A pipeline which is 4m in diameter contains a gate valve. The pressure at centre of pipe is 19.6 N/`(cm)^2`. If pipe is filled with oil of specific gravity 0.87, Find force exerted by oil upon gate and position of center of gravity.

SOLUTION:

Given,

Diameter of pipe,`d=4 m`

Area of pipe,`A=(pi*d^2)/4=4pi m^2`

Pressure at the center of the pipe,`P=19.6**10^4`N/`m^2`

Now,

The pressure head at the centre is

`=p/(rho*g)=(19.6**10^4)/(870**9.81)=22.988 m`

Thus, the height of equivalent free surface of oil from the center of pipe = 22.988 m

Also, the depth of C.G of gate valve from free oil surface, `bar (h....Show More

SOLUTION:

Given,

Diameter of pipe,`d=4 m`

Area of pipe,`A=(pi*d^2)/4=4pi m^2`

Pressure at the center of the pipe,`P=19.6**10^4`N/`m^2`

Now,

The pressure head at the centre is

`=p/(rho*g)=(19.6**10^4)/(870**9.81)=22.988 m`

Thus, the height of equivalent free surface of oil from the center of pipe = 22.988 m

Also, the depth of C.G of gate valve from free oil surface, `bar (h)=22.988 m`

Now,

Force on the gate is,

`F=rho*g*A*bar (h)=870**9.81**4pi*22.988`

`F=2465500 N`

Also,

the point of application of this force is,

`h^** =I_G/(A*bar (h))+bar (h)`

Where,

`I_G=`moment of inertia of pipe

For circular disc,

`I_G=(pi*d^4)/64`,

Thus, `h^**=((pi*d^4)/64)/((pi*d^2)/4*bar (h))+bar (h)`

`=0.043+22.988`

`h^**=23.031 m` 

Thus the centre of pressure is acting at distance of 23.031 m from the imaginary free oil surface.