Question

A cylindrical container with a cross-sectional area of 60.2 cm^2 holds a fluid of density 826 kg/m^3. At the bottom of the container the pressure is 127 kPa. Assume Pat = 101 kPa. What is the depth of the fluid?

Answer 1

Answer:

The depth of the fluid is: 3.208m

Explanation:

The pressure inside a fluid at rest is obtained with the following equation:

P= fluid density × acceleration of gravity  × fluid column height

The pressure at the base of the container only due to the column of the fluid will be the pressure at the base of the container minus the atmospheric pressure:

P= 127KPa - 101KPa= 26KPa= 26000 (kg/(m×s²))

The pressure at the base of the vessel due to the height of the fluid column will be:

P= 26000 (kg/(m×s²)) = 826 (kg/m³) × 9.81 (m/s²) × hight

Therefore, clearing the height of the fluid shaft, the height of the container is obtained which will be:

Fluid column heigh= 26000(kg/(m×s²)) × (1/826) (m³/kg) × (1/9.81) (m/s²)

Fluid column heigh= 3.208m