Saved Bookmarks
| 1. |
Discuss the variation of pressure with depth or pressure produced due to fluid depth h and density of fluid rho. |
Answer» Solution :Suppose a fluid of density `rho` is in a static equilibrium in a container as shown in figure. Consider a cylindrical element of fluid having area of base A and height H. The pressure at points 1 and 2 are `P_(1)andP_(2)` respectively. The force ACTING on liquid column is as below : (1) The forces on point `-1F_(1)=P_(1)A` (downward) (2) The force on point `-2F_(2)=P_(2)A` (upward) (3) Weight of fluid column `W=mgW=Ahrhog` (downward) Fluid column is in equilibrium hence downward forces = upward forces `F_(1)+W=F_(2)` `thereforeF_(2)-F_(1)=W` `thereforeP_(2)A-P_(1)A=Ahrhog`....(1) `thereforeP_(2)-P_(1)=hrhog` `thereforeP_(2)=P_(1)+hrhog` ...(2) Equation (1) shows that pressure difference depends on the vertical distance (h) between the points (1and 2),mass density of the fluid `(rho)` and acceleration due to gravity (g) but it does not depends on the area of cross SECTION. If the effect of gravitation is neglected `P_(2)-P_(1)=hrhog=0` `thereforeP_(2)=P_(1)` It indicates that if the effect of gravitationis neglected , the pressure at every point in fluid is same. Let there be point 1 on the surface of container. The pressure on the surface is atmospheric pressure `P_(a).P_(1)=P_(a)andP_(2)=PthereforeP-P_(a)=hrhogorP=P_(a)+hrhog` ... (3) Pis CALLED absolute pressure. Equation (3) shows that the pressure P at depth below the surface of liquid OPEN to the atmosphere is greater than atomspheric pressure by an amount `hrhog`. The excess of pressure , `P-P_(a)`, at depth h is called a gauge pressure at that point. |
|