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For the showroom, heat is transferred by the mechanisms of
conduction, convection, and radiation.
Heat is conducted to the outer surface of the wall and windows
of the showroom in winter, heat is
convected away by the cold outdoor air while being radiated to
the cold surroundings. Thus, it may be
necessary to keep track of the energy interactions at the
surface, and this is done by applying the
conservation of energy principle to the surface. A surface, in
this case brick and glass, contains no
energy. Therefore, a surface can be viewed as a fictitious
system whose energy content remains
constant during a process (just like a steady-state or
steady-flow system). For the analysis of the
showroom, steady conditions are assumed, the net rate of energy
transfer to a fluid in a control volume
is equal to the rate of increase in the energy of the fluid
stream flowing through the control volume. In
this case, the showroom is the control volume.
Under steady conditions, the rate of heat transfer through any
section of a building of Habitat for
Humanity wall or roof can be determined by knowing the
temperature indoor and outdoor air
temperatures from that a relievable weather website. The walls
and roof for the Habitat for humanity
building are assumed to consist of various layers of materials.
The first step in order to analyze the heat
transfer is to find the overall R-value by determining the
thermal resistances of the individualcomponents using the thermal
resistance network. The overall thermal resistance of a structure
can be
determined most accurately in a lab by actually assembling the
unit and testing it as a whole, but this
approach is usually very time consuming and expensive, we only
want to keep things simple due to
economical and time constrains. The analytical approach
described for this paper is fast and
straightforward, and we expect that the results will be in good
agreement with the real life values.
Now, let us start by analyzing the brick wall, a typical New
England wall will be assumed as shown in the
following figure.
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()()
Then, we are going to find the resistance inside the wall
(( )()
The Rwallwas already found to be 5.72 h.ft2.F/Btu, however we
have to convert it into the unit area for
this this wall which is . Thus, we obtain that Rwall=
0.01664
Then, we found the resistance of outside
(( )()
Therefore the R total without the windows will be
Then, we find the thermal resistances through the wall with
windows:
Awindows= 4 (3*5) = 60ft2
Atotall - Awindows Awall = 343.85ft2 - 60ft2 = 283.85 ft2
( )
