Past paper question about rates

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The discussion revolves around calculating the velocity of water exiting a tank and determining the tank's steady-state fullness. Given that water enters the tank at a rate of 0.2 m³/s and exits through a hole with an area of 0.05 m², the relationship between flow rate, velocity, and area is crucial. Participants emphasize using dimensional analysis to equate the inflow and outflow rates, leading to the equation v * 0.05 = 0.2, where v represents the exit velocity. The uncertainty about whether the inflow and outflow rates are equal in steady state is also addressed. Ultimately, understanding these principles is key to solving the problem effectively.
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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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