How powerful is your heating element? If we assume that your kettle is a closed system (a reasonable first-order assumption), then heat_in must equal heat_out. Assuming all other factors remain constant (e.g., radiant heat from the kettle should be the same regardless of how vigorous the boil, nozzle flow rate, etc.) then the change in water temperature should determine the change in heat_out. So delta_heat_out = delta_T * flow_rate * joules/degree = 60F * 9 gph * joules/degree. Converting to standard units delta_heat_out = 15.5C * 9 gallons/hour * 3785 grams/gallon * 1 hour/3600 seconds * 4.182 joules/gram*C = 613 J/s = 613 W
If delta_heat_in = 5% * heating_element_power, then this suggests you have a 12,267 W heating element, which is larger than any that I am familiar with. Do you have two 5500 W elements? That would likely be within the margin of error of my calculation.
Of course it is also possible that my calculation/assumptions are flawed. Maybe others can chime in.