Answer
$(a)\space 13.87\space m/s$
$(b)\space 0.97\space m^{3}/s$
Work Step by Step
(a) Let's apply equation 11.9 $A_{1}v_{1}=A_{2}v_{2}$ for get a combination between speeds.
$A_{1}v_{1}=A_{2}v_{2}$ ; Let's plug known values into this equation.
$(0.05\space m^{2})v_{1}=(0.07\space m^{2})v_{2}$
$v_{1}=1.4v_{2}-(1)$
Let's apply Bernoulli's equation to the system.
$P+\frac{1}{2}\rho v^{2}+\rho gh=constant$
$P_{1}+\frac{1}{2}\rho v_{1}^{2}+0=P_{2}+\frac{1}{2}\rho v_{2}^{2}+0-(2)$
(1)=>(2),
$P_{1}+\frac{1}{2}\rho (1.4v_{2})^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$
$v_{2}=\sqrt {\frac{2(P_{2}-P_{1})}{0.96\rho }}$ ; Let's plug known values into this equation.
$v_{2}=\sqrt {\frac{2(120\space Pa)}{0.96\times 1.3\space kg/m^{3} }}=13.87\space m/s$
(b) Flow rate = Av = $(0.07\space m^{2})(13.87\space m/s)=0.97\space m^{3}/s$
