I have a tank that is filled with a viscous liquid (~3000 CPs). At the bottom of the tank there is a pipe that leads to a gear pump. There is 22" of pipe straight down from the tank followed by a slight bend (~ 20 degrees) and then 7" to the pump. The liquid level in the tank is variable (H).

It is very important that the pump is not starved on the inlet side. The flow-rate out of the pump is 3000 grams per minute. The liquid has a specific gravity of 0.93.

I want to know at what viscosity will I starve the pump. Conceptually, I can imagine that as the viscosity of the liquid increases there will be more and more resistance to it flowing from the tank through the pipe to the pump. To compensate for this the pump will have to work harder to pull the liquid through the pipe to achieve the same mass flow rate. At some point the pump will not be able to pull hard enough to overcome the resistance and it will starve. I found the following equations which seem to be the right way to go:

$$Q = \frac{\Delta P \pi r^4}{8 \mu L}$$

The problem I have with the equation above is I do not know how to calculate the delta in the pressure. If I just use gravity ($\Delta P = \rho g \Delta H$) the flow rates I get are too small. I assume this is because it is not taking into consideration the negative pressure produced by the pump.

Is there a way for me to calculate the hydraulic vacuum force of the pump? Can I put it in terms of pressure? Can I assume there is a pure vacuum inside the pump and say the pressure difference is -1 atm + the contribution from gravity? Is this something I should be able to look up in the manual for the pump?

Any help or suggestions are appreciated. Thanks!

  • $\begingroup$ The mass balance must be satisfied, so if you are pumping 3kg/min out of the tank, then the flow into the pump must also be 3 kg/min (assuming no nuclear reactions are taking place). With that flowrate and physical properties, calculate the Reynolds number. The Q equation you cite is the Hagen-Poiseuille Law and is only valid for laminar flow. $\endgroup$
    – J. Ari
    Sep 19, 2019 at 14:17
  • $\begingroup$ For my clarification, why are you interested in the viscosity dependence? Does the liquid change viscosity with time, are you looking to use the equipment for a more viscous fluid, etc? $\endgroup$
    – J. Ari
    Sep 19, 2019 at 14:19
  • $\begingroup$ The system was designed for a liquid with a viscosity of 3000 CPs. Now that we have the itin place, R&D is investigating other liquids they may want to try. The manufacturer of the pump informed me that as we start to increase the viscosity we need to worry about starving the pump, but they did not have an exact viscosity at which that would be a concern. I thought it would be a fun mental exercise to try and calculate it, with the added benefit that I could give the information to my R&D folks if I figured it out. How will calculating the Reynold's number help me here? $\endgroup$
    – JBack
    Sep 19, 2019 at 20:03
  • $\begingroup$ You need to make sure you don't use the equation you wrote for regions that are above laminar flow, therefore you need to calculate the Reynolds number to confirm what flow regime you are operating in. $\endgroup$
    – J. Ari
    Sep 20, 2019 at 21:07
  • $\begingroup$ What are the pipe inner diameters? $\endgroup$
    – J. Ari
    Sep 20, 2019 at 21:14

1 Answer 1


What you are investigating is - from the perspective of pumping system design - similar to cavitation. for your purpose $\Delta P$ along the feed pipe is static pressure at pipe inlet minus the minimum allowable pressure at the pump inlet. The first known, from atmospheric pressure, level in tank and specific gravity. Look up NPSH.

The second is the vapor pressure of your medium at the specific temperature, if you fall below this pressure vapor bubbles will form at your pump inlet and the pump will run dry.

The other thing that could happen is that the pump is not able to deliver to sufficient head on the pressure side, if the available head on the suction side is too low, then the pump may stall.

So minimum allowable head on the suction side is the bigger of the two possible heads mentioned above.

Edit to adress question for clarification: Forget, for a moment, the idea of suction. Only positive pressure exists. Fluids flow from high to low pressure. What does it mean for a pump to run dry? It means the flowrate through the pump is higher then the flowrate to the pump. In case of a gear pump, flow rate through the pump is mostly a function of RPM.

Now, you could assume the absolute pressure at the inlet of the feed pipe and put that into the form above, This would be the flowrate if in place of your pump there's be a vacuum. This means, if you pump at or above this flowrate, there's actually be a vacuum - pump runs dry.

However, all fluids will form vapor at low enough pressure, from the perspective of your pump vapor is as bad as vacuum. That's why you want the pressure at the pump inlet above the vapor pressure at that temp.

  • $\begingroup$ Ok so if I understand correctly what you're saying is the pump should not be damaged (i.e. run dry), so long as the pressure at the inlet of the pump is above the vapor pressure of the liquid I am pumping. Is that correct? Couldn't the pump get starved before that point if the liquid is viscous enough? $\endgroup$
    – JBack
    Sep 20, 2019 at 13:50
  • $\begingroup$ Also, isn't this assuming the pump has the capacity to provide enough suction to get the pressure all the way down to the vapor pressure? Couldn't the pump fail before that point? $\endgroup$
    – JBack
    Sep 20, 2019 at 14:55
  • $\begingroup$ @JBack I've update the question $\endgroup$
    – mart
    Sep 20, 2019 at 19:38

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