Ventilation in Poultry Sheds by Muhammad Arslan

 B.Sc (HONS.) Animal Sciences, Poultry Researcher

Broiler Experimental House, UVAS Pattoki

Ventilation is one of the most important stakes in poultry rearing. It has a major role in the health status of reared birds and ultimately it has a great effect on the production and performance of a flock. Ventilation can be defined as “Intake of fresh having a suitable amount of Oxygen essential for the peaceful breathing of birds and going out with suitable amount of Nitrogenous and other gasses causing breathing difficulties to the birds” Summers &  Winters in Pakistan are very hot and dry and it's difficult for a farmer to maintain ventilation in the house. Especially the areas of Sindh Province, Southern Punjab are very harder areas and farmers had to spend a lot of money and resources to maintain ventilation in the house. During summer due to lack of proper ventilation many flocks die partially and production reduces to even lower which cause huge losses to the farmer and national GDP as well.

While discussing ventilation systems in dear homeland Pakistan we should keep in mind about weathers of Pakistan.

We can classify Pakistan into three kinds of weather.

§     Summer/Severe Hot weather

§     Winter/Severe Cold weather

§     Normal weather (Autumn & Spring).

 

o        Ventilation is also divided into three types based on the Pakistani environment.

§     Minimum Ventilation/Winter ventilation

§     Tunnel Ventilation/Summer ventilation

§     Transitional ventilation /Spring Ventilation.

1)Minimum Ventilation /Winter ventilation.

This system of ventilation is adopted when the house temperature is higher than the average temperature of the open environment. In this ventilation, all pads and openings are closed only side ventilators are opened as a source of fresh air intake furthermore, side fans should be preferred for throwing of air outside of the house.

§  Cfm calculations when using low air velocity.

We use 0.4 – 0.5 cfm/pound bodyweight.

OR

0.8-0.9 cfm /pound bodyweight.

§  At higher air velocity

We use 0.5 cfm/ pound bodyweight.

Thumb role for ventilators openings.

1 Sq. inch opening is required for 4cfm. Vents should be opened in front of each other and minimum opening of the vent should be 1.5 inches.

Example of Minimum Ventilation.

Suppose we have a flock size 30k on the first day having bodyweight of 40 gm generally.

40 gm or =0.04kgs

Total weight of flock = 0.04*30000

= 1200 kg.

Required cfm?

Cfm per kg =0.9

1200?

1200*0.9= 1080 cfm

1 side fan provides 8000 cfm.

No. of fans needed?

= Total required CFM/ CFM of one fan

1080/8000 = 0.135

We need 0.135 fans to operate for proper ventilation

What means by 0.135 fans?

We make a cycle of 300 seconds to overcome this confusion.

0.135*300 =40.5 seconds.

The fan will be working for 40-41 seconds and will be switched off in remaining time 300 seconds.

Now we have another thing to do for fresh air intake so we have to calculate side ventilators.

 

Total CFM required /04 this formula to calculate the number of side ventilators.

1080/04 =270 sq. Inches (1 inch of the opening gives 4 cfm, that’s why we divided 1080 with 4)

270 inches are suitable for proper ventilation but How 270 inches should be managed?

1 vent is = 22*11 inches (11 L 22 W)

Minimum opening of a vent should be 1.5 inches

22*1.5 =33 sq. inches for the single vent.

270/33 =8.1 vents from each side are enough for proper ventilation.

2)Tunnel Ventilation /Summer ventilation

Tunnel Ventilation is applied when the required temperature in the house is lower than the average temperature of the open environment. In this ventilation, all pads are used and side vents are closed. We also used water on pads for evaporative cooling. In this type of ventilation, we give 5 Cfm per kg body weight.

1 fan requires 6 cooling pads.

Pad dimension:

L= 6ft

W=2ft

Diameter =0.5ft

Example to calculate Tunnel Ventilation

Required cfm =05 per kg

Total cfm=75000 weight of a flock =15000, 15000 *5=75000)

Cfm of 1 fan =20000

No. Of fans =Total required CFM/CFM of 1 fan.

75000 /20000=3.75 fans.

Means 3 fans will be working 24 hours and 4^th the fan will be working 225 seconds in 05 minutes cycle.

Water calculations for summer ventilation.

0.018*T*Total CFM /8747 (T= Environment temperature – Required Temperature)

8747=British Thermal Unit(BTU) required to convert 01 US gallon into vapours.

Now Suppose.

Environment temperature =35C

Required temperature = 25C

=35*1.5+32 =95 F

=28*1.5+32=82F

=0.018*13*135000/8747

=3.61 US Gallons

Converting it to litres

3.61*3.8=13.72 litres.

1.5 inches pipe in 01 minutes passes 125 litres of water (100 litres generally).

100/13.72= 7.28 cycles for a motor.

1/3 inches pipe =75 litres per minute

=1.25 inches pipe =125 to 150 litres per minute

3) Transitional Ventilation:-

Transitional ventilation is applied when the house temperature is equal to the temperature of an open environment. In this system of ventilation, we use front fans and vents for air intake. We give 02 cfm/kg body weight. We also use pads to fulfil the needs of air.

 

Example of Transitional Ventilation.

Suppose.

Flock size of 3000 having bodyweight of 500 gm(0.5kg) at 16 day calculate Transitional ventilation.

Total weight = average weight *flock size

0.5*30000=15000kg.

Required CFM/kg body weight =2 cfm

15000?

=15000*2=30000cfm required

NO. of fans = total cfm/cfm of 1 fan.

30000/20000=1.5.

1 fan will be working continuously and the other fan will work 2.5 minutes in a cycle of 5 minutes

Now calculation of Vents.

01 sq. inch =4 cfm

30000/4 =7500 inches required for 30000cfm.

We have 24 vents on each side of the house. These vents are opened approximately 6 inches

22*6= 132 inches from 1 vent

132*24 =3168 inches from vents.

7500 – 3168=4332 inches still remains

Now we have to use pads for fulfilling of air amount.

4332 is the area we have to open from pads to fulfill 30000cfm.

24 pads (12 on each wall) each pad with 2ft

Convert pads into the inches

24*12=288 inches on each wall

Total on both walls 288*288=576 sq. inches.

Pad area= 4332/576 = 7.5 inches on each side pads will be opened.

If we have to open just one side then.

4332 /288=15.04 inches.

 

About the Author:

Muhammad Arslan is an animal scientist and belongs from the Institute of Animal & Dairy Sciences, Faculty of Animal Husbandry, UAF. His research interest is poultry sciences and a bonified member of Allama Iqbal Dairy Scientists Association.