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Imagine a thin element of fluid at a depth h , as shown in [link] . Let the element have a cross-sectional area A and height Δ y . The forces acting upon the element are due to the pressures p ( y ) above and p ( y + Δ y ) below it. The weight of the element itself is also shown in the free-body diagram.

Figure A is a schematic drawing of a cylinder filled with fluid and open to the atmosphere on the top. A disk of mass Delta m, surface area A identical to the surface area of the cylinder, and height Delta y is placed in the fluid. A fluid of height h is located above the disk. Figure B is a schematic drawing of the force Delta m x g expressed by the disk, p (y) x A applied by the fluid above the disk, and p (y + Delta y) x A applied by the fluid below the disk.
Forces on a mass element inside a fluid. The weight of the element itself is shown in the free-body diagram.

Since the element of fluid between y and y + Δ y is not accelerating, the forces are balanced. Using a Cartesian y -axis oriented up, we find the following equation for the y -component:

p ( y + Δ y ) A p ( y ) A g Δ m = 0 ( Δ y > 0 ) .

Note that if the element had a non-zero y -component of acceleration, the right-hand side would not be zero but would instead be the mass times the y -acceleration. The mass of the element can be written in terms of the density of the fluid and the volume of the elements:

Δ m = | ρ A Δ y | = ρ A Δ y ( Δ y > 0 ) .

Putting this expression for Δ m into [link] and then dividing both sides by A Δ y , we find

p ( y + Δ y ) p ( y ) Δ y = ρ g .

Taking the limit of the infinitesimally thin element Δ y 0 , we obtain the following differential equation, which gives the variation of pressure in a fluid:

d p d y = ρ g .

This equation tells us that the rate of change of pressure in a fluid is proportional to the density of the fluid. The solution of this equation depends upon whether the density ρ is constant or changes with depth; that is, the function ρ ( y ).

If the range of the depth being analyzed is not too great, we can assume the density to be constant. But if the range of depth is large enough for the density to vary appreciably, such as in the case of the atmosphere, there is significant change in density with depth. In that case, we cannot use the approximation of a constant density.

Pressure in a fluid with a constant density

Let’s use [link] to work out a formula for the pressure at a depth h from the surface in a tank of a liquid such as water, where the density of the liquid can be taken to be constant.

A schematic drawing of the beaker filled with fluid to the height h. Fluid exhibits pressure P0 equal to zero at the surface and pressure P at the bottom of the beaker.

We need to integrate [link] from y = 0 , where the pressure is atmospheric pressure ( p 0 ) , to y = h , the y -coordinate of the depth:

p 0 p d p = 0 h ρ g d y p p 0 = ρ g h p = p 0 + ρ g h .

Hence, pressure at a depth of fluid on the surface of Earth is equal to the atmospheric pressure plus ρgh if the density of the fluid is constant over the height, as we found previously.

Note that the pressure in a fluid depends only on the depth from the surface and not on the shape of the container. Thus, in a container where a fluid can freely move in various parts, the liquid stays at the same level in every part, regardless of the shape, as shown in [link] .

A photo of a few glass containers of different shape connected by a glass tube at the bottom and filled with red fluid. The fluid is at the same height at the different glass containers.
If a fluid can flow freely between parts of a container, it rises to the same height in each part. In the container pictured, the pressure at the bottom of each column is the same; if it were not the same, the fluid would flow until the pressures became equal.

Variation of atmospheric pressure with height

The change in atmospheric pressure with height is of particular interest. Assuming the temperature of air to be constant, and that the ideal gas law of thermodynamics describes the atmosphere to a good approximation, we can find the variation of atmospheric pressure with height, when the temperature is constant. (We discuss the ideal gas law in a later chapter, but we assume you have some familiarity with it from high school and chemistry.) Let p ( y ) be the atmospheric pressure at height y . The density ρ at y , the temperature T in the Kelvin scale (K), and the mass m of a molecule of air are related to the absolute pressure by the ideal gas law, in the form

Questions & Answers

a particle projected from origin moving on x-y plane passes through P & Q having consituents (9,7) , (18,4) respectively.find eq. of trajectry.
rahul Reply
definition of inertia
philip Reply
the reluctance of a body to start moving when it is at rest and to stop moving when it is in motion
An inherent property by virtue of which the body remains in its pure state or initial state
why current is not a vector quantity , whereas it have magnitude as well as direction.
Aniket Reply
the flow of current is not current
bcoz it doesn't satisfy the algabric laws of vectors
The Electric current can be defined as the dot product of the current density and the differential cross-sectional area vector : ... So the electric current is a scalar quantity . Scalars are related to tensors by the fact that a scalar is a tensor of order or rank zero .
what is binomial theorem
Tollum Reply
hello are you ready to ask aquestion?
Saadaq Reply
what is binary operations
What is the formula to calculat parallel forces that acts in opposite direction?
Martan Reply
position, velocity and acceleration of vector
Manuel Reply
*a plane flies with a velocity of 1000km/hr in a direction North60degree east.find it effective velocity in the easterly and northerly direction.*
hello Lydia.
What is momentum
A rail way truck of mass 2400kg is hung onto a stationary trunk on a level track and collides with it at 4.7m|s. After collision the two trunk move together with a common speed of 1.2m|s. Calculate the mass of the stationary trunk
Ekuri Reply
I need the solving for this question
is the eye the same like the camera
I can't understand
same here please
I think the question is that ,,, the working principal of eye and camera same or not?
yes i think is same as the camera
what are the dimensions of surface tension
why is the "_" sign used for a wave to the right instead of to the left?
why classical mechanics is necessary for graduate students?
khyam Reply
classical mechanics?
principle of superposition?
Naveen Reply
principle of superposition allows us to find the electric field on a charge by finding the x and y components
Two Masses,m and 2m,approach each along a path at right angles to each other .After collision,they stick together and move off at 2m/s at angle 37° to the original direction of the mass m. What where the initial speeds of the two particles
2m & m initial velocity 1.8m/s & 4.8m/s respectively,apply conservation of linear momentum in two perpendicular directions.
A body on circular orbit makes an angular displacement given by teta(t)=2(t)+5(t)+5.if time t is in seconds calculate the angular velocity at t=2s
2+5+0=7sec differentiate above equation w.r.t time, as angular velocity is rate of change of angular displacement.
Ok i got a question I'm not asking how gravity works. I would like to know why gravity works. like why is gravity the way it is. What is the true nature of gravity?
Daniel Reply
gravity pulls towards a mass...like every object is pulled towards earth
An automobile traveling with an initial velocity of 25m/s is accelerated to 35m/s in 6s,the wheel of the automobile is 80cm in diameter. find * The angular acceleration
Goodness Reply
(10/6) ÷0.4=4.167 per sec
what is the formula for pressure?
Goodness Reply
force is newtom
and area is meter squared
so in SI units pressure is N/m^2
In customary United States units pressure is lb/in^2. pound per square inch
Practice Key Terms 4

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