FLYWHEEL ENERGY Figure 9-24 shows a flywheel , designed as a flat circular disk, attached to a motor shaft that might also be the camshaft. The motor supplies a torque magnitude TM that we would like to be as constant as possible, i.e., to be equal to the average torque Tavg . The load (the follower system), on the other side of the flywheel, demands a torque TL which is time varying, as shown in Figure 9-17 (p. 242). The kinetic energy in a rotating system is:

where I is the moment of inertia of all rotating mass on the shaft. This includes the I of the motor rotor and of the cams and camshaft plus that of the flywheel. We want to determine how much I we need to add in the form of a flywheel to reduce the speed variation of the camshaft to an acceptable level. We begin by writing Newton’s law for the free-body diagram in Figure 9-24.

The left side of equation 9.14c represents the change in energy E between the maximum and minimum shaft ’s and is equal to the area under the torque-time diagram * between those extreme values of . The right side of equation 9.14c is the change in energy stored in the flywheel. The only way we can extract energy from the flywheel is to slow it down, as indicated by equation 9.13 (p. 253). Adding energy will speed it up. Thus it is impossible to obtain exactly constant camshaft velocity in the face of changing energy demands by the follower load. The best we can do is to minimize the speed variation ( max – min ) by providing a flywheel with sufficiently large I .

* There is often confusion between torque and energy because they appear to have the same units of lb-in ( inlb ) or N-m (m-N) . This leads some students to think that they are the same quantity, but they are not: torque energy. The integral of torque with respect to angle, measured in radians, is equal to energy. This integral has the units of in-lb-rad . The radian term is usually omitted since it is in fact unity. Power in a rotating system is equal to torque x angular velocity (measured in rad/sec ), and the power units are then (in-lb-rad)/sec . When power is integrated versus time to get energy, the resulting units are in-lb-rad , the same as the integral of torque versus angle. The radians are again usually dropped, contributing to the confusion.

EXAMPLE 9-5

Determining the Energy Variation in a Torque-Time Function.

Given: An input torque-time function that varies over its cycle; Figure 9-25 shows the input torque curve from Figure 9-23. The torque is varying during the 360 ° cycle about its average value.

Problem: Find the total energy variation over one cycle.

Solution:

1 Calculate the average value of the torque-time function over one cycle, which in this case is 70.2 lb-in.

2 Note that the integration on the left side of equation 9.14c is done with respect to the average line of the torque function, not with respect to the  axis. (From the definition of

the average, the sum of positive area above an average line is equal to the sum of negative area below that line.) The integration limits in equation 9.14c are from the shaft angle  at which the shaft  is a minimum to the shaft angle  at which  is a maximum.

3 The minimum  will occur after the maximum positive energy has been delivered from the motor to the load, i.e., at a point () where the summation of positive energy (area) in the torque pulses is at its largest positive value.

4 The maximum  will occur after the maximum negative energy has been returned to the load, i.e., at a point () where the summation of energy (area) in the torque pulses is at its largest negative value.

5 To find these locations in  corresponding to the maximum and minimum ’s and thus find the amount of energy needed to be stored in the flywheel, we need to numerically integrate each pulse of this function from crossover to crossover with the average line. The crossover points in Figure 9-25 have been labeled A, B, C, and D. (Program DYNACAM does this integration for you numerically, using a trapezoidal rule.)

6 The DYNACAM program prints the table of areas shown in Figure 9-25. The positive and negative pulses are separately integrated as described above. Reference to the plot of the torque function will indicate whether a positive or negative pulse is the first encountered in a particular case. The first pulse in this example is a positive one.

7 The remaining task is to accumulate these pulse areas beginning at an arbitrary crossover (in this case point A ) and proceeding pulse by pulse across the cycle. Table 9-3 shows this process and the result. There is some numerical error in the integration. The last value should be zero.

8 Note in Table 9-3 that the minimum shaft speed occurs after the largest accumulated positive energy pulse (+200.73 in-lb) has been delivered from the driveshaft to the system.

This delivery of energy slows the motor down. The maximum shaft speed occurs after the largest accumulated negative energy pulse (–60.32 in-lb) has been received back from the system by the driveshaft. This return of stored energy will speed up the motor. The total energy variation is the algebraic difference between these two extreme values, which in this example is –261.05 in-lb. This negative energy coming out of the system needs to be absorbed by the flywheel and then returned to the system during each cycle to smooth the variations in shaft speed.