Question

Oil (density 900 kg/m' and absolute viscosity 0.2 kg/m-s) flows through a pipe 20-m-long and 4-cm wide in inner diameter a- The flow rate is 47 liters/min. when the pipe is laid horizontally. Calculate the pressure needed at the pipe inlet. (7 points) b- If the exit end of pipe is raised by 0.2 m, please find the flow rate at this slope while the inlet pressure remains the same as in the situation (a). (8 points) (a) Air 0.2m (b)

Answer 1

Answer:-

$\rho =900kg/m^3$

D = 4cm

$Q=47l/min$

Velocity will be

V = Q/60 D2 0.047/60 *0.042

$V=0.6233m/sec$

u= 0.2kg/ms

$L=20m$

a) Renolds no.

$Re=\frac{\rho VD}{\mu }$

$Re=\frac{900*0.6233*0.04}{0.2 }$

$Re=112.20$

So flow is laminar.

The pressure drop can be given by

$\Delta P=\frac{32\mu VL}{D^2}$

$\Delta P=\frac{32*0.2*0.6233*20}{0.04^2}$

$\Delta P=49864Pa$

$\Delta P=P_1-P_2=49864Pa$

Here P₂ =101.325kPa atmospheric pressure.

So

$\Delta P=P_1-101.325=49.864$

$P_1=151.189kPa$ Absolute.

b)

From the above calculation we can write

$h_f=\frac{P_1-P_2}{\rho g}=\frac{151.189}{\rho g}$

In new situation the one end of the pipe raised by 0.2 meter, so the piezometric head will be

$h_f=\left \{ \frac{P_1}{\rho g}+H_1\right \}-\left \{ \frac{P_2}{\rho g}+H_2 \right \}$

Here H₁=0 and H₂ =0.2m So

$h_f=\left \{ \frac{P_1-P_2}{\rho g}-H_2\right \}$

$h_f=\left \{ \frac{151189-101325}{900*9.81}-0.2\right \}$

$h_f=5.4477m$

Now the velocity can be calculated as

$h_f=\frac{32\mu VL}{\rho gD^2}$

$5.4477=\frac{32*0.2*V*20}{900*9.81*0.04^2}$

$V=0.6012m/s$

Flow rate

$Q'=AV$

$Q'=\frac{\pi}{4}*0.04^2*0.60122$

$Q'=7.55*10^{-4}m^3/sec$

Or

$Q'=45.3315l/min$