Filling a volume (pool) at a given rate

[whoknows123]

TO fill a child's inflatable wading pool, you use a garden hose with a diameter of 2 cm. Water flows from this hose with a speed of 1.5 m/s. How long will it take to fill the pool to a depth of 25 cm if it is circular and has a diameter of 2.1 m?

how would I solve this problem?

 

[Curious3141]

Volume of cylinder = \pi r^2 h = \pi h \frac{d^2}{4}

Volume of water required to fill child's pool to that depth = ...

Speed of water coming out of hose = 1.5 m/s

Length of cylindrical water column over one second interval (height of cylinder of water) = ...

Volume of cylindrical water column coming out of hose over one second = ...

Number of these cylindrical water columns it would take to fill the pool to the required volume = ...

Hence time taken = ... (answer)

Work out all the "..." and you'll get it.

 

[whoknows123]

Volume of Cylinder = 86.59
Volume of water required? is is the same as the volume of cylinder?
speed of water coming out = 1.5 m/s
Length of cylindrical water column over one second interval=?? how would i find this?

 

[nrqed]

Volume of Cylinder = 86.59
Volume of water required? is is the same as the volume of cylinder?
speed of water coming out = 1.5 m/s
Length of cylindrical water column over one second interval=?? how would i find this?

Curious gave all the steps but I will be a bit more explicit.
There are two distinct parts.

A) How much water do you need to fill the pool. This is the volume of water needed to fill the pool to the depth needed. Can you calculate this? (this does not require anything related to the hose).

B) Now, you must calculate how much water comes out of the hose in one second (for this part, the size or depth of the pool is irrelevant).
Think of the hose as a very long (let's say infinite) cylinder. Since the water flows at 1.5 m/s, you know that in one second, all the water within 1.5 meter of the muzzle of the hose will come out . Therefore, the volume of water expelled by the hose in 1 second is the volume of water contained in a cylinder 1.5 meter long and with a diameter of 0.02 m (it is safer to put everything in meters).


Now, you know how much water (that is how many meter cube of water) is expelled by second and you know the total number of meter cube needed to fill the pool. A simple ratio will give you how long it will take.

 

[whoknows123]

a) would it be the area * velocity, which is 129.885?

 

[whoknows123]

area = pi*height*(d^4/2) = pi*23cm*(2.1m^4/2)?

 

[nrqed]

a) would it be the area * velocity, which is 129.885?

I am confused.. Are you answering part a or part b? Part a does not require the speed at all (or anything related to the hose).
Also, ALWAYS include the units in your answer. AN dmake sure that you are consistent in the unites (use cm everywhere or meters which is usually safer).

But the amount of water ejected by the hose per second comes out to be the area of the opening of the hose times the ejection speed, indeed. But I tried to explain where this comes from...but it is the correct formula indeed.

 

[nrqed]

area = pi*height*(d^4/2) = pi*23cm*(2.1m^4/2)?

Always be consistent with the units (everything in cm or everything in meters) aAlso, nothing is raised to the fourth power!

It would be Pi * .25 m * (2.1 m/2)^2 or Pi* 0.25 m * (2.1 m)^2/4

 

[whoknows123]

Area of the hose would be = .00031459 m^2, multiply this by 1.13 m/s and i get .000355

area of the pool is .8659 m^3

then I divide the two?

 

[nrqed]

Area of the hose would be = .00031459 m^2, multiply this by 1.13 m/s and i get .000355

I thought you said it was 1.5 m/s?
Also, always write your units. Your answer will be in m^3/s

area of the pool is .8659 m^3

then I divide the two?

You mean *volume* of the pool.

Once you write your units, it will be obvious what to do in order to get a result in second. That's one reason why it is important to write the units.

 

[whoknows123]

yes. Thank you, and yes units are important!