Increase Water flow through sealed tank How can I increase the movement of water through a tank that is being supplied in the front and being drawn to the back of the tank with a pump?  I am trying to dissipate heat and need more water movement.
Thank you.
 A: In addition to Gert's theoretical analysis I would like to point out that if you have a closed loop typical of most cooling setups, the value of orienting the cooling tank vertically rather than horizontally becomes almost nil, as there is some downstream return of water from a lower height to a higher height, and this exerts a backpressure on the motor. This applies even when it is not obvious, for example if I run a pipe backwards from the output bottom of a hydroelectric dam to the floor of the input dam reservoir on the other side, the low position of the pipe in the reservoir doesn't relieve you of the need to push a column of water all the way back to the top of the reservoir, because water is largely incompressible.
The basic things you're looking for as an experimentalist trying to increase flow through some sort of heat dissipation system are:


*

*Make sure there are absolutely no air bubbles (or any fluid that is not your coolant) anywhere in the system, make sure that everything is well-connected so that no air can get into the system, make sure that the system is absolutely as full as it can be. Like if you've got a big complex diffuser with metal fins inside the fluid or something like that which air bubbles could get caught on, it's up to you to plug one tube on one side of this, let that be on the "bottom", fill the thing with water up through the other tube, and then bang the diffuser several times against a firm surface to dislodge air bubbles so that they drift up the upper tube. Then reorient the thing and bang on it again, and again. The same might be true of the pump, though it will kind of "bang on itself" as it does its thing. 

*Increase the size of the tubes or any other thin channels that the fluid flows through. Thin channels means friction with the walls which means "head loss" (loss of pressure) as you go through those channels. You want as little resistance as possible.

*Understand the difference between volume flow and fluid velocity and choose which one you need more. Usually you want more volume flow, so you want a bigger reservoir. Sometimes you do need more fluid velocity and so you actually want to transition flow from some wide pipe to some thin pipe and you're willing to accept the extra friction losses.

*Upgrade your pump. 

*If you are still stuck, possibly change fluids.

*If you are still still stuck, you might just need to refrigerate the system that the reservoir is dumping heat into or otherwise increase heat flow on that other end of the reservoir. Like, all of the above is completely pointless if the reservoir interacts with a fan-heatsink combination and the heat sink turns out to be totally blocked with dust or the fan has been shoved against a wall so it's not moving any air.
A: I imagine the situation to be as follows:

Where:


*

*the $P_i$ are the respective pressures

*the $v_i$ are the respective flow velocities

*the $S_i$ are the respective cross-sections

*$\Delta h$ is the height difference between points $1$ and $2$

*$Q_v$ is the volumetric throughput through the system. It is the papameter we want to maximise.


For incompressible, inviscid flow we can apply Bernoulli's Principle, which relates variables pertinent to flow beween two points on the same flowline. The flowline chosen here is the thin red line. The Principle states:
$$\frac12 \rho v_1^2+P_1+\rho gh_1=\frac12 \rho v_2^2+P_2+\rho gh_2\tag{1}$$
In addition, because water is incompressible, we have a continuum equation:
$$S_1v_1=S_2v_2=Q_v\tag{2}$$
At the point $1$ I'll assume the pressure is atmospheric pressure:
$$P_1=P_{atm}$$
So that:
$$\Delta P=P_1-P_2=P_{atm}-P_2$$
Similarly for the respective heights:
$$\Delta h=h_1-h_2$$
So we can re-write $(1)$ as:
$$\frac12 \rho (v_2^2-v_1^2)=\Delta P+\rho g \Delta h\tag{3}$$
Using $(2)$:
$$v_1=\frac{S_2}{S_1}v_2$$
Then with some reworking:
$$\frac12 \rho \Big(v_2^2-\frac{S_2^2}{S_1^2}v_2^2\Big)=\Delta P+\rho g \Delta h$$
$$\frac12 \rho v_2^2\Big(\frac{S_1^2-S_2^2}{S_1^2}\Big)=\Delta P+\rho g \Delta h$$
$$v_2=\sqrt{\frac{2S_1^2(\Delta P+\rho g \Delta h)}{\rho(S_1^2-S_2^2)}}$$
$$Q_v=S_1S_2\sqrt{\frac{2(\Delta P+\rho g \Delta h)}{\rho(S_1^2-S_2^2)}}$$
$$Q_v=S_1S_2\sqrt{\frac{2(\Delta P+\rho g \Delta h)}{\rho(S_1-S_2)(S_1+S_2)}}\tag{4}$$
So $(4)$ is an expression for $Q_v=f(S_1,S_2,\Delta P, \Delta h)$.
We can now evaluate how these variable influence $Q_v$, which we want to maximise.


*

*Influence of $S_1,S_2$:


$$Q_v \propto \frac{S_1S_2}{\sqrt{(S_1-S_2)(S_1+S_2)}}$$
Although not eminently evident from this expression, a little function analysis shows what's intuitive: increasing both cross-sections increases $Q_v$. It's like removing bottle necks to the flow. It boils down to increasing the pipe diameter of all pipes concerned.


*Influence of $\Delta P$:


Clearly increasing $\Delta P$ increases $Q_v$, albeit non-linearly (in the case $\Delta h=0$, then $Q_v \propto \sqrt{\Delta P}$). This comes down to deploying a more powerful pump, of course.


*Influence of $\Delta h$:


Clearly increasing $\Delta h$ increases $Q_v$, albeit non-linearly (in the case $\Delta P=0$, then $Q_v \propto \sqrt{\Delta h}$). This may be or not be a practical thing to implement.
In addition to CR Drost's excellent answer, if we look at the problem from a heat transfer PoV, then increasing turbulence in the tank will help carry off heat of whatever you're cooling in there (like cooling worm). Increased turbulence increases the heat transfer coefficient. This can be achieved by building baffles into the tank.
