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Q: air conditioning- the sizing of split systems for residential applications ( No Answer,   1 Comment )
Question  
Subject: air conditioning- the sizing of split systems for residential applications
Category: Business and Money > Small Businesses
Asked by: kwts-ga
List Price: $2.50
Posted: 27 Jun 2002 05:56 PDT
Expires: 27 Jul 2002 05:56 PDT
Question ID: 34074
the easiest to determine the size of an air conditioning unit for
residential applications, this is usually termed the heat load
expressed in kilowatts of heating or cooling
Answer  
There is no answer at this time.

Comments  
Subject: Re: air conditioning- the sizing of split systems for residential applications
From: alienintelligence-ga on 06 Jul 2002 03:13 PDT
 
Many factors effect a home’s heating or cooling requirement or "load."
A good estimator will measure walls, ceilings, floor space, and
windows to determine the room volumes, and will assess the R-value of
the home’s insulation, windows, and building materials. A close
estimate of the building’s air leakage is also necessary. A blower
door test is the best way to measure air leakage.

A good estimate will also include an inspection of the size, condition
of seals on joints and insulation, and location of the distribution
ducts in forced air systems. The placement of supply and return
registers, should be appropriate for the system type and size.

The orientation of the house also effects heat gain and heat loss
through windows. Overhangs can reduce solar gain through windows.

[INACCURATE METHODS]
How many square feet of living space? This is called
"sizing by square footage" and is the most commonly used
inaccurate method of sizing. A typical value used for air
conditioners is one ton (12,000 Btu/hour) per 500 square
feet (46 m2). This does not take into account differences
among house orientation, insulation levels, design,
construction, and energy efficiency or intended use of the
system.

Another rough method for sizing heating systems involves a
prepared chart such as the one below. You use the chart in
the following way. First, determine the floor area of all
the heated rooms, and the levels of insulation in the
floors, walls, and ceilings. Next, find the category that
best describes the house. Then, multiply both the higher
and lower numbers for heat loss in Btu per hour per square
foot (from the table) by the floor area of the home to give
you a rough range for the heating load.

Home Type or Characteristics ..... Heat loss (Btu/hr/ft2)


1) No insulation in walls, ceilings, or floors; no storm
windows; windows and doors fit loosely .... 90 to 110

2) R-11 insulation in walls and ceilings; no insulation in
floors over crawl spaces; no storm windows; doors and
windows fit fairly tight. ..... 50 to 70

3) R-19 insulation in walls, R-30 in ceilings, and R-11 in
floors; tight-fitting storm windows or double pane windows.
..... 29 to 35

4) "Superinsulated" house with R-24 wall insulation, R-40
in ceilings, and R-19 in floor; tight-fitting storm windows
or double pane windows; vapor barrier sealed carefully
during construction. ..... 21 to 25

5) Earth-sheltered house with little exposure; well
insulated. ..... 10 to 13

For example, if a home’s energy-saving features are best
described by #2, and the home has a heated space of 1,500
square feet (139.35 m2), then 1,500 ´ 50 and 1,500 ´ 70 is
the heating load range. Roughly 75,000 to 105,000 Btu/hour
(18,900 to 26,460 kilocalories/hour.)

Although a chart like this looks official, not all houses
fit the profile given. There is also no accounting for the
other factors mentioned above.

Online calculation info:
http://www.heatload.com/

Program associated with the calculations, in a professional
arena:
http://www.wwwebworks.com/hvac/newfeatures.htm
http://www.mrhvac.com/products/software/elc/elc.htm

Extra links:
http://www.ashrae.org/

Any of this near what you needed? If not, please
clarify. Like, if you wanted US to tell you heat
load, we are going to have to have ALOT more info
;0)

-AI

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