# Question #5fd0d

Dec 28, 2017

The speed is $= 6.4 m {s}^{-} 1$. The flow rate is $= 16.29 {m}^{3} {h}^{-} 1$

#### Explanation:

The flow rate $Q$ is constant

$Q = {v}_{1} {A}_{1} = {v}_{2} {A}_{2}$

The speed ${v}_{1} = 1.6 m {s}^{-} 1$

The cross section area ${A}_{1} = \pi {d}_{1}^{2} / 4 = \pi \cdot {\left(0.06\right)}^{4} / 4 {m}^{2}$

The cross section ${A}_{2} = \pi {d}_{2}^{2} / 4 = \pi \cdot {0.03}^{2} / 4 {m}^{2}$

$\pi \cdot {0.06}^{2} / 4 \cdot 1.6 = \pi {0.03}^{2} / 4 \cdot {v}_{2}$

${v}_{2} = {0.06}^{2} / {0.03}^{2} \cdot 1.6 = 6.4 m {s}^{1}$

The flow rate is

$Q = \pi \cdot {0.06}^{2} / 4 \cdot 1.6 = 0.0045 {m}^{3} {s}^{-} 1 = 16.29 {m}^{3} {h}^{-} 1$