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Estimating the actual heat output of a 8KW diesel heater. The following is a calculation summery. 1 Calculate heat output based on air flow and temperature rise Output air temperature = 172oF Input air temperature = 54oF Temperature rise 118oF Measured inlet air diameter = 2.728” Area of inlet air = πr2 =π(2.728/2)2=5.845 in2 Measured air speed = 2000 ft/min Cubic feet of air per minute = (2000 ft/min )(5.845 in2/144 in2/ft2)=81.2 CFM Estimated heat output =(1.08)(CFM)(Temperature rise) =1.08(81.2CFM)(1180F)=10,348 BTU/Hr 2. Estimate fuel consumption rate in liters per hour Fuel level at start of test = 270 ml Fuel level at end of test = 45 ml Fuel consumed during test= 225 ml Length of test period = 30 min 49 sec = 30.817 min Estimated fuel consumption in liters per hour ((60 min/Hr)/30.817min)(225 ml)=438 ML/Hr = .438 L/Hr 3. Estimated burner efficiency The heat content of diesel is listed as 137,380 BTU/gal (BTU per Gallon). When diesel is burned steam is created which removes some of the heat energy so the usable heat is reduced to 128,450 BTU/gal. There is also lost energy from heating the combustion air. 1 US Gallon = 3.785 liters The total amount of heat released by the heater including combustion gas loss is as follows; ((.438 L/Hr)/3.785 L/gal)(128,450 BTU/gal)=14,864 BTU/Hr Efficiency =( (10,348 BTU/Hr)/(14,864 BTU/Hr))(100) =69.6% 4. What does the 8Kw unit rating mean? When I examined the owner’s manual for the diesel heater, I found listed in a table of Technical Parameters that the unit had a “Calorific value (W)” equal to 8000. When I looked up the definition of Calorific value I found the following; “Calorific value is the amount of heat energy present in food or fuel which is determined by the complete combustion of specified quantity at constant pressure and in normal conditions. It is also called calorific power. The unit of calorific value is kilojoule per kilogram i.e. KJ/KG.” If the 8KW is a calorific value then it is not a indication of the heater output in terms of units of heat per unit of time. I would expect the heat energy to be in KJ/Kg units but there is no reason the calorific value could not be expressed as watts/KG or KW/Kg. So perhaps the owner’s manual should have listed the calorific value of the unit as follows; Calorific value = 8000 w/Kg We know that one watt is approximately equal to 3.412 BTU’s. So now we can express the Calorific value as; Calorific value = 8000 w/Kg x 3.412 BTU/w= 27,296 BTU/Kg On a Google lookup I found 1 kilogram of diesel is equal to 1.1628 liters. Now if we divide the calorific value by 1.1628 we have colorific values in BTU’s and liters. Calorific value= (27,296 BTU/Kg)/1.1628L/Kg= 23,474.4 BTU/L Finally the owner’s manual says the heater consumes a max of 0.55 L/Hr. If we multiply the calorific value by 0.55 we should obtain the expected heat output of this heater in BTU’s/Hr. Expected heat output =( 23,474.4 BTU/L)(0.55 L/Hr)= 12,910.9 BTU/Hr But wait! In my tests the heater consumed 0.438 L/Hr not 0.55 L/Hr. My expected heat output will be as follows; Expected heat output =(23,474.4 BTU/L)(0.438 L/Hr) = 10,281.8 BTU/Hr. This number above is very close to the estimate of heat output that was done previously using the air flow rate and temperature rise of the air as it passed through the heater. That test measured an output heat of 10,348 BTU/Hr. If my assumptions are correct, when the heater manufacturer indicates his unit is rated at 8KW this number is useless for telling you the heat output per unit time until you are also given the fuel consumption rate.