CONSTRUCTION SOLUTIONS WITH THERMAL INSULATION
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Mallor, products for building elements with base of high density expanded polystyrene
combined with two layers of waterproof mortar and fiberglass of 100gr/m2, set that gives an
ideal thermal insulation and, at the same time, an ideal support for ceramic coatings,
microelements and thematic mortars.
ENERGY SAVING WITH THE COATING SYSTEM OF MALLOR’S SWIMMING POOL-SPA.
In the energy study related to the consumption of heat generation and loss in vertical and
horizontal closures and in the water surface, we obtain a comparative for two different cases:
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Reinforced concrete HA-25 basin, without perimeter insulation to avoid heat loss by
transmittance of closures, with layer of 3mm thickness of waterproofing, layer of grip
mortar of 3 mm thickness ready to receive ceramic coating, outside coating of
ceramic mosaics.
Basin with insulation of 45 kg/m3 of 15cm, waterproofing with mortar based on
cement PB Imper of 2 mm thickness, layer of grip mortar of 3mm thickness ready to
receive ceramic coating, outside coating of ceramic mosaics.
Accoring to Appedix E of the Basic Document of Energy Saving exposed in the Technical Code
of building, it is exposed that:
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20
Transmittance of basin of CONCRETE ------------ 2,32 w/m k
20
Transmittance of base with COATING ---------- 0,197 w/m k
20
Water surface -----------------------------------------0,51 w/m k
EXAMPLE:
For a basin with the following size: 5,5 x 3,2 meters and a depth of 0,90 meters we obtain a
comparative where we can see the heat loss that one or another basin undergoes, obtaining
this way an idea of the energy saving.
Surface exposed to the heat loss
NET LOSS
CONCRETE BASIN
COATED BASIN
WATER SURFACE
TOTAL LOSS:
CONCRETE
COATED
33,76 m
20
2
2
2,32 w/m k X 33,76 m = 77,16 w/ºk
20
2
0,197 w/m k x 33,76 m = 6,67 w/ºk
20
2
0,51 w/m k x 17,60 m = 8,98 w/ºk
77,16 + 8,98 = 86,14 w/ºk + 10% loss in pipe net = 94,75 w/ºk
6,67 + 8,98 = 15,65 w/ºk + 10% loss in pipe net = 17,21 w/ºk
Then if we consider that the ideal temperature for a SPA is 34ºC we obtain an energy expense
for 8-hour work period:
CONCRETE
COATED
3,7 kw/h heat pump with 15kw power x 8 hours = 29,6 kw
17,21 x 34ºC/1000 = 0,58 kw x 8 hours = 4,68 kw
Energy saving:
CONCRETE 29,6 kw x 0,148766€ = 4,40€/day x 30 days = 132 €
COATED
4,68 kw x 0,148766€ = 0,69€/day x 30 days = 20,90 €
This is the expense to keep 8 hours the basin at a constant temperature of 34ºC. We will have
to take into account further treatment to keep the temperature of the water the rest of the day.
Recovery
Side coating of swimming pool: 19,33€/m2 x 378,74 = 7321,05 RRP
Expense Concrete
132€/month
Expense Coated
20,90€/month
Diference: 111,11€
7321,05/111,11 = 66 months = 5,5 years
CONCLUSION
With MALLOR COATING we will need much less power of boiler for the maintenance of the
temperature inside the basin because the loss by heat transmission of the closure is less than
the concrete basin.
PROCEDURE TO DETERMINE IF A SWIMMING POOL IS EFFICIENT.
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Exterior diameter of the pipe
Liquid temperature, maximum or minimum.
Surrounding conditions where the pipe is installed, like dry temperature, minimum or
maximum respectively, air speed.
Thermal conductivity of the insulating material, both for pipes and for the pool basin,
that is pretended to use to the average temperature of working of the liquid.
COMPLEMENTARY ACTIONS FOR AN EFFICIENT SWIMMING POOL.
We should take into account the placing of a cover to prevent heat loss throughout the water
surface, it must be protected with thermal barriers against the heat loss of water by evaporation
for the time being in service. This cover will give us a saving in conditioning and moisturizing.
The distribution of heat for water heating and surrounding conditioning of the swimming pool
would be independent from that of other facilities.
In covered pools, a part of the energy needs of the water heating would be made by means of
the incorporation of collecting, storing and use of solar power systems, fulfilling that way the
exigency of the Basic Document about Energy Saving of the Technical Code of building.
Another aspect to be taken into account is the insulation of the pipe net that the pool has. To
determine this insulation, we must check the Regulation of Thermal Installation of Buildings,
where there will be a series of charts of minimum thickness according to the basin situation.
JUSTIFICATION OF TRANSMITTANCES CALCULATION.
CONCRETE
Concrete wall
Waterproofing
Grip layer
Ceramic coating
β = 1,2 w/mºk
β = 1,2 w/mºk
β = 1,2 w/mºk
β = 0,7 w/mºk
Rt = 0,04 + 0,3/1,2 + 0,003/1,2 + 0,003/1,2 + 0,003/0,7 + 0,13 = 0,43
20
U = 1/Rt = 2,32 w/m k
COATED INSULATING
Concrete wall
β = 1,2 w/mºk
Insulation
β = 0,032 w/mºk
Waterproofing
β = 1,2 w/mºk
Grip layer
β = 1,2 w/mºk
Ceramic coating
β = 0,7 w/mºk
20
Rt = 0,04 + 0,3/1,2 + 0,15/0,032 + 0,003/1,2 + 0,003/1,2 + 0,003/0,7 + 0,13 = 0,197 w/m k