A triangular pipe network is shown. Head loss in each pipe is $h_f=rQ^{1.8}$ in consistent units. Pipe AB has $r=1$ and BC has $r=2$. A supply of 100 at A is split equally into AB and AC; withdrawals are 70 at B and 30 at C. Find the resistance coefficient $r$ of AC, to two decimal places.
Equal division gives $Q_{AB}=Q_{AC}=50$. B withdraws 70, so it also receives 20 from C through BC. Thus $H_A-H_B=50^{1.8}$ and $H_C-H_B=2(20^{1.8})$. Subtracting gives $H_A-H_C=50^{1.8}-2(20^{1.8})=r_{AC}50^{1.8}$. Hence $r_{AC}=1-2(20/50)^{1.8}=0.6156$, or 0.62.
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