If h 50ft find The gauge pressure at point 1 108 psig The g

If h = 50ft, find:

The gauge pressure at point 1. [10.8 psig]

The gauge pressure at point 2. [10.8 psig]

The magnitude of the resultant force acting on the valve. [19,500 lb]

The diameter of the float needed to open the valve when the float is half submerged. [10.8 ft]

h/2 h/4 Figure 1: Reservoir

p = g h

where

p = pressure in fluid (N/m², Pa, lb_f/ft², psf)

= density of liquid (kg/m³, slugs/ft³) = 1000 kg/m³

g = acceleration of gravity (9.81 m/s², 32.17405 ft/s²)

h = height of fluid column, or depth in the fluid at which the pressure is measured (m, ft)

Point 1&2 height from open surface = h/2 = 50/2 = 25ft = 7.62m

Hence, p=1000*9.81*7.62 = 74,752..2 Pa = 10.8 psig

Here Point 1&2 both are at same level so gauge pressure will be same for both point.

Total Resultant force = R = g h A

Where A is the area of plate = 0.05h*b = 0.05*50*b = 2.5*b ft²

R = 1000*9.81*15.24*A = 1,13,922.35b N = 33609.92A lb

Here A is not given so for the answer of 19500 lb, A must be = 19500/33609.92 = 0.5801 ft²

So, A= 2.5 b =0.5801 Hence b will be = 0.23 ft

Here tension in the string will be 19500 lb

So, flot has to displace V volume of water for this boyance force.

Hence 19500 = * V * g

So, V = 19500 / 32.17 ( is negligible) = 606.15 ft³

Hence volume of sphare = 4/3 r³ = 606.15

hence r = 5.25 ft

so, d= 2*r = 10.5 ~ 10.8 ft