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Relationship between torque and density in agitator
An agitator is essentially a centrifugal load. This means that the powertospeed ratio is typically a square function. In most agitators, the actual relationship between torque and density is:
Torque = (Shaft Power) * (Fluid density) * (Shaft speed)^2 * (Impeller diameter)^5 / (2 * pi)
So ... as the density increases by 15 percent, the required torque will also increase by 15 percent. Which, given the formula above, means the shaft power will also increase by 15 percent  and require at least 15 percent higher current (since the efficiency and power factor go down as loading increases above 100 percent of nameplate rating due to thermal effects).
It sounds to me like someone on the agitator side didn't do their homework on the fluid properties  or didn't choose the correct diameter impeller for the agitator. On the motor side, if the machine is running slightly fast (i.e. lowerthanexpected cage resistance in the rotor, for example) the power draw  and, consequently, the current draw  will also be higher than expected.
As far as being able to operate continuously in the aboveunity service factor region: machines are rated in one of two ways. Most often, the rating is specified as an alowable rise at 1.0 service factor, with noninjurious heating at the aboveunity condition. Other instances have a slightly higher allowable rise at the continuous overload condition, which automatically means the machine will have lesser rise at unity service factor.
A class B rise according to NEMA MG 1 corresponds to an 80 C rise above ambient at unity  or a 90 C rise at continuous overload. Note that only only one of the two can be actually specified! This is because temperatures are more or less related to the square of the current. Therefore as the current increases (by 15 percent), the heat generated will increase by (1.15^2) = 1.323 or approximately 3032 percent. For a machine already operating at the allowable temperature rise (say 80 C over ambient), this would mean an increase in operating temperature of roughly 25 C. This would put the machine well past the maximum allowable temperature for class B and into the class F operating range.
Having the higher class H insulation in the machine is helpful, certainly. But it is not the real solution. It just means the machine will last a little longer, as the thermal degradation of the insulation proceeds over the life of the equipment. A far better solution is to get the right size equipment (from the original equation above), and install it  whether that is the motor or the impeller or both.
Regardless whether the choice is to "get it right" or just "run the machine into the ground", the relay will need to change to a higher current rating.
Torque = (Shaft Power) * (Fluid density) * (Shaft speed)^2 * (Impeller diameter)^5 / (2 * pi)
So ... as the density increases by 15 percent, the required torque will also increase by 15 percent. Which, given the formula above, means the shaft power will also increase by 15 percent  and require at least 15 percent higher current (since the efficiency and power factor go down as loading increases above 100 percent of nameplate rating due to thermal effects).
It sounds to me like someone on the agitator side didn't do their homework on the fluid properties  or didn't choose the correct diameter impeller for the agitator. On the motor side, if the machine is running slightly fast (i.e. lowerthanexpected cage resistance in the rotor, for example) the power draw  and, consequently, the current draw  will also be higher than expected.
As far as being able to operate continuously in the aboveunity service factor region: machines are rated in one of two ways. Most often, the rating is specified as an alowable rise at 1.0 service factor, with noninjurious heating at the aboveunity condition. Other instances have a slightly higher allowable rise at the continuous overload condition, which automatically means the machine will have lesser rise at unity service factor.
A class B rise according to NEMA MG 1 corresponds to an 80 C rise above ambient at unity  or a 90 C rise at continuous overload. Note that only only one of the two can be actually specified! This is because temperatures are more or less related to the square of the current. Therefore as the current increases (by 15 percent), the heat generated will increase by (1.15^2) = 1.323 or approximately 3032 percent. For a machine already operating at the allowable temperature rise (say 80 C over ambient), this would mean an increase in operating temperature of roughly 25 C. This would put the machine well past the maximum allowable temperature for class B and into the class F operating range.
Having the higher class H insulation in the machine is helpful, certainly. But it is not the real solution. It just means the machine will last a little longer, as the thermal degradation of the insulation proceeds over the life of the equipment. A far better solution is to get the right size equipment (from the original equation above), and install it  whether that is the motor or the impeller or both.
Regardless whether the choice is to "get it right" or just "run the machine into the ground", the relay will need to change to a higher current rating.