1. LECTURE UNIT NO. 3
FLUIDS at REST: PRESSURE CONSIDERATIONS
Fluid Statics deals with systems in which the fluid is either at rest or in uniform motion. In either case
there is no sliding of adjacent layers of fluid relative to each other. As a result, shear stresses do not exist
and thus viscosity does not play a vital role in the behavior of the fluid.
Pressure at a Point: Pascal’s Law
Pressure acts equally in all directions at any point in a static body of fluid.
Consider a small cube of fluid of dimension h located at arbitrary point O within a large static body of fluid.
Summation of forces in the horizontal direction
Summation of forces in vertical direction
Hence, the pressure at any point in a fluid is the same in all directions.
Pressure Variation in Liquids: Pressure-Depth Relationship
Consider a cylindrical volume of liquid with cross-sectional area A and whose height is h.
(P + dP)A – PA – dW = 0
PA + dP(A) – PA – γ (dV) = 0
A(dP) – γ (A x dh) = 0
Thus; dP = γ (dh)
If we integrate, ΔP = γ Δh
P2 – P1 = γ (h2 – h1)

2. Thus, the change in pressure between any two points equal the specific weight times the change in
depth.
Illustration:
Find the intensity of the pressure P at a depth h below the surface of a liquid of a weight density γ
at the free surface is zero. Assume γ is constant.
If h1 = 0 so that P1 = 0 gage (P1 is located in free surface of liquid), and h2 = h = an arbitrary
depth from the free surface.
P=γh
Note:
It is more convenient to express pressure in terms of height of column of fluid rather than in
pressure per unit area.
h=P
γ
Observation of Pressure-Depth Relationship for a Liquid
1. The pressure in a static body of liquid increases linearly with depth, where the density is the slope
of the liquid.
2. As the depth increases, so does the pressure, and vice versa.
3. Depth is measured positively in downward direction from the free surface of the liquid. Thus as the
depth increases, the elevation decreases, and vice versa.
4. The greater the density of the liquid, the greater the change in pressure for a given change in
depth.
5. The equation derived cannot be used for a gas since due to its compressibility the density is not
constant.
6. The head h of liquid in the container is called a pressure head. In general, a pressure head h is that
vertical height of a column of liquid having a specific weight γ that will cause a pressure P.
7. The pressure at any point in a static body of liquid does not depend on the size or shape of the
container holding the liquid. The pressure depends only on the depth of the point and specific
weight of the liquid.
h
1 2 3 4 5
P1 = P2 = P3 = P4 = P5

3. Barometer
The absolute pressure of the atmosphere is measured with a barometer. This device consists of a
tube (filled with mercury) that is inverted and its open end is placed into a container of mercury.
This means that atmospheric pressure supports a column of mercury 760 mm high.
Example 1:
An open tank contains 5.7 m of water covered with 2.8 m of kerosene having a unit weight of 8kN/m 3. If
the diameter of the tank is 1 m. (a) find the pressure at the interface of water and kerosene (b) Find the
pressure at the bottom of the tank (c) Find the total force at the bottom of the tank
Example 2:
From the figure shown:
Air A Air
0.4 m C
0.4 m
Oil
0.5 m (0.90)
B
1.0 m
water
D
(a) Compute the pressure at A (b) compute the pressure at C (c) compute the pressure at D

4. Seatwork:
1. The container shown is filled with air, oil and water. Gage A reads 70 kPa and Gage B reads 14 kPa
less than Gage C. If the unit weight of air is 0.0036 kN/m 3. (a) compute the specific gravity of oil
(b) compute the gage reading on B (c) compute the gage reading on C.
A
Air
0.90 m
0.60 m
Oil
B
0.60 m
H2O 0.90 m
C
2. A pressure in a given tank reads 277 mm of Hg. (a) Determine the equivalent height of column of
water (b) Determine the equivalent height of column of kerosene sp.gr. = 0.82 (c) Determine the
equivalent height of column of nectar sp.gr. = 2.94
3. A pressure gauge at elevation 8 m at the side of a tank containing a liquid reads 80 kPa. Another
gauge at elevation 3 m reads 120 kPa. Compute for the (a) Specific weight (b) Density (c) Specific
gravity