Fluid mechanics - Water flowing through a pipe

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Homework Help Overview

The discussion revolves around fluid mechanics, specifically the flow of water through a pipe and the calculation of mass flow rate using velocity profiles. Participants are examining the method of calculating average velocity and its implications on the results obtained.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between mass flow rate and average velocity, questioning the validity of using a simplified average over the radius instead of the cross-sectional area. There is a discussion about the implications of velocity distribution across different sections of the pipe.

Discussion Status

Some participants have offered clarifications regarding the definition of average velocity in the context of the problem, emphasizing the need to consider the entire cross-sectional area rather than just radial averages. Multiple interpretations of the velocity distribution are being explored, indicating a productive exchange of ideas.

Contextual Notes

There appears to be a misunderstanding regarding the calculation of average velocity and its dependence on the geometry of the pipe, which is under discussion. The original poster's method and assumptions are being critically examined.

theBEAST
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Homework Statement


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The Attempt at a Solution


I am interested in why my method is flawed:
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Alright I know how to do this problem. Essentially you solve mass flow rate = rho * double integral of u dA. HOWEVER, I decided to do this differently. I know that I can solve for the average u from the equation mass flow rate = rho * u_avg * A. I also know that the average u is just the integral of u over the integral of 1. I tried this and I get 1.528 m/s which is clearly not one of the answers. Why does my method not work?
 
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In the expression: ##\dot{m} = \rho u_{avg}A##,
##u_{avg}## needs to be the average over the cross-sectional area of the pipe, not the average over the radius.
 
TSny said:
In the expression: ##\dot{m} = \rho u_{avg}A##,
##u_{avg}## needs to be the average over the cross-sectional area of the pipe, not the average over the radius.

Hmmm, according to the equation doesn't u only depend on r? In other words u is constant for some circular ring r = k?
 
Consider two intervals of radius, one near the center of the pipe and one out at the edge. Suppose the interval dr is the same for both. There are more patches of area in the same interval dr out near the edge. So, the value of the velocity out at the edge contributes more to the average over the cross–section than the value near the center.
 

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