Question

A barrel contains a 0.105-m layer of oil floating on water that is 0.275 m deep. The density of the oil is 630 kg/m3

a) What is the gauge pressure at the oil-water interface?

b) What is the gauge pressure at the bottom of the barrel?

Answer 1

Part A.

Gauge pressure is given by:

P_gauge = rho*g*h

rho = density of oil = 630 kg/m^3

h = height of oil layer = 0.105 m

g = 9.81 m/sec^2

So, At the oil-water interface

P_gauge = 630*9.81*0.105 = 648.9 Pa

Part B.

At the bottom of barrel:

P_gauge = Pressure due to oil layer + pressure due to water

Pressure due to water = rho_w*g*H

rho_w = 1000 kg/m^3

H = height of water = 0.275 m

Pressure due to water = 1000*9.81*0.275 = 2697.8 Pa

So, At the bottom of barrel

P_gauge = P_oil + P_water

P_gauge = 648.9 + 2697.8 = 3346.7 Pa

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