Cooling an ice rink - Load calculation!! 4

[Vicsidhu]

I am designing an ice rink. There is a race track in a stadium (Oval) which is made of materials which has high U-values (blacked out good glass as walls). I have calculated the cooling load, which is quite high as expected. What I have not incorporated is the effect the chunk of ice on the ice track will have to the space. Obviously this will be a negative load to the space.

My question is: How can i determine the negative load this ice race track provides to the room? The referigeration plant is separate and my only concern is the cooling or heating in summer and winters respectiely.

So the parameters for this are:
Ice surface temp: -7°C
Space temp: 17°C
Ice race track Surface area: 5000m²

I need: Load for ice track in kW ot Btu/h

thnx!!

 

[Vicsidhu]

Forgot to mention that the relative humidity in the space will be maintained at 30% RH.

 

[MintJulep]

As a first cut:

UAdT with U being composed of only the convection coefficient between ice and air.

Essentially the ice is an infinite capacity heat sink at -7C temperature.

The ice people must have the opposite problem, right? How much heat from the space goes toward melting the ice. So if the ice system is already designed, and they weren't completely guessing, just ask them.

 

[Vicsidhu]

Thanks MintJulep,

My Total cooling is 2700kW, 1800kW of which is sensible.

So with U=8.33 W/m².°C (RSI=0.12 m².K/W)

I get:

Q=(8.33 W/m².°C)x(5000 m²)x(24° TD)/(1000 W/kW)
= 1000 kW

This seems huge.

 

[waross]

for a reality check, look at the size of the ice plant and try to determine the loading on it.
Yours

 

[MintJulep]

I was going to suggest the same reality check.

 

[Vicsidhu]

Thnx. I will check with the referigeration guys.

Certain other losses come to mind:
-Radiation loss due to temperature difference
-Sublimation (evaporation directly frmo ice)

 

[Warpspeed]

The ice plant will probably be vastly oversized to pull out the latent heat to get the thing frozen to begin with.

But as already mentioned, the steady average power required to keep it running should give you a pretty good idea of the heat balance. Would it be practical to recover some of the rejected heat from the ice plant for space heating ?

 

[imok2]

The effective heat load (including the latent heat effect of convective
mass transfer) is given by the following equation:
Qcv = h(ta – ti) + [K(Xa – Xi)(1226 Btu/lb)(18 lb /mol)]
where
Qcv = convective heat load, Btu/h·ft2
K = mass heat transfer coefficient
ta = air temperature, °F
ti = ice temperature, °F
Xa = mole fraction of water vapor in air, lb mol/lb mol
Xi = mole fraction of water in saturated ice, lb mol/lb mol
When the mole fraction of air is calculated using a relative
humidity of 80% and a dry bulb of 38°F, Xa is approximately
6.6 × 10–3, and Xi for saturated ice at 100% and a temperature of
21°F is 3.6 × 10–3. On the basis of the Chilton/Colburn analogy,
K ? 0.17 lb/h · ft2 (DOE/TIC 1980).
In locations with high ambient wet-bulb temperatures, dehumidification
of the building interior should be considered. This process
lowers the load on the icemaking plant and reduces condensation
and fog formation. Traditional air conditioners are inappropriate
because the large ice slab tends to maintain a lower than normal drybulb
temperature.
From ASHRAE 2002 HANDBOOK, CHAPTER 34, ICE RINKS

 

[DickRussell]

Don't ignore heat transfer downward to ground. Also, if there is no insulation beneath the tubes (such as extruded polystyrene boards), long-term ground freezing problems could be a problem. In one instance I know of, frost heaves below the surface eventually created such a serious problem with leveling and puddling at one end that the rink had to be shut down, piping removed, some excavation done, and the problem fixed. I don't recall if the problem was fixed by insulation or a layer of heating coils well below the refrigeration coils.

 

[sterl]

Year round Ice Rinks all have Sub Floor Heating unless they are built on a ventilated air space...

Rate of heat transfer to ground will not be your difficulty, though. Treat the Ice Surface as a 32-deg. Dew Point impervious membrane for the purposes of doing enthalpy calculations for air stream....

Occupancy and ventilation rate for your buildng will be the dominant factor in AC load...But in terms of contributed moisture, the radiant load (mainly lighting) on the ice surface can cause a series of issues, including contributed humidity. If you don't anticipate dense public skating crowds, or TV level lighting, size your AC and air exchange like any other big-cavity auditorium and take a small benefit for the dew point depression due to the ice slab.

 

[pennpoint]

To all
Ice rinks are not flooded and frozen in one cast. Too much heave (swell). They are frozen in thin layers of water to maintain the near crystal clear effect. Graphics are added at intermeadiate levels. The Zamboni shaves the ice to maintain a flat chip free surface at halftime. It never reaches the graphics level until a new game must be played.




Best regards
pennpoint

 

[Vicsidhu]

I did some calcs, but still have some questions.

My convective heat negative heat load form Ice Track is 654kW. In the same passage in ASHRAE book of refrigeration (Page 34.2), there are some other loads of which radiant seems to be applicable in this situation. There will be radiant loss from the internal loads to ice surface, thus this will be a negative load to my side.

Basically I should be looking at convective and radiant loads, which are 28% and 25% of the total loads. By calculating backwards and considering only these 2 loads, I get radiant load to be 818kW, making the total to be 1472kW.

I wonder if I am right before I put an argument forward to the Refrigeration consultant. Please comment.
----------------------------------------------------
Calculations are included here for reference.
----------------------------------------------------

Calculation for Convective Load:
Qcv = h(ta – ti) + [K(Xa – Xi)(1226 Btu/lb)(18 lb /mol)]
h = 0.6 + 0.00318(25fpm) = 0.6795 Btu/hft².°F

Qcv = 0.6795(62.6°F-19.4°F) + [0.17((6.6x10(-3))-(3.6x10(-3))(1226)(18)
= 29.35 + 11.25
= 40.6 Btu/h. ft²
Qcv = (40.6 Btu/h. ft²) x (55,000ft²)
= 2,233,000 Btuh = 186 tons = 654 kW

Now 654kW will make 44.44% of the total, leaving 55.56% for radiant, which comes to 817.5kW and total is 1,472kW.

 

[waross]

I have been in many rinks were the ice is removed or boarded over in the summer and the building used for anything from concerts to flea markets. This is usually during the season of greates cooling demands. You may consider adding a note to this effect on your worksheets and ignoring the contribution of the ice in your calculations.
respectfully

 

[Vicsidhu]

I have considered different scenarios. Ice Load is considered only from jan to march. Before and after the building has a different pattern, which is more conventional.

Jan-Feb Scenario: 8000 people and 24W/m² lighting.
Rest of the year: 2500 people and 12W/m² lighting