The piping network shown below uses smooth cast iron pipes (C:130). The diameter and length of pipes in the network are pipe \(1:\) dia. \(=0.31 \mathrm{~m}\), length \(=609.6\) \(\mathrm{m}\); pipe 2: dia. \(=0.203 \mathrm{~m}\), length \(=609.6 \mathrm{~m}\); pipe \(3:\) dia. \(=0.1524 \mathrm{~m}\), length \(=\) \(914.4 \mathrm{~m}\); pipe \(4:\) dia. \(=0.1524 \mathrm{~m}\), length \(=1219.2 \mathrm{~m}\); pipe \(5:\) dia. \(=0.203 \mathrm{~m}\), length \(=304.8 \mathrm{~m} ;\) pipe \(6:\) dia. \(=0.203 \mathrm{~m}\), length \(=914.4 \mathrm{~m} ;\) pipe \(7:\) dia. \(=0.203\) \(\mathrm{m}\), length \(=609.6 \mathrm{~m}\). Perform the Hardy Cross analysis and calculate the flow rate in each line after one iteration. Note that \(K_{1}=4.721\) when \(Q\) is in cfs. Use the following initial flow rates: \(Q_{1}\) (top loop) \(=1.811 \mathrm{cfs}, Q_{1}\) (bottom loop) \(=-1.811\) cfs, \(Q_{2}=0.686 \mathrm{cfs}, Q_{3}=-0.314 \mathrm{cfs}, Q_{4}=-0.314 \mathrm{cfs}, Q_{5} \mathrm{cfs}, Q_{6}=0.875 \mathrm{cfs}\), \(Q_{7}=-1.125\) cfs.

Take \(1 \mathrm{cfs}=0.02832 \mathrm{~m}^{3} / \mathrm{s}\).