dr. naeem tahir

BY AYESHA AIMAN

ECH Fan Operation in Broiler Management to Control

Humidity, Temperature in Winter / Summer / Monsoon.

By

Dr. Naeem TahirASM Jadeed Sales Office Faisalabad.

2

The bird!

Homoeothermic

Body temperature---------------105-107˚F(41-42˚C)

Upper lethal body temp.-----------------115˚F(46˚C)

Environment comfort zone--------70-75˚F(21-23˚C)

3

Where is the most heat coming from?

18% ceiling

2%side wall

80%bird

4

Methods of heat loss:

Conduction Convection

Radiation

5

Higher wind speeds increase sensible heat loss and reduce latent heat loss in broilers.

Objective of Fan SystemA System that delivers fresh air into the poultry house and removes obnoxious gases from the poultry house

Fan Operation At a GlanceFan System is For

Moving FRESH AIR INTO a house and moving STALE AIR OUT of the house;

Sending UNWANTED heat, EXCESS moisture, ammonia, OUT of the house; and

AIR MIXING to get heat, moisture, ammonia INTO THE AIR that leaves the house.

8

Objectives of Fan Operation

o Maintain uniform temperatureo Provide fresh airo Remove undesirable gaseso Remove moistureo Keep the litter dryo Keep uniform condition in the house

What is ventilation ?Ventilation means moving the right

amount of air at the right time and in such a way to manage temperature, humidity and other environmental factors for optimum bird performance

Types of ventilation:

Minimum ventilation

Transitional ventilation

Tunnel ventilation10

VENTILATION PARAMETERS: Tunnel fans capacity. Cooling pads area. Minimum ventilation fans. Air inlets. Climate controller. Insulation.

11

VentilationFor ventilation purposes weather can be divided into three types

Weather types

1. Winter2. Summer / Monsoon3. Moderate

Types Of Ventilation

1. Winter => Minimum ventilation2. Summer / Monsoon =>Tunnel ventilation3. Moderate weather =>Transitional ventilation

Relative HumidityMoisture holding capacity of

air at certain temperature.OR

A ratio of the quantity of water vapor in the air compared with the total that can be held at a given temperature

Important Terminology

Relative Humidity

Warm air significantly hold more water than cold air

Water holding capacity of air doubles with each 11oC increase in temperature

Important Terminology

Dew Point TemperatureTemperature at which water vapors are transformed back to liquid

Static PressureWhen we run the exhaust fans a

negative pressure is created inside the house. To measure the difference of outside and inside pressure we use the term ‘Static Pressure’. It is mentioned in Pascal or inches.

CFM

Volume of air being moved by a fan or entering an air in let

Air inlet & Air jet A controllable opening generally located

at the eave controlling the velocity of air

Air that is allowed to travel adjacent to a smooth surface, generally a ceiling or side wall.

Throw The distance an air jet will travel before

its maximum speed is decreased to 75 feet per min.

Less than 75 fpm the air moves aimlessly

Measures of Poultry House

Length 450 ft Width 45 ft Height 8 ft

Fan Calculation

Volume of House Cross Section of House Desire Air Velocity Fan Capacity

Fan Calculation

Volume of House = Length x Width x Height

= 450 x 45 x 8

= 162000 cub. ft

Fan CalculationCross Section of House

Width X Height

45 X 8 = 360 sq. ft

Fan CalculationDesire Air VelocityLength of the houseMostly we recommend (length + 50-75)500 ft/min

Fan Capacity18000 cfm22500 cfm

Fan Calculation (1)

Cross Section x Air Velocity ÷ Fan Capacity = No of Fans

360 x 500 ÷ 22500 = 8

Fan Calculation (2)Volume of House ÷ Fan Capacity = No. Fans

162000 ÷ 22500 = 7.2

= 8

Air Exchange ≤ 1

Air Exchange = House Vol. ÷ Tot. Fan Capacity

= 162000 ÷ 180000

= 0.9-min

Air Velocity Calculation

Air Velocity = Tot. Fan Capacity ÷ Cross Section

= 180000 ÷ 360

= 500 ft/ min

Pad Area Calculation

Pad Size (2``, 4``, 6`` )

Total Fan Installed

Air Velocity for Pad

Pad Area Calculation

Pad Size Air Velocity

2`` 150 ft/min

4`` 250 ft/min

6`` 350-400 ft/min

Pad Area CalculationPad Area = Tot. Fan Capacity ÷ Air Speed

= 180000 ÷ 250/400

= 720 sq.ft ( 4``)

= 450 sq.ft ( 6`` )

MINIMUM VENTILATION / WINTER VENTILATION

Fan Calculation for Minimum Ventilation

No. of Fans = _______________________Avg. body weight × No of birds× CFM requirement per kg

CFM of Fan

Example

Avg body weight = 2.5 kg Cfm required = 0.5 /kg Cfm of fan = 9000 No. of Fans = (2.5 x 30000 x 0.5)/9000 No. of Fans = 4.16 5 fans will be needed for winter

ventilation

Operating time for Fan

Operating Time = _______________Total CFM RequirementTotal Fan Capacity

Example

Total cfm requirement = 37500 Total fan capacity = 9000 x 5 = 45000 Operational time = 37500/45000=0 .83

= 0.83 x 60 sec= 49.9 sec

It means that in 1 min fan should be on for 50 sec and off for 10 sec.

To get reasonable operational time, we multiply by 450 x 4 = 200

It means in 4 min fan should be on for 3 min & 20 sec and off for 40 sec

Inlet Calculation

Inlet area (sq. in.) = Total cfm required ÷4 = 37500/4 = 9375 sq inch = 65.10 sq ft

No. of Vents = Area Required/Area of Vent = 65.10/ 1m x 0.4 m = 65.10/ 3.2808 x 1.3123 = 65.10/4.30 = 15. 13

16 inlets will be required

Inlet Calculation on basis of CFMNo. of inlets = capacity x average weight x

0.5cfm(per kg)/2000cfm

= 30000 x 2.5 x0.5 /2000 =18.75

This formula suggests 19 inlets should be used

Transitional Ventilation Fans

Vol. of House x 40 % ÷ Fan Capacity = No.of Fans

162000 x 40 % ÷ 22500 = 2.88

= 3

Transitional Inlets Calculation

For each 10000 cfm we require 15 sq.ft inlet area Size of each vent

Length Width Area

46`` 6`` 1.91 sq.ft

46`` 8`` 2.55 sq.ft

Transitional Inlets Calculation

No.of fans x capacity ÷ 10000 x 15 = Inlet Area

3 x 22500 ÷ 10000 x 15 = 100.5 sq.ft

Inlet Area ÷ Size of Inlet = No.of Vents

100.5 1.91 52.61 (54) 100.5 2.55 39 (40)

Minimum Ventilation Fan

Number of birds in the houseBody weightCfm per Kg.Fan Capacity

Minimum Ventilation Fan

No.of birds X B.W. X cfm/kg ÷ Fan capacity

30000 X 1.8 X 0.5 ÷ 9000 = No. of fans

= 3

= 3

Side wall Fan spacing

Length of house ÷ (No. of fans X 2) = feet from wall

400 ÷ ( 3 X 2 ) = 67 ft 67 X 2 = 134 ft

Side wall Fan spacing

67 134 134 67

A B CCC

FansPads

Cool Air

Estimated effective temperature reduction when air temp is 85-90˚F:

Air velocity Bird age

Ft/min 1 week 4 weeks 7 weeks100 -4 -2

200 -12 -5 -2

300 -22 -10 -4

400 -15 -7

500 -18 -10

50

Air Velocity Bird Age

Minimum Ventilation Rates per bird

Week Days Cfm/Bird1 1-7 0.102 8-14 0.253 15-21 0.354 22-28 0.505 29-35 0.656 36-42 0.707 43-49 0.808 50-56 0.90

Age (days) Max Airspeed Max Airspeed (m/s) (fpm)

0 – 14 days 0 - 0.5m/s 0 - 100fpm

14 – 21 days 0.5 – 1.8m/s 100 – 350fpm

Air speed for Young Birds

How a bird releasesheat?

A bird releases excess body heat in 2 ways: 1. To the air around it – Sensible Heat – 11BTU/kg or 5BTU/lb. The cooler the air the greater the amount of heat loss. The warmer the air, the

smaller the amount of heat loss 2. Through evaporation of moisture from respiratory system – Latent Heat – 15BTU/kg or 7BTU/lb. the amount of heat a bird loses through the

evaporation of moisture off of its respiratory system depends on the relative humidity of

the air it breathes

3 Areas Where Broilers Release Heat

1. Head,2. Legs,3. Under wing

Heat Loss & Air Speedo Without airspeed evaporative cooling is potentially dangerous and as air velocity increases, the importance of relative humidity to a bird decreases

o If the temperature of the air moving over a bird is equal to its body temperature essentially no heat will be lost to the air. It is not an either/or situation.

o A producer needs to utilize both air movement and evaporative cooling during hot weather to keep birds comfortable and productive.

Wind Chill Effect – Effective Temperature

Any wind-chill curve is an estimation!!– Effective temperature (what a bird perceives thetemperature to be) is a function of:1. Air temperature2. RH3. Bird Age4. Stocking Density5. Wind Speed6. Amount of radiant Heat – roof or side walls

As a result it is very difficult to come up with a chart/formulathat accurately predicts effective temperature!

How much does the air heat up?

1. Amount of heat added to the air in a house the more heat added to the house, the hotter the house will be2. How quickly we exchange the air in the house. the faster the air exchange rate, the cooler a house will be

Remove heat from the house by: rapidly exchanging the air in the house. If we don’t exchange the air rapidly large temperature differences can occur between the inlet and exhaust ends of the house.

Fan Timer Setting Cfm per bird X No. of birds ÷ Fan capacity

0.5 30000 ÷ 27000

= 0.55 (55%)

= 0.55x60= 33 sec.or0.55x300=

165 sec. On and 135 sec. Off

Tunnel Ventilation

Tunnel Ventilation

Transitional Ventilation

Minimum Ventilation

Ventilation with end fans side inlets

Side inlets and side fans

Potential problems:

Not uniform temperature in the house: Proper insulation of the house. First go on tunnel ventilation, then

cooling.

Increased heating cost: Center house brooding. Proper insulation.

Potential problems: Uneven ventilation: Excessive leakage near the pads area, doors, windows, and air inlets. Litter caking :

During Winter: Improper minimum ventilation. Requires proper air inlets opening along with proper no. of 36”

fans.During Summer:

Low air movement. Proper ventilation. (First go on tunnel ventilation, then cooling).

86

Thank You!87