Question 7.P.23: A pipe line 200 mm dia. and 4000 m long connects two reservo......

h_{f}=\frac{f L u^{2}}{2g D},\,u=\frac{4Q}{\pi D^{2}}\,,\,u^{2}=\frac{16Q^{2}}{\pi^{2}D^{2}}\,,\,h_{f}=\frac{8\,f L Q^{2}}{\pi^{2}\;g D^{5}}

Considering the two sections, (total drop)

60=\frac{8\times0.024\times1500\times Q_{1}^{\ 2}}{\pi^{2}g\times0.2^{5}}+\frac{8\times0.024\times2500\times Q_{2}^{\ 2}}{\pi^{2}g\times0.25} \\ =9295.5~Q_{1}^{2}+15492.54~Q_{2}^{~2}

but            {{Q}_{2}}^{2}=({{Q}_{1}}-{0.05})^{2}, Substituting and simplifying

60=9295.52~Q_{1}^{~2}+1549.25(Q_{1}^{~2}+0.05^{2}-2Q_{1}\times0.05)

or              24788.05~Q_{1}^{~2}-1549.25~Q_{1}-21.268=0

Q_{1}={\frac{1549.25\pm[(-\,1549.25)^{2}+4\times21.268\times24788.05]^{0.5}}{2\times24788.05}} \\ ={\frac{1549.25\pm2123.45}{2\times24788.05}}=0.074082\ \mathrm{m^{3}/s},

∴           \mathbf{Q}_{2}=0.024082\mathrm{~m}^{3}/{\mathrm{s}}

The other solution is negative.

Check: For the first section

h_{f\!}=\frac{8\times0.024\times1500\times0.074082^{2}}{\pi^{2}\times9.81\times0.2^{5}}=51.015~\mathrm{m}

For the second section

h_{f\!}=\frac{8\times0.024\times2500\times0.024082^{2}}{\pi^{2}\times9.81\times0.2^{5}}=8.985\;{\mathrm{m}},          Total head = 60 m