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    HEATING DEGREE DAYS

    ANKUSH GAUR

    13M701

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    # Heating degree day(HDD) is a measurement designed to reflectthe demand for energy needed to heat a building.

    # It is derived from measurements of outside air temperature.

    # The heating requirements for a given structure at a specificlocation are considered to be directly proportional to the number ofHDD at that location.

    # A similar measurement, cooling degree day (CDD), reflects theamount of energy used to cool a home or business.

    What is HDD ?

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    # Heating degree days are defined relative to a base temperature

    # Base Temperature=the outside temperature above which a building needsno heating.

    # The most appropriate base temperature for any particular building dependson the temperature that the building is heated to, and the nature of thebuilding (including the heat-generating occupants and equipment within it).

    # The base temperature is usually an indoor temperature which is adequatefor human comfort(internal gains increase this temperature by about 1 to2C).

    # For calculations relating to any particular building, HDD should be selectedwith the most appropriate base temperature for that building.

    # However, for historical reasons HDD are often made available with basetemperatures of 60 F (16 C) or 65 F (18 C)base temperatures that are

    approximately appropriate for a good proportion of buildings.

    How is HDD calculated ?

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    # There are a number of ways in which HDD can be calculated.

    # The more detailed a record of temperature data, the more accurate theHDD that can be calculated.

    # HDD are often calculated using simple approximation methods that use

    daily temperature readings instead of more detailed temperature recordssuch as half-hourly readings.

    FORMULA 1 :

    # This method works satisfactorily if the outside air temperature does notexceed the base temperature. In climates where this is likely to occur fromtime to time, there are refinements to the simple calculation which allowsome 'credit' for the period of the day when the air is warm enough forheating to be unnecessary.

    HDD CALCULATION

    DD = (Tb Tav)

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    HDD CALCULATION contd.

    # This more accurate algorithm enables results to be computed intemperate climates (maritime as well as continental) throughout the year(not just during a defined heating season) and on a weekly as well asmonthly basis

    CONDITION FORMULA USED

    Tmin>Tbase Dh=0

    (Tmax+Tmin)/2>Tbase Dh=(Tbase-Tmin)/4

    Tmax>=Tbase Dh=(Tbase-Tmin)/2-(Tmax-Tbase)/4

    Tmax

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    0

    5

    10

    15

    20

    25

    30

    35

    40

    45

    50

    0 100 200 300 400

    MAX TEMP

    MIN TEMP

    AVG TEMP

    Base Temperature

    Data Source: NasaPower.larc.gov

    HDD HAMIRPUR 2011

    HDD

    Using Formula 1=399.61

    Using Formula 2=405.15

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    0

    10

    20

    30

    40

    50

    60

    0 100 200 300 400

    MAX TEMP

    MIN TEMP

    AVG TEMP

    Base Temperature

    Data Source: NasaPower.larc.gov

    HDD HAMIRPUR 2012

    HDD

    Using Formula 1=412.78

    Using Formula 2=424.59

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    # 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 NewYork City is 5,050 whereas that for Barrow, Alaska is 19,990. Thus, one can saythat, for a given home of similar structure and insulation, around four times theenergy would be required to heat the home in Barrow than in New York.

    HDD ANALYSIS

    # However, this is a theoretical approach as the level of insulation of abuilding affects the demand for heating.

    # For example temperatures often drop below the base temperature duringnight (daily low temperature in diurnal variation), but because of insulation

    heating is unnecessary.

    # Also, buildings include heat mass such as concrete, that is able to storeenergy of the sun absorbed in daytime. Thus, even if the heating degreedays indicate a demand for heating sufficient insulation of a building canmake heating unnecessary.

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    # HDD provides a simple metric for quantifying the amount of heating that

    buildings in a particular location need over a certain period (e.g. aparticular month or year).

    # In conjunction with the average U-value for a building they provide ameans of roughly estimating the amount of energy required to heat thebuilding over that period.

    # One HDD means that the temperature conditions outside the buildingwere equivalent to being below a defined threshold comfort temperatureinside the building by one degree for one day. Thus heat has to beprovided inside the building to maintain thermal comfort.

    # Say we are given the number of heating degree days D in one year andwe wish to calculate the energy required by a building.

    # We know that heat needs to be provided at the rate at which it is beinglost to the environment.

    HDD USE

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    # This can be calculated as the sum of the heat losses per degree of each

    element of the buildings' thermal envelope (such as windows, walls, androof)

    # OR as the average U-value of the building multiplied by the area of thethermal envelopeof the building, or quoted directly for the whole building.

    # This gives the buildings' specific heat loss rate Pspecific, generally given inWatts per Kelvin (W/K).

    # Total energy in kWh is then given by:

    Q = Pspecific24D/1000 [kWh]

    # Note that as total energy consumption is in kWh and heating degree daysare [no. days degrees] we must convert W/K into kWh per degree per dayby dividing by 1000 (to convert W to kW), and multiplying by 24 hours in aday (1 kW = 1kWh per hour). Since one degree temperature difference inCelsius and Kelvin scale are the same, they get cancelled and no

    conversion is required.

    HDD USE contd.

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    # Calculations using HDD have several problems.

    # Heat requirements are not linear with temperature, and heavily insulatedbuildings have a lower "balance point".

    (The balance point is the outdoor temperature at which the heat gain equalsthe total heat losses.)

    # The amount of heating and cooling required depends on several factorsbesides 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=> The amount of wind outside,=> What temperature the occupants find comfortable.

    PROBLEMS

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    # Another important factor is the amount of relative humidity indoors; this isimportant in determining how comfortable an individual will be.

    # Other variables such as precipitation, cloud cover, heat index, buildingalbedo, and snow cover can also alter a building's thermal response.

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

    countries.

    PROBLEMS contd.

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    Present degree day formulaFrom the ASHRAE Handbook, 1980 Systems Volume, the generalequation for the modified degree-day method is:

    IMPROVEMENTS

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    # The key to improving the accuracy of the degree day method is theadoption of "system simulation" (i.e., evaluating the heat losses as met bythe building's heating system).

    # To develop the degree-day equation one must calculate the net effect ofthe heat gains and heat losses.

    # We have to evaluate the transmission heat loss, the ventilation heat lossand the internal heat gain simultaneously.

    IMPROVEMENTS contd.

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    Occupied vs. Unoccupied hrs./week

    # The first step in obtaining an accurate degree-day calculationprocedure which evaluates the effects of the internal heat gains is the

    separation of the various occupancy periods.

    # For example, most buildings are only utilized 40 hours per week, whichmeans that often the internal heat gain only affects the transmission heatlosses 40 hours out of 168 hours., or 24 percent of the week. Therefore, theheat loss is unaffected by any internal heat gain 76 percent of the week.

    # We can use this ratio with reasonable accuracy toproportion the degree-days number into "occupied" and "unoccupied'' components:

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    Refences:

    # Wikipedia.org Degree Days, Heating Degree Days

    # Nasapower.larc.gov.inData Source

    # Sustainable Architectural Science - Szokolay

    # A Simplified Degree-day Method for Commercial and Industrial Buildings-ALFRED GUNTERMANN

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    THANK YOU