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TABLE 7.2 Area Sums for Example 2
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FLYWHEELS 7.7
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7.2.3 Coefficient of Energy Variation The torque-angle relationship for an engine depends on the fuel, gas pressures, reciprocating masses, speed, and engine geometry [7.2]. The large variation that is possible between different engine designs shows that dynamic measurement or kinematic analysis is necessary to determine the torque fluctuation. It is often necessary, however, to come up with a rough estimate for preliminary design purposes or for checking the reasonableness of calculated values. For these purposes, the energy variation for an internal-combustion engine can be estimated by U = Cu KP (7.10)
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where K = 33 000 lb ft rpm/hp [2 J rad/(W s)]. The coefficient of energy variation Cu can be approximated for a two-stroke engine with from 1 to 8 cylinders using the equation Cu = 7.46 (Nc + 1)3 (7.11)
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and for a four-stroke engine with from 1 to 16 cylinders using the two-branched equation Cu = 0.8 0.015 |Nc 1.4|1.3 (7.12)
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Example 3. A 150-hp four-cylinder, four-stroke engine has a flywheel speed of 1000 rpm. Estimate the flywheel necessary for a 2 percent speed variation with a uniform load at an engine speed of 3000 rpm, neglecting the flywheel effect of the other rotating parts. Using Eq. (7.12), Cu = Then from Eq. (7.10), U = 0.22 33 000(150) = 363 lb ft 3000 (7.14) 0.8 0.015 = 0.22 |4 1.4|1.3 (7.13)
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so that from Eq. (7.2), with = 2 (1000)/60 = 105 rad/s, J= 363 = 1.6 lb s2 ft 1052(0.02) (7.15)
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7.2.4 Angular Fluctuation Certain machines, such as electric generators and magnetic digital storage systems, must maintain their angular position within a close tolerance of the constant-speed position. If the torque is known as a function of time, it can be integrated to deter-
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FLYWHEELS 7.8
MACHINE ELEMENTS THAT ABSORB AND STORE ENERGY
mine the angular velocity, and then the angular velocity can be integrated to give the angular position:
(t) =
T( ) d + 0 J ( ) d + 0t + 0
(7.16) (7.17)
(t) =
where the 0t + 0 term represents the constant-speed position. In the more usual instance, the torque is known only as a function of angle. For small values of Cs , however, the torque-time curve is indistinguishable from the torque-angle curve with the angle coordinate divided by avg. Example 4. A generator with the input torque given in Fig. 7.3a must maintain an angular position within 0.25 degrees of the uniform 200-rpm position. Assuming a uniform load, what flywheel inertia is necessary For illustration purposes, the machine cycle will be divided into 10 intervals of t = 0.03 s each, as shown in Fig. 7.3a. For an accurate solution, the problem would be programmed with perhaps 20 intervals. The torque at each step is tabulated (column 3 in Table 7.3), and then the average torque in each interval is placed in column 4. This value, if multiplied by t, would be the area below the curve using the trapezoid rule. Adding these average torques (column 5) and dividing by 10 intervals gives the average torque for the curve, 902 lb ft (column 6), shown as the dashed line in Fig. 7.3a. Subtracting this average, the constant loading torque, from column 4 gives column 7, the average excess of supplied torque in each interval. The running sum of these values (column 8) performs the integration, to give J / t (see Fig. 7.3b). The relative speed at the end of each interval is therefore the value in column 8 times t/J. The procedure is repeated for the second integration, giving columns 9 through 13. Column 13 is then J /( t)2 (Fig. 7.3c), so that the relative angular position is the value in column 13 times t 2/J. The maximum range in column 13 is 6915 ( 7725) = 14 640 lb ft. The maximum angular deviation from the mean position is calculated from half the maximum range, so that max = ( t)2 (14 640) J(2) (7.18)
For max = 0.25 degrees = 0.004 36 rad deviation, this gives J= 0.032(14 640) = 1511 lb s2 ft 0.004 36(2) (7.19)
The speed variation is determined as a by-product of the process. The maximum range in column 8 is 9878 0 = 9878 lb ft. The maximum speed variation is then max min = t(9878) 0.03(9878) = = 0.196 rad/s J 1511 (7.20)
For avg = 2 (200)/60 = 20.94 rad/s, the coefficient of speed fluctuation is then, from Eq. (7.1), Cs = 0.196 = 0.009 36 20.94 (7.21)
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