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This is the first video of Didactic Journey 3, Energy Assessment of Thermal Distribution Systems.
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Specifically, in this video, we are going to talk about underfloor heating.
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Underfloor heating is a heating system that consists of supplying hot water at about 40 degrees Celsius
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through a circuit of pipes embedded in the floor and covered with cement mortar to favor thermal inertia.
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Hot water can be produced with any production system.
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For example, low-temperature boilers, solar panels, or even geothermal energy.
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The heat emitted by the pipes is absorbed by the floor and is emitted into the room by radiation and convention.
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If instead of hot water, cold water is circulated, the underfloor heating could become a cooling floor and be used to air-condition a room in summer.
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Now we are going to carry out a series of simple exercises that will help us to become familiar with the calculation formulas for underfloor heating.
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This first exercise asks us to determine the maximum heat that some underfloor heating can dissipate per unit of time and surface area
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if the interior design temperature is 20°C and the maximum temperature of the floor is 29°C.
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Take into account the transmission coefficient, h equal to 11.5 watts per square meter kelvin.
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The formula that we are going to have to use is the formula that appears on the screen.
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In this formula, we will take into account that Q is the power per unit area in watts
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per square meter, TMS is the average temperature of the floor's surface, Ti is the interior
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temperature of the premises that has been taken into consideration in the design and H represents
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the heat transmission coefficient of the floor by radiation and conduction which usually ranges
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between 10 and 12 watts per square meter Kelvin. Therefore this first exercise is very easy since
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we simply have to replace each of the unknowns by the values in our statement. H that is the
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transmission coefficient is 11.5. The interior design temperature is 20 degrees Celsius and
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the maximum floor temperature is 20 degrees Celsius. If we put this data in order, we will
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get that Q is equal to 11.5 multiplied by the difference between 29 and 20, which will give us
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a value of 103.5 watts per square meter. This second exercise we are going to do is a little
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more complex. The statement says the underfloor heating of a room of 30 square meters has a
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thickness of 1.5 centimeters of stoneware tile whose transmission coefficient is lambda 1 equal
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to 1.15 watts per meter. If the transmission coefficient is h equal to 11.5 watts per square
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meter kelvin and interior design temperature is 23 degrees celsius which will the average
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water temperature be if the thermal load to be covered is 2300 watts? The first thing to know
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is that the thermal demand of the room can be achieved by modifying the supply water temperature
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appropriately. Normally, we will consider that there is a difference of 10 degrees Celsius between
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the supply water and the return water. The average water temperature can be found from the
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from the thermal demand of the design temperature and the transmission coefficient
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Ka, the surface capacity of the floor. This value can be determined if we know the conductivity of
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the material and the transmission coefficient of the floor H. That is, we are in this case,
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we need to calculate Ka, and for this we are going to need x1 and lambda1, which are two
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data from this statement. Since x1 is the thickness of the layer and λ1 is the conductivity
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coefficient of the material, as we already knew, h is the transmission coefficient of the floor by
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convention and radiation. Therefore, to calculate the average temperature, we can calculate it with
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this formula. Q is equal to Ka, that is the transmission coefficient, multiplied by the
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average water temperature minus the flow temperature and from here we would clear the
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average water temperature. Let us now carry out the exercise analytically. The first thing we have
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to do is to clear the transmission coefficient Ka from this formula as shown on the screen.
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Therefore our formula will be as follows. One divided by the thickness of the tile which if we
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notice is in centimeters in the statement but it always has to be in meters divided by the
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transmission coefficient in watts divided by meters plus one divided by the transmission
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coefficient which is 11.5 watts per square meters kelvin and this will give us a result of 10
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watts per square meter Kelvin. Once we have calculated the transmission coefficient we can
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move on to the following formula. In this formula what we are interested in clearing as this is the
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the question asked in the exercise is the average temperature of the water. Q is equal to the
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transmission coefficient multiplied by the average temperature of the water minus the flow temperature.
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The thermal load that we need to cover is 2,300 watts.
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However, it tells us that the room has a surface area of 30 square meters and therefore we will have to divide 2,300 watts by the 30 square meters.
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The coefficient is the one we had calculated previously.
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The average temperature of the water is the unknown we are looking for and the interior design temperature is 23 degrees Celsius indicated in this statement.
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Therefore, if we clear the average temperature of the water, it will be 76.66, which is the result of the division that appears here by 10 plus 23,
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which will give us an average temperature of the water of 30.66 degrees Celsius.
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With this, we have finished the second exercise.
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The next exercise asks us to determine the water flow rate that is needed if the thermal load to overcome is 2,300 watts.
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To determine the water flow rate required for underfloor heating, the heat emitted per second and per square meter must be known.
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And this value has to be able to meet the thermal demand from the thermal gap between the supply and return water.
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If we work in the international system, Q will be in kilobats.
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The specific heat of the water will be 4.18 kJ per kg Celsius and the mass flow rate will be obtained in kilograms per second.
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The thermal gap will be 10 degrees Celsius.
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this way we can state that if we divide the thermal load in kilobats by 41.8
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we can obtain the flow rate in kilograms per second.
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With this simple formula we will be able to calculate the water flow rate of the exercise
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and therefore the result of the exercise that appears on the screen will be 2.3 kilobats divided by 41.8
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will give us a mass flow rate of 0.055 kilograms per second.
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This is the last exercise we are going to see in this video.
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We are told a flat is to be heated without the floor heating.
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The dining room is of 20 square meters.
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The large bedroom is 12 square meters.
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The small bedroom is 10 square meters and the bathroom is 4 square meters.
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We are asked to calculate the length of the water circuit in each room.
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As we know that the thickness is 20 cm and the average distance is 5 m,
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to calculate the length of the water circuit, we will have to use the formula that appears on the screen.
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L is equal to A, which is the surface area, E, which is the thickness, plus 2 times L,
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which is the distance in meters between the collector and the area to be heated.
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As a guideline, if the thickness is 20 cm and the length is 5,
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The above formula can be simplified as L is equal to the area divided by 0.2 plus 10.
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Therefore, let's carry out the exercise.
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To calculate the dining room, we know that the area is 20 meters, the thickness 20 centimeters,
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which I remind you that it must be converted into meters, plus twice length 5.
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Therefore, the dining room gives us a length of 110 meters.
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The bedroom will be the area in meters divided by 0.2 plus 10, 70 meters.
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The small bedroom, 10 square meters, 0.2 plus 10, 60 meters.
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And the bathroom, 4 square meters divided by 0.2 plus 10, 30 meters.
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All these exercises contain formulas which will be essential to know when carrying out a complex underfloor heating exercise
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like the one we will see in the next video.
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I hope you enjoyed this video.
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See you in the next one.
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- Idioma/s:
- Subido por:
- Nuria Esther J.
- Licencia:
- Reconocimiento - No comercial - Sin obra derivada
- Visualizaciones:
- 6
- Fecha:
- 20 de mayo de 2024 - 17:38
- Visibilidad:
- Clave
- Centro:
- IES ANTONIO MACHADO
- Duración:
- 09′ 35″
- Relación de aspecto:
- 1.78:1
- Resolución:
- 1280x720 píxeles
- Tamaño:
- 33.36 MBytes
