Two reservoirs are connected by two pipes P and Q. The pipes have the same diameter and length and are placed in parallel. If the friction factor of P is 9 times that of Q, then the discharge in P to that in Q is

Concept:Head loss \(h_l=\frac{fLV^2}{2gd}\)where,f = FRICTION factor, L = length of pipe, V = velocity of flow, and d = diameter of pipe.When two pipes connected in parallelhl = hl1 = hl2Q = Q1 + Q2Calculation:Given:dP = dQ, LP = LQ, hLP = hLQ, AP = AQFP = 9FQTwo pipes are connected parallel, so head loss will be same.\(\BEGIN{array}{l} {{\RM{h}}_{{{\rm{L}}_{\rm{P}}}}} = \frac{{{{\rm{F}}_{\rm{P}}}{{\rm{L}}_{\rm{P}}}{\rm{V}}_{\rm{P}}^2}}{{2{\rm{g}}{{\rm{d}}_{\rm{P}}}}},{\rm{\;}}{{\rm{h}}_{{{\rm{L}}_{\rm{Q}}}}} = \frac{{{{\rm{F}}_{\rm{Q}}}{{\rm{L}}_{\rm{Q}}}{\rm{V}}_{\rm{Q}}^2}}{{2{\rm{g}}{{\rm{d}}_{\rm{Q}}}}}\\ \Rightarrow \frac{{9{{\rm{F}}_{\rm{Q}}}{{\rm{L}}_{\rm{P}}}{\rm{V}}_{\rm{P}}^2}}{{2{\rm{g}}{{\rm{d}}_{\rm{P}}}}} = \frac{{{{\rm{F}}_{\rm{Q}}}{{\rm{L}}_{\rm{Q}}}{\rm{V}}_{\rm{Q}}^2}}{{2{\rm{g}}{{\rm{d}}_{\rm{Q}}}}} \end{array}\)\( \Rightarrow 9{\rm{V}}_{\rm{P}}^2 = {\rm{V}}_{\rm{Q}}^2\)⇒ VQ = 3VP ⇒ VP/VQ = 1/3QP = APVP and QQ = AQVQ\(\frac{{{{\rm{Q}}_{\rm{P}}}}}{{{{\rm{Q}}_{\rm{Q}}}}} = \frac{{{{\rm{A}}_{\rm{P}}}{{\rm{V}}_{\rm{P}}}}}{{{{\rm{A}}_{\rm{Q}}}{{\rm{V}}_{\rm{Q}}}}} = \frac{1}{3} = 0.33\)