I just got back from a booster pump re-and-re. The new pumps are 3 25 HP 575V pump assembly with control - each pump is labeled with a 22.2A draw. The engineer drawings are specifying that we are to supply 3-#6 to the 3-pump assembly fused with a 80A time delay fuse. Looking immediately at table 13 - an 80A fuse is too large for #6 T90. Yet then there is 28-200(d) in the CEC. Here is my question - given that you can up a fuse size to 225% of the motor FLC if it does not start, does the conductor size have to be increased accordingly? It does not mention this a far as I can see in the CEC.

I realize on the face of it, that it appears the engineer didn't consult the code - I would size the conductor normally (22.2 * 1.25) + 22.2 + 22.2. = cable #4. It appears that one motor is a backup only and may not be used at all unless another is manually disconnected.

On a second look at the CEC all appears fine as long as only two motors are connected to operate at once - following 28-204(c) up to 300% of the conductor rating is permitted.

Davo,
What sort of system is this if it supplies a 575V motor?
Just inquisitive there mate.

I will reply to Trumpy first. 347/600 is the nominal system voltage. Very common in Canada and usually used where 277/480 is used in the US.
So to Davo 28-204 is the right rule and you are applying the right conditions. 125% if the largest motor with the FLC of each additional motor that must be able to run at the same time = wire size. The overcurrent rules allow a breaker as big as necessary provided it does not exceed %300 of the ampacity of the wire feeding the lot.
So 50 amp wire could have a breaker up to 150 amps. If you use #6 r90 then up to 190 amps, not 200. You probably have to start with a max of 100 amp breaker unless it trips, which lets you go up from there.