What Is the Depth and Pressure of Fluid in a Cylindrical Container?

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SUMMARY

The discussion focuses on calculating the depth of fluid in a cylindrical container with a cross-sectional area of 65.2 cm² and a fluid density of 806 kg/m³. The pressure at the bottom of the container is given as 116 kPa. Using the formula P = ρgh, the depth of the fluid is determined to be 14.67 m. Additionally, the discussion addresses the pressure change when an extra 2.05 x 10⁻³ m³ of fluid is added, assuming no spillage occurs.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically hydrostatic pressure.
  • Familiarity with the equation P = ρgh for calculating pressure in fluids.
  • Knowledge of unit conversions, particularly between cm² and m².
  • Basic skills in algebra for manipulating equations and solving for variables.
NEXT STEPS
  • Research the effects of additional fluid volume on pressure in closed containers.
  • Learn about the implications of fluid density on pressure calculations.
  • Explore the concept of hydrostatic equilibrium in different fluid systems.
  • Investigate the applications of fluid mechanics in engineering and design.
USEFUL FOR

Students studying fluid mechanics, engineers working with hydraulic systems, and anyone involved in pressure calculations in cylindrical containers.

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Homework Statement



A cylindrical container with a cross sectional area of 65.2 cm[tex]^{2}[/tex] holds a fluid of density 806kg/m[tex]^{3}[/tex]. 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 [tex]\times[/tex] 10[tex]^{-3}[/tex] m[tex]^{3}[/tex] 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[tex]^{2}[/tex]
[tex]\rho[/tex] = 806 kg/m[tex]^{3}[/tex]
P(bottom) = 116 kPa = 1.16 [tex]\times[/tex] 10[tex]^{5}[/tex] Pa

closed manometer [tex]\rightarrow[/tex] P = [tex]\rho[/tex]gh

h = [tex]\frac{P}{\rhog}[/tex] = [tex]\frac{1.16 \times 10^{5}}{806 \times 9.81}[/tex]
= 14.67 m
 
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sorry i didn't mean to finish it yet there is a later post with more in it
 

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