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The previous descriptions of how ideal machines work were simplifications and are only approximately true. The discussions of hydraulic and electric motors require some re-interpretation because of the effects of adding the losses to the machine models.
For the hydraulic motor connected to a pump, flow from the pump will equal the flow into the motor. However, actual flows are not the same as the displaced (ideal) flow. The ideal equation will yield too high a motor speed because it does not account for internal (volumetric) leakage of the pump and of the motor.
N = 60 Q/D
Where N is the shaft speed in rpm,
Q is the hydraulic flow in volumetric units per second, and
D is the volumetric displacement per revolution.
A similar situation exists with the electrical motor. Recall that electrical power is the product of potential and current (E × I). With armature current, there will be a voltage drop (loss) across the armature resistance. The result is that steady-state speed will be slightly less than would be calculated under ideal conditions.
N = (60/2π) × E/(kmg × IF)
where N is the no-load speed in rpm,
E is the applied voltage to the motor input (armature) terminals in V,
Kmg is the motor conversion constant, and
IF is the field current.
However, more can be gleaned from these ideal equations. Furthermore, some important similarities must be pointed out. The first is that the displacement of the hydraulic motor is perfectly analogous to the field current in the electric motor.