A verbal ODE problem I can't get through..

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calculator13

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The height of a water tank, shaped like a circular cylinder, is 3m and the diameter of the bottom is 5m. When emptying the container through a hole, shaped like a round, with a diameter of 5cm, at the bottom, water velocity in the hole approx 2.5*sqrt(h) m/s has been observed, where h is the depth of the water in the tank. How long does it take for the tank to empty?
 
calculator13 said:
The height of a water tank, shaped like a circular cylinder, is 3m and the diameter of the bottom is 5m. When emptying the container through a hole, shaped like a round, with a diameter of 5cm, at the bottom, water velocity in the hole approx 2.5*sqrt(h) m/s has been observed, where h is the depth of the water in the tank. How long does it take for the tank to empty?
Click to expand...
What does VERBAL ODE problem mean? Are you not allowed to use pencil/paper/computer to solve the problem? Or is it another way of saying "word problem involving DE"?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
Subhotosh Khan said:
What does VERBAL ODE problem mean? Are you not allowed to use pencil/paper/computer to solve the problem? Or is it another way of saying "word problem involving DE"?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
Click to expand...
It is a word problem involving differential equations. I have been able to write that the water coming out of the tank is equal to the water velocity multiplied by time. And that the velocity can be written as a derivative of place in respect to time. So what I have is

h'= 2.5*sqrt(h)
h'*t= 2.5*sqrt(h)*t

My problem is in order to find out the time it takes for the tank to be emptied i need something else that equals h' or h'*t. But with the information I have been given I can't find a solution for this that works. I have calculated that the volume of the tank is 75/4*pi, but I don't know how I can relate that to the depth of the water or time..
 
calculator13 said:
...So what I have is

h'= 2.5*sqrt(h)
Click to expand...

Not quite. The expression 2.5*sqrt(h) just gives the velocity of the water leaving the 5cm diameter hole (in m/s). NOT the velocity of the water in the larger tank.

However, you can use this expression to calculate the volume of water leaving the tank every second (in terms of h). Once you have this, use it to write an expression for how fast the level in the big tank drops (dh/dt written in terms of h)
 
calculator13 said:
It is a word problem involving differential equations. I have been able to write that the water coming out of the tank is equal to the water velocity multiplied by time. And that the velocity can be written as a derivative of place in respect to time. So what I have is

h'= 2.5*sqrt(h)
h'*t= 2.5*sqrt(h)*t

My problem is in order to find out the time it takes for the tank to be emptied i need something else that equals h' or h'*t. But with the information I have been given I can't find a solution for this that works. I have calculated that the volume of the tank is 75/4*pi, but I don't know how I can relate that to the depth of the water or time..
Click to expand...
You have to set up the DE considering the Physics Concept. We will consider a small differential time 'δt' .The concept is:

During a given time period, the VOLUME lost in the cylinder = Volume expelled through the hole​

Volume expelled through the hole, during δt = A * V * δt = π/4 * (d2)*2.5*[√(h)] * δt ............................what are the units of 'd' and h

Volume reduction in cylinder (with constant diameter of 3 5 m) due to drop in height of δh = π/4 * (52) * δh ...............................[edited]

Now equate those and form your ODE.
 
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calculator13 said:
The height of a water tank, shaped like a circular cylinder, is 3m and the diameter of the bottom is 5m. When emptying the container through a hole, shaped like a round, with a diameter of 5cm, at the bottom, water velocity in the hole approx 2.5*sqrt(h) m/s has been observed, where h is the depth of the water in the tank. How long does it take for the tank to empty?
Click to expand...
Did you get your ODE?
 
Subhotosh Khan said:
You have to set up the DE considering the Physics Concept. We will consider a small differential time 'δt' .The concept is:

During a given time period, the VOLUME lost in the cylinder = Volume expelled through the hole​

Volume expelled through the hole, during δt = A * V * δt = π/4 * (d2)*2.5*[√(h)] * δt ............................what are the units of 'd' and h

Volume reduction in cylinder (with constant diameter of 3 5 m) due to drop in height of δh = π/4 * (52) * δh ...............................[edited]

Now equate those and form your ODE.
Click to expand...
Professor Khan, I have a comment. Isn't the volume of a cylinder is \(\displaystyle \pi r^2 h\), so why did you write \(\displaystyle \frac{\pi}{4}r^2 h\)?
 
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