HEAT LOAD ESTIMATING USING THE SHORT COMPREHENSIVE METHOD © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Hello, I would like to show you how I estimate heat loads using the short comprehensive method.  The short comprehensive method for heat load estimating is used where the construction of the structure or use of the space is considered different to the expected average. For example, a living area where there is an unusually large amount of glass in the walls, or where an area will have a large number of people in it for long periods of time. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 To use this method I use some charts and forms. Copies of these are in the Training Room.  This is a chart for calculating heat through the ceiling and the floor. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 This is a chart for calculating heat through the walls. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 This is a chart for calculating heat through the windows. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 This is a form for calculating the comprehensive heat load. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 I have been asked to calculate the heat load for bedroom 1.  Here it is on the plan. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 The building survey gave this information. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Let’s have a look at my calculations. There are a few steps. Here’s the first one  Step 1 is calculating the heat load through the ceiling and floor.  If the floor area is 4 m x 5 m, total floor area equals 20 m².  Use line A on the chart. It shows that the heat load through the floor and ceiling is 200 watts. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 200 watts is the first sub-total on the form for calculating comprehensive heat load. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Step 2 is calculating the heat load through the internal and external walls.  The internal wall length is 4 m + 5 m, equalling 9 m  Use line A on the chart. It shows that the heat load through the internal walls is 360 watts.  Use line A on the chart. It shows that the heat load through the internal walls is 360 watts.  The external wall length is 4 m + 5 m, equalling 9 m.  Use line D on the chart. It shows that the heat load through the external walls is 600 watts. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 360 watts is the second sub-total on the form for calculating comprehensive heat load. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 600 watts is the third sub-total on the form for calculating comprehensive heat load. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Step 3 is calculating the heat load through the windows.  Total window area for the bedroom is 1.8 x 1.2 x 2, equalling 4.32 m².  If the windows had been either west or south facing, then multiplying or dividing factors would need to be applied. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Use line C on the chart. It shows that the heat load through the windows is 1350 watts © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 1350 watts is the fourth sub-total on the form for calculating comprehensive heat load. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Step 4. The room to be air conditioned is a bedroom. Therefore 400 watts is added to the load to compensate for the heat produced by the people sleeping in the room.  400 watts is the fifth sub-total on the form for calculating comprehensive heat load. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 So the total cooling capacity for this application is the sum of all of the sub-totals calculated.  Total heat load= totalling 2910 watts.  This means that an air conditioning system, which has a cooling capacity of 2910 watts, will provide satisfactory performance in this application. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 Once the heat load has been estimated, the next step is to select an appropriately sized air conditioner. To do this I need information from the air conditioner manufacturers. Most manufacturers will have a brochure that contains the specifications of a range of systems. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 We have to be careful when recommending the right size air conditioning system.  When a heat load estimate is calculated it provides a guide to the required cooling capacity for a system which will create satisfactory conditions within the air conditioned space. This calculation is made for a given set of conditions, and usually assumes an ambient temperature of about 32 °C and a relative humidity of about 65%.  An over sized system will not always provide satisfactory performance. In some cases, a slightly smaller system may be more desirable. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence

 An over sized system will rapidly cool the area without allowing sufficient time for de-humidification (moisture removal). Frequent cycling of the system on and off by the thermostat can also result in more wear and tear than if the system runs steadily for longer periods.  On the other hand an under sized system will be incapable of providing comfort during times of extreme heat. © Commonwealth of Australia 2010 | Licensed under AEShareNet Share and Return licence