georges's profile - activity

Merci beaucoup Julien.

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.

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)?

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.

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.

Hi.

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.

Thank you Neal.

Hi Neal. Do you happen to know at which depth does Christensen and Burch formula calculate the water mains temperature? Thank you.

Thank you Neal. Do you know whether or not there is an hourly version of the Christensen and Burch formula? I can not use the daily one.

Hi Neal. Thank you for the reply. Would you please take a look at the edited version of my initial reply? Thank you and I apologize once again for not clarifying the issue from the very start.

Thank you for the reply David. Maybe I should have been more precise from the very start: I am actually looking for a ground temperature so that I could calculate the cold (inlet) water temperature for domestic hot water loads. I am quite new to EnergyPlus, in DOE2, this variable is defined as: cold (inlet) water temperature. Does this change your reply? From what I have seen by the link you provided these are mostly "Ground Heat Transfer in EnergyPlus" questions.

EDITED:

Hi, I am quite new to EnergyPlus, and have been using DOE up until now. I want to find out, how does EnergyPlus calculate the Cold (inlet) water temperature? I googled a bit, and looks like the term EnergyPlus uses is: "Water Mains temperature" instead of "Cold water temperature". I found an equation for EnergyPlus 8.3 Water Mains temperature in here: http://bigladdersoftware.com/epx/docs...

The thing that I am confused of, is lack of pipes depth input.

DOE2 uses the following equation to calculate the Cold (inlet) water temperature ("Carslaw and Jaeger semi-infinite medium conduction equations"): www.energy.ca.gov/2013publications/CE...

It has an input for both the pipes depth and soil type. It is also used to calculate the hourly cold water temperature while EnergyPlus one, seems to give only the daily value.

I would be very grateful for any clarification on this would.

Thank you.

EDITED on 10pm (UTC -5:00).