Question

(a) Calculate the absolute pressure at an ocean depth of 1250 m. Assume the density of sea water is 1028 kg/m³ and that the air above exerts a pressure of 101.3 kPa.
. Pa

(b) At this depth, what force must the frame around a circular submarine porthole having a diameter of 28.0 cm exert to counterbalance the force exerted by the water?
N

Answer 1

given data

atmospheric pressure=101.3*10^3 Pa

=101.3*10^3 N/m^2

density of water=1028 kg/m^3

acceleration due to gravity=9.8 m/s^2

depth=1250 m

Calculate the absolute pressure.

Absolute Pressure at the depth 'h' , P = Po + hρg

= 101.3 X 10 ³ + 1250 * 1028 * 9.8

= 101.3 X 10 ³ + 12593 X 10 3

P = 12694.3 kPa

(B)

Diameter of the circular hole d=28 cm

d = 0.28 m

Calculate area

A =pi*(diameter/2)^2

= pi*(0.28/2)^2=0.0615 m^2

Calculte the pressure fotce.

Force F = P*A

F =12694300*0.0615

= 780700 N