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My background is in electronics, not mechanical engineering and fluid dynamics, so please keep that in mind. Also, this is not a homework or academic assignment.

Q: A dewar is partially filled with liquid nitrogen (LN2) and is open to the atmosphere. If a container, open on the top, has a hole in the bottom and is lowered into the liquid, how quickly will the liquid fill the container?

Specifics: The dewar has an inside diameter of 14" and has 8" of LN2 ( volume ~1200 cubic inch). Second container is 4" outside diameter, 3.75" inside diameter, 7" height. A small hole, ~40 mils, is drilled in the bottom of the container. Container will be lowered into the LN2 to a depth of 6". Atmospheric pressure: Sea level.

I need to determine the rate of flow into the secondary container so I know how quickly it will fill to a 6" depth (~66 cubic inch).

Elflord
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    Why would you want to fill a container in this method? I used to fill liquid nitrogen carts on the flightline all the time. We used a pressure build up coil that boiled off some nitrogen to build up pressure to push the liquid into the cart. The fluid dynamics is not difficult, but the complication of the cryogenic liquid boiling at contact with the container complicates things. If you cool the container before placing it in the nitrogen it would be simpler. To give you an idea, the density of liquid nitrogen is less than water so it should fill a bit slower as I think the viscosity is similar – Gwydionforge Apr 19 '17 at 15:01
  • Just as a sidenote. You should use fewer abbreviations. For non-native or non-experts your question is difficult to understand. – MrYouMath Apr 19 '17 at 15:06
  • Gwydionforge - There will be electronics inside the secondary container; the purpose of slowing filling it is to slowly lower the temperature of the circuits. Too fast a temperature change will result in damage to the parts due to differing thermal coefficients of the integrated circuit packages and solder bonds. – Elflord Apr 19 '17 at 16:10

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I am at work and cannot perform the calculations for you, but if you solve the following equation you will find the velocity $v$ being the square root of 2 times Earth gravity times depth $x$:

$$v = \sqrt{2 \cdot 9.81 \, \mathrm{m/s^2} \cdot x}$$

If I remember correctly, from the velocity you should be able to calculate the volume rate $\dot{V}$ by multiplying the area of the opening $A$ with the velocity $v$:

$$\dot{V} = A \cdot v$$ $$\mathrm{\frac{m^3}{s} = m^2 \cdot \frac{m}{s}}$$

then just convert to the volume units of your choice.

http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html Read the example on Bernoulli's equation and the vented tank example.

Gwydionforge
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