雅田主营:杭州工业风扇,工业吊扇,浙江大型工业风扇,工业节能风扇

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  • 大型工业风扇降温的原理
  • 本站编辑:杭州雅田机电荣鼎国际官网发布日期:2019-06-02 16:59 浏览次数:

大型工业吊扇运行在理想的情况下,您可以把超大型工业风扇吹出的风想象成柱状的气流,柱状气流的外表面就会与空气产生摩擦,那 阻力=K1(阻力常数)*周长 通风量 =k2(速度常数)*截面积。拿一个直径 为 730mm的风扇和直径为 7300mm 的超做比较:  

1、直径 730mm 的风扇:周长=0。73 × 3。14=2 。29m 截面积=0。36 × 0。36 × 3。14=0。407 平方米

 2、直径 7200mm 的风扇 :周长=7.2 × 3.14=22.9m 截面积=3.6 × 3.6 × 3.14=40.7 平方米 

由此可见,通过同样面积的气流的阻力是小风扇的 0.1 倍  

原理:大的气流流体运动所受的摩擦力相对小,从而使大风扇能够驱动大量的空气却比小风扇所需的能量相对要少! 一旦空气的惯性得到克服,大量的空气流动的自身动能使之只需很小的后续动力即可维持空气的持续流动。而小型的风扇的气流摩擦比较大,能量大部分都损耗在摩擦过程中。

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The large industrial ceiling fan is ideally operated, and you can imagine the wind blown out by the Super large industrial fan as a columnar air flow, and the outer surface of the columnar air flow will rub against the air. That resistance = K1(resistance constant) * perimeter ventilation = K2(velocity constant) * cross-sectional area. Take a 730mm diameter fan and a 7300mm diameter super large industrial fan for comparison:
1, 730mm diameter fan: perimeter = 0.73 × 3.14 = 2. 29M cross-sectional area = 0.36 × 0.36 × 3.14 = 0.407 square meters
2, 7200mm diameter fan: perimeter = 7。2 × 3。14 = 22。9 M cross-sectional area = 3。6 × 3。6 × 3。14 = 40。7 square meters
It can be seen that the resistance of large fans passing through the same area of air flow is 0.1 times that of small fans.
Principle: The friction force of the large airflow fluid movement is relatively small, so that the large fan can drive a large amount of air but the energy required by the small fan is relatively less! Once the inertia of the air is overcome, the self-movement of a large amount of air flow can enable it to maintain the continuous flow of air with only a small amount of subsequent power. The friction of the small fan is relatively large, and most of the energy is lost during the friction process.

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