How can I discharge a fully charged 450V 47uF capacitor?
The capacitor stores 0.5 * V * V * C = 200v * 200v * 0.000047

Energy = 40,000 / 20,000 = 2 watt seconds. 
(oops forgot the 0.5 === ONE  watt second)

Thus a 1 watt resistor should easily survive, of any value.

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Consider a 10 ohm. The time constant will be C * R = 47 * 10 = 470 microsecond.

The current will be 200 volts / 10 ohms = 20 amps peak.

The power (peak, at start of the discharge) will be 200v * 20a = 4,000 watts.

But the discharge tails off very rapidly.

You can write the integral of the energy, time = 0 to time = infinity.