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#1 |
Member
naman doshi
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I am simulating desert cooler in cfx,-steady state
i have a desert cooler inside a room (just a plane section of 0.5*0.5m), (there is no cooler body) and a vent as opening with the ambient condition of air (temp, mass fraction) Q1. now how should i give the condition of the cooler? Q2. how do i give the source for the cooler (to account for the mass of water added in cooler to air)? velocity needed =5m/s (normal to plane direction) temp of air(output of cooler)- known mass fraction of air - known I have calculated mass flow of total air(dry air+moist)= velocity (5m/s)*area of cooler(0.5*0.5m) *(density of moist air at that output condition of cooler) please help |
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#2 |
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Gert-Jan
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This is completely unclear to me. What is a desert cooler?
Do you want to cool the Sahara? Or an icecream? Then it has to a dessert cooler. What question are you trying to answer using CFD? Velocity profile? Temperature? Condensation? Moisture concentration? And a sketch would help a lot... |
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#3 |
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Glenn Horrocks
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In Australia we know this as a Coolgardie safe - https://en.wikipedia.org/wiki/Coolgardie_safe
If you want to model the cooling action by simple empirical relations then this is a simple simulation where you can use a source term (or thermal conditions on a wall) to model the cooling. If you want to make it a little more complex, but still reasonably straight forward you can also track the water vapour in the air. But if you want to model the evaporation process directly, the water flow in the fabric and all those details this will be a very challenging simulation to do in CFX. Definitely not for beginners. So back to Gert-Jan's question: "What question are you trying to answer using CFD?"
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#4 |
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Gert-Jan
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I only know dessert coolers. And then call them fridge.
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#5 | |
Member
naman doshi
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Quote:
An evaporative cooler (also swamp cooler, desert cooler and wet air cooler) is a device that cools air through the evaporation of water. it os used to cool room. like air conditioner but uses water to cool air and then this cool air is sent out inside room. ![]() ans2. i want to get steady condition inside room velocity at each point, water mass fraction, and temperature distribution. |
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#6 | |
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naman doshi
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Quote:
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#7 |
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Glenn Horrocks
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That's easy then. Just use a source term to absorb some heat and another source term to add some humidity to the air (which I presume is modelled as an additional variable or maybe a multicomponent mixture).
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#8 |
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naman doshi
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#9 | |
Member
naman doshi
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Quote:
sir, Value of mass fraction is coming much more than it can be. for eg at 100%relative humidity,40'C ,mass fraction of water(Humidity Ratio or specific humidity) is 0.0488. But I am getting around 0.9 which is impossible. fluid used air and water vapour (both ideal gas) |
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#10 |
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Glenn Horrocks
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Then you are not setting the simulation up correctly. But without any details of what you are doing there is no way for us to know what is wrong. Please attach your output file, and image of what you are modelling and what you expect the results to look like.
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#11 | |
Member
naman doshi
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Quote:
https://drive.google.com/file/d/1NKc...ew?usp=sharing cooler back-taken as opening with only normal velocity 5m/s(black arrows) cooler body(small box) - sources as suggested by you (water (5kg/hour=.0014kg/sec ) + energy(2500 (latent heat of vapourisation of water) *.0014)(kJ/sec) (-ve)) wall- as ambient condition (313k) constant temp cooler walls(sides walls top and bottom) -adiabatic both vents as openings with opening temp and mass fraction to that of the environment (313k and .0187 respectively)(blue arrows) cooler body is the part green highlighted https://drive.google.com/file/d/19-m...ew?usp=sharing contour of mass fraction (updated) https://drive.google.com/open?id=1v8...ZNMccJ10Xhb2eG contour of temp updated https://drive.google.com/file/d/1skx...ew?usp=sharing Last edited by namandoshi; April 18, 2018 at 21:35. |
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#12 | |
Member
naman doshi
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Quote:
sir, I think I m doing some mistake in modelling moist air, I have made a new material basic setting- variable composition mixture of air ideal gas and water vapour ideal gas. thermodynamic state gas mixture properties ideal mixture in default domain - imported my new material basic settings- continuous fluid fluid models heat transfer -thermal energy turbulence- sst transport eq -on for water ideal gas air ideal gas- constrained Should i change to water vapour 25'C? |
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#13 |
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Glenn Horrocks
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You did not appear to attach an image of the mass fractions.
Please attach your output file.
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#14 | |
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naman doshi
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Quote:
mass fraction |
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#15 |
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Glenn Horrocks
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That image is showing a maximum mass fraction of about 0.1, not 0.9. Still, 0.1 is higher than saturation so I suspect that is higher than you are looking for.
Are you sure your mass fraction rate is correct? For instance are you sure all the water is evaporated in the fabric? Most of the coolers I have seen like this have water dripping off the bottom - which suggests not all the water evaporates. Also, could some water be swept away in droplet form and not evaporate?
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#16 | |
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naman doshi
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Quote:
but still, air must not be able to hold mass fraction of .1 at 306K. or should i solve my prb taking 1)water vapour at 25'C and air ideal gas? 0r 2)both air and water vapour @ 25'C |
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#17 |
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Glenn Horrocks
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The different materials will just change the material properties a small amount. So it will not matter too much much of those material models you use. But you should try both and see what difference it makes to check my comment.
You need to think about the physics of this device. If you put more water into the device than the air can evaporate then what happens? Does the air come out saturated and any left over water just drips to the bottom? Then you should make your source terms model this situation. You can define a source term to make the mass fraction 0.0488 (or whatever the saturated mass fraction is), and adjust the heat source term to remove the amount of heat based on the amount of water evaporated. And if you want to go further you should take into account the change in saturation mass fraction due to temperature changes. As you can see, when you start looking into the details of an apparently simple system it starts getting complex.
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#18 | |
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naman doshi
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Quote:
As you can see, when you start looking into the details of an apparently simple system it starts getting complex. i have started simulations with above 2 mixutres will post results soon. |
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#19 | |
Member
naman doshi
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Quote:
mass fraction https://drive.google.com/file/d/1AUF...ew?usp=sharing temp https://drive.google.com/file/d/1jaY...ew?usp=sharing seems like absurd results i have only change materials no boundary conditions were changed. |
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#20 |
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Glenn Horrocks
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You can't just try models at random and hope to get the correct answer. If you want to see the difference between the material models, first look at the density, viscosity, specific heat and thermal conductivities the different models generate. The difference models are just changing these properties.
As for simulation accuracy, that is an FAQ: https://www.cfd-online.com/Wiki/Ansy..._inaccurate.3F In short, you have to validate and verify your model if you want any hope of it being accurate.
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