Environmental Control

Systems I- Temperature and

Humidity, Arch 353

Lecture # 6

Dr. Hussain Alzoubi

1

Emissivity e : The ratio of the radiation intensity of a nonblack body

to the radiation intensity of a blackbody

Absorption : The process in which incident radiant energy is

retained by a substance by conversion to some other form of energy

reflectance r :is the ratio of the radiant energy reflected from the surface

to the total radiation falling on the same surface area.

-All incident energy must be either absorbed or reflected, and e+r=1.0

-a good reflector is poor emitter, and a poor reflector is a good emitter

for example: white color plaster is good reflector and bad emitter

2

Absorbance and reflectance factors for different surfaces

3

How to calculate radiation from surfaces?

R= eT4

Where R= radiant energy

= the radiation constant, 1.36x10-11 Kcal/(sec)(m2)(K4)[radiation in kcal/sec)(m2)

OR = the radiation constant, 5.67x10-8 W/(m2)(K4)

e = the emissivity of the surface

T =the absolute (Kelvin) temperature

for two opposite surfaces

The net radiation :

R= eA (T14 – T2

4 )

T2

T1

Surface 1 Surface 2

4

Example1:

A person with a total body surface area of

1.35 m2 and an average skin temperature of 32 C is standing unclothed

in a room in which the walls, ceiling and floor are at a temperature

of 15 C. Find the net rate of Heat loss from the body (radiation only)

in watts.

Solution:

From table 3-4 in the previous slides emissivity of skin e = 0.96

Converting the temperature to Kelvin

T1 = 32 + 273 = 305 K

T2 =15 + 273 = 288 K

Rnet = eA (T14 – T2

4 )

= 0.96 * 5.67x10-8 W/(m2-K4)*1.35*(3054 – 2884)

= 130 watt………………answerTherefore a human body in a room of 15 C loses 130 joule per second

If the walls and ceiling have the same temperature degree of what human body

has, the human body would not lose energy by radiation.

The net heat loss per square meter is 130/1.35 = 96.3 Watt

5

How to calculate heat in spaces

1- Calculate heat gain or loss from windows

2-Calculate heat gain from occupants

3-Calculate heat gain from appliances, devices,

lighting fixtures..etc

4-Calculate heat gain or loss from walls

5-Subtract heat gain from heat loss to find heating

loads.

Examples:

A room has the following appliances run

for 6 hours:

1-Two computers with power 200 watt for each.

2-Three lighting fixtures with 250 watt for each

3-One secretary whose work is typing

Heat loss through walls is 2000 watt or (joule/second)

Solar heat gain from windows is 600 watt or(joule/second)

Find the required heat to keep the same temperature

Degree in the room.(Assume all the input power of

the devices is converted into heat energy 6

Solution:

1-rate of heat gain from computers= N X power

= 2x 200 = 400 watt

N: number of computers

2-rate of heat gain from lighting = 3x250 = 750 watt

(assume all lighting energy is converted to heat energy )

3-Secretary (from table ) 500 (Btu)x 0.293(watt) = 147 watt(1 Btu/hour [I.T.] = 0.293 071 07 watt)

Total heat from heat generators = 400x6+750x6+147x6

Total heat from heat generators = 2400+4500+879=7779watt-hour

Required heat = solar heat + appliances heat – heat loss

From walls

= 600x6 + 7779- 2000x6 = -621watt-hour

The needed power = -621/6 = -103.5 watt.

The room needs a small heater with a power of 103.5 watt

to balance the heat loss.

7

Humidity and human comfort

8

9

10

11

Source of heat

12