Tunnel ventilation (moving air at high speed over the total length of the house) is based on the principle of increasing heat loss by increasing air velocity. This allows to remove more heat from the birds at higher enviromental temperatures.
To create a desired level of air velocity, we have to calculate the dimensions of the house and the fan capacity. The objective is to allow the air that goes in through the inlet and removed by the exhaust fans, will pass through along the length of the house.
If the cross section of the house (width x height) is 15 x 3 meter, the surface area of this cross-section is 45 m2. If 8 fans are installed, each with a capacity of 10.000 m3/hr, the total air movement through the house will be 80.000 m3/hr or 22 m3/sec. This means that every second, 22 m3 of air is passing the cross section of 45 m2 in the house. This results in an air velocity of 22 m3 / 45 m2 =0.49 m/sec
If the cross section of the house is reduced by lowering the ceiling to 2 meter, the cross section is 30 meter and the resulting air velocity will be: 22 m3 / 30 m2 = 0.73 m/sec
If the ventilation capacity is increased to 200.000 m3/hr (56 m3/sec), the resulting air velocity will be: 56 m3 / 45 m2 = 1.2 m/s for a house of 3 meter high and 56 m3 / 30 m2 = 1.9 m/s for a house of 2 meter high
We normally consider an air speed of 2 m/s as an acceptable air speed for broilers. For a house with a cross-section surface of 45 m2, this would mean a ventilation capacity of:
45 m2 x 2 m/s = 90m3/s or 324.000 m3/hr for a house with a cross section surface of 45 m2 or
30 m2 x 2 m/s = 60 m3/s or 216.000 m3/hr for a house with a cross section surface of 30 m2.
To have a more effective cooling, we can increase the air speed to 3 m/s, which will require:
45 m2 x 3 m/s = 135 m3/s or 486.000 m3/hr for a house with a cross section surface of 45 m2 or
30 m2 x 3 m/s = 90 m3/s or 324.000 m3/hr for a house with a cross section surface of 30 m2.