epgep cooling load example 2012

COOLING LOAD CALCULATION

CLTD/GLF method

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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