# Water flow between reservoirs

I've got a homework problem that looks like this and i've been given the following data, $$d=1\,\text{mm}$$ and kinematic viscosity $$\nu=1.06\times10^{-6}\,\text{mm}^2/s$$ and it also states that I should consider the flow to be laminar. The problems asks to compute the flow in the tube. My approach to solving this problem, given that I have been given height differences was like this.

Initially I computed the pressure at datum 3 against datum 2 like this $$p_3 =p_2 + \rho\,g\,\Delta h \approx 1.007 \,\text{bar}$$ After this, I can compute a velocity component using Bernoulli's theorem for datums 1 and 3. I wrote the equations like this, $$\frac{p_1}{\rho\,g} + \frac{v_1^2}{2g}+h_1 = \frac{p_3}{\rho\,g} + \frac{v_3^2}{2g}+h_3$$ Considering datum 3 as zero, and velocity at datum 1 to be zero (large tank assumption thing) I was able to write the equation as follows $$\frac{p_1}{\rho\,g}+h_1=\frac{p_3}{\rho\,g}+\frac{v_3^2}{2g}$$ After this it was a matter of solving for $v_3$ which comes out to be $$v_3 \approx 2.8\,\text{m}/\text{s}$$ Ultimately I compute the flow as follows $$Q = v_3\,A_3 \approx 0.002\,\text{L}/\text{s}$$ Now my issue is two fold,

1. Am I correct in considering datum 2-3 to be hydrostatics?
2. Why have I been given the kinematic viscosity and the assumption that the flow is laminar?

It seems that I could use this formula to compute the velocity at the pipe $$\text{Re} = \frac{v_3\,d}{\nu}$$ but then I would have to make the assumption that $Re\approx2300$, which doesn't seem quite fair.

I also believe that we can neglect losses, so what's up with that?

Bernoulli equation is not appropriate to use for this problem, which is dominated by viscous stresses. If you do a force balance on the fluid in the tube, you obtain:$$\rho g AL+(p_{in}A-p_{out}A)=\tau_{wall}\pi D L$$where A is the cross sectional area of the tube, $p_{in}$ is the inlet pressure to the tube (determined from the hydrostatics of the upper tank), $p_{out}$ is the outlet pressure from the tube (determined from the hydrostatics of the lower tank), D is the tube diameter, $\tau_{wall}$ is the viscous shear stress at the tube wall, and L is the length of the tube. The first term on the left is the weight of the fluid, the term in parenthesis on the left describes the pressure forces acting on the fluid at the inlet and outlet ends of the tube, and the term on the right is the viscous frictional force at the tube wall. If we divide the above equation by the tube area A, we obtain:$$\rho g L+(p_{in}-p_{out})=\frac{4L}{D}\tau_{wall}$$ The shear stress at the wall is related to the shear rate $\gamma_{wall}$ by:$$\tau_{wall}=\eta \gamma_{wall}$$ where $\eta$ is the dynamic viscosity. In addition, the shear rate at the wall is related to the average flow velocity v by:$$\gamma_{wall}=\frac{8v}{D}$$If we combine the three previous equations, we obtain:$$\rho g L+(p_{in}-p_{out})=\frac{32\eta Lv}{D^2}=\frac{128\eta LQ}{\pi D^4}$$where Q is the volumetric flow rate. The term on the right should be recognizable from the Hagen Poiseulle Law.
I think you are supposed to assume Poiseulle flow in the connecting pipe (which requires that flow be laminar), in which a constant parabolic velocity profile develops throughout the pipe (away from its ends). You need to calculate $\Delta p$ between pipe ends, and to do so it's alright to make hydrostatic assumption since the fluid levels in reservoir are changing only gradually.