so, let's crunch the numbers in the original question. your load is a 10 kVA alternator, and your input is about a 1 kVA motor, and you want to store energy in the flywheel. you didn't mention whether the drive motor remains connected to the flywheel after the alternator is engaged. in either case, the flywheel and alternator will spin down, so you will only get 8 kVA out of the alternator for a very short period of time, a little bit longer with the drive motor still connected, but it will still spin down. with the drive motor still connected, it will spin down to the point where 1 kVA in will deliver about 0.7 to 0.8 kVA from the alternator. with the motor disconnected, it will obviously run down and stop. you have several ways energy leaks out of the system, friction in the motor and alternator (bearings and such), aerodynamic losses in the motor, alternator, and especially the flywheel, and resistance losses in the motor and alternator windings.
while you can store a lot of kinetic energy in the flywheel, and the bigger the flywheel the better, the fact is that once you begin to use that energy, the only way of maintaining the flywheel's kinetic energy is for the input energy to exceed the energy delivered to the load.
Diver300 posted a formula that should be useful, as well as a valid warning that the flywheel material has to be able to withstand the g-forces at the rim. i have seen videos of flywheels failing, some of them almost explosively. there's also a Mythbusters episode where they test the failure of CD and DVD disks at high speeds, and there were at least a couple of explosive failures at something like 50,000 rpm