epgep cooling load example 2012
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COOLING LOAD CALCULATION
CLTD/GLF method
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CLTDs
GLFs
Cooling LoadTemperatureDifferences
Glass Load Factor(include transmission and
solar radiation)
Heat gain through thewalls, floor, and
ceiling
Heat gain throughthe window
+ If there is shading
ShadingCoefficient (SCs)
(GLF) x (SC) !!!
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Cooling load calculation of a single family house using CLTD/GLF method
Floor Plan of the Single Family House
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Example House Characteristics
Roof constructionConventionel roof-attic-ceiling combination U = 0.28 W/(m2·K)
Wall constructionBrick, insulation, gypsum wallboard U = 0.34 W/(m2·K)
Partition wall U = 0.4 W/(m2·K)
DoorsWood, solid core U = 1.82 W/(m2·K)
WindowsClear double-pane glass in wood frames3 mm thick.
U = 2.84 W/(m2·K)The window glass has a 600 mm overhang at the top.Assume closed, medium-color venetian blinds.
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Outdoor design conditionsTemperature of 36 °C dry bulb with a 13 K daily ran geHumidity ratio is 0.0136 kg vapour/kg dry air (23.7 °C wet bulb)
Indoor design conditionsTemperature of 24 °C dry bulbRelative humidity ratio is 50%
OccupancyFour people
Appliances and lights470 W for the kitchen and 50% in the living room
Find the sensible, latent and total cooling load.
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SolutionThe cooling load must be made on a room-by-room basis to determine the proper distribution of air.
For walls, roof and door
)CLTD(UAQ ⋅⋅=&
whereCLTD – Cooling Load Temperature Difference, K
ASHRAE Fundamentals 2001, Ch. 28, Table 1 and 2 according to– orientation– outdoor design temperature– daily temperature range
]W[
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Solution
For windows
whereGLF – Glass Load Factor, K
ASHRAE Fundamentals 2001, Ch. 28, Table 3 and 4 according to– window orientation– type of glass– type of interior shading– outdoor design temperature.
))SC(()GLF(AQ ⋅⋅=& ]W[
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