Calculating Water Speed in a Pipe: Solving a Fluid Dynamics Problem

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Homework Help Overview

The problem involves calculating the speed of water flowing through a pipe, given the diameter of the pipe and the time taken to fill a bathtub. The context is fluid dynamics, specifically focusing on flow rates and volume calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between flow rate, area, and velocity, questioning how to calculate the volume of water flowing into the bathtub over time. There are inquiries about the appropriate units and the setup of the problem, including the conversion of volume and area.

Discussion Status

There is active engagement with various interpretations of the problem. Some participants have offered guidance on using the area of the pipe and the concept of flow rate, while others are seeking clarification on specific calculations and unit conversions. The discussion is productive, with participants exploring different aspects of the problem.

Contextual Notes

Participants are navigating the conversion of units from liters to cubic meters and the implications of time in their calculations. There is a focus on ensuring that all variables are correctly defined and understood in the context of the problem.

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Homework Statement


Water flowing through a 2.68 cm diameter pipe fill a 284 L bathtub in 8.2 minutes. What is the speed of the water in the pipe?



Homework Equations


p+pgh +1/2pv^2=p2 +pgh + 1/2pv2^2


The Attempt at a Solution


I know that p=pat for both cases but the water does not start from any height and does not end up at any height. I don't know how to start that one. Thank you for your help.
 
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If the area of the pipe opening is A, and the speed of the water flowing through it is v, what is the volume of water that flows into the bathtub in 1 second?
 
volume=q/a. So is the volume 284 L in this case?
 
Also, for the area, would I just consider the cross section since there is no height in this case?
 
Yes area is the cross section. If the area of the cross section is A, and the speed of flow is v, then a volume Av will flow into the pipe per second. Do you see why?
 
No I'm trying but I don't. Could you explain it to me please.
 
Also,if that is the case would not velocity be in L/m^2? would I need to solve it for 8.2 minutes?
 
If the circular opening of the pipe has area A, then a small cylindrical volume of length v(dt) of water will leave in the small time dt. The volume of this small cylinder is A(v)(dt) (area of cross section times length). So dV = A(v)(dt), or dV/dt = Av.
 
If the rate at which the water is flowing into the bathtub is Av, how much volume will flow into the tub in 8.2 minutes in terms of Av?
 
  • #10
Av*492 s. But wouldn't Av have to be in m/s?
 
  • #11
Av will have units of m^3/s. (if you write A in units of m^2, and v in units of m/s)
 
  • #12
Now just solve Av(492) = 284.
 
  • #13
How do I find v since it's not given?
 
  • #14
OH I get it! Thank You!
 
  • #15
I solved for V and my units were in m/s but that did not work. Why is it wrong still?
 
  • #16
What was your value for A? It should be 5.64 x 10-4 m2.
 
  • #17
OH I got it! I had to convert L into m^3 then the final answer is in m/s. Thank you very much!
 

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