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
Mass is mass. If you add something which has mass to the bucket, the bucket now has more mass. It doesn't matter if it was more dense or less dense. If you add 1kg to a bucket, you add 1kg.
Now there are two areas where this falls apart. One is in the case where the object you are adding is actually less dense than the air (not just less dense than the water, but actually less dense than air, like a helium balloon). In this case, we can forget about the minor detail of whether the object is floating on the water or not, we can focus on the entire bucket which is immersed in air. This object displaces a larger mass of air than its own mass, so it will actually add lift to the bucket. The mass of the bucket will still be bucket+water+object, but the bouyancy forces will make the bucket feel lighter. Indeed, this is precisely the mechanism used by hot air balloons.
The other corner case is the case where the bucket is already filled to the brim with water. Adding a single drop of water would cause a drop to have to spill over the brim. Now when we add our toy boat the situation gets a little more complicated. The boat will displace a mass of water equal to the mass of the boat. This displacement will cause the water to rise and spillover. If you tally up all of the masses in this case, you will find that the bucket+water-in-the-bucket+toy boat will have the same mass as the bucket+water did before you added the toy boat. Where did the extra mass go? There's a puddle on the ground outside of the bucket, whose mass is exactly equal to that of the toy boat.