How Can I Solve This Fluid Dynamics Homework Problem?

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Homework Help Overview

The discussion revolves around a fluid dynamics problem involving the application of Bernoulli's equation. The original poster presents a series of calculations related to pressures at different points in a fluid system, attempting to solve for unknown pressures based on given velocities and heights.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Bernoulli's equation to calculate pressures at different points, but expresses confusion regarding the correct application of the equation. Some participants question the understanding of the equation and suggest reevaluating the pressure terms and assumptions made in the calculations.

Discussion Status

Participants are actively engaging with the original poster's calculations, providing feedback and clarifications. There is an exploration of the impact of significant figures on the results, and some participants suggest that the question's phrasing may lead to ambiguity in the answers.

Contextual Notes

There are discussions about the accuracy of the values provided in the problem statement, with some participants noting that the lack of precision in the given measurements could affect the final answers. The original poster also expresses uncertainty about their calculations and seeks further guidance.

Adam Rabe
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Homework Statement


Difficult pressure at pipes question.PNG


Homework Equations


AV = AV
P + 1/2*p*v^2 + pgh = P + 1/2*p*v^2 + pgh

The Attempt at a Solution


For a.) ... divide 3 by 0.1 because of (AV = AV) = 30 ms-1. answer is right.
For b.) I am stuck on this one. I tried using the 2nd relevant equation above but i can't seem to find the answer and I am sure i need to use that one. Heres my working out.

Pressure A: P + (0.5 *1000 *3^2) = 5 *10^5... Therefore P = 495500 Pa
Pressure B: P = 495500 + (0.5 *1000 *30^2) + (1000 *10 *1) = 955500 Pa
Answer is 6.5 *10^4... (65000)

For c.) point is located at the same height and same area as A, therefore pressures are equal = 5 * 10^5 Pa. Answer is right

Any help would be greatly appreciated. Thanks in advance
Adam
 
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Adam Rabe said:
Pressure A: P + (0.5 *1000 *3^2) = 5 *10^5... Therefore P = 495500 Pa
You don't seem to understand the Bernoulli equation. P is the pressure at the given point. At A it is 5*10^5 Pa. The other terms are then added to this to get a quantity that is the same throughout the streamline.
 
haruspex said:
You don't seem to understand the Bernoulli equation. P is the pressure at the given point. At A it is 5*10^5 Pa. The other terms are then added to this to get a quantity that is the same throughout the streamline.
Hello sorry for the late reply.
i think i see where i went wrong...
Using your advice, i added the pressure change due to height and subtracted the pressure difference between the two speeds (as fluid speeds up pressure must drop)

5*10^5 + (1000 *10) - ((0.5 * 1000 * 30^2) - (0.5 * 1000 * 3^2)) = 64500

But I am still 500 Pa off from the right answer, am i on the right track?
 
Adam Rabe said:
64500
That's what I get. Looks like the answers have been rounded to 2 significant figures.
The question statement is poor. It gives values like "1m2", "5x105Pa". Standard interpretation says these are only accurate to one decimal place. If you follow that through the calculation you find that the answer is indeterminate. E.g. just change the 3m/s to 3.1m/s and you get a much smaller answer. It should quote the values as 3.00m/s etc. to justify 2 sig figs in the answer.
 

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