You want to make sure your pipe is sized such that the flow, Q, out of tank 1 is greater than the flow into tank 1.
Or, at some level, you want to know the maximum flow, Q, that the system will allow. You can solve this kind of problem using the Bernoulli Equation for conservation of energy with the Darcy Weisbach equation to account for frictional losses equation:
$\frac{P_1}{\gamma}+z_1+\frac{v_1^2}{2 g}=\frac{P_2}{\gamma}+z_2+\frac{v_2^2}{2 g}+f\frac{L}{D}\frac{v^2}{2g}$
If we assume both tanks are under atmospheric pressure, the fluid in both tanks is not moving, and we neglect minor losses from the bends in the pipe and the change in pipe diameter then the equation reduces to:
$z_1-z_2=f\frac{L}{D}\frac{v^2}{2g}$
Where the term $z_1-z_2$ is what you're calling head. Actually, because both tanks are full, it does not matter if the pipes are at the top or botom of the tanks, As long as both pipes are submerged the change in head (from free surface to free surface) is what's important. so the variables we have are:
- $head$: the difference in water surface elevations
- $f$:a friction factor that depends on what the pipe is made of, and how fast the fluid is moving in it,
- $L$: the pipe length,
- $D$: the pipe diameter,
- $v$: the fluid velocity,
- $g$: is gravity, which is a constant
Your system is a little more complicated because we have two pipe diameters, so the equation should look like,
$z_1-z_2=f_A\frac{L_A}{D_A}\frac{v^2}{2g}+f_B\frac{L_B}{D_B}\frac{v^2}{2g}$, although i'm going to focus on the concept here, so i'm going to stick with the simpler equation.
to find the friction factor, $f$, a moody diagram is typically used... although because you're having backflow at times, i'm going to assume your flow is low enough that we can estimate the friction factor as:
$f=\frac{64}{Re}=\frac{64 \nu}{v*D}$
where $\nu$ is the kinematic viscosity of water.
This makes the equation:
$z_1-z_2=\frac{64 \nu}{v*D}\frac{L}{D}\frac{v_A^2}{2g}$
solving for the flow rate $Q$ ($Q=V A$) gives:
$Q=\frac{\pi D^4 g (z_1-z_2)}{128 L \nu}$
So you can see that the maximum flow is strongly influenced by pipe diameter! The bigger the pipe, the more flow from tank 1 to tank 2 will be possible. Additionally, contracting from 1" pipe to 0.5" pipe is restricting the flow even more, do to energy losses at the connection. Shortening the length of pipe used will also help you, but not nearly as much as increasing the pipe diameter. Doubling the pipe diameter will give you 16 times as much flow! The change in elevation of the free-sufrace of both tanks (head) is for sure important, but if you can't change it, that's ok. Finally, the pipe material is an important factor at higher flow rates, but maybe not for your current situation at a low flow rate.
Best Answer
The static pressure $p$ depends solely on the height $h$. If $p_0$ is the atmospheric pressure, then:
$$p=p_0+\rho gh$$
Where $\rho$ is the liquid's density and $g\approx10\:\mathrm{ms^{-2}}$.
So for the left hand side tank the pressure would be slightly higher because the tank is taller (not because it has higher or lower capacity).
Flow through the pipe always causes some viscous pressure loss though.
Answering OP's concern (comment section):
If you have easy access to the stream, a tank is not required.
The pressure discussed in my answer is called hydrostatic pressure and for good reason: if one blocks the end of the pipe, the pressure $p$ at that point would be exactly as indicated above ($p=p_0+\rho gh$, where $h$ is the height difference between the water's open surface and the pipe's end).
But when flow is allowed a certain amount of pressure is lost due to friction in the pipe. In that case the corrected pressure can be noted as:
$$p'=p_0+\rho gh-\Delta p$$ Without going into full details of pipe flow theory we can say with certainty that $\Delta p$:
A short, wide, smooth, straight pipe operated at low flow speed will deliver pressure almost identical to the hydrostatic pressure $p$.
But using a long, narrow (etc) pipe can lead to almost all of the hydrostatic pressure being lost to friction ($\Delta p \approx p_0+\rho gh$).
Further reading: Darcy-Weisbach.