What is Heat Load Calculation?
449590
there are alot of factors involved in heat loads.go to ASHRAE{1981}conditions include storage size,outside surface area,inside dimensions,volume,insulation.ect ect go to what i said here you should  find all you want to know
434523
* The heat content of a chemical system is called the enthalpy (symbol: H)\n    * The enthalpy change (H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.\n    * H = H(products) - H(reactants)\n    * H is specified per mole of substance as in the balanced chemical equation for the reaction\n      The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)\n      1 calorie (1 cal) = 4.184 joules (4.184 J)\n    * Energy changes are measured under standard laboratory conditions\n      25oC (298K) & 101.3kPa (1 atmosphere) \n\nEnthalpy changes are measured in KJ per mole, but experiments on measuring enthalpy changes always measure a temperature change. The temperature change is then converted to enthalpy change using the following formula:\n\nDH    =    m x c x DT\n\nm = mass in grams\n\nc = specific heat capacity in Joules/Kelvin/gram\n\nDT = change in temperature in degree Kelvin\n\nThe experiment  is done by simple calorimeter experiments which involve carrying out the reaction in a heat resistant vessel and measuring the change in temperature. \n\nThis is also a common practical assessment, and you should be aware of the sources of error in your measurements such as heat loss to surroundings, incorrect readings, inaccurate thermometers etc.
848677
I am a HVAC trainer in the USA, and as such I have to explain how to calculate what amounts of energy it takes to do a certain heating jobs.  Looking at your question, I am of the impression that you are asking how much energy it takes to bring the temperature of 100 gallons of water from 74deg F to 90deg F.  I will explain  how to do this calculation, plus ask and answer a question that I feel might be more appropriate to an application. \n\nMost HVAC technicians define a BTU as the amount of heat energy it takes to raise one pound of water one degree Fahrenheit.  Tables will state that there are approximately 8.34 pounds of water in one gallon of water.  Taking this information, we can calculate that there are 834 pounds of water in 100 gallons of water (100 * 8.34 = 834).   So, if we raised 100 gallons of water 1deg F, we will have put 834 BTU’s into that water.  The temperature difference you talk about is equal to the difference between 74deg and 90deg which is 16deg F (90 – 74 = 16).  If one degree requires 834 BTU’s, then 16 degrees will require 16 times that amount, or 13,344 BTU’s (834 * 16 = 13,344).\n\nNow for the real issue, if you have a flow of water entering the slab at 90deg F, and it comes out of the slab at 76deg F, and if the flow rate is …… lets say 8 gallons per minute, than how much energy is the water putting into the slab?  To answer this, we use the same concept as above except we figure the flow rate in pounds per hour and multiply that times the temperature difference.  The process goes like this: (8 gallons per minute) * (8.34 pounds per gallon) * (60 minutes per hour) = 4003.2 pounds per hour.  Now multiply the pounds per hour flow rate times the temperature difference, and we get the energy going into the slab (4003.2 * 16 = 64051.2 BTU’s per hour).  So, the heat pump would have to have a capacity of a little over 5 tons (1 ton is 12,000 BTU per hour) to meet that load. \n\nNow, the energy one puts into a slab should be based on the heat loss of the structure.  It might be that you only have to put 18,000 BTU’s into the slab to offset the heat loss.  The flow rate and temperature difference can be adjusted based on the heat loss.
963248
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.
1186504
A.   thermal efficiency=work done/total amount of heat taken= 3200/1.55x10^4= 0.206\n\nB.  heat discarded=heat taken-work done=15500-3200=12300 J\n \nC. Heat evolved= 15500 J\nthe mass of gasoline is heat/heat evolved per gm.\n=15500/46000\n= 0.3369\n\nD. power output = 63x 3200=201600 J
366921
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
170755
Specific heat 0.13 J/gK
238117
If by "calculate" you mean process the results of experimental meaurements of speed, force, etc., then you measure the initial total energy (potential, kinetic) of the test object, measure all known energy input that occurs along the path (integral of applied external force times velocity dot product), and subtract from the total of those two the final total energy. The result is the energy lost to dissipative processes like friction and air resistance.\nIf instead you use calculate in the sense of predicting the dissipative losses, you need a model of the object (its drag coefficient and the coefficient of friction). If the path is constant-slope and speed and air density are constant, you can calculate the end state analytically because drag force is constant and so is normal force and thus friction. Otherwise you have nonlinearities which require a time-stepped simulation to solve.
476810
http://ushotstuff.com/Heat.Scale.htm
412854
I don't know what you plan on using this information for but I'm sure it it very similar to other steels. For many calculations small errors in the number have little or no practical effect on the answer.\n\nMy heat transfer book gives the following values:\n\nMild steel 36 Btu/(hr)(sqft)(degF/ft) @ 32 degs F\nMild steel 33 Btu/(hr)(sqft)(degF/ft) @ 212 degs F\n\n1% Carbon steel 26.2 Btu/(hr)(sqft)(degF/ft) @ 32 degs F\n1% Carbon steel 25.9 Btu/(hr)(sqft)(degF/ft) @ 212 degs F
909507
The change in energy is change in heat + change in work (dE = q + w)\n\nIn this situation, the system is releasing heat to the surroundings, , so it is exothermic and q < 0.  q = -350 J.\n\nIn this situation, the system is doing work on the surroundings, w < 0.  To calculate work, W = PdV.  In PV work, since J = kg-m^2/s^2 = Pa-m^3, you need to make sure you convert pressure units from atm to Pascals (1 atm = 101,325 Pa), and volume units from liters to m^3 (1000 L = 1m^3).  \nw = -(1.00 atm)(20.36 L - 1.20 L)(101,325 Pa/1 atm)(1 m^3/1000 L) = -1941 J\n\nThe total energy change is dE = q + w = (-350) + (-1941) = -2291 J
725119
I kept reading anyway.\n\nThe heater is already telling you the power consumption.\n900W = 0.9 kW\nTherefore running the heater for one hour uses 0.9 kWH.\n\nYour formula is wrong. Watt/1000  * cost/kWH = cost per hour\n\nFor heaters, it is true that Volts * Amps = Watts.\nTherefore if you know the supply is 120V and the power is 900W, you can calculate that the current load is 7.5 A.
963007
multiply the mass of the water by the specific heat and the (temperature required minus the ambient temperature).that will give you the energy in joules.if you know in how much time you want this temperature to be achieved you can calculate the power as power is nothing but the rate at which the energy is dispensed
671610
How did you get a degree in engineering? I am amazed you have to ask someone how to do this.\n\nIt is energy out divided by energy in. \n\nYou determine the amount of energy in by the amount of gas being burned and its heating value. You determine the amount of energy out by the amount of heat absorbed by the fluid being heated by the heater.
678335
calculate the heating and cooling loads.Calculate the air flow rate knowing inlet and exit temperatures.From the air flow rate size the duct such that the air speed does not exceed 6ft/minute. Look up the temperatures in the comfort zone on your air psychometric chart. Knowing the pressure drop choose your grills such that you lose just enough of the pressure not to influence the flow rate or the comfort air speed mentioned above. Return air plus make up air should be constant approximately. Good luck!
790311
To get the most out of your room air conditioner, you'll need to consider the size of the unit that best suits your needs.\nThe cooling capacity of room air conditioners is measured in British thermal units per hour, or Btu/h. To figure out the maximum cooling capacity for a room, you will first need to calculate the size of the room where the air conditioner will be placed.\nTo do this, simply multiply the width of the room by the length of the room.  Use the following information and formulars to help you with calculating your air conditioning requirements:\n\n\nHeat Load\n\nThe amount of heat generated is known as the heat gain or heat load. Heat is measured in either British Thermal Units (BTU) or Kilowatts (KW). 1KW is equivalent to 3412BTUs.  The heat load depends on a number of factors, by taking into account those that apply in your circumstances and adding them together a reasonably accurate measure of the total heat can be calculated. Factors include: \n\nThe floor area of the room\n The size and position of windows, and whether they have blinds or shades \nThe number of room occupants (if any) \nThe heat generated by equipment \nThe heat generated by lighting \nArea of Room\n\nThe amount of cooling required depends on the area of the room. To calculate the area in square metres:\n\n\nRoom Area BTU = Length (m) x Width (m) x 337\n\n\nWindow Size and Position\n\nIf, your room has no windows, you can ignore this part of the calculation. If, however there are windows you need to take the size and orientation into account.\n\n\nSouth Window BTU = South Facing window Length (m) x Width (m) x 870\nNorth Window BTU = North Facing windows Length (m) x Width (m) x 165\nIf there are no blinds on the windows multiply the result(s) by 1.5.\nAdd together all the BTUs for the windows.\n\nWindows BTU = South Window(s) BTU + North Window(s) BTU\n\n\nOccupants\n\nYou will have to take that into account people who normally working in the space. The heat output is around 400 BTU per person.  \n\nTotal Occupant BTU = Number of occupants x 400\n\n\nEquipment\n\nClearly most heat in a room is generated by the equipment. This is trickier to calculate that you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is probably safer to overestimate the wattage than underestimate it.\n\n\nEquipment BTU = Total wattage for all equipment x 3.5\n\n\nLighting\n\nTake the total wattage of the lighting and multiply by 4.25.\n\n\nLighting BTU = Total wattage for all lighting x 4.25\n\n\nTotal Cooling Required\n\nAdd all the BTUs together.\n\nTotal Heat Load = Room Area BTU + Windows BTU + Total Occupant BTU + Equipment BTU + Lighting BTU\n\nThis is the amount of cooling required so you need one or more air conditioning units to handle that amount of heat.\n\n--------------------\n\nGeneral thumb rule \n\n100 sqr ft area with roof height of 9 - 10 ft will need 1 Ton AC unit.
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.