$4\mathrm{.}4\mathrm{.}1.$ The total discharge from $A$ to $B$ in Figure $P4\mathrm{.}4\mathrm{.}1$ is $0\mathrm{.}936\frac{m^3}{s}.$ Pipe $1$ is $100m$ long with a

diameter of $40cm,$ and pipe $2$ is $100m$ long with a diameter of $50cm.$ Determine the head loss

between A and $B($junctions$)$ and the flow rate in each pipe using $($a$)$ Hardy$-$Cross principles and $($b$)$

the method of equivalent pipes. Assume $e_1=2\mathrm{.}0mm$ and $e_2=1\mathrm{.}25mm$ for the highly tuberculated

pipes, $20C$ water, and minor losses are negligible. Hint: A direct solution is possible using the

conservation of mass and energy principles.

Figure P$4\mathrm{.}4\mathrm{.}1$