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I'm trying to figure out what kind of pump I should use to boil water at low temperature. Here is my idea of the setup:

enter image description here

The dark blue water comes in, the pump sucks down the light-blue water. This will reduce the pressure in the top, forcing the dark blue water start boiling.

The water then condenses and gets pumped out.

An important condition is of course the temperature of the incoming water. This can be calculated using the Antoine equation,

$$p =10^{\left(8.07131 – \frac{1730.63}{233.426+T}\right)}$$

where $p$ is pressure in mmHG, and $T$ is temperature in degrees Celsius.

I'm trying to figure out what how I can translate the boiling of water to psi. I get confused because as some water gets pumped away, that will be replaced by newly condensed water, thus increasing the pressure again.

I'm assuming I should use a positive-displacement pump. Please let me know if this is a bad idea for some reason.

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  • $\begingroup$ How are you condensing the water after it has boiled? Are you heating the water? To get water to boil at ambient temperatures you need to drop the pressure to ~15mmHg or expand the gas space by a factor of 50. And the dark blue water will be able to rise ~10m to prevent the formation of such a vacuum. $\endgroup$
    – CleptoMarcus
    Apr 22, 2016 at 10:46
  • $\begingroup$ @CleptoMarcus, the idea is that the dark blue water will cool down the light blue water, transferring the heat back. The water is not at ambiant temperature, i'm looking at somewhere between 40-80 degrees. How do you get the water-rise? Thanks $\endgroup$
    – Himmators
    Apr 22, 2016 at 11:16
  • $\begingroup$ Why do you want to boil water? What is your ultimate goal here? $\endgroup$
    – Carl Witthoft
    Apr 22, 2016 at 11:24
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    $\begingroup$ The dark blue water at the bottom is at atmospheric pressure at best, ~101000Pa, the pressure at the top (the vacuum) is at ~7000Pa (@40°C) so the static head of water is dP/(rho * g) or (101000-7000)/(1000kg/m³ * 9.8m/s²) = 9.6m $\endgroup$
    – CleptoMarcus
    Apr 22, 2016 at 11:25

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You can use a water pump to evaporate water. But you shouldn't.

The evaporation chamber needs to be closed, except for the suction pipe. Otherwise when you lower pressure water (or air) will flow in, raising pressure again. So you can create small batches of vapor.

The water will evaporate where pressure is lowest, this is likely to be your pump inlet. Not what you want. You need to make sure that the ambient pressure in the evaporation chamber is lower than vapor pressure at whatever temeprature your water has. At the same time, ambient pressure + hydraulic head between level in chamber and pump inlet needs to be higher by a margin, else you have evaporation in your pump. Not good.

Then, you have low pressure, low temperature vapor. What do you want to do with it and how do you transport it? If you use a blower, the same blower could also create the vacuum for your evaporation and save you all the pump headaches.

So you could create vapor with a water pump in a batch process, but I fail to see an aplication where this makes sense.

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  • $\begingroup$ Great answer! I'm researching solar powered water desalination. Boiling water at a lower temperature would decrease power consumption while also reducing the need for insulation (because the temperature-difference would be less).The idea would be to use a heat exchange to the new water to cool the water to a lower temperature before pumping it out, however the other problem is much greater and makes this approach not practical. $\endgroup$
    – Himmators
    Apr 22, 2016 at 12:21
  • $\begingroup$ Water desalination has btw been done in batch, called multi-stage flash desalination, I now understand why. ;) $\endgroup$
    – Himmators
    Apr 22, 2016 at 12:23
  • $\begingroup$ Sugestion: From your comment, I'm not sure you understand multi-stage flash desalination. Understand heat pumps, a lot of the relevant thermodynamics are there. There are many different solar desalination schemes, it's likely that the world needs actual implemantation more than yet another idea. I also think that you have some way to go before you can improve on the state of the art. So go you, and good luck! $\endgroup$
    – mart
    Apr 22, 2016 at 20:48

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