1
Chapter 4
4-1. Use Equation 4-50 in the textbook:
(1)/(1)
2
1
c
moo
go
gM
QCAP
RT





4
7.42 10 kg/s
m
Q

4-2. Chlorine:
3
1380 kg/m (density)
2



342
1380 kg/m 2.81 m/s 3.14 10 m
1.22 kg/s
m
m
Q

b. For liquid discharge thru a hole, use Equation 4-7,
2
mo cg
QAC gP
For turbulent liquid discharge thru a hole, 0.61
C
3
252
42 3
1 kg m/s 10 N/m
14.2 kg/s
m
Q


4-3.
Sideways View:
8.5 m
0.79 m
Tank
Dike
1 m
Top View:
20 m
10 m 20 m
a. Total volume of liquid leaked = total volume of liquid in dike.
22
22
1
4
3.14
Vl Dh






4
b. Total volume of liquid in dike = total volume leaked out of tank
2
2
3
4
3.14
t
D
Vh

c. Use Equation 4-18 in textbook to calculate time of leak, with 0
g
P for a vented
tank:

L
The hole area and tank area are as follows:

222
3.14 0.1 m 0.00785 m
44
D
A

 


5
4-4. a. Air flowing thru a 100-m long pipe from a source at 100 bara to 1 bara.
The pipe is 0.5 cm in diameter. For air, from Table 4-3 in the text, 1.40
.
The velocity head loss, K, is due entirely to the pipe length, 4
f
f
L
Kd
.
For new, commercial steel pipe, from Table 4-1, the pipe roughness is 0.046 mm
.
Assuming fully developed turbulent flow, Equation 4-34 is used to calculate the friction
factor.
2
2
2
15 mm
4log 3.7 4log 3.7 10.42
0.046 mm
1/10.42 9.60 10
d
f
f

  
 
  

  


0.005 m
f
For our specific case, the pressure drop ratio is,
100 1 0.99
PP

a. Since the velocity head loss is high, we can use the asymptotic solution using Equation 4-
6
4-5. Liquid cyclohexane flowing thru a hole:
Benzene Liquid
Equation 4-7 applies for discharge thru a hole:
mcg
For this case:
2
2
3
778.1 kg/m
10 barg
g
P
Substituting into Equation 4-7:

52
52 9 2 42
4.79 10 m 1.56 10 kg / m s
= 1.89 kg/s
 
 
Cyclohexane
7
4-6. Liquid cyclohexane draining out the bottom of a tank:
1 cm diameter hole
10 m
2 m
0 barg
8


52
2
4
7.85 10 m
0.61 9.8 m/s
= 8.37 10 s = 23.25 hr


Same answer as other method.
4-7. Liquid cyclohexane flowing thru a pipe:
100 m
This may require a trial and error solution since we need the velocity to determine the
9
Solve for u, the liquid velocity:
We know everything but the friction factor, f. Substitute in everything we have.


3
2 100 m 778.1 kg/m
f
f
Assume that the flow is fully developed turbulent flow. We can check this later.

2
2
9.73 10 0.00948
f
 
The velocity is then,
0.00947
Let’s check the Reynolds number and see if our assumption of fully developed turbulent
flow is valid.

3
22
1 N/m 1 kg m/s
0.00102 Pa s Pa N
 
 
 
From Figure 4-7 for this Reynolds number and a pipe roughness of 0.01, we can see that
The flow rate is determined from:
10
 
2
33.14 0.03 m
2.07 kg/s

4-8. Oxygen gas flowing thru a hole.
For natural gas, from Table 4-3, M = 32.1 and 1.40
.
a. Use Equation 4-49
0
0.528
P

Thus, any source pressure above 1.89 atm will result in choked flow. Since our
source is at 2 atm, the flow will be choked.
b. Use Equation 4-50:

1/ 1
2
c
gM



For our case,
01.0C for choked flow,
2
2
Then it follows that:


2
1/ 1 2.40/0.40
6.00
kg m/s
kg-mole K
22
0.833 0.334
12.40




 





Putting it all together,

52 5 2
0.0392 kg/s
mchoked
Q
4-9. Natural gas flowing thru a pipe:
10 bara 1 bara
100 m
ID: 0.5 cm
f
L
0.005 m
f
For our specific case, the pressure drop ratio is,
100 1 0.99
PP

4-10. Use Equation 4-50 to solve for the area, A:


1/ 1
2
c
moo
choked
gM
QCAP




12
Then,
From Equation (4-50),
2
82
b
ar 100,000 N/m kg
5.33 10 m






The diameter of the hole is,
82
4
3.14
2.60 10 m = 0.260 mm
d

4-11. H2S stored at 100 psig = 114.7 psia.
13
QRT
Substituting into the equation:
3oo6
For H2S gas, from Table 4-3, 1.30