Question

(a) A large, open at the top, upright cylindrical tank contains fuel oil with a density of 0.890 x 103 kg/m². If the air pressure is 101.3 kPa, determine the absolute pressure (in Pa) in the fluid at a depth of 32.5 m. 283465x The absolute pressure at any point in the fluid is the sum of atmospheric pressure and the pressure due to the depth of the fluid. Did you remember to account for atmospheric pressure? Pa (b) Determine the force (in N) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 4.00 cm. ON

Answer 1

Air pressure on cylinder top = Po=10l-3 kPa =) po = 102.3 xicoo Pa = 101300 Pa Depth of fluid = d = 32.5m density of fluid = = 0.890 X 10² Kg 1m² = 890 kg / m3 (a) Absolute pressure in fluid at depth d=p=Pot syd - where g is acceleration due to gravity = 9.8 m/s2 =) p = 101300 + (890 x 9.8x32.5) Pa =) P = 101300 + 283465 Pa = P = 384765 Pa (6) Pressure at depth d due to fluid only =P,= gg d =) P, =89089.8832.5 Pa = P, = 283465 Pa diameter of window =D = 4cm = 4x10-2m radius of window' = x= ļ = 2 x 10-2 m Area of window = A=79²=R x 4x10-4 m2 Force exerted by fluid = F = Pressure X Area = F =P, A = 283465 * *xu xlo un = F = 356.21 N