How Do You Calculate Pressure at Different Points Inside a Fluid-Filled Barrel?

[MAPgirl23]

A barrel contains a layer of oil of thickness 0.110 m floating on water of depth 0.245 m. The density of the oil is 630 kg/m^3

a) What is the gauge pressure at the oil-water interface?
b) What is the gauge pressure at the bottom of the barrel?

**
a) Atmospheric Pressure + Pressure caused by oil = Total Pressure at oil-water interface
Pressure caused by oil = density * depth * g
Pressure caused by oil = 630 kg/m^3 * 0.11 m * 9.81 = 679.14 Pa
Total Pressure = 101325 Pa + 679.14 Pa but still gives me the wrong answer

***The 101325 is atmospheric pressure which is exerted in addition to the pressure caused by the oil.

b) Pressure at bottom = Pressure at oil-water interface + pressure caused by water
Pressure caused by water = density * depth * g
Pressure caused by water =1000 * 0.245 m * 9.8 = 2401
P = 102004.833 Pa + 2401 Pa but still gives me the wrong answer

***The 102004.833 is the atmospheric pressure plus the pressure of the oil which is exerted on the bottom of the container in addition to the pressure caused by the water

Please help

 

[OlderDan]

A barrel contains a layer of oil of thickness 0.110 m floating on water of depth 0.245 m. The density of the oil is 630 kg/m^3

a) What is the gauge pressure at the oil-water interface?
b) What is the gauge pressure at the bottom of the barrel?

**
a) Atmospheric Pressure + Pressure caused by oil = Total Pressure at oil-water interface
Pressure caused by oil = density * depth * g
Pressure caused by oil = 630 kg/m^3 * 0.11 m * 9.81 = 679.14 Pa
Total Pressure = 101325 Pa + 679.14 Pa but still gives me the wrong answer

***The 101325 is atmospheric pressure which is exerted in addition to the pressure caused by the oil.

b) Pressure at bottom = Pressure at oil-water interface + pressure caused by water
Pressure caused by water = density * depth * g
Pressure caused by water =1000 * 0.245 m * 9.8 = 2401
P = 102004.833 Pa + 2401 Pa but still gives me the wrong answer

***The 102004.833 is the atmospheric pressure plus the pressure of the oil which is exerted on the bottom of the container in addition to the pressure caused by the water

Please help

In your previous problem you did not include atmospheric pressure in the calculation. There is a reason for that. Pressure guages usually read the difference between the measured pressure and antmospheric pressure. In the current problem, the first reading should be just the pressure increase above 1 atmosphere due to the layer of oil. The second part should follow correctly from the first when you add the extra pressure from the water.

 

[MAPgirl23]

The density of the oil is 630 kg/m^3 = 0.63 g/cm^3

The pressure from the oil is 63g/cm^2 per meter depth
At 0.11 m deep the pressure from the oil at the interface is 0.11 x 63 = 6.93 g/cm^2

Water pressure is 100g/cm^2 per meter depth
At 0.245 m deep the pressure from the water is 0.245 x 100 = 24.5 g/cm^2

Pressure at the bottom of the barrel is 6.93 + 24.5 = 31.43 g/cm^2

The units should be in Pa
What do I do now?

 

[OlderDan]

The density of the oil is 630 kg/m^3 = 0.63 g/cm^3

The pressure from the oil is 63g/cm^2 per meter depth
At 0.11 m deep the pressure from the oil at the interface is 0.11 x 63 = 6.93 g/cm^2

Water pressure is 100g/cm^2 per meter depth
At 0.245 m deep the pressure from the water is 0.245 x 100 = 24.5 g/cm^2

Pressure at the bottom of the barrel is 6.93 + 24.5 = 31.43 g/cm^2

The units should be in Pa
What do I do now?

Why did you convert the density units to g/cm^3 in the first place? If you work with the kg/m^3 units, your answer will come more naturally. Unit conversion is not difficult, but you should not need it in this problem.