Question 6: Water at 25 °C ($\rho$= 1000 kg/m$^3$; $\mu$=10$^{-3}$ N.s/m$^2$) is continuously fed to a tank which is open to the atmosphere. A pump draws water at very low volumetric flow rate from the tank and pumps water to a point 2 (open to atmosphere, too) at a steady state, shown in figure below. The specific shaft work, supplied to the pump, is +80 J/kg. The pump is 100% efficient. The average velocity at the point 2 is 0.5 m/s. Please calculate Z1 by making necessary and valid assumptions (1 J = 1 N.m and g=9.81 m/s$^2$).
The flow inside the tube is in laminar regime; thus, the Fanning friction factor is given by the equation below:
$f_F = \frac{16}{Re}$
It is believed that the frictional losses in straight pipe sections can be given using the following equation:
$\sum F_{straight} = \left(\frac{4f_F L}{d_{tube}}\right)\left(\frac{v^2}{2}\right)$
And minor losses due to fittings, valves etc..
$\sum F_{minor} = \left(\sum K\right)\left(\frac{v^2}{2}\right)$
$\sum K = K_{elbows} + K_{valves} + K_{exit} + K_{enterance} + K_{expansion/contraction} + K_{pipe \ exit \ etc.}$
Steady-state Mechanical Energy Balance for incompressible flow (modified Bernoulli equation)
$g(Z_2 - Z_1) + \left[\left(\frac{v_2^2}{2}\right) - \left(\frac{v_1^2}{2}\right)\right] + \left[\left(\frac{P_2}{\rho}\right) - \left(\frac{P_1}{\rho}\right)\right] - \bar{W}_s + \sum F = 0$
where
$\sum F = \sum F_{straight} + \sum F_{minor}$
Data: K$_{45^\circ \ elbow}$= 0.35; K$_{tank \ to \ pipe \ exit}$= 0.5; K$_{pipe \ exit \ at \ point \ 2}$= 1
