1 00:00:10,740 --> 00:00:16,539 This is the first video of Didactic Journey 3, Energy Assessment of Thermal Distribution Systems. 2 00:00:17,179 --> 00:00:20,760 Specifically, in this video, we are going to talk about underfloor heating. 3 00:00:21,379 --> 00:00:27,100 Underfloor heating is a heating system that consists of supplying hot water at about 40 degrees Celsius 4 00:00:27,100 --> 00:00:33,880 through a circuit of pipes embedded in the floor and covered with cement mortar to favor thermal inertia. 5 00:00:35,140 --> 00:00:38,719 Hot water can be produced with any production system. 6 00:00:38,719 --> 00:00:44,899 For example, low-temperature boilers, solar panels, or even geothermal energy. 7 00:00:46,240 --> 00:00:53,359 The heat emitted by the pipes is absorbed by the floor and is emitted into the room by radiation and convention. 8 00:00:54,780 --> 00:01:03,500 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. 9 00:01:03,500 --> 00:01:12,500 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. 10 00:01:12,500 --> 00:01:21,500 This first exercise asks us to determine the maximum heat that some underfloor heating can dissipate per unit of time and surface area 11 00:01:21,500 --> 00:01:29,500 if the interior design temperature is 20°C and the maximum temperature of the floor is 29°C. 12 00:01:29,500 --> 00:01:39,780 Take into account the transmission coefficient, h equal to 11.5 watts per square meter kelvin. 13 00:01:39,780 --> 00:01:43,739 The formula that we are going to have to use is the formula that appears on the screen. 14 00:01:43,739 --> 00:01:48,560 In this formula, we will take into account that Q is the power per unit area in watts 15 00:01:48,560 --> 00:01:56,280 per square meter, TMS is the average temperature of the floor's surface, Ti is the interior 16 00:01:56,280 --> 00:02:01,280 temperature of the premises that has been taken into consideration in the design and H represents 17 00:02:01,280 --> 00:02:05,980 the heat transmission coefficient of the floor by radiation and conduction which usually ranges 18 00:02:05,980 --> 00:02:12,879 between 10 and 12 watts per square meter Kelvin. Therefore this first exercise is very easy since 19 00:02:12,879 --> 00:02:19,060 we simply have to replace each of the unknowns by the values in our statement. H that is the 20 00:02:19,060 --> 00:02:26,340 transmission coefficient is 11.5. The interior design temperature is 20 degrees Celsius and 21 00:02:26,340 --> 00:02:31,580 the maximum floor temperature is 20 degrees Celsius. If we put this data in order, we will 22 00:02:31,580 --> 00:02:39,599 get that Q is equal to 11.5 multiplied by the difference between 29 and 20, which will give us 23 00:02:39,599 --> 00:02:45,659 a value of 103.5 watts per square meter. This second exercise we are going to do is a little 24 00:02:45,659 --> 00:02:52,360 more complex. The statement says the underfloor heating of a room of 30 square meters has a 25 00:02:52,360 --> 00:02:59,219 thickness of 1.5 centimeters of stoneware tile whose transmission coefficient is lambda 1 equal 26 00:02:59,219 --> 00:03:08,080 to 1.15 watts per meter. If the transmission coefficient is h equal to 11.5 watts per square 27 00:03:08,080 --> 00:03:14,419 meter kelvin and interior design temperature is 23 degrees celsius which will the average 28 00:03:14,419 --> 00:03:21,340 water temperature be if the thermal load to be covered is 2300 watts? The first thing to know 29 00:03:21,340 --> 00:03:26,879 is that the thermal demand of the room can be achieved by modifying the supply water temperature 30 00:03:26,879 --> 00:03:32,840 appropriately. Normally, we will consider that there is a difference of 10 degrees Celsius between 31 00:03:32,840 --> 00:03:39,919 the supply water and the return water. The average water temperature can be found from the 32 00:03:39,919 --> 00:03:45,219 from the thermal demand of the design temperature and the transmission coefficient 33 00:03:45,219 --> 00:03:51,939 Ka, the surface capacity of the floor. This value can be determined if we know the conductivity of 34 00:03:51,939 --> 00:03:57,060 the material and the transmission coefficient of the floor H. That is, we are in this case, 35 00:03:57,060 --> 00:04:04,939 we need to calculate Ka, and for this we are going to need x1 and lambda1, which are two 36 00:04:04,939 --> 00:04:11,479 data from this statement. Since x1 is the thickness of the layer and λ1 is the conductivity 37 00:04:11,479 --> 00:04:17,839 coefficient of the material, as we already knew, h is the transmission coefficient of the floor by 38 00:04:17,839 --> 00:04:23,220 convention and radiation. Therefore, to calculate the average temperature, we can calculate it with 39 00:04:23,220 --> 00:04:30,480 this formula. Q is equal to Ka, that is the transmission coefficient, multiplied by the 40 00:04:30,480 --> 00:04:35,860 average water temperature minus the flow temperature and from here we would clear the 41 00:04:35,860 --> 00:04:41,480 average water temperature. Let us now carry out the exercise analytically. The first thing we have 42 00:04:41,480 --> 00:04:47,019 to do is to clear the transmission coefficient Ka from this formula as shown on the screen. 43 00:04:47,680 --> 00:04:53,980 Therefore our formula will be as follows. One divided by the thickness of the tile which if we 44 00:04:53,980 --> 00:05:01,279 notice is in centimeters in the statement but it always has to be in meters divided by the 45 00:05:01,279 --> 00:05:09,420 transmission coefficient in watts divided by meters plus one divided by the transmission 46 00:05:09,420 --> 00:05:16,839 coefficient which is 11.5 watts per square meters kelvin and this will give us a result of 10 47 00:05:16,839 --> 00:05:24,139 watts per square meter Kelvin. Once we have calculated the transmission coefficient we can 48 00:05:24,139 --> 00:05:30,819 move on to the following formula. In this formula what we are interested in clearing as this is the 49 00:05:30,819 --> 00:05:37,220 the question asked in the exercise is the average temperature of the water. Q is equal to the 50 00:05:37,220 --> 00:05:43,300 transmission coefficient multiplied by the average temperature of the water minus the flow temperature. 51 00:05:43,300 --> 00:05:47,639 The thermal load that we need to cover is 2,300 watts. 52 00:05:48,040 --> 00:05:59,139 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. 53 00:05:59,639 --> 00:06:02,680 The coefficient is the one we had calculated previously. 54 00:06:03,100 --> 00:06:11,399 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. 55 00:06:11,399 --> 00:06:22,379 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, 56 00:06:22,560 --> 00:06:27,819 which will give us an average temperature of the water of 30.66 degrees Celsius. 57 00:06:28,319 --> 00:06:30,519 With this, we have finished the second exercise. 58 00:06:30,519 --> 00:06:38,199 The next exercise asks us to determine the water flow rate that is needed if the thermal load to overcome is 2,300 watts. 59 00:06:38,199 --> 00:06:45,040 To determine the water flow rate required for underfloor heating, the heat emitted per second and per square meter must be known. 60 00:06:45,500 --> 00:06:50,779 And this value has to be able to meet the thermal demand from the thermal gap between the supply and return water. 61 00:06:51,220 --> 00:06:54,379 If we work in the international system, Q will be in kilobats. 62 00:06:54,699 --> 00:07:03,579 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. 63 00:07:04,000 --> 00:07:06,240 The thermal gap will be 10 degrees Celsius. 64 00:07:06,240 --> 00:07:13,040 this way we can state that if we divide the thermal load in kilobats by 41.8 65 00:07:13,040 --> 00:07:16,800 we can obtain the flow rate in kilograms per second. 66 00:07:17,680 --> 00:07:23,079 With this simple formula we will be able to calculate the water flow rate of the exercise 67 00:07:23,079 --> 00:07:32,540 and therefore the result of the exercise that appears on the screen will be 2.3 kilobats divided by 41.8 68 00:07:32,540 --> 00:07:38,379 will give us a mass flow rate of 0.055 kilograms per second. 69 00:07:38,720 --> 00:07:41,420 This is the last exercise we are going to see in this video. 70 00:07:41,699 --> 00:07:45,519 We are told a flat is to be heated without the floor heating. 71 00:07:45,860 --> 00:07:48,040 The dining room is of 20 square meters. 72 00:07:48,199 --> 00:07:50,180 The large bedroom is 12 square meters. 73 00:07:50,319 --> 00:07:54,079 The small bedroom is 10 square meters and the bathroom is 4 square meters. 74 00:07:54,579 --> 00:07:58,019 We are asked to calculate the length of the water circuit in each room. 75 00:07:58,019 --> 00:08:03,399 As we know that the thickness is 20 cm and the average distance is 5 m, 76 00:08:03,759 --> 00:08:08,959 to calculate the length of the water circuit, we will have to use the formula that appears on the screen. 77 00:08:09,399 --> 00:08:15,959 L is equal to A, which is the surface area, E, which is the thickness, plus 2 times L, 78 00:08:16,620 --> 00:08:21,639 which is the distance in meters between the collector and the area to be heated. 79 00:08:21,759 --> 00:08:26,279 As a guideline, if the thickness is 20 cm and the length is 5, 80 00:08:26,279 --> 00:08:34,299 The above formula can be simplified as L is equal to the area divided by 0.2 plus 10. 81 00:08:34,620 --> 00:08:36,639 Therefore, let's carry out the exercise. 82 00:08:37,179 --> 00:08:42,860 To calculate the dining room, we know that the area is 20 meters, the thickness 20 centimeters, 83 00:08:43,840 --> 00:08:48,740 which I remind you that it must be converted into meters, plus twice length 5. 84 00:08:48,980 --> 00:08:52,620 Therefore, the dining room gives us a length of 110 meters. 85 00:08:52,620 --> 00:09:01,039 The bedroom will be the area in meters divided by 0.2 plus 10, 70 meters. 86 00:09:01,759 --> 00:09:07,960 The small bedroom, 10 square meters, 0.2 plus 10, 60 meters. 87 00:09:07,960 --> 00:09:13,779 And the bathroom, 4 square meters divided by 0.2 plus 10, 30 meters. 88 00:09:14,299 --> 00:09:21,899 All these exercises contain formulas which will be essential to know when carrying out a complex underfloor heating exercise 89 00:09:21,899 --> 00:09:24,580 like the one we will see in the next video. 90 00:09:25,220 --> 00:09:27,139 I hope you enjoyed this video. 91 00:09:27,659 --> 00:09:29,220 See you in the next one.