Filling a volume (pool) at a given rate

In summary, to fill a child's inflatable wading pool using a garden hose with a diameter of 2 cm and a water flow speed of 1.5 m/s, it would take approximately 129.885 seconds to fill the pool to a depth of 25 cm if it is circular with a diameter of 2.1 m. This can be calculated by finding the volume of water required to fill the pool, and then using the speed and diameter of the hose to calculate the volume of water expelled in one second. The total volume of water required is then divided by the volume expelled per second to determine the time needed to fill the pool. It is important to use consistent units throughout the calculations.
  • #1
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?
 
Physics news on Phys.org
  • #2
Volume of cylinder = [tex]\pi r^2 h = \pi h \frac{d^2}{4}[/tex]

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.
 
  • #3
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?
 
  • #4
whoknows123 said:
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.
 
  • #5
a) would it be the area * velocity, which is 129.885?
 
  • #6
area = pi*height*(d^4/2) = pi*23cm*(2.1m^4/2)?
 
  • #7
whoknows123 said:
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.
 
  • #8
whoknows123 said:
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
 
  • #9
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?
 
  • #10
whoknows123 said:
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.
 
  • #11
yes. Thank you, and yes units are important!
 

1. How is the rate of filling a pool determined?

The rate of filling a pool is determined by dividing the volume of the pool by the time it takes to fill it. This will give you the volume per time, commonly measured in liters per minute or gallons per hour.

2. How do you calculate the time it takes to fill a pool?

The time it takes to fill a pool is calculated by dividing the volume of the pool by the rate at which it is being filled. This will give you the time in minutes, hours, or days.

3. Can the rate of filling a pool change?

Yes, the rate of filling a pool can change depending on factors such as the size of the pool, the size of the filling mechanism, and any external factors that may affect the flow of water into the pool.

4. How do you maintain a constant rate of filling a pool?

To maintain a constant rate of filling a pool, you can use a flow control valve or adjust the speed of the filling mechanism. It is important to regularly check and adjust the rate to ensure that the pool is filling at a steady pace.

5. How does the temperature affect the rate of filling a pool?

The temperature can affect the rate of filling a pool as warmer water is less dense and therefore takes up more space. This means that it will take longer to fill a pool with warm water compared to cold water, assuming that the rate of filling remains constant.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
Replies
20
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
9K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top