## Essential University Physics: Volume 1 (3rd Edition)

$v_1 = \sqrt{\frac{2\Delta p}{\rho(b^2-1)}}$
We simplify the Bernoulli equation using the known conditions to find: $p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ We call $\Delta p=p_1 -p_2$ to find: $\Delta p + \frac{1}{2} \rho v_1^2 = \frac{1}{2}\rho v_2^2$ We know that $\Delta p = \frac{1}{2}\rho b^2 v_1^2 - \frac{1}{2}\rho v_1^2$. Simplifying using the known conditions as well as math, we find: $v_1 = \sqrt{\frac{2\Delta p}{\rho(b^2-1)}}$