What is the effect of changing the input torque of an alternator with excitation kept constant? and what is the effect of changing its excitation with a constant torque?

  • \$\begingroup\$ How can you keep constant torque or change it? If your system can deliver constant torque, then it will spin the alternator until they will reach the equilibrium, means if no load is connected it will increase the speed to the infinity. \$\endgroup\$ – Marko Buršič Jul 14 '16 at 7:12

Ignoring losses...

excitation * speed * k = voltage
excitation * current * k = torque

where k is some constant (the same in each equation) dependent on the geometry of the alternator

It's not clear how you intend to 'increase torque' without specifying the characteristics of the load. To cut you some slack, let's assume the load is something for which we can easily see what will happen, something like a rectifier charging batteries with a substantially constant terminal voltage, accepting whatever charge current you throw at them. For a torque-limited driver, they will control the speed of the alternator by increasing the loading rapidly as the alternator voltage exceeds the battery voltage. In this case, increasing the torque will increase the charge current linearly with the torque, while the terminal voltage and alternator speed will change little. Other loads will behave differently.

Increasing the excitation is trickier to visualise, as the k constant rises. Other things being equal, increased excitation is like putting a gear-box on the shaft, reducing the speed and increasing the torque of the alternator. How the speed and load voltage reacts to that depends on the characteristics of the driver and the load.

If we multiply out to get power

mechanical power = speed * torque
electrical power = voltage * current

we can see that the excitation and the geometry constant have cancelled, and for no losses, input and output power are equal.

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  • \$\begingroup\$ @Abdu: Your question doesn't make much sense. If you increase the input torque for a given load the speed will go up. The electrical output power will increase too until it balances the mechanical input. \$\endgroup\$ – Transistor Jul 14 '16 at 6:26

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