December 2, 2020

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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-

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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

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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

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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.

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Substituting into the equation:

For H2S gas, from Table 4-3, 1.30