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ISAT 413 - Module III: Building Energy Efficiency

Topic 7: Transient Heating and Air Conditioning Loads

Thermal Admittance

Intermittent Heating

Air Conditioning

Examples of Air Conditioning

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Under steady state conditions the heat loss from a building depends on the thermal transmittance and the ventilation rate. A true steady state is never attained since the outside climate varies on a 24-hour cycle as well as from day-to-day.

When a material is heated up or cooled down the energy change depends on the density and specific heat of the material as well as on the thermal conductivity.

Transient Building Loads

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Comparison of Materials for Heat Transfer and Energy Storage

Heavyweight concrete needs a much greater thickness than expanded polystyrene for the same insulating effect but that it needs 3110 times as much energy to heat or cool that thickness.

Material Density (kg/m3)

Specific heat (J/kg.K)

Thermal conductivity (W/m.K)

t (mm) E (MJ)

Heavyweight concrete

2100 840 1.00 1000 3.110

Brick 1700 840 0.84 840 1.200 Fiberboard 300 1000 0.05 50 0.015 Expanded polystyrene

25 1000 0.03 30 0.001

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In the electrical circuit the impedance, Z , is introduced as the ratio of the voltage to the current; the reciprocal of the impedance, 1/Z is called the admittance.

The thermal admittance is given the symbol Y and since it is a reciprocal of a resistance-type term then it has the same units as thermal transmittance (i.e. W/m2.K).

It can be defined as the rate of heat flow per unit surface area between the internal surface and a space for each degree of swing in the temperature of the space.

Thermal Admittance, Y

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Thermal Admittance and Transmittance for some typical constructionConstruction Y

U (W/m2.K) (W/m2.K)

External brick, 105 mm thick 4.2 3.3External brick, 335 mm thick 4.7 1.7Asbestos cement sheet, 5 mm thick 6.5

6.5external brick, 105 mm thick with plaster 13 mm thick on inside 3.3 2.6As above but with an air gap of 25 mm 3.5

1.4 With reference to the brickwork values, the thermal transmittance, U, of a structure reduces with thickness but the admittance, Y, tends to a constant value as thickness increases above a certain value.

Adding a second layer of brick with an air cavity, although substantially reducing the value of U, has a very small effect on the admittance, Y, which is largely determined by the layer of plaster on the inside surface.

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For the more common case of intermittent heating rather than steady-state cases, the inside temperature will vary throughout the 24-hour period. A simple on-off system with n hours on and (24-n) hours off would give the temperature and heat flow variations shown in Figures below.

Intermittent Heating

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Fluctuating Heat Input For intermittent heating the fluctuating heat input replace mean heat input, and the admittance, Y, is used instead of the thermal transmittance, U.

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The input which cause the need for air conditioning are as follows:

Air Conditioning

Solar Radiation

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Example of Air ConditioningA room 5 m by 4 m by 3 m high has one external wall ( 5m by 3 m) which faces south and which has a window, 3 m by 2 m. The window is single-glazed with normal exposure and white venetian blind fitted internally. There are two occupants who do light office work for 8 hours per day. For a typical sunny day in June using the data below calculate the required air conditioning load to maintain the inside dry resultant temperature at 18oC, Assume that the rooms above, below, and at the sides of the room are at the same conditions as the room. Assume also that the room is well-sealed.

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DataMean total solar irradiance = 155 W/m2

Peak total irradiance at solar noon = 535 W/m2

24-hour mean external temperature = 16.5oC

Mean sol-air temperature = 20oC

External air temperature at 13:00 = 20.5oC

Decrement fatcor for external wall = 0.31

Time lag for glazing = 1 hour

Time lag for external wall = 9 hours

Sol-air temperatures at 04:00 = 10oC

Heat gains for light office work per person = 140 W

Heat gains from office equipment = 2000 W

Average period of use of equipment = 4 hours

below. table the in as and of values the Take

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Item A (m2) U (W/m2.K)

UA (W/K) Y (W/m2.K)

YA (W/K)

External wall

9 0.73 6.57 3.6 32.4

Window 6 5.60 33.60 5.6 33.6

Internal walls

39 - - 3.5 136.5

Floor 20 - - 2.9 58.0

Ceiling 20 - - 6.0 120.0

A=94 (UAo)=40.17 (YA)=380.5

Example of Air Conditioning (continued)

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Example of Air Conditioning (continued)

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Example of Air Conditioning (conclusion)

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