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18/01/2016 LECTURE – 5

Pressure distribution in a :

 There are many instances where the fluid is in stationary condition. That is the movement of (or ) is not involved.  Yet, we have to solve some problems involving stationary (similar to what you have studied in of the course Engineering ).  The condition of fluid, for such case is termed as hydrostatic.  Recall, a fluid at rest cannot resist shear.  Therefore, the normal on any plane through a fluid element at rest is called fluid , p. (Recall you have studied normal stress & in Mechanics course).  This normal stress, when fluid is at rest, is taken positive for compression (conventionally taken through & across various course).  To describe about this pressure, let us describe through equilibrium of forced in a small wedge of fluid at rest. ( again recall principles of statics)

Δs Δy

b θ Δx

 For a body in static condition you should satisfy ∑Fx = 0

∑Fy = 0

∑Fz = 0  The above fluid element is also in static equilibrium.

 As this element is isolated from its surroundings, we need to incorporate corresponding (Free body diagram).  By convention the compressive forces may be acting on the wall of the fluid element, i.e., forces act into the plane in respective directions.  Let px be the per unit area acting on the plane b∆z.  Let pz be the force per unit area acting on the plane b∆x.  Let py be the force per unit area acting on the plane ∆x∆z.  Let ps be the force per unit area acting on the plane b∆s.

For ∑Fx = 0;

pxs b z  p b  ssin  0 ……………………… (1)

For ∑Fz = 0;

1 p b x  p b  scos  g  x  zb  0 ………. (2) zs 2

For ∑Fy = 0;

11 p  x  z  p  x  z  0………………….. (3) 22yy From (1),

pxs b z  p b  ssin  0

pxs b z  p b  z

ppxs

From (2),

1 pzs b x  p b  x  g  x  zb  0 2 1 p p  g  z zs2

For limit (∆z → 0; ∆x → 0) pz → ps

So, pz = ps = px = py = p = pressure

At a static point, pressure is a scalar property without any orientation. P(x,y,z,t) Pressure force on fluid element: Due to pressure, forces act on the respective plane of interest. Let us consider a rectangular elemental prism of volume ∆x∆y∆z

Net x force on an element due to pressure variation (Source: Fluid Mechanics by F.M. White)

 Force due to pressure is called pressure force.  At any mathematical point pressure is described as p(x,y,z,t) as it is a scalar quantity.  Therefore, let us suggest that in x direction there are two planes normal.  On the left plane, let p be the pressure. p  On the right plane, pressure will be px  x So, net force due to pressure acting in the x direction will be:

p F  p  y  z ( p   x )  y  z px x p = - x  y  z x Similarly, net force in y direction:

p Fx    yz  py y

Similarly, net force in z direction due to pressure

p Fx    yz  pz z Force is a vector & the net pressure force is given by,

ˆˆˆ Fp   Fpx i   F py j   F pz k

p  p  p [  ˆi -  ˆj  k] ˆ x  y  z x  y  z

As the volume ∆x∆y∆z is arbitrary & choosen by us, the net pressure force per unit volume

p  p  p f [  ˆi -  ˆj  k] ˆ p x  y  z =p

= of pressure i.e., it is the gradient of pressure that cause force in the fluid.

Gage Pressure & Pressure: In the classes, you might have already seen how pressure is expressed.  The units are N/m2 or Pa or kPa etc.  Might have heard about 1. Absolute pressure, 2. Gage pressure, 3. Vacuum pressure

Hydrostatic Pressure Condition:

 If a fluid is in rest, it will not have any shear force.  Similarly, in a static liquid, there will not be any viscous force.  For the same rectangular prism element of the fluid in rest (or static): We should apply principles of statics ∑F = 0

 As the fluid is in static, the forces acting on it will be  pressure force  gravity force

ρg

g00 iˆˆ  j  gkˆ p i.e. ∑Fx = 0 means  x  y  z  0 x p Or,  0 x p Similarly, ∑Fy = 0 implies  0 y

p Similarly, ∑Fz = 0 means  x  y  z  g  x  y  z  0 x z p Or, g z z

It can be also expressed as,

∑⃗⃗⃗⃗⃗퐹⃗⃗ = 0 , p  x  y  z  g  x  y  z  0

pg

p i.e., g z or

In hydrostatic condition, it is now clear that,

p p  0 and  0 x y That is, there won’t be variation of pressure in horizontal direction for the same fluid.

p dp So, we can hence write as in static conditions. z dz dp g dz z dp gz dz B

BB dp  g dz z A AA z2

z2 z pB p A   g z () z21  z

pA p B  g z () z21  z

If B is water surface, then PB = = 0 (gage pressure)

So, pA = ρgz(z2 – z1) = ρgh, where h = height of water surface level from surface.