The device itself runs on around 2-5uA when asleep. When awake draw is aroun 15-30mA. The device is awake for 1 second every 15 minutes.

OK, let's do some calculations.

I'm going to assume that your circuit will be powered directly from the supercap at 2.7V and can tolerate a supply voltage as low as 2.5V.

1 second every 15 minutes is a duty cycle of 1/(15*60) = 1/900 = 0.1111%. That percentage of 30 mA is 33 µA. Plus 5 µA continuous is 38 µA mean current drain.

If the supercap is fully charged at 2.7V and can discharge to 2.5V, that's a dV of (2.7 - 2.5) = 0.2V. You can calculate the amount of time a 10F supercap can power that circuit for, using the formula dT = C dV / I

= 10 * 0.2 / 38e-6

= 52,600 seconds

= 14.6 hours.

That's assuming worst case current consumption.

If I've made any incorrect assumptions, you'll need to correct them and rerun the calculations.

The second part i would like to have a go at is charging the device by wireless energy transfer.I am wondering if it is possible to transfer enough energy to keep the capacitor charged so power is available when required.

Mean power consumption is voltage multiplied by mean current, which is 2.7 * 38e-6 which is about 100 µW.

The transmitter for the wireless power transfer will be around 3-4 meters from the device to be powered.

This is outside my experience. Is it possible to transfer enough power, magnetically I assume, to produce 100 µW at the receiver at a distance of 3~4 metres? I can't answer that question.

If you have a continuous charging design, you don't need such a big capacitor; you just need one to hold the charge for 15 minutes or a bit longer. C = I dT / dV = 38e-6 * 15*60 / 0.2 = 0.17 farads.

Can you use some other method of getting the power into the device? Light, perhaps?