3.8.  ILLUSTRATIVE PROBLEMS AND SOLUTIONS

This section provides a set of illustrative problems and their solutions to supplement the material in Chapter 3.

I3.1. The torque–speed curve of a two-phase ac servomotor is illustrated where it is assumed that this characteristic can be represented by a straight line:

Figure I3.1

(a)  Derive an equation for this torque–speed relationship, relating T and ω.

(b)  The motor-shaft output power is the product of the motor’s derived torque, T, and its speed, ω:

P = Tω.

Derive the speed, ω_max, where the motor output power is a maximum.

(c)  Derive an expression for the corresponding maximum motor-shaft output power, P_max.

(d)  Determine the maximum motor-shaft output power, P_max, if the no-load motor speed, ω_NL, is 5000 rev/min and the stall torque, T₀, is 4 in-oz.

SOLUTION: (a)                      

(b)  Substituting the equation for T derived in part (a) into the equation for the motor-torque output power

P = Tω

we obtain the following:

To find ω_max, we differentiate P with respect to ω as follows:

Therefore,

To check that this is a maximum, we take the second derivative of P as follows:

Since the second derivative is negative, ω_max is a maximum.

(c)  Substituting the expression found for ω_max in part (b) into the expression for motor-shaft output power,

P = Tω

we obtain the following:

Substituting the expression for ω_max found in part (b) into the expression for T in part (a), we obtain the following:

T_max = T₀/2.

Substituting this expression for T_max into the expression for P_max′ we obtain the following:

(d)      

Therefore,

P_max = 2.726 watts

and P_max occurs at a speed of 2500 rev/min.

I3.2. A control-system engineer needs to use an 8 : 1 gear reduction for a two-phase ac servomotor which is to be used for a positioning control system. The 8 : 1 gear reduction is to be made of two gear passes. The stock room has a limited number of gears available which can permit one of the following two configurations:

Figure I3.2

Each configuration has the same 8 : 1 gear reduction. The number of teeth on each gear is shown. At the instant of starting, the motor develops a torque T_D. The control-system engineer wants to use the gear configuration which will provide the higher initial load accelertion.

(a)  Write one parametric equation expressing the developed motor torque, T_D, in terms of the inertias, the gear reductions, and the load acceleration, a_L, with everything reflected to the motor shaft. (One parametric equation can represent both systems.)

(b)  Assuming that J_L = 16J₂ = 4J_m, determine which system has the higher initial load acceleration.

SOLUTION: (a) Reflecting a_L and all inertias to the motor shaft, we obtain the following:

where N₁ represents the gear ratio between J_m and J₂, and N₂ represents the gear ratio between J₂ and J_L.

Simplifying this equation, we obtain the following:

(b)  Setting the initial developed torques, T_D, of system A and system B equal to one another, substituting J_L = 16J₂ = 4J_m, and using the gear ratios indicated, we obtain the following:

Factoring out J_L and then canceling it from both sides of this equation, we obtain the following:

a_LA(2 + 0.132 + 0.125) = a_LB(2 + 0.1955 + 0.125).

Therefore,

a_LA(2.257) = (2.3205)a_LB

and

a_LA = 1.00281a_LB.

Therefore, we conclude that System A has the higher initial load acceleration.