Fluids Pressure and Depth
SUBJECT: Aeronautics
TOPIC: Hydrostatic Pressure
DESCRIPTION: A set of mathematics problems dealing with hydrostatics.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman  August 1996
A fluid is a substance that flows easily. Gases and liquids are
fluids, although sometimes the dividing line between liquids and
solids is not always clear. Because of their ability to flow, fluids
can exert buoyant forces, multiply forces in a hydraulic systems,
allow aircraft to fly and ships to float.
The topic that this page will explore will be pressure and depth.
If a fluid is within a container then the depth of an object placed
in that fluid can be measured. The deeper the object is placed in the
fluid, the more pressure it experiences. This is because is the
weight of the fluid above it. The more dense the fluid above it, the
more pressure is exerted on the object that is submerged, due to the
weight of the fluid.
The formula that gives the P pressure on an
object submerged in a fluid is:
P = r * g * h
where
 r (rho) is the density of the fluid,
 g is the acceleration of gravity
 h is the height of the fluid above the
object
If the container is open to the atmosphere above, the added
pressure must be included if one is to find the total pressure on an
object. The total pressure is the same as absolute pressure on
pressure gauges readings, while the gauge pressure is the same as the
fluid pressure alone, not including atmospheric pressure.
P_{total} = P_{atmosphere} +
P_{fluid}
P_{total} = P_{atmosphere} + ( r * g * h
)
A Pascal is the unit of pressure in the metric system. It
represents 1 newton/m^{2}
Example:
Find the pressure on a scuba diver when she is 12 meters below the
surface of the ocean. Assume standard atmospheric conditions.
Solution:
The density of sea water is 1.03 X 10^{ 3} kg/m^{3}
and the atmospheric pressure is 1.01 x 10^{5}
N/m^{2}.
P_{fluid} = r g h = (1.03 x10 ^{3} kg/m^{3})
(9.8 m/s^{2}) (12 m) = 1.21 x 10^{5}
Newtons/m^{2}
P_{total} = P_{atmosphere} + P_{fluid} =
(1.01 x 10^{5}) + (1.21 x 10^{5} ) Pa = 2.22 x
10^{ 2} kPa (kilo Pascals)
Exercises :
 What is the pressure experienced at a point on the bottom of a
swimming pool 9 meters in depth? The density of water is 1.00 x
10^{3} kg/m^{3}.
(answer)
 The interior of a submarine located at a depth of 45 meters is
maintained at normal atmospheric conditions. Find the total force
exerted on a 20 cm by 20 cm square window. Use the density of sea
water given above.
(answer)
 How many atmospheres is a depth of 100 meters of ocean
water?
(answer)
 If the weight density of pure water is 62
pounds/ft^{3}, find the weight of water in a swimming pool
whose dimensions are 20 ft by 10 ft by 6 feet.
(answer)
 An airplane in level flight whose mass is 20,000 kg has a wing
area of 60 m^{2}. What is the pressure difference between
the upper and lower surfaces of its wing? Express your answer in
atmospheres.
(answer)
Related Pages:
Aeronautics Activities
Aerospace Activities Page
Aerodynamics Index
Air Pressure
Air Density
