What's wrong with my work?

The correct answer IS $41\mathrm{.}14$ but I can't get my work there and I don't know what else to do$.1$

$\frac{1}{1}$ point

Water flows in a cast iron pipe with diameter $D=0\mathrm{.}61[m]$ from an upstream point A to a downstream point B $.$ The distance between point A and B is $x=2\mathrm{.}1[km]$ and the elevation drop between A and B is $z=28\mathrm{.}9[m].$ The discharge in the pipe is $Q=1\mathrm{.}3[\frac{m^3}{s}].$ Ignoring minor losses, compute the total headlosses along the pipe in using Darcy$-$Weisbach. What is the drop in pressure between points A and $B$ in $[$psi$]?$ Assume normal conditions.

$D=0\mathrm{.}61m$

$x=21\mathrm{.}0km=2100m$

$\frac{1\mathrm{.}3m\frac{p}{s}}{\frac{(0\mathrm{.}61(m){)}^2}{4}}=4\mathrm{.}45\frac{m}{s}$

$z=28\mathrm{.}9m$

$Q=1\mathrm{.}3\frac{m^3}{s}$

$h_L=?$

$\frac{p_1}{}+z_1=\frac{p_2}{}+z_2+h_L$

$e=0\mathrm{.}26mm$

$\frac{p_1}{}-\frac{p_2}{}=z_2-z_1+h_2$

$=0\mathrm{.}26{10}^{-3}m$

$\frac{P}{}=(-z+h_L),f=0\mathrm{.}016$

$=(-z+f(\frac{x}{D})\frac{y^2}{2g})$

$=(-28\mathrm{.}9m+(0\mathrm{.}016)\frac{2100(m)}{0\mathrm{.}61(m)}*\frac{(4\mathrm{.}45\frac{m}{s}{)}^2}{2(9\mathrm{.}81\frac{m}{s^2})})$

$P=(26\mathrm{.}7)9810$

$55\mathrm{.}6m$

$=\frac{261927Pa}{6895}$

$p=41\mathrm{.}1$