# Fluid Mechanics Essay

P_2

A

P_1

Force due to P_1 on area A acting up = P_1 A

Force due P_2 on area A acting down = P_2 A

Force due to the weight of the element =mg= ρA(z_2- z_1 )g

Since the fluid is at rest, there can be no shear forces and hence no vertical forces on the side of element due to surrounding fluid. Considering upward as positive, sum of all forces = 0,

We have

P_1 A-P_2 A - ρA(z_2- z_1 )g=0

Or

P_2- P_1= - ρ(z_2- z_1 )g

So, under the influence of gravity , pressure decreases with the increase height.

Equality

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hence the ∂p/∂z cannot vary horizontally. This implies ∂p/∂z= -ρg = constant

Hence for equilibrium ρ must be constant over any horizontal plane.

Therefore the conditions for equilibrium under gravity are The pressure at all points on a horizontal plane must be the same The density at all points on a horizontal plane must be the same The change of pressure with elevation is given by

The pressure variation with respect to height is given by dp= -∫_(z_1)^(z_2)▒〖ρg dz〗

The integration can be performed, if the relation between density and pressure is known.

Variation of pressure with altitude in a fluid of constant density

Generally liquids come under this category.

Integrating the above equation, taking density out , we have p_2-p_1= -ρg(z_2-z_1)

Variation of pressure with altitude in a gas at constant temperature (Isothermal)

From gas equation, we have ρ=p/RT And we have general relation for the variation of pressure in vertical direction dp/dz= -ρg= -pg/RT dp/p= -g/RT dz integrating from p=p_1 when z=z_1 and p=p_2 when z=z_2 ln〖(p_2/p_1 )=e^((-g/RT (z_2-z_1 ) ) ) 〗

Variation of pressure under adiabatic conditions p/ρ^γ =constant=p_1/(ρ_1^γ ) ρ=ρ_1 (p/p_1 )^(1/γ)

Substituting in dp/dz= -ρg= -(ρ_1 g)/(p_1^(1/γ) ) p^(1/γ)

z_2-z_1=(p_1^(1/γ))/(ρ_1 g) [p^((γ-1)/γ)/((γ-1)/γ)]

Between p=p1 and p=p2 and substitute p1/ρ_1 =