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Answer 1 · 2017-12-28T10:18:48

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

The flow rate $Q$ $Q$ is constant

$Q = v_{1} A_{1} = v_{2} A_{2}$ $Q=v_1A_1=v_2A_2$

The speed $v_{1} = 1.6 m s^{- 1}$ $v_1=1.6ms^-1$

The cross section area $A_{1} = π \frac{d_{1}^{2}}{4} = π \cdot \frac{{(0.06)}^{4}}{4} m^{2}$ $A_1=pid_1^2/4=pi*(0.06)^4/4m^2$

The cross section $A_{2} = π \frac{d_{2}^{2}}{4} = π \cdot \frac{{0.03}^{2}}{4} m^{2}$ $A_2=pid_2^2/4=pi*0.03^2/4 m^2$

$π \cdot \frac{{0.06}^{2}}{4} \cdot 1.6 = π \frac{{0.03}^{2}}{4} \cdot v_{2}$ $pi*0.06^2/4*1.6=pi0.03^2/4*v_2$

$v_{2} = \frac{{0.06}^{2}}{{0.03}^{2}} \cdot 1.6 = 6.4 m s^{1}$ $v_2=0.06^2/0.03^2*1.6=6.4ms^1$

The flow rate is

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