Qassim University
Fluid mechanics
Faculty of Engineering
CE230
Level 6
Time allowed 60 min
1432 – 1433 H
First midterm exam
Second semester
___________________________ ________________________________________________
Clear and clearance of your letter will be much appreciated.
Name:
PIN: Model Answer
Try the following questions
--------------- ----------------------- --------------------------- ------------------------------ ----------- --1. Find the capillary rise of water in a tube of diameter 0.3mm. Take surface tension
of water as 0.07N/m, specific gravity of water as 9810N/m3 and =0.0.
Answer of question No 1
4 cos
 d
40.07 cos(0)
= 0.095m= 95 mm
h
98100.0003
h
1
2. The 2 m square gate AB in figure below is supported by two support one at A and the
other at B. Find the hydrostatic force action on the gate and its position of action.
3.0m
Water
y'
ycp
A
Cg
Cp
1.60 m
e
B
1.20 m
Answer of question No 2

the hydrostatic force can be calculated using the following equation
P   . y '.A
A  2 * 2  4m 2
  9810 N / m3
1.6
 3.80m
2
P  9810 3.84  149112N
y'  3.0 

Action position of the force may be calculated as follow:
I cg. sin 2 ( ) 2 * 23 / 12 0.82
=0.056m
e

 
A. y '
43.8
The hydrostatic force is located at a distance y cp from the water surface.
ycp  e  y '
ycp  0.056  3.8  3.856 m
2
2- Multiple choice questions
1. Dynamics viscosity is defined as equal to
a) Kinematics viscosity / density
b) Kinematics viscosity x density
c) Pressure x density
d) Kinematics viscosity x pressure
2. Kinematics viscosity has the dimension as
a) ML-2 T-1
b) ML2 T-1
-2 -1
c) L T
d) L2 T-1
3. Surface tension has the units of
a) force/unit length
b) force/unit area
c) force/unit volume
d) none of above
4. Newton's law of viscosity is given by the relation
du
dy
1 du
c)  
 dy
a)   
b)   
2
du
dy
d) none of above
5. Capillary rise of liquid in a thin tube is given by:
4 cos
 d
4 cos
c) h 
 d
a) h 
b) h 
4 cos
 r
d) none of above
6. The point of application of hydrostatic pressure on a surface is
a) centre of pressure
b) intensity of pressure
c) centroid
d) none of these
7. The pressure of 2.7m of head of water is equal to
a) 26.476 kN/m2
b) 26.476 N/m2
2
c) 264.76 N/m
d) 26476 kN/m2
8. centre of pressure below water surface for a vertical immersed surface is given by:
a)
I cg
 y'
A. y'
I cg
c) y '
A. y '
b)
d)
I cg
A. y'
 y'
A. y'
 y'
I cg
9. the total pressure on an immersed surface is
a)  . y'.A
b)  .hcg
c)  . y ' / A
d)
 .A
y'
10. The value of normal atmospheric pressure is
a) pressure due to 760 mm of mercury
b) 2.01325 bar
c) all of the above
d) None of the above
1
b
Solution of question No 3
2
3
4
5
D
a
A
a
6
a
3
7
a
8
a
9
a
10
a
4. In Figure below, a mercury manometer is connected to enclosed water
tank. If the following data is available: specific gravity of the oil = 0.9,
specific gravity of mercury = 13.6, h1=0.5 m, h2 =0.4m and h3=0.45m.
Compute the enclosed air pressure at point 1.
6
7
5
4
3
Solution of Problem No 4
p3  p2   mercury .h3
p3  0.0  13.6 * 9810 .0.45 =60037N/m2
p4  p3
p5  p4   oil .h2
p5  60037  0.9 * 9810 0.4 =56506N/m2
p7  p6  p5
pair  p1  p7   water .h1
pair  56506  9810 0.50=51600N/m2
__________________________ _____________________________________
Good luck Dr. Yousry Ghazaw
4