Filling a volume (pool) at a given rate

  • Thread starter Thread starter whoknows123
  • Start date Start date
  • Tags Tags
    Rate Volume
Click For Summary

Homework Help Overview

The problem involves calculating the time required to fill a child's inflatable wading pool using a garden hose. The pool has a circular shape with a specified diameter and depth, while the hose has a given diameter and water flow speed. The discussion focuses on understanding the volume of water needed and the rate at which water is expelled from the hose.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the volume of the pool and how to calculate it using the formula for the volume of a cylinder. There are questions about the relationship between the volume of water needed and the volume expelled by the hose. Some participants explore how to determine the volume of water coming out of the hose per second and the necessary calculations for both parts of the problem.

Discussion Status

Participants are actively engaging with the problem, raising questions about calculations and units. Some guidance has been offered regarding the need to calculate the volume of water required to fill the pool and the volume of water expelled by the hose. There is an ongoing exploration of the necessary formulas and the importance of unit consistency.

Contextual Notes

There are discussions about ensuring consistent units throughout the calculations, as well as clarifications on the distinction between the volume of the pool and the volume of water expelled by the hose. Some participants express confusion regarding the calculations and the steps involved.

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
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.
 
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?
 
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.
 
a) would it be the area * velocity, which is 129.885?
 
area = pi*height*(d^4/2) = pi*23cm*(2.1m^4/2)?
 
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.
 
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
 
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!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K