Trying to answer but YA taking breather. Or is it???
No it just doesn't want to display my answer. Try again...
I may be reading your question wrong, but it looks like you want to charge the capacitor from the power supply, in such a way that if the power goes off the voltage on the capacitor will hold the relay closed for 30-60 seconds, until the capacitor discharges.
That would work, but it is clunky, and you will need a big capacitor.
Assuming that the relay will continue to hold closed until the voltage drops to about 60% of normal,
C = 2 * T / R where R = the coil resistance of the relay
Let's say you have a very small, low power relay with coil resistance 1200 ohms. For 30 seconds:
C = 60 / 1200 = 0.05F = 50,000uF
You would probably go for an aluminium electrolytic cap 47,000 uF, 16V.
If your relay has a lower resistance coil, which it probably does, you will need a bigger capacitor, or several in parallel.
You may need to charge the cap through a rectifier diode like 1N4001, so that when the power fails it doesn't discharge back through the power supply.
The way I would do it is to use a darlington transistor, FET or solid-state relay to switch the line. Then you can do the same thing but with a much smaller capacitor, and you can run it from a battery or power direct from mains through a capacitor / rectifier circuit.
Don't try that latter idea unless you know what I'm talking about, it's very easy to kill yourself...