1. HEAT GAIN CALCULATIONS

2. Heating Degree Day (HDD)
Heating degree day (HDD) are quantitative indices designed to reflect the demand for energy needed to heat a home or business.
These indices are derived from daily temperature observations, and the heating requirements for a given structure at a specific location are considered to be directly proportional to the number of HDD at that location.
A similar index, cooling degree day' (CDD), reflects the amount of energy used to cool a home or business.

3. HDD are defined relative to a base temperature - the outside temperature above which a building needs no heating.
The most appropriate base temperature for any particular building depends on the temperature that the building is heated to, and the nature of the building (including the heat-generating occupants and equipment within it).

4. For calculations relating to any particular building, HDD should be selected with the most appropriate base temperature for that building.
However, for historical reasons HDD are often made available with base temperatures of 65°F (18°C), or 60°F (15.5°C) - base temperatures that are approximately appropriate for a good proportion of buildings.

5. There are a number of ways in which HDD can be calculated: the more detailed a record of temperature data, the more accurate the HDD that can be calculated.
However, most HDD are calculated using simple approximation methods that use daily temperature readings instead of more detailed temperature records such as half-hourly readings.
One popular approximation method is to take the average temperature on any given day, and subtract it from the base temperature. If the value is less than or equal to zero, that day has zero HDD. But if the value is positive, that number represents the number of HDD on that day.

6. HDD can be added over periods of time to provide a rough estimate of seasonal heating requirements.
In the course of a heating season, for example, the number of HDD for New York City is 5,050 whereas that for Barrow, Alaska is 19,990.
Thus, one can say that, for a given home of similar structure and insulation, around four times the energy would be required to heat the home in Barrow than in New York. Likewise, a similar home in Los Angeles, California, whose heating degree days for the heating season is 2,020, would require around two fifths the energy required to heat the house in New York City.

7. Example
For a typical New York City winter day with High = 40F and Low = 30F, the Average Temperature = 35F.
For that day HDD = ( ) = 30.
A month of thirty similar days might accumulate HDD = 900.
A year (including summer average temperatures above 70F) might accumulate an annual HDD = 5000

8. Example of Use
HDD provide a simple metric for quantifying the amount of heating that buildings in a particular location need over a certain period (e.g. a particular month or year). In conjunction with the average U-value for a building they provide a means of roughly estimating the amount of energy required to heat the building over that period.

9. Method
One HDD means that the temperature conditions outside the building were equivalent to being below the temperature required for thermal comfort inside the building by one degree for one day
Thus heat has to be provided inside the building to maintain thermal comfort.
The rate at which heat needs to be provided is the rate at which it is being lost to the environment. The rate at which heat is being lost to the environment for one degree temperature difference is simply the U-value of the dwelling (as calculated by averaging over all components) multiplied by the area of the thermal envelope of the dwelling.

10. For a ten-degree temperature difference it is ten times this amount
For a ten-degree temperature difference it is ten times this amount. If we multiply the rate at which the building is losing heat by the time (in hours) over which it is losing heat we get the amount of heat lost in Wh (or BTU) and, therefore, the amount of heat that has to be provided by the heating sources.
To convert Wh to kWh we simply divide by Therefore if the area of the dwelling’s thermal envelope (walls, roof and floor) is A, its average U-value is U and the number of degree days is D then the amount of heat required in kWh to cover the period in question is just:
A×U×D×24/1000

11. where the factor of 24 is needed to get the value in kWh rather than kW days or simply BTUs in the case of US units.
The same method is used (except you don't need to divide by 1000) for US units with the result being in BTUs rather than in kWh

12. Problems
Calculations using HDD have several problems. Heat requirements are not linear with temperature, and heavily insulated buildings have a lower "balance point".
The amount of heating and cooling required depends on several factors besides outdoor temperature:
How well insulated a particular building is
the amount of solar radiation reaching the interior of a house
the number of electrical appliances running (e.g. computers raise their surrounding temperature)
the amount of wind outside
and individuals' opinions about what constitutes a comfortable indoor temperature .

13. Another important factor is the amount of relative humidity indoors; this is important in determining how comfortable an individual will be.
Other variables such as wind speed, precipitation, cloud cover, heat index, and snow cover can also alter a buildings thermal response

14. Another problem with HDD is that care needs to be taken if they are to be used to compare climates internationally, because of the different baseline temperatures used as standard in different countries and the use of the Fahrenheit scale in the US and the Celsius scale almost everywhere else. This is further compounded by the use of different approximation methods in different countries

16. Cooling Degree Day
Technically, a cooling degree-day is calculated when there is a 1-degree Fahrenheit (F) difference between 65 degrees F and a mean outdoor air temperature of 66 degrees F, on any given day.

17. Heating degree days: Definition: Difference between
outside and inside temp.
averaged over the day.
If the average temperature is 55 degrees (a hour span, from 12:00AM – 11:59PM)…
And the temperature inside of a house is 65 degrees…
What is the difference between outside and inside temp?

18. Calculations: Assume a 65 degree indoor temp. year round
If the avg. temp. in January (31 days) was 30 degrees, what is number of the heating degree days?
Hint: Multiply the difference in indoor and outdoor by the number of days.
Answer:
(65 – 30) * 31 days 
35 degrees * 31 days 
1085 HDD

19. Calculations: Assume a 65 degree indoor temp. year round
If the avg. temp. in June (30 days) was 85 degrees, what is number of the heating degree days?
Hint: Multiply the difference in indoor and outdoor by the number of days.
Answer:
(65 – 85) * 30 days 
-20 degrees * 30 days 
- 600 HDD

21. 6500 HDD zone: What does this mean?
The average temperature each day is subtracted from the indoor air temperature (65 degrees).
A heating degree day is calculated for each day over an entire year.
The total heating degree days over an entire year is roughly 6500 in Mass.

22. Why is HDD important?
Engineers, architects, designers want to know the HDD of a particular climate…
The greater the HDD, the larger a heating system must be.
South has less HDD, therefore only a small heating system is necessary
Most systems are oversized.
Home builders and owners should know about HDD for replacing an older system (which may have been larger than needed)

23. HEAT GAIN AND COOLING LOAD

24. Space Heat Gain and Space Cooling Load
Space heat gain is the rate at which heat enters a space, or heat generated within a space during a time interval.
Space cooling load is the rate at which heat is removed from the conditioned space to maintain a constant space air temperature.
The figure shows the difference between the space heat gain and the space cooling load.
The difference between the space heat gain and the space cooling load is due to the storage of a portion of radiant heat in the structure.
The convective component is converted to space cooling load instantaneously.

25.                                                                                                                   
Differences between Space Heat Gain and Space Cooling Load

26. Cooling Load Temperature Difference (CLTD) and Cooling Load Factor (CLF)
Cooling load temperature difference and cooling load factor are used to convert the space sensible heat gain to space sensible cooling load

27. Cooling Load Temperature Difference
The space sensible cooling load Qrs is calculated as:
where A = area of external wall or roof
U = overall heat transfer coefficient of the external wall or roof.
For fixed conditions of outdoor/indoor temperatures, latitudes, etc. Corrections and adjustments are made if the conditions are different

28. Cooling Load Factor The cooling load factor is defined as:
CLF is used to determine solar loads or internal loads.

29. Space Cooling Loads
Space cooling load is classified into three categories:
1 External Cooling Loads
External cooling loads have the following components:
Solar Heat Gain through Fenestration Areas, Qfes
where As = un-shaded area of window glass
Ash = shaded area of window glass
max. SHGFsh = maximum solar heat gain factor for the shaded area on window glass (Table 4)
max. SHGF = maximum solar heat gain factor for window glass (Table 5)
SC = shading coefficient

30. The corresponding space cooling load Qfs is:

31. 1.2 Conduction Heat Gain through Fenestration Areas, Qfe
The space cooling load due to the conduction heat gain through fenestration area is calculated as:
where A = fenestration area
U = overall heat transfer coefficient for window glass CLTD = cooling load temperature difference

32. 1.3 Conduction Heat Gain through Roofs (Qrs) and External Walls (Qws)
The space cooling load due to the conduction heat gain through roofs or external walls is calculated as:
where A = area for external walls or roofs
U = overall heat transfer coefficient for external walls or roof
CLTD = cooling load temperature difference

33. 1.4 Conduction Heat Gain through Interior Partitions, Ceilings and Floors, Qic
The space cooling load due to the conduction heat gain through interior partitions, ceilings and floors is calculated as:
where A = area for interior partitions, ceilings or floors
U = overall heat transfer coefficient for interior partitions, ceilings or floors
Tb = average air temperature of the adjacent area
Ti = indoor air temperature

34. Internal Cooling Loads
Electric Lighting
Space cooling load due to the heat gain from electric lights is often the major component for commercial buildings having a larger ratio of interior zone. Electric lights contribute to sensible load only. Sensible heat released from electric lights is in two forms:
(i) convective heat from the lamp, tube and fixtures.
(ii) radiation absorbed by walls, floors, and furniture and convected by the ambient air after a time lag.

35. The sensible heat released (Qles) from electric lights is calculated as:
where Input = total light wattage obtained from the ratings of all fixtures installed
Fuse = use factor defined as the ratio of wattage in use possibly at design condition to the installation condition
Fal = special allowance factor for fluorescent fixtures accounting for ballast loss, varying from 1.18 to 1.30
The corresponding sensible space cooling load (Qls) due to heat released from electrical light is:

36. CLF is a function of
(i) number of hours that electric lights are switched on (for 24 hours continuous lighting, CLF = 1), and
(ii) types of building construction and furnishings.
Therefore, CLF depends on the magnitude of surface and the space air flow rates

37. People
Human beings release both sensible heat and latent heat to the conditioned space when they stay in it. The space sensible (Qps) and latent (Qpl) cooling loads for people staying in a conditioned space are calculated as:
where n = number of people in the conditioned space
SHG = sensible heat gain per person
LHG = latent heat gain per person

38. CLF for people is a function of
Adjusted values for total heat is for normal percentage of men, women and children of which heat released from adult female is 85% of adult male, and that from child is 75%.
CLF for people is a function of
(i) the time people spending in the conditioned space, and
(ii) the time elapsed since first entering.
CLF is equal to 1 if the space temperature is not maintained constant during the 24-hour period

39. Power Equipment and Appliances
In estimating a cooling load, heat gain from all heat-producing equipment and appliances must be taken into account because they may contribute to either sensible or latent loads, and sometimes both

40. Loads from Infiltration and Ventilation
Infiltration load is a space cooling load due to the infiltrated air flowing through cracks and openings and entering into a conditioned room under a pressure difference across the building envelope.
The introduction of outdoor ventilation air must be considered in combination with the infiltrated air.
Infiltration and ventilation loads consist of both sensible and latent cooling loads.