How to calculate the air volume of the fan required by the product

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How to calculate the air volume of the fan required by the product
Shijiazhuang Jingjin Mechanical Equipment Installation Co., Ltd. Zhou Yanhui 13833456715
For different customers in different regions, the manufacturer has the obligation to guide customers how to select appropriate air volume. The air volume selection method is introduced as follows:
First, you must know some known conditions:
1.1 Calories is equal to 1g of water weighing 0 ℃, which is the heat required to raise its temperature by 1 ℃.
2。 One watt for one second is equal to one joule
3。 One card equals four. 2 Joules
4。 Specific heat (Cp) of air at constant pressure (10mmAq)=0. 24（Kcal/Kg℃) 
5。 Standard air: the moist air with temperature of 20 ℃, atmospheric pressure of 760mmHg and humidity of 65% is the standard air, and the weight of air per unit volume (also called specific weight) is 1200g/M * 3
6。 CMM and CFM refer to the volume of air discharged per minute, and the former unit is m3/min; The latter is in cubic feet per minute. 1CMM=35.3CFM。 
2. Formula calculation
1、 It is known that the total heat output of the fan (H)=specific heat (Cp) × Weight (W) × Allowable temperature rise of vessel (△ Tc)
Because: weight W=(CMM/60) × D=volume between units (per second) multiplied by density
=（CMM/60)·1200g/M*3
=（Q/60）  × 1200g/M*3 
Therefore, total heat (H)=0. 24（Q/60） ·1200g/M＊3·△Tc 
2、 Electric appliance heat (H)=(P [power] t [seconds])/4.2
3、 From the first and second, we know that: 0.24 (Q/60) · 1200g/M * 3 · △ Tc=(P · t)/4.2
Q=（P × 60）/1200·4.2·0。 24·△Tc
Q=0。 05P/△Tc……………………………………………… （CMM)
=0。 05·35.3 P/△Tc=1。 76 P/△Tc…………………………（CFM）
4、 The converted Fahrenheit degree is: Q=0.05.1. 8 P/△Tf=0.09 P/△Tf………………………（CMM） 
=1。 76·1.8 P/△Tf=3.16 P/△Tf…………………………（CFM） 
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3. Example
Example 1: A computer consumes 150 watts of power and a fan consumes 5 watts. When the summer temperature reaches 30 ℃, the CPU is allowed to work at 60 ℃. The required fan air volume is calculated as follows:
P=150W+5W=155W； △Tc=60—30=30
Q=0。 05 × 155/30=0。 258CMM=9。 12CFM (air volume required for work)
Therefore, the fan with actual air volume of Qa shall be selected
Example 2: A SWITCHING power supply consumes 250 watts of power, and a fan consumes 20 watts. The local summer temperature is 55 ℃ at the highest level. If the supply is allowed to work at 95 ℃, the required fan air volume is calculated as follows:
P=250W+20W=270W; △Tf=95-55=40
Q=0。 09 × 270/40=0。 6075CMM=21。 44CFM (air volume required for work)
Therefore, the fan with actual air volume of Qa shall be selected
Air flow Q=m ³/ h
Q=sectional area * wind speed
Wind speed=air volume (m ³/ h) /(sectional area (m2/h) ＊ 3600)
Wind speed (m/s)
Air volume (m ³/ h）
Sectional area (m2/h)
Fan outlet wind speed formula
Six fans are used to build a wind tunnel. The pressure of each fan is 354pa, the flow is 30000 m3/h, and the wind speed at the outlet of the wind tunnel is required to be 9m/s. What is the outlet area of the wind tunnel?
Air flow Q=m ³/ h
Q=sectional area * wind speed
Wind speed=air volume (m ³/ h) /(sectional area (m2/h) * 3600)
Wind speed (m/s)
Air volume (m ³/ h)
Sectional area (m2/h)
One degree equals one thousand watt hours
One watt equals one joule per second
1 degree equals 1000 joules per second * hour (3600 seconds)
One degree equals 3600000 joules