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How to calculate heating load based on heating degree hours?

asked 2015-07-28 17:18:59 -0500

georges's avatar

updated 2015-08-03 19:55:30 -0500


I am interested in explanation on how to calculate the heating load of a single room, if we know the annual heating degree hours? But doing this by hand, not using EnergyPlus, DOE or similar energy analysis application. I just want to understand the principle for now.

Is this is the right approach:

heating load (per hour) = total heat loss (per hour) * heating degree hour

For the sake of simplicity let's say that our single room is made of bricks (both four walls, ceiling and floor), that it has no windows, nor doors. The room is a brick of dimensions 3x3x3 meters. It has no rooms around it, basically it's one room house. The thermal transmitance coefficient (U) for brick wall = 0.0026 kW/m2*C Let's also assume that there is no heat loss through ventilation nor heat gains.

heatLoss through each room wall/ceiling/floor is = 3*3 * 0.0026 = 0.0234 kW/C. For all six "sides" of the room, it is = 6 * 0.0234 = 0.1404 kW/C

Let's assume that for some particular hour during a year, the heating degree hour is 20 C (in winter for example).

Does that mean that for that particular hour, the heating load of our room is: 0.1404 * 20 = 2.808 kW?

Did I understand this correctly?

Thank you for the reply.

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@georges, did you see this question yet? How to describe climate data / calculate degree days? Please search before you post. If the answer is not there, it helps others if you can reference the older post and explain why you need a new question.

Neal Kruis's avatar Neal Kruis  ( 2015-07-28 17:34:56 -0500 )edit

Thank you for your reply Neal. You seemed to misunderstand my question. I am not looking for a way to calculate the heating/cooling degree days. I already have heating (or cooling does not matter) degree hours, and based on this I would like to calculate a heat loss of a certain room, for a particular hour.

georges's avatar georges  ( 2015-07-28 17:53:59 -0500 )edit

@georges My apologies. I didn't read the question thoroughly.

Neal Kruis's avatar Neal Kruis  ( 2015-07-29 09:53:41 -0500 )edit

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answered 2015-07-29 03:38:07 -0500

updated 2015-07-29 04:16:11 -0500

You can indeed approximate heating and cooling loads (losses) by doing a manual calculation, but know that this won't be very accurate since you'll be missing a lot of parameters (for example solar heat gains, radiation, latent loads...).

Anyways, your basic loads are as follows:

  • Envelope losses - from floors, walls, ceilings and windows - that can be calculated using q = UA.dT.
  • Infiltration losses, which basic equation is the sensible heat equation q = cp.ρ.Q.dT (where Q is the volumetric airflow rate m3/s, cp is the specific heat capacity of air)
  • Potential ventilation loads (same basic equation as infiltration)

It's always very useful to know those basic equations by heart, so that you can double check any more complicated model (such as a modeling software) outputs. For example, double checking the heating and cooling energy consumption of constant volume dedicated outside air system (DOAS) is really easy using heating and cooling degree days: using a very basic dimensional analysis you can figure out how you're supposed to plug the heating degree days in there to derive kWh/year (I can help if you want). It's also a perfect application since it should be fairly accurate.

I suggest you read up Chapter 17 and 18 of the ASHRAE Handbook of Fundamentals (they are on a four year cycle so anything like 2009, 2013), and you can also see the Manual J calculations from from ACCA.

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Thank you for the reply Julien. I am not working on an actual project. This only an exercise to see whether or not I have understood the basic principles. I do have the ASHRAE fundamentals 2009,but nor chapters 17, or 18 are of much of the help as they explain the whole heating/cooling load process in detail with tens of factors, coefficients, buffers, which I do not not need,at the moment.They also do not explain the way calculation of heating load could be calculated by the use of heating degree hours. So please can you read my initial reply, and tell me whether or not it is correct? Merci.

georges's avatar georges  ( 2015-07-29 04:56:49 -0500 )edit

I gave you the answer. Q = U.A.dT where dT = (Tin-Tout) in case of heating. So if you have calculated your heating degree hour with the correct Tin, then yes, this is your CONDUCTION loss through the exterior surfaces.

Julien Marrec's avatar Julien Marrec  ( 2015-07-29 11:53:55 -0500 )edit

Thank you Julien. I think I got confused because deltaT in your formula is Tindoors-Toutdoors, while in case of HDH (deltaT=HDH) would be TbaseTemperature -Toutdoors. Where TbaseTemperature is a few degrees less than Tindoors.

Does this mean that for this simplified example, the conduction loss through the exterior surfaces is basically the heating load for that room (again if we neglect ventilation, heat gains, and other mentioned)?

georges's avatar georges  ( 2015-07-29 12:59:32 -0500 )edit

No, as I said, you need to at least include infiltration, as all buildings even without ventilation have significant infiltration losses. For example, you should try to find typical air changes per hour for your type of buildings, there's plenty of literature out there. If sizing is your goal, you should check out manual J, and it's associated excel spreadsheet. It's a poor way to do it but used a lot and exposes the basic idea...

Julien Marrec's avatar Julien Marrec  ( 2015-07-29 16:42:11 -0500 )edit

Ok,let's say use rule of a thumb air changes of 0.5 per hour. infiltrationHeatLoss = roomVolume x deltaT x 0.5 x 0.018 .So does this mean that the heating load of a room for some particular hour would equal the heat loss of that room for that particular hour? heatingLoadOfAroom = heatLossOfAroom = sumOfConductionLosses + infiltrationHeatLoss?sumOfConductionLosses = surface1CondutionLoss + surface2ConductionLoss ... where: surface1ConductionLoss = surfaceArea x U x Heating Degree Hour? Again,I only need the calculation of heating load by the use of Heating Degree Hours.

georges's avatar georges  ( 2015-07-29 17:44:42 -0500 )edit

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Asked: 2015-07-28 17:18:59 -0500

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Last updated: Jul 29 '15