Pressure of fluid in container

[faoltaem]

Homework Statement

A cylindrical container with a cross sectional area of 65.2 cm$$^{2}$$ holds a fluid of density 806kg/m$$^{3}$$. At the bottom of the container the pressure is 116 kPa
a) What is the depth of the fluid?
b) Find the pressure at the bottom of the container after an additional 2.05 $$\times$$ 10$$^{-3}$$ m$$^{3}$$ of this fluid is added to the container. Assume no fluid spills out of the container.

Homework Equations

The Attempt at a Solution

a) A = 65.2 cm$$^{2}$$
$$\rho$$ = 806 kg/m$$^{3}$$
P(bottom) = 116 kPa = 1.16 $$\times$$ 10$$^{5}$$ Pa

closed manometer $$\rightarrow$$ P = $$\rho$$gh

h = $$\frac{P}{\rhog}$$ = $$\frac{1.16 \times 10^{5}}{806 \times 9.81}$$
= 14.67 m

 

[faoltaem]

sorry i didn't mean to finish it yet there is a later post with more in it

 

[m2003]

The depth of the fluid is 14.67 m.

b) Since the additional fluid does not spill out of the container, the volume of the container remains the same. Therefore, the pressure at the bottom of the container will remain the same as well. So, the pressure at the bottom of the container after adding the additional fluid will also be 116 kPa or 1.16 \times 10^{5} Pa.

I would like to add that the pressure at any point in a fluid is equal in all directions and is caused by the weight of the fluid above that point. This is known as Pascal's Law and is an important principle in fluid mechanics. In this problem, we used the equation \rhogh to calculate the depth of the fluid, assuming the fluid is incompressible and the pressure at the top of the fluid is atmospheric pressure. Additionally, it is important to note that the units used in this problem are consistent, as pressure is typically measured in Pascals (Pa) and density in kilograms per cubic meter (kg/m^{3}).