# Room air conditioning (pressure increase problem)

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 October 8, 2011, 09:20 Room air conditioning (pressure increase problem) #1 New Member   Join Date: Mar 2011 Posts: 5 Rep Power: 8 Hi, I'm trying to simulate heat transfer from house to envoirment and find heat losses through walls, door, window etc..I'm using FloEFD. There is an air conditioner which has inlet volume flow rate with high temperatrure(30oC). I've also added outlet volume flow with same rate. I got 2 issues; 1) When I run the analyze, pressure inside the house drastically increases (to nonsense values). If I increase the outlet flow rate like 2 times higher, pressure inside the house drastically decreases. Then I changed the outlet flow with environment pressure and the issue solved but I'm not sure about the environment pressure has no effect on heat losses. What else can i do to fix this problem? 2) I've also added some radiators and add volume sources(with heat generation rate) to them. Heat conduction in solids active, problem is external(to take into account the effects of wind), time dependent but the sources have no effect on temperature distrubution. I've checked toggle for the source.. Thank you Last edited by unas; October 8, 2011 at 09:23. Reason: grammar

 October 17, 2011, 07:05 #2 Senior Member   Boris Marovic Join Date: Jul 2009 Posts: 502 Rep Power: 16 Hi unas, 1. It is the correct way to define an outlet pressure condition. The pressure condition does work exactly as the real life condition. The heat loss is only by the fluid leaving through the opening with the calculated temperature. This is the way to set it up. The calculation always takes into account to have a balanced mass. That means what enters the simulation has to leave again, but only in terms of mass. If you define a volume flow rate for 30°C inlet and another one as outlet with the same amount of volume flow then you don't consider the same mass as the temperature at the outlet can be different and with a given volume flow rate the mass is different than at the inlet, thus haveing the mass balance not fulfilled and you suck either more or less mass out of your building causing a lower or higher pressure because of the different density of air for the inlet and outlet temperature according to the equation: Mass=Volume*desnity And we all know the desity of air is lower for hotter temperatures. With a pressure opening you satisfy the needs for a mass balance and giving the system a "releasing valve" as the pressure can be released and is not forced to build up by the limitation on how much has to flow out. You could also define inlet and outlet mass flow rate but the recommended way is to use an outlet pressure condition. 2. The heat sources should have an effect. But do you need time dependency? Because this process can take quite long in real life already. The best way to test it is using really high heat source so you get a very high internal temperature and a very low external temperature and neglect the wind in this test. For example having 500°C inside and -100°C outside and you should see the temperature gradient through the wall and then the same for 60°C inside and 0°C outside. All in steady state for the test. Do you want to see how long it takes to cool down or heat up or why are you unsing transient? The steady state should show you that it works and if you need transient it takes longer (not always suitable for a quick test) and make sure to manually specify the manual time step in the calculation control options as you would otherwise calculate very tiny steps per interation and therfore probably don't see much changes over a long calculation time. Usually you get time steps automatically around miliseconds or smaller and with such processes taking up to hours it would cause an extreme calculation time. I hope this helps, Boris

April 11, 2012, 22:52
#3
New Member

lihui
Join Date: Jun 2010
Posts: 12
Rep Power: 9
Quote:
 Originally Posted by Boris_M Hi unas, 1. It is the correct way to define an outlet pressure condition. The pressure condition does work exactly as the real life condition. The heat loss is only by the fluid leaving through the opening with the calculated temperature. This is the way to set it up. The calculation always takes into account to have a balanced mass. That means what enters the simulation has to leave again, but only in terms of mass. If you define a volume flow rate for 30°C inlet and another one as outlet with the same amount of volume flow then you don't consider the same mass as the temperature at the outlet can be different and with a given volume flow rate the mass is different than at the inlet, thus haveing the mass balance not fulfilled and you suck either more or less mass out of your building causing a lower or higher pressure because of the different density of air for the inlet and outlet temperature according to the equation: Mass=Volume*desnity And we all know the desity of air is lower for hotter temperatures. With a pressure opening you satisfy the needs for a mass balance and giving the system a "releasing valve" as the pressure can be released and is not forced to build up by the limitation on how much has to flow out. You could also define inlet and outlet mass flow rate but the recommended way is to use an outlet pressure condition. 2. The heat sources should have an effect. But do you need time dependency? Because this process can take quite long in real life already. The best way to test it is using really high heat source so you get a very high internal temperature and a very low external temperature and neglect the wind in this test. For example having 500°C inside and -100°C outside and you should see the temperature gradient through the wall and then the same for 60°C inside and 0°C outside. All in steady state for the test. Do you want to see how long it takes to cool down or heat up or why are you unsing transient? The steady state should show you that it works and if you need transient it takes longer (not always suitable for a quick test) and make sure to manually specify the manual time step in the calculation control options as you would otherwise calculate very tiny steps per interation and therfore probably don't see much changes over a long calculation time. Usually you get time steps automatically around miliseconds or smaller and with such processes taking up to hours it would cause an extreme calculation time. I hope this helps, Boris
yes,absolutly correct.
Li hui

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