Is this the correct replacement part ?
https://www.maplin.co.uk/p/2a-time-delay-glass-20mm-fuse-10-pack-gl62s
Yes, the Maplin fuse should be fine.
Obviously we're talking 12 volts but what rate would you charge it at ?
The recommended constant voltage for charging a sealed lead acid battery varies with temperature (and from manufacturer to manufacturer). At 20°C, the standby/float charge voltage should, typically, be 2.26V to 2.31V per cell or 13.56V to 13.86V for a 12V battery. To prevent battery damage during charging (e.g. from excessive gas production), the charge current should be limited to less than a quarter of the battery's capacity (Ah/4).
'I have previously replaced the battery in our alarm a couple of times with no issues but I installed yours today and although everything looked ok initially and the Low Battery warning disappeared, the alarm sounded after about an hour and the keypad did not illuminate or respond to key presses. I disconnected the new battery and reconnected the old one and someone was able to reset the alarm with the keypad. I don't know when the keypad started working again as I was nowhere near it when the code was entered.', if that gives any clues as to what originally happened ?
Is there a date code on the battery?
If the new battery was old stock (left on the shelf for a long time without a top-up charge), lead sulphate crystals could have formed on its plates - acting as a barrier to recharging. Initially, that would cause the charge current to be very low and the charge voltage high. That would be perceived by the alarm's controller as a fully charged battery (hence, the warning disappeared). As the resistive sulphation layer broke down, the charge current would increase. Unless your alarm's controller has the means to limit that current, the flat battery might have drawn sufficient current to starve the keypad, and other parts of the alarm system, of power.
A battery that has suffered mild sulphation can, usually, be recovered by charging it at higher voltage for 12 hours - with the current limited to
one tenth of the battery's capacity.
... I suppose that I could use my DP832. However, I've seen plenty of people on here blowing up their PSUs (including the DP832) when attempting to charge batteries/
You could protect your power supply by fitting a diode in series with its output. The charge voltage would need to be increased by the magnitude of the diode's forward voltage drop.