# Bernoulli's Thereom

#### Cass

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.

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#### Cass

Here is the picture of the problem

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#### topsquark

Math Team
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.
$$\displaystyle P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$

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