Question
What is the gauge pressure at the bottom of the cylinder?
oil= 0.88m
brine= 1.11m
p oil= 790 kg/m^3
p brine= 1025 kg/m^3
oil= 0.88m
brine= 1.11m
p oil= 790 kg/m^3
p brine= 1025 kg/m^3
Answers
GPT 3.5
To find the gauge pressure at the bottom of the cylinder, we need to calculate the pressure exerted by both the oil and brine columns.
The gauge pressure at a certain depth is given by the formula:
P = ρgh
Where:
P is the pressure
ρ is the density of the fluid
g is the acceleration due to gravity
h is the depth
For the oil:
P_oil = ρ_oil * g * h_oil
= 790 kg/m^3 * 9.8 m/s^2 * 0.88 m
= 6787.44 Pa
For the brine:
P_brine = ρ_brine * g * h_brine
= 1025 kg/m^3 * 9.8 m/s^2 * 1.11 m
= 11298.15 Pa
The total pressure at the bottom is the sum of the pressures from the oil and brine:
P_total = P_oil + P_brine
= 6787.44 Pa + 11298.15 Pa
= 18085.59 Pa
Therefore, the gauge pressure at the bottom of the cylinder is 18085.59 Pa.
The gauge pressure at a certain depth is given by the formula:
P = ρgh
Where:
P is the pressure
ρ is the density of the fluid
g is the acceleration due to gravity
h is the depth
For the oil:
P_oil = ρ_oil * g * h_oil
= 790 kg/m^3 * 9.8 m/s^2 * 0.88 m
= 6787.44 Pa
For the brine:
P_brine = ρ_brine * g * h_brine
= 1025 kg/m^3 * 9.8 m/s^2 * 1.11 m
= 11298.15 Pa
The total pressure at the bottom is the sum of the pressures from the oil and brine:
P_total = P_oil + P_brine
= 6787.44 Pa + 11298.15 Pa
= 18085.59 Pa
Therefore, the gauge pressure at the bottom of the cylinder is 18085.59 Pa.