Title Rhodes Solutions Ch8 127.4 KB 21
##### Document Text Contents
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SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT

EXERCISE 8.1:
Design a positive pressure dilute-phase pneumatic transport system to carry 500 kg/hr
of a powder of particle density 1800 kg/m3 and mean particle size 150 m across a
horizontal distance of 100 metres and a vertical distance of 20 metres using ambient
air. Assume that the pipe is smooth, that four 90˚ bends are required and that the
allowable pressure loss is 0.7 bar.

SOLUTION TO EXERCISE 8.1:
Design in this case means determine the pipe size and air flowrate which would give a
total system pressure loss near to, but not exceeding, the allowable pressure loss.
The design procedure requires trial and error calculations. Pipes are available in fixed
sizes and so the procedure is to select a pipe size and determine the saltation velocity
from Text-Equation 8.3. Saltation velocity is important since, in any system with
horizontal and vertical pipelines, the saltation velocity is always greater than the
choking velocity - so if we avoid saltation, we avoid choking. The system pressure
loss is then calculated at a superficial gas velocity equal to 1.5 times the saltation
velocity (this gives a reasonable safety margin bearing in mind the accuracy of the
correlation in Text-Equation 8.3). The calculated system pressure loss is then
compared with the allowable pressure loss. The pipe size selected may then be
altered and the above procedure repeated until the calculated pressure loss matches
that allowed.

Step 1 Selection of pipe size: Select 50 mm internal diameter pipe.

Step 2 Determine gas velocity

Use the Rizk correlation of Text-Equation (8.3) to estimate the saltation
velocity, USALT. Text-Equation (8.3) rearranged becomes:

1
1

f

2
22

p
SALT

Dg10M4
U

where = 1440x + 1.96 and = 1100x + 2.5.
In the present case = 2.176, = 2.665 and USALT = 9.21 m/s.
Therefore, superficial gas velocity, U = 1.5 x 9.211 m/s = 13.82 m/s.

SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT Page 8.1

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Step 3 Pressure loss calculations
a) Horizontal Sections
Starting with Text-Equation 8.15 and expression for the total pressure loss in the
horizontal sections of the transport line may be generated. We will assume that all
the initial acceleration of the solids and the gas take place in the horizontal sections
and so terms 1 and 2 are required. For term 3 the Fanning friction Equation is used
assuming that the pressure loss due to gas/wall friction is independent of the presence
of solids. For term 4 we employ the Hinkle correlation (Text-Equation 8.17). Terms
5 and 6 became zero as = 0 for horizontal pipe. Thus, the pressure loss, PH, in the
horizontal sections of the transport line is given by:

pH
f HUfH

2

2
p(1 H )U

2

2
2f g fU

2LH
D

2f p p 1 H UpH
2 LH

D

where the subscript H refers to the values specific to the horizontal sections.

To use this Equation we need to know H, UfH and UpH. Hinkle’s correlation gives
us UpH:
UpH U (1 0.0638 x

0.3 p
0.5) = 11.15 m/s

From continuity, G . p(1 H )U pH

Solids flux, G = Mp/A =
500
3600

1

4
(0.05)2

70.73 kg / m2.s

thus

pHp
H U

G
1 = 0.9965

and UfH
U

H

13.82
0.9965

13.87 m / s

Friction factor fp is found from Text-Equation (8.19) with CD estimated at the relative
velocity (UfH - UpH), using the approximate correlations given below, (or by using an
appropriate CD versus Re chart [see Chapter 2])

Rep < 1 : CD = 24/Rep
1 < Rep < 500 : CD = 18.5 Rep

0.6

500 < Rep < 2 x 10
5 : CD = 0.44

SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT Page 8.2

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In the present case = 3.4, = 3.6 and USALT = 10.94 m/s.
Therefore, superficial gas velocity, U = 1.5 x 10.94 m/s = 16.41 m/s.

Step 3 Pressure loss calculations
a) Horizontal Sections
Starting with Text-Equation 8.15 and expression for the total pressure loss in the
horizontal sections of the transport line may be generated. We will assume that all
the initial acceleration of the solids and the gas take place in the horizontal sections
and so terms 1 and 2 are required. For term 3 the Fanning friction Equation is used
assuming that the pressure loss due to gas/wall friction is independent of the presence
of solids. For term 4 we employ the Hinkle correlation (Text-Equation 8.17). Terms
5 and 6 became zero as = 0 for horizontal pipe. Thus, the pressure loss, PH, in the
horizontal sections of the transport line is given by:

pH
f HUfH

2

2
p(1 H )UpH

2

2
2fg f U

2LH
D

2f p p 1 H UpH
2 LH

D

where the subscript H refers to the values specific to the horizontal sections.

To use this Equation we need to know H, UfH and UpH. Hinkle’s correlation gives
us UpH:

UpH U(1 0.0638 x

0.3 p
0.5) = 12.24 m/s

From continuity, G . p(1 H )U pH

Solids flux, G = Mp/A =

700
3600

1

4
(0.04)2

154.7 kg / m2.s

thus

pHp
H U

G
1 = 0.9874

and UfH

U

H

16.41
0.9874

16.62 m / s

Friction factor fp is found from Text-Equation (8.19) with CD estimated at the relative
velocity (UfH - UpH), using the approximate correlations given below, (or by using an
appropriate CD versus Re chart [see Chapter 2])

SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT Page 8.10

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Rep < 1 : CD = 24/Rep
1 < Rep < 500 : CD = 18.5 Rep

0.6

500 < Rep < 2 x 10
5 : CD = 0.44

Thus, for flow in the horizontal sections, Rep =
xUU pHfHf

for ambient air f = 1.2 kg/m

3 and = 18.4 x 10-6 Pas, giving

Rep
1.2 16.62 12.24 1 10 3

18.4 10 6
285.5

and so, using the approximate correlations above, CD = 18.5 Re

-0.6 = 0.622

Substituting CD = 0.622 in Text-Equation 8.19 we have:

2

3p 24.12
24.1262.16

101
040.0

622.0
1000

2.1
8
3

f = 0.00143

To estimate the gas friction factor we use the Blasius correlation for smooth pipes,

. The Reynolds number calculated based on the superficial gas

velocity:

f g 0.079 Re
0.25

Re

0.04 1.2 16.41
18.4 10 6

42800, which gives fg = 0.0055.

Thus the components of the pressure loss in the horizontal pipe from Text-Equation
8.15 are:
Term 1 (gas acceleration):

f H
UfH

2

2
1.2 0.9874 16.622

2
163.6 Pa.

Term 2 (solids acceleration):

p
1 H UpH

2

2
1000 1 0.9874 12.242

2
946.8 Pa.

Term 3 (gas friction):

2f g fU

2LH
D

2 0.0055 1.2 16.412 80
0.04

7096 Pa.

SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT Page 8.11

Page 20

Knowing the required pressure gradient, the packed bed voidage and the particle and
gas properties, Text-Equation 8.26 can be solved for Urel , the magnitude of the

relative gas velocity:

2
rel

sv

f
rel2

2

2
sv

U
1

x
75.1U

1
x

150
H

p

2

rel6rel2

2

26

5

U
47.0

47.01
10220
0.2

75.1U
47.0

47.01
)10220(

102
1509000

17940 Urel
2 78819 Urel 9000 0

noring the negative root of the quadratic,Ig Urel 0.1114 m / s

We now adopt a sign convention for velocities. For standpipes it is convenient to take

rom the continuity for the solids (Text-Equation 8.11),

downward velocities as positive. In order to create the pressure gradient in the
required direction, the gas must flow upwards relative to the solids. Hence, Urel is
negative: Urel = -0.1114 m/s

F

solids flux,
Mp Up 1 p A

The solids fl g/ux is given as 300 k m2s and so:

p
300

1 0.47 2000
0.283 m / s U

olids flow is downwards, so Up = + 0.283m/s

he relative velocity,

S

T Urel Uf Up

Hence, actual gas velocity, Uf 0.1114 0.283 0.1716 m / s (downwards)

ownwardsTherefore the gas flows d at a velocity of 0.1716 to the

s velocity is therefore:

m/s relative
standpipe walls.
The superficial ga

Uf 0.47 0.1716 0.0807 m / s U

SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT Page 8.20

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From the continuity for the gas (Text-Equation 8.12) mass flow rate of gas,
Mf Uf fA

= 0.0114 kg/s.

So for the standpipe to operate as required, 0.0114 kg/s of gas must flow from upper
vessel to lower vessel.

SOLUTIONS TO CHAPTER 8 EXERCISES: PNEUMATIC TRANSPORT Page 8.21