# Bernoulli's equation

#### mathsaj

Hi all,
Can someone explain clearly where these numbers are coming from and some advice on mathematics tools I need to solve similar manipulation of problems questions, this would be much appreciated....

Using continuity, U1 X A1 = U2 X A2 therefore u1 = u2 x 50sqr /100sqr = U2/4

So substituting back into Bernoulli's equ:

p1/row x grav + u1sqr/2 x grav = u2sqr / 2 x grav the other variables have been cancelled due to the actual problem.

We get pressure 1 or P1/ rough x gravity + U2 sqr/32 = u2sqr / 2 x grav and so p1/row x grav = 15xu2sqr / 32 x grav.

Can anyone help? Where does this 15 come from?

#### skipjack

Forum Staff
Write 15 as 16 - 1 so that you get two fractions on the right-hand side $$\displaystyle ^{(\frac{15}{32}\,=\,\frac12\,-\,\frac{1}{32})}$$. There seems to be a typing error ("x grav" omitted) on the left-hand side of the previous equation. You can use LaTeX to make your equations more readable.

$$\displaystyle u_1\,=\,\frac{u_2}{4}\text{ so }\frac{u_1^2}{2}\,=\,\frac{u_2^2}{32}.$$

$$\displaystyle \frac{p_1}{\rho\Psi}\,+\,\frac{u_1^2}{2\Psi}\,=\,\frac{u_2^2}{2\Psi}\text{ so }\frac{p_1}{\rho\Psi}\,+\,\frac{u_2^2}{32\Psi}\,=\,\frac{u_2^2}{2\Psi}.$$

$$\displaystyle \text{Hence }\frac{p_1}{\rho\Psi}\,=\,\frac{u_2^2}{2\Psi}\,-\,\frac{u_2^2}{32\Psi}\,=\,\frac{15u_2^2}{32\Psi}.$$

#### mathsaj

Thank you very much.... I have a lot of work to do... in your opinion what maths skills do you think I need to solve these types of questions?

#### mathsaj

Hi another one for you....

p1/row g + u1sqr /2g +z1 - {u2sqr / 2g +z2 = head loss

p1 /row g = 5 and z1 - z2 = -0.5m

hl = 5- 0.5 +u1sqr/2g - u2sqr/2g

could you explain how the next equ happens : hl = 4.5 - u2sqr/2g[1 - u1sqr/u2sqr]

then using continuaty equ u1a1=u2a2, the equ changes to u1sqr = u2sqr d2 to the power of 4 over d1 to the power of 4, I am aware that area equals pi d sqr over 4 but how did the power of 4 appear at bottom and top.

What maths techniques will help me solve these by myself?

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