Transcript ME31B: CHAPTER SIX

ME31B: CHAPTER SIX
ENVIRONMENTAL CONTROL
FOR AGRICULTURAL OR SMALL
BUILDINGS
6.1 INTRODUCTION
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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
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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
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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.
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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
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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.
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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
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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
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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.
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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.
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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 m 3
/ min?
21oC wb
B
19oC
Solution: ha = 60.4 kJ/kg;
A
32oC
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
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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
5oC
C
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
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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
FARM BUILDING
Exhaust Fan
C
EVAPORATIVE
COOLER
A
Moist pad
air
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
28oC
A
32oC 35oC
Solution
26oC
24oC
C
B
D
28oC
A
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
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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
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An important function of buildings is to
provide an environment of controlled
temperature and relative humidity.
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Therefore,
design
requires
an
understanding of heat generation and
heat transfer.
6.3.1
Heat Conduction Through
Flat Bodies
Wall of a single homogenous material having heat conductivity (K), has
heat flow through it given as:
Q

A
K
t1  t 2
L
K is the thermal conductivity for the material 1 m thick (Watt/m. 1 o K)
Q is the heat flow through the wall in Watt
A is the area of wall in m2
t1 and t2 are temperatures of opposite faces of the wall in o C
L is the thickness of the wall in metre.
6.3.1
Heat Flow Through Walls
and Insulators
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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 nonhomogenous 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.
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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
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The Coefficients are grouped in three
Categories:
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(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.
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(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.
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(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
If U represents the combined heat transmission coefficient of the
entire wall then the heat flow for 1 m2 area, Q = U (ti – to).
In general, it can be shown that:
1
U
 R 
1 L1 L2 1
 

f i K1 K2 f o
U is the overall heat transfer coefficient
R is the resistance for heat transfer
fi is the surface conductance inside
f0 is the surface conductance outside
Example: For the hot weather situation below, find the overall heat
transfer coefficient of the wall.
Outside
Air
O
33 C
Inside
1.25 cm
insulation
Concrete
a
b
c
d
e 23OC
Given: fi = 9.42 Watt/m2. oK ;
f o = 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.
Solution
Outside
Air
O
33 C
Inside
1.25 cm
insulation
Concrete
a
b
c
d
e 23OC
Given: fi = 9.42 Watt/m2. oK ;
f o = 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.
1
U



1
L
1
1
1
 1 


f i K1 Cair Cconcrete f o
1
0.0125
1
1
1




9.42
0.05
6.28 514
.
34.26
011
.  0.25  016
.  019
.  0.03  0.74
U is then 1/0.74 = 1.35 Watt/m2 1oK
Note
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As has been done for the wall, the U
values for all elements of the building
should be calculated.
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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
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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
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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)
T 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.
Solution Concluded
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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
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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
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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.
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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.
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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.