marke Posted September 11, 2006 Report Share Posted September 11, 2006 I often see statements from "reputible" sources suggesting that the use of oversized motors causes energy to be wasted. In many cases, this is not true, infact there can be an advantage in using an oversized motor. For example, from a manufacturers data sheet (attached) I see the following: 11KW motor. 100% load 89.2% eff 15KW motor 75% load 89.7% eff 22KW motor 50% load 90.7% eff So, if I have an 11KW load, I would be most efficient to use the 22KW motor at half load, followed by the 15KW motor at three quarter load and least efficient to use the 11KW motor at full load. The oversized motors would run cooler, and therefore would last longer as well. The disadvantages of using an oversized motor would be that the start current will be higher and the capital cost is also higher. I find the same if I compare the 30, 45 and 55KW motors, and others as well. As I see it, the use of an oversized motor does not waste energy, it actually can save energy!! Best regards,4_pole_motors.pdf Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
GGOSS Posted September 11, 2006 Report Share Posted September 11, 2006 I would tend to agree with most of that, however expect that you're comment regarding starting current would apply to DOL starting only. The amount of torque required to start the load would be the same irrespective of motor size, therefore it should be possible to start at the same level of current (actual amps) via for example a soft starter that allowed for start current adjustment in 1% FLC increments, in other words upper end soft starter products. Regards, GGOSS Link to comment Share on other sites More sharing options...
marke Posted September 11, 2006 Author Report Share Posted September 11, 2006 Now theres a trap!! If we have a 11KW load and we need to have 50% start torque, then with the 11kw motor, we would have a start current Is to develp that torque. If now we double the size of the motor, then the start torque relative to the motor is now 25%. The rated current is double the 11KW, but torque is proportional to current squared. If we had the same start current (Amps) then it would a lower start torque per amp and so we would have less start torque. With the larger motor (twice the size, we will need a start current that is 41% higher than the 11KW motor (start current = rt2 times the start current) Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
GGOSS Posted September 11, 2006 Report Share Posted September 11, 2006 You're absolutely correct. Don't know what I was thinking. Maybe it's time to open that Fish 'n' Chip shop. Regards, GGOSS Link to comment Share on other sites More sharing options...
mariomaggi Posted September 11, 2006 Report Share Posted September 11, 2006 Dear marke, you wrote some interesting considerations. But these considerations are related to standard motors with standard efficiency. If you compare a standard efficiency motor with an high-efficiency motors (for example, Class Eff1 or Eff1+), you will see better results. In other words, if the motor is not so stressed by heat due to low quantity of copper used, and induction on magnet steel is normal and not too high, motor efficiency will be better. You can find the same conditions in a bigger motor at 50% of load. Please note that using a bigger motor the power factor will decrease, and you will be obliged to make a compensation. Furthermore, for frequently started motors, the higher inertia will be a further cost in energy for acceleration. Regards Mario Mario Maggi - Italy - http://www.evlist.it - https://www.axu.it Link to comment Share on other sites More sharing options...
waisak Posted January 2, 2010 Report Share Posted January 2, 2010 I agree with Mr.Mario, it is true that we must comparing the efficiency of oversized motor with energy-efficient motor at operating load. therefore we can obtain the energy saving. Considering motor is just not only about efficiency but the value of power factor which is decreasing sharply at use less than 50% of load. The graphic on the site below may be can help us more understand: http://www.p2pays.org/ref/40/39569.pdf Regard's Waisak Student of Food and Technology Department Bogor Agricultural University Indonesia Link to comment Share on other sites More sharing options...
bob14 Posted October 25, 2021 Report Share Posted October 25, 2021 Hi All, In the case of grossly oversized delta connected induction motor, would wiring the motor in star save energy? In star , the shaft power available would be reduced by 3. Let's take a simple example, a 30 k W delta connected installed motor consumes 8 k W. Would rewiring the motor to star save energy, obviously assuming that the starting torque in star would suffice to start the load? Is it an issue to wire delta designed motor to star? Thanks Bob Link to comment Share on other sites More sharing options...
marke Posted October 25, 2021 Author Report Share Posted October 25, 2021 Hi Bob The question that I first look at, is how much energy is being wasted by the motor being oversized? Typically, the only recoverable energy that is being wasted, is the iron loss and that is small relative to the KW rating of the motor. Take for example, a modern motor with a full load efficiency of say 93%. I would expect that the iron loss probably amounts to around 40% to 60% of the total losses, so with total losses of 7% of the motor rating, we could expect that the iron loss is in the order of 3% - 4% of the motor KW rating. - very small. Reducing the voltage, will reduce the iron loss, but it will also reduce the flux in the iron, so will increase the slip loss. As the voltage applied to the motor is reduced, the work component of the current will increase, increasing the copper loss which may result in a net increase in losses. If the load is small enough, the increase in copper loss will be less than the reduction in iron loss yielding a net energy reduction. - This is the principle that the NASA (Nola) energy saving algorithm works on, and also the "energy optimiser" function in many VFDs. Very small motors have a high iron loss and tend to be overfluxed, so can achieve good savings. Reconnecting a delta induction motor in star, is effectively reducing the voltage to root 3 so effectively dropping from 400V to 230V in a 50 Hz world. Effectively, the motor KW rating reduces to one third by connecting the delta designed motor in star, so provided that the KW load is less than one third of the rating of the motor, it is OK to connect it in star. Any higher load will cause the rotor to overheat. Best regards, Mark Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
bob14 Posted October 25, 2021 Report Share Posted October 25, 2021 Thanks Mark. Link to comment Share on other sites More sharing options...
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