Question

What is the fluid pressure (in units of ܲܽPascals) a distance 3.2 meters below the surface of Lake Superior?

Answer 1

fluid pressure = 132590 pascal

Step-by-step explanation:

Fluid pressure can be calculated by using following relation:

`P_o - P_g = g\time p* (Z_b - Z_a )`

Where

Pb = fluid pressure at a distance of 3.2 m below from surface

Pa = fluid pressure at surface = atmospheric pressure = 101325 Pascal

g = gravitational constant = 9.8 m/sec2

`\rho = density of fluid = 997 kg/m³`

Z = location depth from surface of lake

Therefore ,
`Z_a = 0 meter`

`Z_b = 3.2 meter`

`P_b - P_a =g*\rhop* (Z_a - Z_b)`

`= 9.8 * 997* (3.2-0)`

`= 31265 kg/m-sec^2`

1 Pascal = 1 Kg/(m.Sec)

`P_b = P_a +31265`

= 101325 + 132590 Pascal

= 132590 pascal