Pipeline carrying water has a diameter of 0.5 m and 2 km long. To increase the delivery another pipeline of the same diameter is introduced parallel to the first pipe in the second half of its length. Find the increase in discharge if the total head loss in both the cases is 15 m. Assume 4 f = 0.02 for all the pipes

Diameter of pipe, D=0.5 m

Length of pipe, L = 2 Km=2000m

Coefficient of friction 4f= 0.02

Head loss in both the cases $ish_f$= 15m

Length of another parallel pipe $L_2$=1000 m

Diameter of another pipe line , $D_2$=0.5 m

Increase in discharge:

Case. I. Discharge (Q) for single pipe of length 2000 m and diameter 0.5m: The head lost due to friction in single pipe is given as:

$h_f=\frac{4fLV^2}{D×2g}$

Where ,V = velocity of flow for a single pipe

15=$\frac{0.02×2000×V^{2}}{0.6×2×9.81}$

V=$(\frac{0.6×2×9.81×15}{0.02×2000})^{1/2}$=2.10 m/s

∴Discharge,Q=AV=$\frac{π}{4} 0.5^{2}×2.10=0.4123 m^{3}⁄s$ ………(i)

Case II. When an additional pipe of length 1000m and diameter 0.5 m is connected in parallel with the last half of the pipe:

Let, $Q_1$=Discharge in first pipe,

$Q_ 2$=Discharge in second parallel pipe,

$Q_p$=Discharge in the main pipe (when pipes are connected in parallel)

Then, $Q_p=Q_1+Q_2$

As the pipes are in parallel have the same diameter and length,

$∴Q_1=Q_2=\frac{Q_p}{2}$

Consider the flow through ABC or ABD:

$[V_BC=\frac{Q_p⁄2}{area}=\frac{Q_p⁄2}{π⁄4×0.5^{2}}=2.54Q_p]$

Substituting these values in (ii),we get

15=$52.82Q^2_{P} +13.15Q^{2}_P$

$Q_p=[\frac{15}{52.82+13.15}]^{1/2}=0.4768 m^3⁄s$

∴Increase in discharge= $Q_p-Q$

=0.4768-0.4123=$0.0645m^3⁄s$ …..(ANS)