3 independent balances (one for each species)

b.

7 unknowns ( m1 , m3 , m5 , x2 , y2 , y4 , z4 )
– 3 balances
– 2 mole fraction summations
2 unknowns must be specified

c.

y2 = 1 − x2

FG kg A IJ = m + b1200gb0.70g FG kg A IJ
HhK
HhK...