ME31B: CHAPTER SIX

ENVIRONMENTAL CONTROL FOR AGRICULTURAL OR SMALL BUILDINGS

6.1 INTRODUCTION Structures for livestock and poultry have the

basic function of climate modification and environmental control.

In tropical areas, protection from high temperature and radiation is needed while for temperate areas, protection from both cold and hot weather is necessary.

Thus, principle for design of effective shades are greatly important as well as the design of closed, insulated, mechanically ventilated buildings.

6.1.1 Environment An animal's environment is the total of

all external conditions that affect its development, response and growth. Factors can be:

a) Physical eg. space, light, sound, pressure and equipment.

b) Social: number of birds or animals per cage or pen

c) Thermal: air temperature, relative humidity, air movement and radiation.

6.1.2 Homeothermy

This is the fixation of animal internal temperature through a balance of heat produced and lost. The allowable range of body temperature is but a few degrees. Typical body temperatures are:

Humans 37 °C Sheep 39 ° C Swine 39 °C

Cattle    38.5 °C Goats 40 ° C Chickens 41.7 °Ç

Horses 38 °C Cats & dogs 38.6 °C

Homeothermy Contd.

Animals struggle to maintain this constant temperature and prefer to be in a restricted temperature range called the comfort zone.

At this comfort zone, the animal can physically adjust to maintain this temperature.

For man, the comfort zone is between 22 ° C to 30 °C.

Animals produce most when their surrounding temperature is at the comfort zone. Their productivity decreases as temperature rises.

6.1.3 Balance of Heat The maintenance of essentially a constant

body temperature by the animal while subjected to a wide range of environmental conditions depends on balancing heat production and loss.

Heat can be transferred by conduction, convection, radiation and evaporation.

The production of heat in the animal is proportional to its weight.

i.e. H = C W n , where n and c are constants; W is weight.

Balance of Heat Contd. For conduction, heat loss depends mainly on

the surface area exposed to the environment. For cattle, total surface area can be empirically related as:

A = 0.12 W0.60 ; A is area in m2 ; W is body weight in kg.

A big animal has a large total surface area but a small surface area per unit weight while a small animal has a big surface area per unit weight. This latter factor controls heat loss from the body of an animal.

Balance of Heat Concluded This follows that: A big animal produces a large amount of heat

but loses a small amount of it, while a small animal produces a small amount of heat but loses a large amount. Also, unlike in a big animal, a small one has its insulating and temperature controlling mechanisms still under development. This means that:

Small animals cannot live in cold environment while larger ones can. Also small animals can live in hot weather unlike the big ones that will be uncomfortable in hot weather.

6.2 PSYCHROMETRY AND USE IN ENVIRONMENTAL CONTROL 6.2.1 Introduction: Recall use of the

psychrometric chart. The chart may be used in the solution of many environmental control problems for livestock and poultry housing. Recall the following definitions:

a) Humidity ratio (w): The mass of water mixed with a unit mass of dry air in grams of water vapour per kg of dry air.

b) Relative humidity = Water vapour pressure at a certain temp. Saturated vapour pressure at the same temp.

Definitions Contd.

c) Specific volume(V): of an air/water vapour mixture is the space occupied by the mixture per unit mass of dry air. It is expressed in cubic metres per kg (m3 /kg).

d) Dew-point temperature: The temperature at which the air is cooled in the atmosphere without change in the humidity ratio during which the moisture condenses.

e) Enthalpy: The total heat in an air/water vapour mixture. This includes both sensible and latent heat. The enthalpy for a unit weight of dry air (kg) is referred to as specific enthalpy.

Pychrometry Contd. h) Wet bulb temperature: The temperature

obtained by vaporizing moisture to bring it to saturation at constant enthalpy.

Providing supplemental artificial heat to livestock and poultry buildings during the cold season is common.

This process of adding heat is represented by a horizontal line since no change in humidity ratio takes place.

Sensible cooling is merely the reverse of heating and can proceed as long as the final temperature is above the dew point of the moist air moisture.

6.2.2 Sensible Heating and Cooling of Moist Air

Example: Moist air at 5 ° C dry bulb and 3 °C wet bulb is brought into an animal house

through a heater at the rate of 3400 m 3 /hr (0.94 m3 /s). The air leaves the heater at 20

° C dry bulb. What is the heat requirement if Joules/hr?

A B

5oC 20oC db.

Solution: With 5 °C dry bulb and 3 °C wet bulb, ha is 14.9 kJ/kg

Since sensible heat is involved, move right from A to B at 20 °C dry bulb,

hb = 30 kJ/kg.

Solution Concluded

The heater fan is located at position A so the specific volume at A

should be determined. Va = 0.792 m3 /kg.

Mass rate of air, Ma in kg/hr is given as:

Ma = Qv/Va ; Qv is the volumetric rate of ventilation in m3 /hr

Va is the specific volume at condition A (m 3 /kg)

Ma = 3400 m3 /hr = 4292.93 kg/hr

0.792 m3 /kg

Heat added, H = Ma (hb - ha) = 4292.93kg/hr (30 - 14.9) kJ/kg

= 64823.2 kJ/hr = 64823.2 x 1000 J = 18006.44 watts = 18 kW.

60 x 60 s

Example: Moist warm air is cooled mechanically from 32 °C dry bulb and 21 °C wet

bulb to 19 °C dry bulb. How much sensible heat must be removed if the rate is 125 m3

/ min?

21oC wb

B A

19oC 32oC

Solution: ha = 60.4 kJ/kg; hb = 47.2 kJ/kg

The specific volume of air at point B (final state of cooling) is 0.842 m 3 /kg.

The fan is also at B.

Mb = 125 m 3 /min x 1/0.842 kg/m = 148.46 kg/min = 8907.36 kg/hr.

Heat removed , H = Mb (ha - hb) = 8907.36 kg/hr (60.4 - 47.2) kJ/kg

= 117,577 kJ/hr = 117,577 x 1000 = 32660.28 watts = 32.66 kW

3600

6.2.3 Combined Heating and Humidification of Moist Air

This is a common process in the environmental control of livestock buildings. Moisture is continually produced in both vapour and liquid forms by animals and poultry.

They also produce sensible heat. Thus, incoming ventilation air is both

heated and humidified as it moves through the building.

Example: Moist air at 5 ° C dry bulb and 80 % relative humidity is brought into a

growing house for hogs. The air is removed with a 300 m 3 /min exhaust fan at 18 °C

dry bulb and 15 °C wet bulb. How much sensible and latent heat is added to the building

per hour?

15oC

B

80% RH Humidity Ratio

A C

5oC 18oC

Solution: ha = 16 kJ/kg; hb = 42 kJ/kg and hc = 29 kJ/kg

Mass of air exchange must be determined based on specific volume at B

because the exhaust fan is moving air at inside environmental conditions

ie. Vb = 0.836 m3 /kg.

Solution Concluded

Mb = 300 m 3 /min x 1/0.836 kg/m = 358.85 kg/min = 21531 kg/hr.

Sensible heat = qac = 21,531 (29 - 16) = 279,903 kJ/hr = 77.8 kW

Latent heat = qcb = 21,531 (42 - 29) = 279,903 kJ/hr = 77.8 kW

Total heat = qab = 21,531 (42 - 16) = 559,806 kJ/hr = 155.6 kW

Note: q ab = qac + qcb.

6.2.1 Evaporative Cooling: The evaporative cooling process, which consists of

converting sensible heat to latent heat can be used for the cooling of agricultural

buildings under certain hot weather conditions. The system is adiabatic (ie. no

heat is gained or lost in the process).

B Insulation

Warm

Exhaust Fan A air

C

Moist pad

FARM BUILDING

EVAPORATIVE COOLER

Example: Outside summer air at 35 °C dry bulb and 40 % relative humidity is drawn

through an evaporative cooler at the rate of 800 m3 /min. The air leaves the cooler and

is introduced into a livestock building at 28 °C dry bulb and 24 °C wet bulb. Sensible

and latent heat from the livestock heat the air to 32 °C dry bulb and 26.5 °C wet bulb

temperature before it is withdrawn from the building with exhaust fans. What is the

amount of exchange of sensible heat for latent heat in the cooler, how much total heat is

added for latent heat in the cooler, and how much total heat is added to the ventilation

air as it goes through the livestock building.

26oC

24oC

C

B

D A

28oC 32oC 35oC

Solution

26oC

24oC

C

B

D A

28oC 32oC 35oC

Note that AB follows the constant enthalpy line, as the evaporative cooling process is

adiabatic. In order to partition AB into latent heat (BD) and sensible heat (AD), locate D

as shown above.

ha = hb = 72.2 kJ/kg; hd = 64.5 kJ/kg. The enthalpy difference between D and B or 7.7

kJ/kg represents the amount of sensible heat withdrawn from each kg of air in cooling it

7 °C from A to D. This sensible heat is thus converted to latent heat through the

evaporative process. BC process is a heating and humidifying process within the

livestock structure as discussed in section 6.2.3.

Solution Concluded Vc = 0.892 m3 /kg; hc = 82.5 kJ/kg Mass of air at C = 800 m3 /min 0.892 m3 /kg = 896.86 kg/min = 53812 kg/hr Sensible heat (qda) = Mc (ha - hd) = 53812kg/hr (72.2 - 64.5) kJ/kg = 414,350 kJ/hr Latent heat (qdb) = same (this is an adiabatic

process) Total heat added in house by animals = Mc (hc - hb) = 53812 (82.5 - 72.7) = 527,358 kJ/hr

6.3 HEAT TRANSFER THROUGH BUILDING SURFACES

An important function of buildings is to provide an environment of controlled temperature and relative humidity.

Therefore, design requires an understanding of heat generation and heat transfer.

6.3.1        Heat Conduction Through Flat Bodies

W a ll o f a s in g le h o m o g e n o u s m a te r ia l h a v in g h e a t c o n d u c t iv ity (K ), h a s

h e a t f lo w th ro u g h it g iv e n a s :

Q A Kt t

L

1 2

K is th e th e rm a l c o n d u c t iv ity f o r th e m a te r ia l 1 m th ic k (W a tt /m . 1 o K )

Q is th e h e a t f lo w th ro u g h th e w a ll in W a tt

A is th e a re a o f w a ll in m 2

t1 a n d t 2 a re te m p e ra tu re s o f o p p o s ite fa c e s o f th e w a ll in o C

L is th e th ic k n e s s o f th e w a ll in m e tre .

6.3.1        Heat Flow Through Walls and Insulators

Walls that enclose a building are rarely homogenous.

Usually, they consist of combinations of different materials and air spaces.

Therefore we have heat flow through non-homogenous walls.

Materials that transmit heat poorly are termed heat insulators e.g. cork, mineral wool, fibre glass, wood shaving or saw dust.

Heat Transfer Through Walls and Insulators Contd. These coefficients are termed

conductance. Its unit is Watt/m2. oK. Conductance differs from conductivity,

which is given per m thickness. If a wall is composed of several materials,

the calculations are simplified by means of experimental coefficients that separate the total heat flow of the three modes (conduction, convection and radiation).

Experimental Heat Coefficients

The Coefficients are grouped in three Categories:

(i) The conductance of non homogenous units: The conductance of non homogenous building block are obtained by experiments. Some of these are given in Table 6.1.

Heat Coefficients Contd. (ii) Surface conductance from building

surfaces: Heat transmission from a surface is a combined process of conduction, convection and radiation.

The rate of heat flow is affected by the temperature and emissivity of the surface, air velocity and temperature difference between the surface and air.

The surface conductance (f) combines the effect of the three modes of heat transfer. Conductance for inside surface (fi) is chosen for zero air speed ( 9.42 Watt/m2. oK ) and the conductance for outside surfaces (fo) for 25 km/hr wind speed (34.26 Watt/m2. oK)

Heat Coefficients Contd. (iii) Conductance through the

interior air spaces of framed wall construction. The conductance (a’) for vertical air spaces increases with temperature.

It is high for narrow spaces but decreases to nearly constant value for spaces greater than 2 cm in width.

A satisfactory average value for vertical spaces is 6.28 Watt/m2. oK

 

6.2 Cold Weather Situation.

L1 L2

Inside Outside

ti

ta

tb

tc to

K1 K2

To estimate the heat flow through a non homogenous wall, it is

necessary to compute its combined heat transmission coefficient.

The heat flow consists of the following phases:

(i) Inside air to wall

(ii) Flow through the length of solid wall L1 with K1

(iii) Flow through the length of solid wall L2 with K2

Wall to outside air.

Heat Transmission Coefficient

I f U r e p r e s e n t s t h e c o m b i n e d h e a t t r a n s m i s s i o n c o e f f i c i e n t o f t h e

e n t i r e w a l l t h e n t h e h e a t f l o w f o r 1 m 2 a r e a , Q = U ( t i – t o ) .

I n g e n e r a l , i t c a n b e s h o w n t h a t :

1 1 11

1

2

2UR

f

L

K

L

K fi o

U i s t h e o v e r a l l h e a t t r a n s f e r c o e f f i c i e n t

R i s t h e r e s i s t a n c e f o r h e a t t r a n s f e r

f i i s t h e s u r f a c e c o n d u c t a n c e i n s i d e

f 0 i s t h e s u r f a c e c o n d u c t a n c e o u t s i d e

Example: For the hot weather situation below, find the overall heat

transfer coefficient of the wall.

Outside Air Inside 1.25 cm 33OC Concrete insulation a b c d e 23OC

Given: fi = 9.42 Watt/m2. oK ; fo = 34.26 Watt/m2. oK ; Conductance

for concrete = 5.14 Watt/m2. oK ; conductance for air space is

6.28 Watt/m2. o K and Conductivity for insulation board is 0.05 Watt/m. o K.

S o l u t i o n

O u t s i d e A i r I n s i d e 1 . 2 5 c m 3 3 O C C o n c r e t e i n s u l a t i o n a b c d e 2 3 O C

G i v e n : f i = 9 . 4 2 W a t t / m 2 . o K ; f o = 3 4 . 2 6 W a t t / m 2 . o K ; C o n d u c t a n c e

f o r c o n c r e t e = 5 . 1 4 W a t t / m 2 . o K ; c o n d u c t a n c e f o r a i r s p a c e i s

6 . 2 8 W a t t / m 2 . o K a n d C o n d u c t i v i t y f o r i n s u l a t i o n b o a r d i s 0 . 0 5 W a t t / m . o K .

1 1 1 1 1

1

9 4 2

0 0 1 2 5

0 0 5

1

6 2 8

1

5 1 4

1

3 4 2 6

0 1 1 0 2 5 0 1 6 0 1 9 0 0 3 0 7 4

1

1U f

L

K C C fi a i r c o n c r e t e o

.

.

. . . .

. . . . . .

U i s t h e n 1 / 0 . 7 4 = 1 . 3 5 W a t t / m 2 1 o K

Note

As has been done for the wall, the U values for all elements of the building should be calculated.

See Table 6.2 for Computation of Coefficient of heat transfer (U) for various walls, roofs and ceilings.

6.3.1        Rate of Overall Heat Loss or Gain from a Building 

Once the U values have been determined for each element of the building (walls, ceiling, windows, doors etc.), the area of each element is determined, and design temperatures for inside and outside are chosen for the location.

Rate of Overall Heat Loss or Gain from a Building  It follows then that for each building element:

Q A x U x T

Where: Q is the total heat transfer rate through an element (W)

A is the area of the building element (m2)

U is the coefficient of heat transfer for the element (W/ m2.o K)

is the temperature differential across element.For the building as a whole, the total heat exchange rate will equal the sum of the Q values.

T

Solution Concluded

Door 1.5 x 15 x 2.4 = 54 W Window 1.0 x 15 x 6.0 = 90 W Total Heat Loss = 1844 W  Metal Roof Roof 27 x 15 x 3.03 = 1227 W Wall 42.5 x 15 x 2.9 = 1849 W Door 1.5 x 15 x 2.4 = 54 W Window 1.0 x 15 x 6.0 = 90 W Total Heat Loss = 3220 W

Comment

It is obvious that much more heat must be supplied to the metal roof house.

A ceiling or other insulators will provide a substantial saving.

6.4 ARTIFICIAL COOLING OF LIVESTOCK In general, livestock are considered to

be depressed by temperature over 25 °C. Some of the various ways of providing some relief for animals under heat stress conditions are:

a) Drinking water: Cooled drinking water has proved to be of value.

Water is cooled to temperature 15 to 20 °C.

(b) Air Movement

Fan movement of air in open feed lots for beef obtained good

results. Large diameter (1 m) slow speed fan can be located

at 20 m spacing around the fence line. The fan's centre line

could be 2 1/2 m above the ground and the fans tilt downwards

at an angle of 7 °. This increases the convective heat loss.

Fan

2.5 m

Artificial Cooling of Livestock Contd. c) Evaporative cooling: The absolute

values of evaporative cooling would vary depending on pad thickness (say 6 to 10 cm), pressure drop and material.

The method is successful in dry areas. With properly designed pads and air flows

below 1 m3/s, the dry bulb temperature of the incoming air can be reduced to within 1 1/2 ° C of the wet bulb temperature.

Artificial Cooling of Livestock Contd. d) Air conditioning or mechanical cooling:

Air conditioning or refrigeration equipment is necessary for mechanical cooling. For that reason, it is an expensive means of cooling and of questionable feasibility for domestic livestock and poultry production.

Air conditioning may become economical in future for many types of domestic animal enterprises e. g. Summer production of meat, milk and eggs and air conditioning may be introduced in other areas where high environmental temperatures make production unsuitable.