1. The document contains example problems related to fluid statics involving calculations of pressure differences, fluid levels, densities, forces, and center of pressure locations using principles of fluid mechanics. 2. Problem 1 involves calculating the pressure difference between two points in a manometer containing water and oil. 3. Problem 2 calculates the height of oil in a tank containing immiscible water and oil layers. 4. Problem 3 determines the force and center of pressure on a submerged gate using fluid properties and gate geometry.

1. Engineering Fluid Mechanics I
Example Problems: Fluid Statics
Problem 1. Pressure Difference in a Manometer
a) Write an expression for the water difference between points A and B, pA-pB, in terms
of γw , the specific gravity of the oil, SGoil, the levels of the fluids z’s at A, B, C, D
and E.
b) Determine the pressure difference between locations A and B in the figure below. The
water has a γw = 9800 N/m3
and the oil specific gravity, SGoil = 0.85.
Solution:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
w oil w oil
w oil oil
3
9800 N/m 0.15 m 0.85 0.10 m 0.20 m 0.85 0.15 m
73.5 Pa =
A B A C C D D E E B
A C C D D E E B
A C C D D E E B
A B
p p p p p p p p p p
z z z z z z z z
z z SG z z z z SG z z
p p
γ γ γ γ
γ
− = − + − + − + −
= − − − − − − − −
⎡ ⎤= − − + − + − + −⎣ ⎦
⎡ ⎤= − × + × − + − + ×⎣ ⎦
= −
Problem 2. In the figure below, the tank contains water and immiscible oil at 20o
C. What
is h in cm if the density of the oil is 898 kg/m3
?
Solution: We identify the level of the top surface of the oil as 1, the interface between the
oil and the water as 2 and the level of the top surface of water as 3. We choose the zero
level to be the bottom of the tank; therefore:
Water
Oil,
SG = 0.85
A
D
E
0.15 m
0.1 m
0.2 m
0.15 m
Water
Oil
B
C

2. 1 2 312 cm + 8 cm = 20 cm = 0.2 m, 8 cm = 0.08 m, 6 cm+12 cm 8 cm = 0.26 m.z h h h z z= + + + = = +
The density of oil is, ρoil = 898 kg/m3
; therefore, its specific weight,
3
oil oil 898 9.8 8800.4 N/m .gγ ρ= = × = The specific weight is:
3
water 9,800 N/m .γ = Now, we integrate
dp
dz
γ= − between 1 and 3:
( ) ( ) ( ) ( )
3 3 2 3
1 1 1 2
3 1 oil water
oil 2 1 water 3 2 oil 1 2 water 3 2 0 .
p z z z
p z z z
dp
dp dz p p dz dz
dz
z z z z z z z z
γ γ γ γ
γ γ γ γ
= − ⇒ = − ⇒ − = − −
= − − − − = − − − =
∫ ∫ ∫ ∫
We set the difference between p1 and p3 to be zero, because both of them are equal to
the atmospheric pressure (both ends are open to the atmosphere). Substituting the
values of the specific weights and the levels, we get:
( ) ( ) ( ) ( )oil 1 2 water 3 2 0 8,800.4 0.2 0.08 9,800 0.26 0.08 0.z z z z hγ γ− − − = ⇒ + − − − = In
this expression, the only unknown is h:
( )
9,800
0.26 0.08 0.12 0.08 m = 8 cm = .
8,800.4
h h= − − =
Problem 3. Gate AB is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric
pressure effects, compute the force F on the gate and its center of pressure position X
from point A (see Figure).
Solution: First, to find the resultant force, we use the following expression:
oil gate (neglect atmospheric pressure, )c aF h A pγ= . The specific weight of oil may be
expressed as: 3
oil oil w 0.82 9.8 kN/m .SGγ γ= × = ×
The height, hc, corresponds to the vertical distance from the oil surface to the centroid of
the gate. This centroid, given that the gate is rectangular, sits in the middle of the gate at
1.2/2 or 0.6 m from A along the gate. hc may then be expressed as:
4 m + (1+0.6) m sin 40 = 5.028 m =c ch h= × °
You have to be careful here to use the right units for the angle. Here, we use degrees; but,
sometimes, the default is to use radians; explicitly state that the angle is in degrees when
computing the sine. The area of the gate is expressed as the length times its width:
2
gate gate1.2 m 0.8 m = 0.96 m =A A= ×

3. The center of pressure sits below the centroid by the followind distance along the plate
incline (note that along the width into the paper, xCP = 0, i.e. no shift for the center of
pressure relative to the centroid in the x-direction):
CP
gate
sinxc
c
I
y
h A
θ
≡
By using the following figure:
Therefore, ( )( )
33 41 1
0.8 m 1.2 m 0.1152 m = .
12 12
xc xcI ba I= = = Note, here that we used
the coordinate system, as shown in the figure above, to determine that a corresponds to
the length and b corresponds to the depth into the paper. The value for yCP is:
( )
( )( )CP CP
gate
0.1152 sin 40sin
0.0153 m =
5.028 0.96
xc
c
I
y y
h A
θ °
≡ = =
Note that if yCP is positive, then the position of the center of pressure is below the
centroid. Now, we evaluate the distance X from A to the center of pressure as:
CP0.6 0.615 m =X y X= + =
Problem 4. The tank shown below is 2 m wide into the paper. Neglecting atmospheric
pressure, find the resultant hydrostatic force on panel BC.
Solution: The resultant force acting normal to panel BC may be expressed as: w cF h Aγ=
(neglect atmospheric pressure, ap ). The centroid depth is determined as follows:
3 3/ 2 4.5 m = .c ch h= + = The area of the gate is determined as follows:
( ) ( ) ( )
2 2 2
panel widthpanel length
3 m 4 m 2 m 10 m =A A= + × =
Therefore, the resultant force is:
( ) ( ) ( )3 2
w 9.8 kN/m 4.5 m 10 m 441 kN =cF h A Fγ= = × × =