/* Add your own MailChimp form style overrides in your site stylesheet or in this style block.
We recommend moving this block and the preceding CSS link to the HEAD of your HTML file. 230 Btu/hour, Seated, Light Work ………………….. 255 Btu/Hour, Heavy Work, Lifting …………………. The program will then calculate how many rooms the heating or cooling system will need to heat or cool. Now we can calculate the heat generation of the fan motors in the evaporator. Cooling Load Calculation Example.

. Q = 1.8 kWh/day. Now we need to calculate the heat load from air infiltration. The walls, roof and floor are all insulated with 80mm polyurethane with a U value of 0.28W/m, U = U value of insulation (we already know this value) (W/m, A = surface area of walls roof and floor (we will calculate this) (m, Temp in = The air temperature inside the room (, Temp out = The ambient external air temperature (, CP = Specific Heat Capacity of product (kJ/kg.°C), m = the mass of new products each day (kg), Temp enter = the entering temperature of the products (°C), Temp store = the temperature within the store (°C), resp = the respiration heat of the product (1.9kJ/kg), time = length of time they spend inside each day per person (Hours), heat = heat loss per person per hour (Watts), lamps = number of lamps within the cold room, wattage = the rated power of the fan motors (Watts), power = power rating of the heating element (kW), cycles = how many times per day will the defrost cycle occur. In ou r example, this block load occurs at 4 p.m., the time when the sum of the space loads for rooms 101 and 102, plus the ventilation load, is the highest. changes = number of volume changes per day, energy = energy per cubic meter per degree Celsius. The last thing we need to do is calculate the refrigeration capacity to handle this load, a common approach is to average the total daily cooling load by the run time of the refrigeration unit. Next we will calculate the cooling load from the product exchange, that being the heat brought into the cold room from new products which are at a higher temperature. This website uses cookies to improve your experience. 4 people x 400 = 1,600. Substituting in the formula. Now If you’re doing this for a real world example then I recommend you use a design software such as the Danfoss coolselector app for speed and accuracy.

. How solved this question, for which application you 1 ton for 1000Sq/ft, Why each cubic meter of new air provides 2kJ/°C ? For this we can the use the formula of: In this cold room evaporator we’ll be using 3 fans rated at 200W each and estimate that they will be running for 14 hours per day. Fruit and vegetables give off carbon dioxide so some stores will require a ventilation fan, this air needs to be cooled down so you must account for this if it’s used. Actually rate to 86.7 kWh/day and not 72.27 kWh/day as stated earlier, Q = U x A x (Temp out – Temp in) x 24 ÷ 1000 During this time energy is used but you will not see a temperature change while the product changes between a state of liquid and ice. CTRL + SPACE for auto-complete. Scroll to the bottom to watch the video tutorial. Q= 3 x 4 hours x 100W / 1000 If you were to calculate for a critical load you should use greater precision. Introduction . Shouldn’t the unit be running for (24 – 3*30mins) 22.5hrs instead of 14 or are there any other factors that may have made the unit run lesser hours that I am unaware of? U = The U-Factor is the reciprocal of the Resistance, Glass exposed to the suns rays will allow heat into the building based on the Area (Ft2) of the window, the SC (Shading Coefficient) and its SCL (Solar Cooling Load Factor), SC = Glass Shading Coefficient (Effected by window blinds). Assumes CLF = 1. Sample of Cooling Load Estimation With HAP. 