{"id":6553,"date":"2023-03-23T09:55:10","date_gmt":"2023-03-23T08:55:10","guid":{"rendered":"https:\/\/www.slingerlandtechniek.nl\/?p=6553"},"modified":"2023-03-23T09:58:00","modified_gmt":"2023-03-23T08:58:00","slug":"understanding-fan-operation-and-speed-control-copy","status":"publish","type":"post","link":"https:\/\/www.slingerlandtechniek.nl\/en\/understanding-fan-operation-and-speed-control-copy\/","title":{"rendered":"Understanding fan operation and speed control"},"content":{"rendered":"\t\t
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Understanding fan operation and speed control with variable speed drives<\/h1>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Some basic mechanical laws come into play in the operation of a fan. But what happens when you change the speed by controlling it back with an AC drive? Understanding fan operation and speed control will help you better manage your fan and optimize its performance.<\/p>

When the fan speed changes, the air volume also changes. This also affects the pressure (\u2206 PStat) that the fan can handle, which is directly proportional to the air volume. This means that the pressure increases or decreases quadratically with speed.<\/p>

In addition, the fan’s power input changes to the 3\u1d49 power output as the speed changes. This can have important implications for energy consumption and efficiency.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Sample exercises - fan operation and variable frequencies<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Here are some exercises that will help you understand fan operation and speed control.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Example 1:<\/strong> Suppose we have a fan with a capacity of 10,000 m\u00b3\/hour at 60 Hz and a power consumption of 3 kW. We want to increase the capacity to 12,000 m\u00b3\/hour. What should the new frequency be?<\/p>

Solution:<\/strong><\/p>

(Hz2 \/ Hz1) x Q1 = Q2<\/p>

(Hz2 \/ 60) x 10,000 = 12,000<\/p>

Hz2 = (12,000 \/ 10,000) x 60 = 72 Hz<\/p>

Answer:<\/strong> The new frequency must be 72 Hz.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Example 2:<\/strong> Suppose we have a fan with a pressure of 300 Pa at 50 Hz and a power input of 1 kW. We want to reduce the pressure to 150 Pa. What should the new frequency be?<\/p>

Solution:<\/strong><\/p>

(Hz2 \/ Hz1) \u00b2 x Pa1 = Pa2<\/p>

(Hz2 \/ 50) \u00b2 x 300 = 150<\/p>

Hz2 = \u221a[(150 \/ 300) x 50\u00b2] = 35.4 Hz (rounded to 1 decimal place)<\/p>

Answer:<\/strong> The new frequency should be about 35.4 Hz.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Example 3:<\/strong> Suppose we have a fan with a speed of 1200 opm and a capacity of 8000 m\u00b3\/hr at 60 Hz. We want to increase the speed to 1500 opm. What should the new frequency be?<\/p>

Solution:<\/strong><\/p>

(Hz2 \/ Hz1) x n1 = n2<\/p>

(Hz2 \/ 60) x 1200 = 1500<\/p>

Hz2 = (1500 \/ 1200) x 60 = 75 Hz<\/p>

Answer:<\/strong> The new frequency must be 75 Hz.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Contactpage<\/a><\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
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Learn about the physics behind air and what the effects are of different wind speed.<\/p>\n","protected":false},"author":2,"featured_media":5496,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"content-type":"","footnotes":""},"categories":[42],"tags":[],"class_list":["post-6553","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-information"],"yoast_head":"\nUnderstanding fan operation and speed control<\/title>\n<meta name=\"description\" content=\"What will happen when we change the speed of fans? 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