Answer to Question #251841 in Mechanical Engineering for Kay

Question

The diameter of an open-topped tank 1.5 m high increases uniformly from 4.2 m at the [20] base to 6 m at the top. Discharge from the base of the tank takes place through a

75 mm diameter pipe which is 3 m in length. The pipe discharges the water to the atmosphere 1.5 m below the base of the tank. Initially, the tank is fully filled with water.

Take ƒ = 0.01. Find the time to empty the tank.

Answer 1

d_p=75 mm= 0.075m

l_p=3m

d₁=4.2m, r₁=2.1m

d₂=6m, r₂=3m

h₁=1.5m

h₂=$\frac{0.7*1.5}{0.3}=3.5m$

Volume of water in tank,

V=$\frac{1}{3}[π{r_2}^{2}(h_1+h_2)-π{r_1}^{2}h_2]$

=$\frac{1}{3}[π*3^{2}*(1.5+3.5)-π*2.1^{2}*3.5]=30.95m^{2}$

Area of pipe,

$=\frac{π}{4}d^{2}=\frac{π}{4}*0.075^{2}=4.416*10^{-3}m^{2}$

Apply Bernoulli's equation;

$\frac{p_1}{\rho g}+\frac{{V_1}^{2}}{2g}+h_1+h_3=\frac{p_2}{\rho g}+\frac{{V_2}^{2}}{2g}+0+h_f$

$h_1+h_3=\frac{{V_2}^{2}}{2g}+h_f$ this is equation (i)

Where $h_f=\frac{fLV^{2}}{2gd}=\frac{0.2*3*{V_2}^{2}}{2*9.81*0.075}=0.02039{V_2}^{2}$

Substituting values to equation (i)

$1.5+1.5=\frac{{V_2}^{2}}{2*9.81}+0.02039{V_2}^{2}$

V₂=6.484m/s

Area of pipe*V₂=$\frac{Volume\ of\ water\ in\ tank}{time}$

$4.416*10^{-3}*6.484=\frac{30.95}{t}$

$t=\frac{30.95*10^{3}}{4.416*6.484}$

t=1080.91s