Answer
(a) $32cm/s$
(b) $43 Pa$
Work Step by Step
(a) We can find the required speed as
$v_2=\frac{A_1}{A_2}v_1$
$\implies v_2=\frac{\frac{\pi}{4}d_1^2}{\frac{\pi}{4}d_2^2}v_1$
$\implies v_2=(\frac{d_1}{d_2})^2v_1$
We plug in the known values to obtain:
$v_2=(\frac{1.1cm}{0.75cm})(15cm/s)$
$v_2=32cm/s$
(b) We can find the pressure drop as follows:
$\Delta P=\frac{1}{2}(v_2^2-v_1^2)$
We plug in the known values to obtain:
$\Delta P=\frac{1}{2}(1060)((0.323)^2-(0.15)^2)$
$\Delta P=43 Pa$
