If you have not any idea about your transformer, you can know your limits from its phisical dimensions. The transformer is either small or big.
If it is small, so you can load it and measue, you can use an oscilloscope to display the output waveform.
Connect the transformer, increase the load, when it reaches the maximum you will notice the output wave start to have flat top, transformer noise will increase, and over heating will occure. Ust 0.75 of that rating or less.
If it is big and the above is hard to do
Expect the output power should be about
Area in cm = sqrt (Power) in watt
i.e. 10 cm iron area is good for 100 Watt
Area is for Iron core not the coil, as it is better to mutiply thr real area of iron by .9 to take the lamination insulation out of effective atea..
Wire Diameter in mm = (.5 to .8) sqr (I) in amp
e.g 9 Amp need .8x3 2.4 mm wire diameter, use next higher standered.
0.5 is taken for outer coil with better ventillation
0.8 is taken for inner coil with less ventillation
0.3 sometimes used for oil immersed or forced air cooling
It is all based on heat dissipation in ohmic resistance
Now you need to know is it a primary/ secondary type or auto transformer (where either the primary or secondary is shared with the other), this auto configuration can increase your power by a ratio due to fact that some power is passed through without transformer action from primary to secondary, the transformer action is needed for the voltage difference between prim and sec. That ratio of increased power handling is
Power auto = power non auto x (v max - V min)/V max
as you see if "in Volt" = "Out Vout", Vmax-Vmin=0 and actually it is a simple coil, no transformer is needed, but if you use a primary/secondary you must transform the full power by the magnitic field for that same case...
Knowing your max rating be safe and use a portion of it
Forgot to add the above is for 50 Hz, for 60 Hz, the power increases by 60/50 or 1.2 for the same core area
Maged A. Mohamed