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Question:

Should the no-load power of the pump be equal to the rated power?

There is a pump, the rated power is 1200W, but only more than 600 W measured!What's the matter, please?I was not measuring no-load, pumping water in the pool, this work environment should be sufficient to pump full load?Also, is the rated power should not change?Thank you

Answer:

We must first clear the meaning of rated power; rated power is called the reference numerical electrical power consumption in the voltage conditions specified, and the actual power consumption of electrical appliances is different in different situations, the pure resistance (such as lighting, electric) as long as the voltage conditions, the actual consumption of electric power close to or equal to the rated power, and for appliances like motor, electric power consumption, not only related with the voltage, but also with the relevant load. For example, someone with 200 pounds of energy, then the person in the condition of small meal, maximum load of 200 pounds, but if you take as long as he is 100 pounds, of course he relaxed, he would not say must bear 200 pounds!You measure the pump power is less than 1200W, first you did not mention your use of voltage is not rated voltage, then you don't want him full load, the suction is far less than 8 meters, there is no mention of whether you send water to 50 meters (Yang Cheng), there may be insufficient voltage, and you have don't need it so much power consumption. Of course, one advantage is that it doesn't burn easily. It is easy to burn if it is used at full load.
Manufacturers in order to make money, the nameplate marked "1200W", in fact, the maximum power of the motor, only 750W.
Not equal to.1. Calculate the shaft power of centrifugal pump:Pa = HQ P /102 KWType H - pump rated lift, m;Q - rated flow of the pump, m3/s;P -- medium density, /m3 kg;Efficiency -- pump rated conditions (ETA = ETA ETA ETA m h V, m -- h -- mechanical efficiency, hydraulic efficiency,ETA V - volume efficiency);Pa - shaft power under rated working condition of pump, KW.2, determine the motor powerP=KPa/.Among them, P--- motors are equipped with powerShaft power under rated condition of Pa-- pumpETA - pump drive efficiency, general pump is using direct drive 1.K-- original motive power margin coefficient (shaft power less than 15KW, 1.25; 15 Pa > 55KW, 55KW > 1.15;, 1.1)

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