fluid dynamics Bernoulli's equation

Question

I really can't seem to prove this.

Using $A_1u_1 = A_2u_2 $ I get: $$\frac{r_a^2}{r_b^2} = 1.$$

Using Bernoulli's equation, I can't figure out what the other terms should be?

$$\frac{1}{2}\rho u_a^{2} + \rho g z_1 + pa = \frac{1}{2} \rho u_2^{2} + \rho g h + P_b$$

What is the $z_1$ at point $A$? and is $z_2$ i.e the height at point $B$ equal to $h$, if I take $z=0$ at the center of the circle?

Answer 1

We have $$ pB = pA – rho * g * h + ½ * rho * ( U_A^2 – U_B^2 ) $$ and $$ S_A * U_A = S_B * U_B -> U_A = ( S_B / S_A ) * U_B $$ or -> $$ U_B = ( S_A / S_B ) * U_A = ( r_A^2 / r_B^2 ) * U_A $$

( U = velocity, S = section, r = radius )

then, $$ pB = pA – rho * g * h + ½ * rho * ( U_A^2 – U_B^2 ) $$

$$ pB = pA – rho * g * h + ½ * rho * U_A^2 * ( 1 – U_B^2 / U_A^2 ) $$

$$ pB = pA – rho * g * h + ½ * rho * U_A^2 * ( 1 – ( U_B / U_A )^2 ) $$

$$ pB = pA – rho * g * h + ½ * rho * U_A^2 * ( 1 – ( r_A / r_B )^4 ) $$

or

$$ pB = pA – rho * g * h - ½ * rho * U_A^2 * (( r_A / r_B )^4 - 1 ) $$

Hope you can understand. I have to learn Latex.