Re: [EnergyPlus_Support] Outdoor swimming pool

Karen

On Mon, Jan 9, 2012 at 2:10 PM, Jim Dirkes <jim@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

Dear Karen et al,
As I think more about this, several things come to mind:

·         It?s a simple matter to calculate the amount of energy lost by evaporation; pool heater people have needed that information for a long time. It does not matter (to them) whether the energy for evaporation is taken from the air or water; they just need to replace it.

·         For the purposes of pool water heating energy calculations, it doesn?t matter where the energy goes either; the pool water heater must replace it.

·         I reached out to a colleague who is the Engineering Manager for one of the major pool dehumidification (air handler) companies to clarify this, since I did actually keep falling asleep in Thermodynamics class. He says that essentially all of the heat of vaporization is removed from the pool?s water if the pool and surrounding air are the same temperature. (EngineeringToolBox.com says much the same, but my friend explained it better.)

·         For an unheated pool, the evaporation would cool the pool?s water and create some sensible temperature exchange between the pool and surrounding air. It will also change the rate of evaporation.

o   If the pool is heated and indoors, and the heater works properly, that won?t be true because the heater will act to maintain the water temperature at a constant value.

o   If the pool is heated and outdoors, the rate of evaporation will change as the outdoor temp and humidity change.

·         The ASHRAE Handbook and my colleague reminded me about radiant heat transfer to the sky (at night) or from the sky (actually solar energy) for an outdoor pool. These are aspects that I have no clue how to calculate. I hope they balance J.

The Building Performance Team
James V. Dirkes II, P.E., BEMP , LEED AP
1631 Acacia Drive NW
Grand Rapids, MI 49504
616 450 8653

From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Jim Dirkes
Sent: Monday, January 09, 2012 6:34 AM

To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: RE: [EnergyPlus_Support] Outdoor swimming pool

Enlightenment (much) appreciated! Perfect practical example! Thank you!

The Building Performance Team
James V. Dirkes II, P.E., BEMP , LEED AP
1631 Acacia Drive NW
Grand Rapids, MI 49504
616 450 8653

From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Karen Walkerman
Sent: Sunday, January 08, 2012 10:11 PM
To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: Re: [EnergyPlus_Support] Outdoor swimming pool

Hi James,

Engineering ToolBox has a better explanation of pool heat transfer than I think I can give:

http://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html

Also, a little thought experiment - when you get out of a pool, and the surface of your skin is wet, as the water evaporates, you feel cool. Why? The evaporation of the water must be removing heat from you, not just from the air around you. In the case of the pool, there is energy transfer from the pool, TO the evaporating water. If the surrounding space is very, hot then the evaporating water could also absorb heat from the air, but this depends on the space temperature, the pool temperature and the vapor pressure in the space. Exactly how to split the cooling effect between the pool and the space, I'm not sure, but based on my understanding, under normal pool circumstances, most of the cooling effect is applied to the body of water, not to the surrounding air.

--
Karen
On Sun, Jan 8, 2012 at 9:08 PM, Jim Dirkes <jim@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

Karen,
I kept falling asleep in Thermodynamics class, so I am not sure about this?.

How could there be energy transfer to the pool water as a result of evaporation? I am having a hard time imagining it, or at least imagining that it would be more than a tiny amount.

Enlightenment welcome!

The Building Performance Team
James V. Dirkes II, P.E., BEMP , LEED AP
1631 Acacia Drive NW
Grand Rapids, MI 49504
616 450 8653

From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Karen Walkerman
Sent: Saturday, January 07, 2012 5:37 PM

To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: Re: [EnergyPlus_Support] Outdoor swimming pool

The other thing to consider is that the cooling due to evaporation will be split between the pool water and the space. Do you have the thermal mass of the pool water included in the space? This will help stabilize the air temperature. In this case you'd also need to include your pool heat source into your zone heat balance.

Karen

On Saturday, January 7, 2012, Jim Dirkes <jim@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>
>
> Dear Ery,
>
> I think that Evaporation is pretty straightforward; if a pound of water evaporates, it must absorb a certain amount of energy from the air into which it evaporates.
>
> There is no ?efficiency? which affects evaporation from a pool surface. Rather, the key is to calculate the amount of evaporation correctly. I am no expert in this. I use a spreadsheet (from an ASHRAE Journal article, I think) that is supposed to calculate evaporation using the Shah method, which in turn is supposed to be a good method.
>
> The Shah method assumes a static room temperature and humidity, which is probably not true in your case. To account for varying air conditions above an outdoor pool (and the direct impact on evaporation), I recently modified that method in another spreadsheet (It?s the engineer?s tool of choice!). The new spreadsheet calculates evaporation with the varying outdoor air conditions found in a TMY weather data set. Who knows if I?ve applied it properly or whether it is even applicable! Nonetheless, it seems that the approach is defensible so I?m using it for a current project since I have nothing better.
>
> I?ve attached both spreadsheets in case they may be helpful.
>
> (Warning: They are not documented very well and you?ll have to become ?Sherlock Holmes? to figure out what I?m doing.)
>
> If you or any other forum members decide to look at them closely and find that I?ve got an error in there, please let me know!
>
> To keep things straight, I have attached:
>
> 1.       ?IndoorPoolCalc?? This is the Shah method pool evaporation calcs for an indoor pool.
>
> 2.       ?TMY-based Hourly Schedule?.? This is the adaptation for outdoor pools. Beware, there are psychrometric functions included, so you must enable macros. Also, I use it for other purposes and have hidden the inapplicable columns.
>
> 3.       ?OutdoorPoolEvap?.? This is the .CSV file used by a Schedule:File to modify the maximum evaporation rate.
>
>
>
> The Building Performance Team
> James V. Dirkes II, P.E., BEMP , LEED AP
> 1631 Acacia Drive NW
> Grand Rapids, MI 49504
> 616 450 8653
>
>
>
> From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Ery Djunaedy
> Sent: Saturday, January 07, 2012 3:43 PM
> To: EnergyPlus_Support@xxxxxxxxxxxxxxx
> Subject: Re: [EnergyPlus_Support] Outdoor swimming pool
>
>
>
>
>
> Thanks Jim.
>
> I tried the OtherEquipment route before. What I found is that the room gets too cool. My gut feeling is that there is an "efficiency" factor on the evaporative cooling effect that needs to be applied so that the room temperature can be "reasonably" cool. I was just lost trying to determine the efficiency factor.
>
> Any thoughts?
>
> Ery
>
>
>
> On 01/07/2012 08:23 AM, Jim Dirkes wrote:
>
>
>
> Dear Ery,
>
> It sounds as though you want to consider the evaporation from the pool surface as an evaporative cooler. I trust that you have determined the rate of evaporation already and now need to apply that to the space.
>
> Since the evaporation rate in pounds / kg of water per hour is known, you also know the equivalent sensible cooling rate (roughly 1000 BTU / pound of water). Armed with this information on an hourly basis, two possible strategies come to mind:
>
> 1.       Define a water cooling coil which mimics your evaporative cooling effect. (This looks messy because there is no apparent way to directly control the cooling capacity and you will also need to define a chilled water loop and delete it?s energy from your totals.)
>
> 2.       I think past posts have noted that you can define objects such as OtherEquipment with a negative energy value (i.e., cooling). If you also define a fractional schedule that mimics your evaporation rate, you should get the temperature effect represented properly.
>
>
>
> The Building Performance Team
> James V. Dirkes II, P.E., BEMP , LEED AP
> 1631 Acacia Drive NW
> Grand Rapids, MI 49504
> 616 450 8653
>
>
>
> From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Ery Djunaedy
> Sent: Saturday, January 07, 2012 2:20 AM
> To: EnergyPlus_Support@xxxxxxxxxxxxxxx
> Subject: Re: [EnergyPlus_Support] Outdoor swimming pool
>
>
>
>
>
> A question and some thoughts on this topic.
>
> I developed a similar method as Jim described below, but for the purpose of my simulation, this is not enough. The method is used primarily to design a heating system for the pool. My problem is totally the opposite: an indoor swimming pool in the tropics. The swimming pool is inside a semi-enclosed naturally ventilated space. Water heating is not required, as it is assumed that the water is constantly replenished with the right temperature.
>
> The method below will calculate the evaporation rate, which will become the latent heat for the space. This is pretty straight forward as Jim mentioned, but it does not give any free sensible cooling to the room. The crux of the problem that I want to achieve is how to calculate the free cooling from the water. So here is what is missing from this method:
>
> 1. Convection heat to/from the pool. The air temperature is higher than the water temperature, so the room should be cooled some what.
>
> 2. On the evaporation rate calculation, the method assumes that all of the heat to evaporate the water is coming from the water, i.e. the reduction in water temperature. As this evaporation happens in the surface, should not some of the heat come from the air, i.e. there should be a red
>
>