I Siphoning through a straight tube

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Research in nuclear physics is focused on pre-cooling a dewar with liquid nitrogen before adding liquid helium to reduce costs. To siphon out the nitrogen, a metal tube is used, requiring higher pressure in the dewar than atmospheric pressure. Calculations indicate that approximately 1.132 bar of absolute pressure is needed to initiate flow, factoring in the weight of the nitrogen. Initially, only 1.092 bar is necessary when the height difference is one meter. The confusion arose from mistakenly including a velocity term in the pressure calculations.
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Hi y'all,
I am currently doing research in nuclear physics dealing with cryogenics. We want to pre cool the inner most part of a dewar with liquid nitrogen before we put liquid helium in since it's quite expensive. Once the nitrogen stops boiling off and reaches somewhat of an equilibrium, the excess liquid needs to be siphoned out of the dewar. Our only option is to use a metal tube sticking straight up into the air. In order to siphon the liquid nitrogen out, we need a higher pressure in the dewar than the atmosphere. I go through my calculations with Bernoulli's equations and I seem to get around 1.8 bars needed inside the dewar. That answer seems too high to me and I don't know where I'm going wrong. Here are some specifics:

Total volume flow rate: 2 liters/hour (.002 m^3/hour)
Diameter of metal tube: 3mm
Diameter of Inner Dewar Compartment: 101.6 mm
Density of Liquid Nitrogen: 807.0 kg/m^3
Total Height of Tube: 1.5 m
Height from surface of liquid nitrogen to top of tube: 1 m

I appreciate any help.

Thanks
 
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The pressure at the bottom of the tube due to the weight of the nitrogen will be ##\rho g h## or 0.11875 bar. Adding the atmospheric pressure, you get an absolute pressure of 1.132 bar. If you maintain a pressure inside the dewar larger than this, the nitrogen will flow out the tube.

At the beginning of the process (when the difference in height is 1 m), you only need 1.092 bar of absolute pressure in the dewar to initiate flow outside of the tube.
 
Ah okay makes sense. I think I was including a velocity term that gave me a higher answer. Thanks!
 
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