QUESTION 6 HEATING AND COOLING LOAD CALCULATIONS 6.1 Explain the need for coolin

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QUESTION 6 HEATING AND COOLING LOAD CALCULATIONS 6.1 Explain the need for cooling and heating load calculations. Explain the differences between conventional cooling and heating load calculations 6.2 Define fenestration and explain the need for fenestration and its effect on air conditioning 6.3 Compute the heat gain for a window facing southeast at 32° north latitude at 10 a.m. central daylight time on August 21. The window is regular double glass with a 13-mm air space. The glass and inside draperies have a combined shading coefficient of 0.45. The indoor design is 25°C, and the outdoor temperature is 37°C. Window dimensions are 2 m wide and 1.5 m high. 6.4 If the window has a 0.5-m overhang at the top of the window as shown in Fig. Q2, determine how far the shadow will extend downward. Fig. Q2: Shading angles and dimensions

Explanation / Answer

Question 6.1:

Solution: (a) Heating and cooling load calculations are carried out to estimate the required capacity of heating and cooling systems, which can maintain the required conditions in the conditioned space. To estimate the required cooling or heating capacities, one has to have information regarding the design indoor and outdoor conditions, specifications of the building, specifications of the conditioned space (such as the occupancy, activity level, various appliances and equipment used etc.)

(b) Heating Load:

Cooling Load:

Question 6.2

Solution: (a) Fenestration refers to any glazed (transparent) apertures in a building, such as glass doors, windows, skylights etc.

(b) Fenestration is required in a building as it provides:

(c) Because of their transparency, fenestrations transmit solar radiation into the building. Heat transfer through transparent surfaces is distinctly different from heat transfer through opaque surfaces. When solar radiation is incident on an opaque building wall, a part of it is absorbed while the remaining part is reflected back. As will be shown later, only a fraction of the radiation absorbed by the opaque surface is transferred to the interiors of the building. However, in case of transparent surfaces, a major portion of the solar radiation is transmitted directly to the interiors of the building, while the remaining small fraction is absorbed and/or reflected back. Thus, the fenestration or glazed surfaces contribute a major part of cooling load of a building. The energy transfer due to fenestration depends on the characteristics of the surface and its orientation, weather and solar radiation conditions. A careful design of fenestration can reduce the building energy consumption considerably.

Question 6.3

Solution: Window Area = 2 m x 1.5 m = 3.0 m2

Heat transfer Coefficient (U) = 3.5 W/m2K

Heat gain due to Transmission:

Q1 = UA(t0-t1)

= (3.5) (3) (37 - 25)

= 126 W

Heat gain due to Solar:

Qs = (SHGFmax)(SC)(CLF)A

Constant Values: For 320North Latitude, Facing SESHGF = 580 W/m2

Facing SE at 10 A.M.

CLF = 0.79 and SC = 0.45

Qs= (580) (0.45) (0.79) (3)

= 618.6 W

Heat Gain = 126 W + 618.6 W

Heat Gain = 744.6 W Ans

Question 6.4:

Solution: From given Fig.

Y (Distance of shadow) = dTan?/Cos? …………… (1)

D (Given) = 0.5 m

For 320 North Latitude, 10 A.M., August

? = 560

? = 600

Facing South East,

? = 450

? =? -? = 60 - 45 = 150

Putting values in (1)

Y (Distance of shadow) = dTan?/Cos? = (0.5) Tan56/Cos15

y = 0.77 m Ans

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