A building Boston, Massachusetts has the following glass wall areas (north wall
ID: 2322541 • Letter: A
Question
A building Boston, Massachusetts has the following glass wall areas (north wall is Group B type wall which has no window and net wall area for the north exposure is 400 ft^2): The glass is 1/4 -in heat absorbing double glass. All windows have medium color Venetian blinds.(Assume U =0.45 for glass). Floor is on an unconditioned space and assume no heat gain from basement. Room temperature is 75 F. The building has a 3,500 ft^2 flat roof (Roof # 9) with 4-in heavy weight concrete: 2-in insulation and a suspended ceiling. Determine the peak month and time and find the peak heat gain through walls, glass, ceiling + roof).Explanation / Answer
Solution:
Assuming outside temp as 90 (peak temp for Massachusetts)
Total Area for all walls : 400 + 900 + 900 + 400
= 2600 ft2
Heat Gain( convection) is given by Qg = U A (T-To)
= 0.45 (2600) (90-75)
= 17750
For flat roof , heat gain (conduction) is given by
Qf = K A dt/dx
= 0.11 (3500) (90-75)/0.33 ( k is 0.11 for concrete & dx is 4/12 ft)
= 17500
With 2 inch insulation there will be no loss of heat from inside and it get trapped &with time keep on increasing.
Peak month is likely to be July and evening time when all heat from day time is trapped inside the room
Peak heat gain is 17750 + 17500
= 35250
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