Electro-dynamics; by Carus-Wilson Charles Ashley 1860-

Author:Carus-Wilson, Charles Ashley, 1860- [from old catalog]
Language: eng
Format: epub
Tags: Electric motors, Direct current
Publisher: London, New York and Bombay, Longmans, Green and co.
Published: 1898-03-25T05:00:00+00:00

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second ; the torque will then rapidly diminish as the car speeds up; we may plot the curve as far as 26 feet per second.

From the records of actual performance we find that the average current at full speed is 70 amperes, giving a retarding torque of 5,760 inch-pounds, equal to a tractive eflTort of 42,000 pounds on 27-inch wheels. This includes all forces opposing the motion. Since the weight of the train is 35 tons, this corresponds to a tractive eflTort of 19*7 pounds per ton on wheels of 33 inches diameter.

Construct a curve B giving torque available for acceleration, by deducting from the horizontal ordinates of curved an amount equal to 5,760 inch-pounds; this curve will cut the axis of torque at ^=7,610 inch-pounds, and the axis of speed at 21*2 t.p.s. This gives us the final speed of the car, which is equivalent to 14'5 miles per hour.

Curve B shows that the accelerating torque is constant from the start up to 16 f p.s., that it then rapidly diminishes, until at 21*2 fp.s. it becomes nothing; the car then ceases to accelerate, and uniform speed is attained.

We here assume that the retarding torque remains constant at all speeds. This is not strictly correct, but it is probably not far from the truth. If the values of the retarding torques for diflferent speeds can be obtained from experiments, the true form of the curve of accelerating torque can be deduced. No experiments of this kind have been made with these motors; we shall therefore take the average value, and assume it to remain constant at all speeds.

Since the weight of the train is 35 tons, each motor

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will have to accelerate 17*5 tons. Using the Equation 84 we find the initial acceleration to be 0*463 f.p.s. per second. As the speed when the rheostat is all out is 16 f.p.s., the acceleration will remain constant for 34*6 seconds. Plot the first portion of curve G showing the speed in f.p.s. on a base of seconds. Let the point p^ represent the time when the rheostat is all out. Draw a line at right angles to op,, passing through the point q^y and cutting the speed base in s^. Draw a horizontal line passing through ^j and p^ cutting the speed base in ty Then q^t^ is the accelerating torque at the point j?i on the acceleration curve, and the tangent at the point p^ is at right angles to the line drawn through the points, q^ on the torque curve and s^ on the speed

We have then a general construction for drawing the acceleration curve. Take the point p,, where the acceleration ceases to be constant, draw a horizontal line to cut the torque curve in q^ and the speed base in t^. Measure ^j«i along the speed base, join g',5p and from p^ draw a line at right angles to q^s^, and continue this line for a period during which the acceleration may be assumed to be constant, say to the point p^.

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