What Is the Minimum Diameter for a Pipe Serving 360 Apartments to Prevent Leaks?

AI Thread Summary
The discussion centers on calculating the minimum diameter for a pipe serving 360 apartments, each consuming 0.05 liters per second, to prevent leaks. The calculations suggest that the required diameter is approximately 0.07 meters or 7.6 cm, based on a discharge speed of 4 m/s. Participants express confusion about the implications of peak water usage times and the relevance of preventing leaks in the context of the problem. There is also uncertainty regarding the interpretation of the problem statement and the significance of the provided figures. Ultimately, the focus remains on determining the appropriate pipe diameter for adequate water supply without leaks.
vinamas
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Homework Statement


An engineer is planning for a network of pipes to a residential tower containing 360 apartments each one of them consumes 0.05 liters/s of water what is the least possible pipe discharge for the main pipe in the network
if the fastest speed of discharge in the pipe is 4m/s then what is the least diameter possible that doesn't cause leaking

Homework Equations


J=Av=V/t

The Attempt at a Solution


I convert from liters to volume try to apply the law of course it doesn't work I have two variables
https://www.physicsforums.com/threads/pipe-discharge-and-diameter.860980/goto/post?id=5402647#post-5402647 my work:
360*0.05*10^-3/s=A*4
A=0.0045
0.0045=(d/2)^2*pi
d=0.07m. I think the 360 apartments are vital to solve this problem but I don't know where to use them
the diameter is 0.2m How do I get this answer?[/B]
 
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0.05 l is a shot glass. Is this per year ?

Could you list off the relevant equations a bit more completely ?

And show your work ?
 
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BvU said:
0.05 l is a shot glass. Is this per year ?

Could you list off the relevant equations a bit more completely ?

And show your work ?
Oh sorry I just edited it its liters per second
 
Can you show your work ?
And how do you find the 0.05 l/s ?
 
BvU said:
Can you show your work ?
And how do you find the 0.05 l/s ?
its given in the problem

my work:
360*0.05*10^-3/s=A*4
A=0.0045
0.0045=(d/2)^2*pi
d=0.07m. I think the 360 apartments are vital to solve this problem but I don't know where to use them
 
360 is the number of users; 4 is m/s I suppose. You get 7.6 cm diameter.

But don't most of these 360 users sleep around the same time and wake up around the same time and want to shower at approximately the same time ? Would they be happy with 3 liters/minute each ?
 
BvU said:
360 is the number of users; 4 is m/s I suppose. You get 7.6 cm diameter.

But don't most of these 360 users sleep around the same time and wake up around the same time and want to shower at approximately the same time ? Would they be happy with 3 liters/minute each ?
Well that's the question lol
 
BvU said:
360 is the number of users; 4 is m/s I suppose. You get 7.6 cm diameter.

But don't most of these 360 users sleep around the same time and wake up around the same time and want to shower at approximately the same time ? Would they be happy with 3 liters/minute each ?
So what is wrong about my method?
 
No idea what's wrong. I calculate what you calculate, but it is highly unlikely that's really what is asked. I also don't understand the mentioning of leaking in the problem statement.
Nor, in fact, the "least possible pipe discharge for the main pipe in the network" ? (even before the 4 m/s is provided).

Any clue in your notes/textbook ?
 
  • #10
BvU said:
No idea what's wrong. I calculate what you calculate, but it is highly unlikely that's really what is asked. I also don't understand the mentioning of leaking in the problem statement.
Nor, in fact, the "least possible pipe discharge for the main pipe in the network" ? (even before the 4 m/s is provided).

Any clue in your notes/textbook ?
You can forget those details the diameter of the main pipe is what the question asks I know its confusing
 

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