Adama Science &Technology University
School of Mechanical, Chemical and Materials
Engineering
Department of Chemical Engineering
FLUID MECHANICS FOR CHEMICAL ENGINEERS
(ChE 2206)
•
Dr. Alemu G.
22/03/2025
Dr. Alemu G.
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Course Information
Course Name: Fluid Mechanics for Chemical Engineers
Course Code: ChE2206
Lesson Title: Practice Problems and Solutions Part I
Instructor Name: Dr. Alemu G.
Peer Reviewers
1.
2.
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Desta Getachew
Etefa Gurmesa
Dr. Alemu G.
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CH- I Recap
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Properties- Recap
1. Density
2. Viscosity & kinematic viscosity
3. Specific volume, Specific weight,
Specific gravity
4. Surface tension
5. Pressure
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Tutorial Q1.1
• A reservoir of carbon tetrachloride (CCl4) has a
mass of 500 kg and a volume of 0.315 m3.
Determine the CCl4:
a) Weight
b) Density
c) Specific weight
d) Specific volume
e) Specific gravity
Take gravitational acceleration, g = 9.81 m/s2
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Solutions
a) Weight, W= mg
= (500 kg)(9.81m/s2)
= 4905 kgm/s2
= 4905 (kgm/s2) x N/(kgm/s2) x (kN/1000N)
= 4.905 kN
b) Density,  =
m
500
kg
kg

x 3  1587 3
V 0.315 m
m
c) Specific weight,  =
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W
4905 kN
kN

x 3  15.57 3
V 0.315 m
m
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Solutions
d) Specific volume,
3
1
1
m
υ  
 6.301 10 -4
ρ 1587x kg
kg
m3
e) Specific gravity, SG
ρ CCl4
1587 kgm 3


x
 1.59
3
ρH2 O 1000 kgm
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Tutorial Q1.2
• The volume of a rock is found to be 0.00015 m3. If
the specific gravity of the rock is 2.60, determine its
mass and weight.
ρrock
ρrock


 2.60
kg
ρH2 O 1000x
m3
S.G
 rock = 2600 kg/m3
Mass, m = V = (2600 kg/m3)(0.00015 m3) = 0.39 kg
Weight, W = mg = (0.39 kg)(9.81m/s2) = 3.826 N
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Tutorial Q1.3
• Nitrogen gas occupies a volume of 100 L at 120 kPa
and 100oC. Determine its specific volume and
specific weight.
Solutions:V = 100 L = 100  10-3 m3
P  ρRT
ρ 
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P
120 kPa
kg


1.08
RT 
m3
kPa  m 3 
 8.314

kmol  K 100  273 K 

 28 kg



kmol


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Solutions
• Specific volume,
1
1
m3
υ  
 0.923
kg
ρ 1.08 x
kg
m3
• Specific weight,  = g
= (1.08 kg/m3)(9.81 m/s2)
= 10.6 N/m3
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Tutorial Q1.4
• Determine the capillary rise of water at 10oC in a
tube if the tube diameter is 1 mm.
What will happen if the tube diameter increases by
50%? Data: water @ 10oC = 0.00742 N/m
Assumption: for water Ɵ = 0
h

 cos 0  1.51 mm
1000 Nm   0.001m 
2 0.00742 Nm -1
-3
if tube diameter increases by 50%
h
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

2 0.00742 Nm -1 cos 0
1000 Nm   0.001 m  1.5 
-3
Dr. Alemu G.
 1.0 mm
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