hydrostatic equilibrium

fluid flow operation sem 3 chemical engineering

1. HYDROSTATIC EQULIBRIUM NAME:- HIREN DHINGANI(131130105007) DEPARTMENT :- CHEMICAL ENGINEERING Mo: 9913413660 SAL COLLEGE OF ENGINEERING

2. DEFINATION OF HYDROSTATIC EQUILIBRIUM hydrostatic balance occurs when compression due to gravity  is balanced by a  pressure gradient force in the opposite direction.

3. HYDROSTATIC EQULIBRIUM In this column of static the static fluid the pressure at any point is the same direction The pressure is also constant at any point is the same in all directions Let, Cross sectional area = A Pressure =p Hight =h Hight of the base= h+dh Pressure= p+dp

4. Force (P+DP)A is acting downwards.....taken as + ve Force to gravity is acting downwards and is equal to mass times acceleration due to gravity=mg= Vƍg = A.dhƍ.g .... taken as - ve . Force PA is acting upwards ....taken as – ve . As the fluid element is in equilibrium the resultant of these three force action on it must be zero. Thus,

5. +PA – A dh.ƍ.G – ( P+dp )A =0 +PA – A dh.ƍ.G –PA – A.dp =0 - A.dh.ƍ.g – A.dp =0 dp+dh.ƍ.g =0 This 1 equationis the desired basic equation that can be used for obtaining the pressure at any height. Let apply it to incompressible and compressible fluids. 1

6. 1. Incompressible fluids:- For incompressible fluids, density is independent of pressure. Integrating equation 1. we get, dp + g.ƍ.dh =0 P + hƍg =constant From equation (2) is clear that the pressure is maximum at the base of the column or container of the fluid and it decrease as we move up the column. 2

7. If the pressure at the base of the column is p1 where h=0 and pressure at any height h above the basic is p2 such that p1>p2 then, (p1-p2) = h.ƍ.g In this equation the pressure difference in a fluid between any two points can be obtain by measuring the height of the vertical column of the fluid, 3

8. 2. Compressible fluids:- Compressible fluids density varies with pressure . For an ideal gas, the density is given by the relation. ƍ = PM RT P =absolute pressure M=molecular weight of gas R=universal gas constant T=absolute temperature. Putting of value of ƍ from equation (4) into equation (1) dp + g(PM/RT)dh =0 4 5

9. Re arranging equation (5) dp + g. M dh =0 P RT Integrating equation (6),we get lnP + g M .h = constant RT Ingrtrating the above equation between two heights h1 and h2 where the pressure acting are p1 and p2,we get ln p2 = -g M(h2-h1) p1 RT Equation (7) is known as the B aromaric equation. 6 7 8

10. THANK YOU