Water at $40C$ flows from tank $A$ to tank $B$ as shown in the figure. Find the volumetric flow $Q,$ neglecting entrance losses to the capillary tube as well as exit losses. Assume laminar flow and use the following energy equation and the head loss through the pipe,

$\frac{p_1}{}+{}_1\frac{V_1}{2g}+z_1+h_p=\frac{p_2}{}+{}_2\frac{V_2}{2g}+z_2+h_t+h_L$

$h_L=\frac{128LQ}{g{D}^4}$

where, $=992k\frac{g}{m^3}$ and $=6\mathrm{.}51{10}^{-4}N*\frac{s}{m^2}$ are the density and viscosity of water, $D$ and $L$ are the diameter and length of the tube, respectively, and $V$ is the mean velocity through the tube. For laminar flow, ${}_1={}_2=2.$