Type
Quiz
Book Title
Chemical Process Safety: Fundamentals with Applications-- 4/e 4th Edition
ISBN 13
978-0134857770

978-0134857770 Chapter 4 Part 1

December 2, 2020
1
Chapter 4
4-1. Use Equation 4-50 in the textbook:
(1)/(1)
2
1
c
moo
go
gM
QCAP
RT





4-2. Chlorine:
3
1380 kg/m (density)
2
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
m
4-3.
Sideways View:
8.5 m
0.79 m
Tank
Dike
1 m
Top View:
a. Total volume of liquid leaked = total volume of liquid in dike.
4
b. Total volume of liquid in dike = total volume leaked out of tank
c. Use Equation 4-18 in textbook to calculate time of leak, with 0
g
P for a vented
tank:
The hole area and tank area are as follows:
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.
For our specific case, the pressure drop ratio is,
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:
For this case:
Substituting into Equation 4-7:
Cyclohexane
7
4-6. Liquid cyclohexane draining out the bottom of a tank:
2 m
0 barg
8
Same answer as other method.
4-7. Liquid cyclohexane flowing thru a pipe:
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.
Assume that the flow is fully developed turbulent flow. We can check this later.
The velocity is then,
Let’s check the Reynolds number and see if our assumption of fully developed turbulent
flow is valid.
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
2.07 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
1.40/0.40
1.40 1
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:
For our case,
01.0C for choked flow,
Then it follows that:
Putting it all together,
4-9. Natural gas flowing thru a pipe:
10 bara 1 bara
100 m
ID: 0.5 cm
For our specific case, the pressure drop ratio is,
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),
The diameter of the hole is,
4-11. H2S stored at 100 psig = 114.7 psia.
13
Substituting into the equation:
For H2S gas, from Table 4-3, 1.30