![]() ![]() Tell your customer to stick to making tyres. In this condition the motor can reach instability if the nature of load is either constant torque or square load and the motor will stall in doing so will inhibit the type of sparking you mentioned. If you go into the 'field weakening area' where the armature voltage is kept constant and the speed is increased by reducing the field flux (speed is inversely proportion to flux) the power available will be fixed and the torque available with progressively reduce with increases in speed. ![]() If you reduce in speed you still have the same level of torque available and the power is within the speed torque envelope. This is called the 'constant torque area' where the available torque is constant and subsequently the power varies with speed (voltage). The operating area in armature voltage control is based on a constant field flux. If the gearing is such that the decrease is from 1000 rpm to 40 rpm then the power available will be very similar to 1500HP less losses. So from 0 rpm to 1000 rpm the power (HP) is directly proportional the speed the speed is (almost) directly proportional to the applied armature voltage if we ignore the effects of armature reaction et al. Your clients's motor is 1500HP based on rated speed of 1000 rpm. RE: gear ratio change - how does it affect the motor ? jbartos (Electrical) 21 May 04 23:25 This means a higher motor output requirement at a lower speed, which as stated above, is not possible due to temp rise issues. If this lower torque is sufficient to run the rubber mill, the proposed system of 800/40 RPM gear box will work with a lower load on the motor.īut if the lower gear box torque (at 800 RPM) is insufficient (more likely given the rubber load on the mill), then to increase the torque, the motor amp-turns (or rather the motor amps) has to increase. The motor output torque stays the same at the lower speed (since torque = k*field flux density *armature amp-turns).īut the gear box output torque will now be reduced by 4/5 (=800/1000). RE: gear ratio change - how does it affect the motor ?Īs the motor speed is reduced (by increasing the field current and assuming there is no saturation), the motor output (in terms of armature amp-turns) is reduced to maintain the winding temp limits. ![]() "Venditori de oleum-vipera non vigere excordis populi" ![]() Realistically though, it can probably do less effective shredding at faster speed: in other words probably no net gain in productivity, and maybe even a loss due to inefficiencies throughout the system! The proper people to consult on this would of course be Banbury, but if it were that simple, wouldn't they have already done it? Think about it, if they could have simply changed the gearbox ratio and offered increased throughput, why would they NOT? They probably already knew something different. If the Banbury can still shred with a 20% reduction in torque, no problem. A Banbury mixer works by "exploding" the rubber slabs between high pressure rollers turning at slightly different speeds. The problem then shifts to the work being performed. At 800:40 or 200:1, the output shaft torque will only go up 200 x the motor torque, 20% less than before (200/250) although when he runs at 1000 rpm the output speed will be 20% faster (40/50). Now however he is changing the torque requrements via the gearboxes. The old speed reduction ratio was 1000:40 or 250:1, so the output shaft torque was 250 x the motor torque (ignore losses for sake of discussion). When you lower the speed through gear reduction however, you AMPLIFY torque by the inverse ratio. At his 20% overspeed he may have lost as much as 50% of the motor's available torque (no I did not get out the calculator), so that may explain the strain on the motor! HP can't go up so you begin to lose torque rapidly. Once you go over base speed however, the max. etc.) As you lower the speed and the torque requirement stays constant, you are lowering the output HP at the same time. The motor has a fixed MAXIMUM HP rating, so that stays essentially constant (other than overload capacity, temperature rises etc. Horsepower is the term used to describe a speed / torque relationship. ![]()
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