Purifying and Air Pipe Method: Energy Waste and Vigor Request for School - Term Paper Example

Ventilation According to the definition given by Energy Saving Trust (2006) ventilation is defined as being the replacement of stale indoor air with fresh air from out door by use of purpose-provided openings and also through cracks and gaps found in the building envelope. The school and gym complex are expected to have good ventilation so as to avoid stale air from building up. In common practice scenario a ventilation rate of between 0.5 and 1.5 air changes per hour (ACpH) is recommended while a higher efficiency value of 0.3 is recommended for the case of Passivehaus. In this taken the lower value of the common practice will be used which is the value 0.5 ACpH. It is recommended by The Energy Saving Trust (2006) that best practice for new dwelling need to use whole house mechanical ventilation with heat recovery (MVHR) system whole label is system 4 in part F in the building regulations. In order for the system to have a high energy saving benefits it will require that the complex design is such that it will have an airtight property of less than 5m3/hr/m2 so as to ensure almost all the air passes via the heat exchanger. In this design MVHR system with dimensions of 1810mm(h) x 1250mm(w) x 480mm(d) weighing 250 kg and having the following recovery performance: supply/extract air flow rate of 350m3/h at 100 Pa; fresh/exhaust air flow rate is 900m3/h; 90% heat recovery rate and fan power of 68w at 200 m3/h Calculating ACpH In order to calculate the ACpH the volumes to be ventilated are to be established from which the ACpH will then be calculated. The ACpH will be 0.5 of the total volume of the complex Volume of school complex The volume is calculated by taking area of the floor times the height of the building. Calculating floor area for the school complex The total school complex area is calculated by neglecting the curved areas School complex width = (9m +9m + 17m + 9m + 8m) = 52m The school complex length up to the medium room = 10m +5m+6m++9m+3m+3m+15m+3m+ 5m = 59m Approximate area of school complex up to medium room = 59m x 52m = 3068m2 Remaining area = {(9m + 8m+ 7m + 8m) x (52)} - {(9m+8m+7m+8m) x (12m +12m+ 12m) = 1664m2 – 1152m2 = 512m2 The total approximated floor area for school = 3068m2 +512m2 = 3580m2 Volume of school complex up to 4m height = 4 x 3580 = 14320 m3 Volume of school complex above 4m height This is the 2 m extra height for the gym area at the school complex Area of the gym area = 20m x20m = 400m2  Volume of the gym above the 4m height = 400m2 x 2 = 800 m3 Total volume of the school complex = 14320 m3+800 m3 = 15120m3 ACpH for school complex = 0.5 x total of school complex volume = 0.5 x 15120 = 7560 m3 Considering that the MVHR system to be used can handle 900m3 per hour The total number required in the school complex = 7560m3/ 900m3 = 8.4 Thus 9 MVHR systems will be needed which will need to be distributed evenly within the complex. Calculating floor area of the sport centre When calculating the area of sport centre the curved area at the waiting area has been neglected while the calculation of other areas is by use of the dimensions given on the drawing. Basket ball court area = 31m x21m = 651m2 Area of rooms adjacent to basket ball court = 13m x (9m+3m+14m+10m) = 13m x 36m = 468m2 Area of rooms adjacent to swimming pool = (6m+6m+8m+3m+6m+7m) x (5m +7m) = 36m x 12m = 432m2 Area of floor surrounding swimming pool = (22m x 31m) – (25m x 15m) = = 682m2 – 375m2 = 307m2 Thus total area of sports centre = 651 +468 + 432 + 307 = 1858m2 Total volume of sports centre = 1858 m2 x 6m = 11148m3 ACpH for sports complex = 0.5 x total of sports complex volume = 0.5 x 11148 = 5574 m3 Considering that the MVHR system to be used can handle 900m3 per hour The total number required in the sports complex = 5574m3/ 900m3 = 6.2 Thus 7 MVHR systems will be needed. Heat Loss Heat loss in different housing categories is determined by calculation of thermal transmittance rate U of the elements interacting with the external environment. U gives the heat amount transmitted in 1 m2 of an element in every 1 °K difference of temperature and the unit given to the quantity being W/m2K. The design of the school and sports complex is such that all the elements meet the best practice requirements given in Part L of the building regulations. In the calculation of the amount of heat lost through an external enclosure of a space the formulae used is The buildings are in the UK and a temperature difference of 240K as to the general practice. In calculating the energy loss through enclosure each external enclosure will have to be analysed separately. The elements to be considered are the flat roofs with U = 0.25, the walls 0.35, glazed windows U = S/G 4.3W/m2K and double glazed (D/G) doors U = 2.5W/m2K, the glass areas enclosing the buildings whose U is taken as 2.5W/m2K (same for doors). Areas of enclosures associated with the school complex Total wall area for school complex including windows and doors Calculating floor area for the school complex School complex width = (9m +9m + 17m + 9m + 8m) = 52m School complex area of wall facing away the sports complex = Full length of complex = 10m +5m+6m++9m+3m+3m+15m+3m+ 5m+9m + 8m+ 7m + 8m = 91m Area up to 4 m height = 91x 4 = 364m2 Also area of the full length of wall facing the sports complex = 364m2 Area of the wall at the side of soft are = total length of wall involved x 4m Total length involved = {(9m + 8m+ 7m + 8m) x2} +52 = 116m  Total area of the wall at the side of soft area = 116m x4m = 464m2 Total area of wall 4m high (including windows and doors) = 364m2 + 364m2 +464m2 = 1192m2 Area of the glass “wall” = School complex width x4m = (9m +9m + 17m + 9m + 8m) x4m = 52m x4m = 208m2 Glass ‘wall’ area for the 2m extra upper area surrounding gym Length around gym = (20x4) = 80m Approximately 7m length of the upper part of gym is not made of glass “wall” Thus glass area = (80m -7m) x 2 = 146m2 The non glass wall area = 7x2 = 14m2 Area of windows and doors From the school complex drawings the total number of windows = 45 windows and 3 doors Height and width of each window is1.5m and 1.8m respectively Thus area of each window = 3.24 Total area of windows in school complex = 3.24 x 45 = 145.8m2 Height and with of each door = 1.8 and 2m respectively Area of door = 3.6m2 Area of the 3 doors in the school complex = 3.6 x 3 = 10.8 m2 Total area of windows and doors = 145.8m2+10.8 m2 = 156.6m2 If total area of wall 4m high (including windows and doors) = 1192m2 Total area of wall 4m high (excluding windows and doors) = 1192 – 156.6 = 1035.4m2 Calculating energy losses the external enclosures for school complex Energy losses from the wall (U wall = 0.35W/m2oK) = Area of external enclosure x U x temperature difference =1035.4 x 0.35 x 24 = 8697.36w Energy loss through glass “wall” (U = 2.5W/m2K) = (146m2 +208m2) x 2.5x 24 = = 354 x 2.5 x 24 = 21240w Energy loss through windows = Area of windows x U x temperature difference =145.8 x4.3x 24 = 15046.56 w Energy loss through doors = 10.8 m2x 2.5x 24 = 648w Energy loss through roof The energy loss through roof is calculated by considering the area of the roof Area of the roof = total area of the school complex floor= 3580m2 Considering U for flat roof = 0.25, and temperature change = 24 Energy loss = 3580 x 0.25 x 24 = 21480w Total energy loss through enclosures = 8697.36w +21240w +15046.56 w+648w+21480w =67111.92w Energy losses in school complex through ventilation Ventilation loss =  The total volume school complex as calculated = 15120m3 The ACpH value can be taken as 1.5 for the whole school complex  Total ventilation loss in school complex =  = 181440w Energy losses at sports complex Calculating floor area of the sport centre In calculating the area of sport centre the curved area of waiting area is neglected and the other areas are calculated using the dimensions given on the drawing. Area of the basket court = 31m x21m = 651m2 Area of rooms adjacent to basket court = 13m x (9m+3m+14m+10m) = 13m x 36m = 468m2 Area of rooms adjacent to swimming pool = (6m+6m+8m+3m+6m+7m) x (5m +7m) = 36m x 12m = 432m2 Area of swimming pool floor and surrounding area = (22m x 31m) = 682m2 = 307m2 Thus total area of sports centre = 651 +468 + 432 + 682 m2 = 2233m2 Heat loss through roof of sports centre complex Area of sports centre roof = total area of sports centre = 2233m2 Considering U for flat roof = 0.25, and temperature change = 24 Energy loss = 2233 x 0.25 x 24 = 13398w Energy losses through door and windows for sports complex From the school complex drawings the total number of windows = 12 windows and 1door Height and width of each window is1.5m and 1.8m respectively Thus area of each window = 3.24 Total area of windows in school complex = 3.24 x 12 = 38.88m2 Height and width of door = 1.8 and 2m respectively Area of door = 3.6m2 Energy loss through windows = Area of windows x U x temperature difference =38.88 x4.3x 24 = 4012.416 w Energy loss through door = 3.6 m2x 2.5x 24 = 216w Calculating area of external enclosures of sports complex Total enclosure length of complex = {(13+21)x2} +[{(9+3+14+10+31) +(36-31)}x2] = 68 +124 = 192m Total enclosure area of sports complex (including windows and door) = 192x 6 = 1152m2 Total enclosure area excluding windows and door = 780m2 – (38.88m2+3.6 m2) = 1109.52 m2 Area of glass enclosure = (9m +3m+ 14m + 10m + 31m) x 6m = 67m x 6m = 402m2 Thus area of wall enclosure = 1109.52 m2 - 402m2= 707.52m2 Energy losses from the wall (U wall = 0.35W/m2oK) = Area of external enclosure x U x temperature difference =707.52x 0.35 x 24 = 5943.17w Energy loss through glass “wall” (U = 2.5W/m2K) = 402 x 2.5x 24 = 24120w Total energy loss through enclosures in sports complex = 13398 + 4012.416+ 216 + 5943.17+ 24120 = 47689.59w Energy losses in sports complex through ventilation Ventilation loss =  The total volume of sports complex as calculated = total area of sports centre x of height sports complex = 2233m2 x 6 = 13398m3 The ACpH value can be taken as 1.5 for the whole sports complex  Total ventilation loss in school complex =  = 160776w Power Requirements The power requirement has been calculated for each property type so as to ascertain the electrical demand for each. As away of ensuring the power demand has been kept to a minimum value all the power consuming appliances will have low energy consumption option. In order to calculate the power demand for each of the categories of power appliances the equation to be used is The total power demand for each area of the sports and school complex will found from the sum of demand for each service. References Cryer S. (2013). Heat loss in building. The royal academy of engineering. Energy Saving Trust Annual Review 2011-2012". Retrieved23 March 2013. Energy Saving Trust Annual Review 2011-2012". Retrieved 21 March 2013. Read More

Areas of enclosures associated with the school complex Total wall area for school complex including windows and doors Calculating floor area for the school complex School complex width = (9m +9m + 17m + 9m + 8m) = 52m School complex area of wall facing away the sports complex = Full length of complex = 10m +5m+6m++9m+3m+3m+15m+3m+ 5m+9m + 8m+ 7m + 8m = 91m Area up to 4 m height = 91x 4 = 364m2 Also area of the full length of wall facing the sports complex = 364m2 Area of the wall at the side of soft are = total length of wall involved x 4m Total length involved = {(9m + 8m+ 7m + 8m) x2} +52 = 116m  Total area of the wall at the side of soft area = 116m x4m = 464m2 Total area of wall 4m high (including windows and doors) = 364m2 + 364m2 +464m2 = 1192m2 Area of the glass “wall” = School complex width x4m = (9m +9m + 17m + 9m + 8m) x4m = 52m x4m = 208m2 Glass ‘wall’ area for the 2m extra upper area surrounding gym Length around gym = (20x4) = 80m Approximately 7m length of the upper part of gym is not made of glass “wall” Thus glass area = (80m -7m) x 2 = 146m2 The non glass wall area = 7x2 = 14m2 Area of windows and doors From the school complex drawings the total number of windows = 45 windows and 3 doors Height and width of each window is1.5m and 1.8m respectively Thus area of each window = 3.

24 Total area of windows in school complex = 3.24 x 45 = 145.8m2 Height and with of each door = 1.8 and 2m respectively Area of door = 3.6m2 Area of the 3 doors in the school complex = 3.6 x 3 = 10.8 m2 Total area of windows and doors = 145.8m2+10.8 m2 = 156.6m2 If total area of wall 4m high (including windows and doors) = 1192m2 Total area of wall 4m high (excluding windows and doors) = 1192 – 156.6 = 1035.4m2 Calculating energy losses the external enclosures for school complex Energy losses from the wall (U wall = 0.35W/m2oK) = Area of external enclosure x U x temperature difference =1035.4 x 0.35 x 24 = 8697.

36w Energy loss through glass “wall” (U = 2.5W/m2K) = (146m2 +208m2) x 2.5x 24 = = 354 x 2.5 x 24 = 21240w Energy loss through windows = Area of windows x U x temperature difference =145.8 x4.3x 24 = 15046.56 w Energy loss through doors = 10.8 m2x 2.5x 24 = 648w Energy loss through roof The energy loss through roof is calculated by considering the area of the roof Area of the roof = total area of the school complex floor= 3580m2 Considering U for flat roof = 0.

25, and temperature change = 24 Energy loss = 3580 x 0.25 x 24 = 21480w Total energy loss through enclosures = 8697.36w +21240w +15046.56 w+648w+21480w =67111.92w Energy losses in school complex through ventilation Ventilation loss =  The total volume school complex as calculated = 15120m3 The ACpH value can be taken as 1.5 for the whole school complex  Total ventilation loss in school complex =  = 181440w Energy losses at sports complex Calculating floor area of the sport centre In calculating the area of sport centre the curved area of waiting area is neglected and the other areas are calculated using the dimensions given on the drawing.

Area of the basket court = 31m x21m = 651m2 Area of rooms adjacent to basket court = 13m x (9m+3m+14m+10m) = 13m x 36m = 468m2 Area of rooms adjacent to swimming pool = (6m+6m+8m+3m+6m+7m) x (5m +7m) = 36m x 12m = 432m2 Area of swimming pool floor and surrounding area = (22m x 31m) = 682m2 = 307m2 Thus total area of sports centre = 651 +468 + 432 + 682 m2 = 2233m2 Heat loss through roof of sports centre complex Area of sports centre roof = total area of sports centre = 2233m2 Considering U for flat roof = 0.

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