Download HEAT GAIN CALCULATIONS

HEAT GAIN CALCULATIONS
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.
• 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).
• 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.
• 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.
• 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.
Example
• For a typical New York City winter day with
High = 40F and Low = 30F, the Average
Temperature = 35F.
• For that day HDD = (65 - 35) = 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
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 Uvalue for a building they provide a means
of roughly estimating the amount of energy
required to heat the building over that
period.
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.
• 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 1000.
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
• 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
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
.
• 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
• 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
• http://www.degreedays.net/
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.
Heating degree days:
Definition:
Difference between
outside and inside temp.
averaged over the day.
• If the average temperature is 55 degrees (a 24
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?
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
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
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.
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)
HEAT GAIN AND COOLING
LOAD
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.
Differences between Space Heat Gain and Space Cooling Load
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
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.
• http://personal.cityu.edu.hk/~bsapplec/
heating.htm
• For fixed conditions of outdoor/indoor
temperatures, latitudes, etc. Corrections and
adjustments are made if the conditions are different
Cooling Load Factor
• The cooling load factor is defined as:
• CLF is used to determine solar loads or
internal loads.
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
• The corresponding space cooling load
Qfs is:
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
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
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
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.
• 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:
• 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
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
• 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
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
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.