Tabreed mechanism and force technique (calculations)

To calculate the surface area of cell pads needed to cool down the production unit depends on many methods. First of all, calculate the volume of air in the unit that is required to be changed once a minute. Then, calculate the number of fans, which are capable to change the air mass once an hour.

Techniques

First technique: It depends on the need per meter from the unit of certain volume of air each hour and it is approximate, where it is required for every meter 200 cm - 250 cm from air in an hour.
Example:
A house of 100 m long and 12 m wide has a surface area of 1200 m² needs 270,000 m³ of air in an hour. (225 x 1200) = 270,000 m³.

Second technique: It depends on the amount of air needed for every kilogram of living body weight that exist in the house per hour. However, it neglects very important facts.
Example:
*A production unit has a surface area of 12 m x 100 m =1200 m². This has the ability to contain:
- 12,000 birds with a rate of 10 birds per m² OR
- 20,400 birds with a rate of 17 birds per m² OR
- 23,000 birds (batteries) with a cage area of 500 cm²/per bird

*Every 1 Kilogram of body weight needs 6 m³ to 9 m³ of air in an hour with good insolated structure. As also, every 1-kilogram from the body weight needs 9 m³ to 15 m³ from the air in an hour with poor isolated structure. Therefore, it is important to calculate accurately the needs of the units; this depends fully on the amount of meat present and purpose of the usage of this unit.

Third technique: This technique is considered the most current, where the volume of air (per unit of production) is calculated. The aim is to change it every minute (or less).
Example:
A unit with measurements of 100 m long, 12 m wide and 3 m high
Size of air mass = 3,600 m³ is changed every minute; it means 216,000 m³ per hour

Fourth technique: This is the ideal method that could be used for houses which depend on REAL TUNNEL ventilation way. In this method the following calculations are made:
- The cross sectional area (Width x Height) regardless of the house length
- The velocity of this cross sectional area, which is required to move inside the house, is selected. Either 1m/sec or 1.5 m/sec or 2 m/sec i.e. 60 m/min or 90 m/min or 120 m/min
- The number of fans that have the capability to move this cross sectional area of air with the required velocity
- The working cell surface area
Example:
A house whose dimensions are as follows:
- Width 12 m and height 3 m; calculate the cross sectional area = (12x3) = 36 m²
- Required air velocity = (1.5 m/sec or 90 m/min), 36 m x 90 min = 3240 m/min, 3240 x 60 min = 194400 cm/hr
- Fan capacity = 42000 m³ (at Zero Pascal) x 80% = 33600 m³/hr (The actual capacity after installation)
- Required Number of fans = 194400 ÷ 33600 = (5.78) or 6 fans
- Cell surface area = 194400 ÷ 5400 = 36 m cell area

Important note:
1- The capability of the fans in the market is calculated at zero Pascal, meaning that the pressure before the fan and afterwards is equivalent. When the fans in the units are exposed to batteries or concrete columns inside the house, the fans airflow capacity is reduced. Therefore, careful consideration must be taken into account that 10% to 25% of the fans capacity is increased to be more precise.
2- Square meter of the cell pad with a thickness of 10 cm has the ability to pass:
1 m³/sec meaning 3600 m³/h (1 x 60 min x 60 sec) OR
1.5 m/sec. Meaning 5400 m/h (1.5 x 60 min. x 60 sec) OR
2 m/sec meaning 7200 m/hr (2 x 60 min x 60 sec)

The cell pads with thickness 15 cm are equivalent to 1½ cell pad with a thickness of 10 cm.

Important considerations
1- The accuracy of calculations and maximum production of the house should be taken into consideration
2- The condition of the house whether it is closed well or if there is any leakage, which has a negative impact on the efficiency of the cooling system, should also be taken into consideration
3- The degree of isolation of the walls and ceiling and the amount of heat leakage is to be considered too
4- The relative humidity that is inside the house should be regarded. Whenever the relative humidity is lower, the efficiency of the cooling system is higher. Also whenever the temperature is at the maximum record, the relative humidity will be at its lowest records (functioning hours)