Past paper question about rates

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SUMMARY

The discussion centers on calculating the water velocity as it exits a tank with a given inflow rate of 0.2 m³/s and an outlet area of 0.05 m². By applying the principle of conservation of mass, the outflow rate can be expressed as the product of the exit velocity (v) and the outlet area. In steady state, the inflow rate equals the outflow rate, leading to the equation 0.2 m³/s = v * 0.05 m², allowing for the determination of the exit velocity and the water level in the tank.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically conservation of mass.
  • Familiarity with dimensional analysis in physics.
  • Basic knowledge of velocity and area relationships in fluid flow.
  • Ability to solve algebraic equations for unknown variables.
NEXT STEPS
  • Calculate the exit velocity using the equation v = inflow rate / outlet area.
  • Explore the concept of steady state in fluid systems.
  • Study the principles of Bernoulli's equation for further insights into fluid flow.
  • Investigate the effects of varying outlet area on fluid velocity and tank levels.
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Students studying fluid dynamics, engineers working on hydraulic systems, and anyone involved in water resource management or tank design.

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Homework Statement


Water is falling gently into a large tank at 0.2m^3s^-1 and leaves through a small hole in the bottom of the tank of area 0.05m^2. What is the water velocity as it leaves? How full is the tank in the steady state?




Homework Equations


Well if the rate of water entering is 0.2m^3s^-1 by dimensional analysis rate=velocity*area
So the rate of water leaving the tank will be v*Area = v*0.05

I'm not sure how to find the velocity of the water as it leaves the tank because surely these two rates will not be the same (except in the steady state??)
 
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captainjack2000 said:

Homework Statement


Water is falling gently into a large tank at 0.2m^3s^-1 and leaves through a small hole in the bottom of the tank of area 0.05m^2. What is the water velocity as it leaves? How full is the tank in the steady state?


Homework Equations


Well if the rate of water entering is 0.2m^3s^-1 by dimensional analysis rate=velocity*area
So the rate of water leaving the tank will be v*Area = v*0.05

I'm not sure how to find the velocity of the water as it leaves the tank because surely these two rates will not be the same (except in the steady state??)
You've almost got it.

What is v * Area? or rather what are the units of v * A?
 
Last edited:
sorry what do you mean
how to do change velocity*area?
 

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