Bernoulli's Thereom

Dec 2014
2
0
Canada
I am trying to locate the formula to calculate pressure (psi) when given pipe diameter and the velocity. This is a plumbing related question. I was using the formula v1p1= v2p2 and the answers are not correct. I uploaded the picture in the replies.
Thanks in advance
 
Last edited:
Dec 2014
2
0
Canada
Here is the picture of the problem
 

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topsquark

Math Team
May 2013
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The Astral plane
I am trying to locate the formula to calculate pressure (psi) when given pipe diameter and the velocity. This is a plumbing related question. I was using the formula v1p1= v2p2 and the answers are not correct. I uploaded the picture in the replies.
Thanks in advance
\(\displaystyle P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2\)

and, as you mentioned, the continuity equation \(\displaystyle A_1v_1 = A_2v_2\)

So, for example, to find \(\displaystyle P_E\) we need \(\displaystyle v_E\):
\(\displaystyle \left [ \pi \left ( \frac{d_D}{2} \right ) ^2 \right ] v_D = \left [ \pi \left ( \frac{d_E}{2} \right )^2 \right ] v_E \)
(Watch the units!) I get \(\displaystyle V_D = 2.25 \text{ in/s}\)

Now apply Bernoulli:
\(\displaystyle P_D + \rho g h_D + \frac{1}{2} \rho v_D^2 = P_E + \rho g h_E + \frac{1}{2} \rho v_E^2\)
We can consider \(\displaystyle h_D = h_E\), so the middle terms drop out, and we know \(\displaystyle P_E\), \(\displaystyle v_D\), and \(\displaystyle v_E\). (And of course the density is the same for both points.)

\(\displaystyle P_D + \frac{1}{2} \rho v_D^2 = P_E + \frac{1}{2} \rho v_E^2\)
I get \(\displaystyle 50 + \frac{1}{2} \rho (2.25)^2 = P_E + \frac{1}{2} \rho (3)^2\)

The bad news is that there isn't enough information to find a value of \(\displaystyle \rho\), so that's as far as we can go for this one. If we knew the pressure at any of the other points then we could solve this. I'd talk with your instructor.

-Dan

Addendum: Also, please post your images right side up.
 
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