Question

A 2m long pipelines tapers uniformly form 10cm diameter to 20cm diameters at its upper end. The pipe centreline slopes upwards at a angle of 30° to the horizontal and the flow direction is from smaller to bigger cross section. If the pressure gauges installed at lower and upper ends of the pipeline read 200kpa and 230kpa respectively. Determine the flow rate and the pressure mid length of the pipeline. Assume no energy losses.

Answer 1

(Refer Image for clarity of question)

$P_C=200 \ Kpa;P_a=230\ KPa$ at gauge pressure.So difference of pressure between these points will be absolute.

$h_c=0;$

$h_a=2\ sin30\degree=\frac{2}{2}=1\ m;$

$h_q=PB \ sin 30\degree =\frac{1}{2}$

$P_a-P_c=30KPa=30,000KPa..........(1)$

Solution:Let area of pipe at $C ,Q,A$ be $A_c,A_q,A_a$ respectively.

As$\ Q$ is midpoint, Diameter at $Q=\frac{D_c+D_a}{2}=\frac{10+20}{2}=15\ cm$

$\frac{D_c}{D_a}=\frac{10}{20}=\frac{1}{2};\frac{D_c}{D_q}=\frac{10}{15}=\frac{2}{3}$

And $\frac{A_q}{A_c}=(\frac{D_q}{D_c})^2=(\frac{2}{1} )^2=4\implies A_q=4A_c ;$

$\frac{A_a}{A_c}=(\frac{D_a}{D_c})^2 =(\frac{3}{2})^2=\frac{9}{4}\implies A_a=\frac{9}{4}A_c ;$

By using continuity equation,$A_av_a=A_cv_c\implies v_a=\frac{A_a}{A_c}v_c=\frac{v_c}{4}......(2)$

And ,$A_av_a=A_cv_c\implies v_q=\frac{A_q}{A_c}v_c=\frac{4v_c}{9}........(3)$ .

Assuming the fluid to be water,$\rho=1000\ Kg/m^3,g=10m/sec^2$ ,Apply Bernaulli Equation between $A$ and $C$

$P_a+\frac{1}{2}\rho v_a^2+\rho gh_a=P_c+\frac{1}{2}\rho v_c^2+\rho gh_c$

$\implies \frac{1}{2}\rho (v_c^2-v_a^2)=(P_a-P_c)+\rho g(h_a-h_c)$

$\implies \frac{1}{2}\times 1000\times (v_c^2-(\frac{v_c}{4})^2)=30000+1000\times 10\times (1-0)$

$\implies v_c^2=\frac{40\times 32}{15}=\frac{1280}{15}\implies v_c=\sqrt{\frac{1280}{15}}=9.237\ m/sec$

$D_c=10\ cm=0.1 \ m;D_q=15\ cm=0.15\ m,$ $D_a=0.2\ m$

Flow rate at $C=A_cv_c=\frac{\pi D_c^2}{4}v_c=\frac{\pi\times 0.1^2}{4}\times 9.237=0.29\ m^3/sec$

As fluid is incompressible,flow rates at all points are equal.

Flow rate at mid point$(Q)$ $=$ Flow rate at $C$ $=0.29m ^3/sec$

To determine pressure at mid point ,apply bernaulli equation at $C$ and $Q$ .

$\implies \frac{1}{2}\rho (v_c^2-v_q^2)=(P_q-P_c)+\rho g(h_q-h_c)$

$\implies \frac{1}{2}\rho (v_c^2-(\frac{4v_c}{9})^2)=(P_q-P_c)+\rho g\times \frac{1}{2}$

$\implies \frac{1}{2}\times 1000\times \frac{65}{81}\times \frac{1280}{15}=(P_q-P_c)+5000$

$\implies P_q-P_c=29238\ Pa=29.238\ Kpa\implies$ $P_q=229.238\ KPa$ (gauge pressure at mid point)