CHAPTER 12

ELECTRICAL PROBLEMS

To solve electrical problems the A and B scales are used to do the working, and the answers read off on what are known as the Efficiency and Voltage Drop scales, but there are instances when a setting is made on these latter scales, and the answer read off the A or B scales.

As with the trigonometrical scales, so with the electrical scales, manufacturers of slide rules arrange the Voltage Drop and Efficiency scales rather differently. One type has these scales in the trough of the rule, and settings are made with a metal indicator attached to the end of the slide, whereas another type of rule has a special pair of A and B scales, and the Voltage Drop scale is read by means of the hair line, and the Efficiency scales by means of an indicating arrow. In principle, both of these types are very similar.

It must be noted that these scales are only applicable to direct current calculations, or for alternating current that is free from induction.

EFFICIENCY OF DYNAMOS

For this purpose the A scale is taken as the kilowatt scale, and the B scale as the horse-power scale, commencing at 10 horse-power.

Set 10 horse-power under 746 (usually marked as a constant), on the Kw. scale, and it will be seen that the efficiency shown by the indicator is, 100%. Thus, with the rule in this position, all ratios between the Kw. and H.P. scales represent 100% efficiency.

As the slide is moved to the left, thus giving lower kilowatt readings per horse-power, the indicator shows a decreasing efficiency.

From the foregoing it will be seen that having made a setting on the Efficiency scale, the dynamo output in kilowatts will be shown on the Kw. scale directly over the horse-power, on the H.P. scale, needed to drive the dynamo.

EXAMPLE: What would be the output with 25 horse-power from a dynamo of 75% efficiency?

Set the indicator to 75% on the Dynamo Efficiency scale, and corresponding with 25 on the H.P. scale, the answer 14.0 kilowatts will be found on the Kw. scale. (Fig. 15.)

EXAMPLE: What is the efficiency of a dynamo giving 13.8 kilowatts with an input of 23 horse-power?

Set the hair line to 13.8 on the Kw. scale and bring 23 on the H.P. scale to correspond with it, and then read off the indicator 80.5% on the Dynamo Efficiency scale.

EXAMPLE: What horse-power would be need to drive a dynamo 75% efficient supplying 40 kilowatts?

Set the indicator to 75% and on the H.P. scale, opposite 40 on the Kw. scale, read the answer as 71.5 horse-power.

EFFICIENCY OF MOTORS

The Motor Efficiency scale is used in a similar manner to the Dynamo Efficiency scale to determine the consumption of a motor of known horse-power. As before, the A and B scales are taken as the Kilowatt and H.P. scales.

EXAMPLE: Find the efficiency of a motor which, with 19 kilowatts, delivers 19.9 horse-power.

Line up these two values on their proper scales, and off the indicator read the answer 78% on the Motor Efficiency scale.

EXAMPLE: Ascertain the power in kilowatts, required by a motor developing 41 horse-power, having an efficiency of 84%.

Set the indicator to 84% on the Motor Efficiency scale and opposite 41 on the H.P. scale read the answer on the Kw. scale as 36.4 kilowatts.

EXAMPLE: What horse-power would a 70% efficient motor deliver with 25 kilowatts?

Set the indicator this time to 70% on the Motor Efficiency scale, and opposite 25 on the Kw. scale read the answer on the H.P. scale as 23.5 horse-power.

VARIATION OF RESISTANCE WITH TEMPERATURE OF A COPPER CONDUCTOR

This is determined by using the small Temperature scale situated sometimes in the centre, and sometimes at the right-hand end of the slide.

British Standards Specification No. 7, accepted by the Standards Association of Australia, determines the effect on the resistance of a copper conductor for temperatures ranging between 40° and 130° F., standard conditions being 60° F.

[Only slide rules with temperature engravings from 40° to 130° F. confirm to this specification. Other rules are either marked in °C. or in Fahrenheit equivalents of 0, 15, 25, 50 and 70° C., a range greater than that covered in B.S.S. No. 7.
The following remarks and those referring to the Voltage Drop scales will only apply to rules correct to B.S.S. No. 7.]

By setting 60° F. on this scale to the resistance in ohms on the Kw. scale, the resistance at other temperatures between 40° and 130° F. can be found. similarly, if the resistance is known at any temperature between 40° and 130° F. its value at the standard temperature of 60° F. can be obtained.

EXAMPLE: The resistance of a copper conductor at 60° F. is 11 ohms. What is the resistance at 80° F.?

Set 60° on the Temp. scale under 11 on the Kw. scale, and opposite 80° read off the answer of 11.5 ohms.

EXAMPLE: What is the resistance at 60° F. of a conductor having a resistance of .5 ohms at 100° F.?

Set 100° on the Temp. scale under 5 on the Kw. scale, then the required resistance of .459 is found opposite 60°.

The Temperature scale can also be used to vary the cross sectional area of a conductor whose resistance at working temperature must not exceed a certain value, when the design of a conductor has been carried out at the standard temp. of 60° F.

EXAMPLE: A conductor of .02 square inches is found necessary at a temperature of 60° F. What would be the area necessary for a working temperature of 50° F., assuming that the resistance is to be kept constant?

As the resistance is reduced for a drop in temperature, a smaller conductor would be satisfactory.

Set 60° to 20, i.e., .02 sq. ins. on the Kw. scale and read off 19.9, i.e., .0196 sq. ins., opposite 50° on the Temperature scale.

VOLTAGE DROP

The voltage drop is equal to the current flowing in amps multiplied by the resistance of the conductor in ohms; this resistance varies directly with the length of the conductor and inversely with the cross-sectional area.

The Voltage Drop scale is so constructed that the drop is given for a copper conductor having a resistance of .0240079 ohms for a length of 1000 yards and a cross-sectional area of one square inch, which is the B.S.S. figure for a temperature of 60° F.

In order to simplify the calculation the current scale has been graduated from 1 to 100 to correspond with currents of from 10 to 1000 amps, the left-hand index being marked 10 amps.

The length and cross-sectional area scale has been graduated in a similar way to correspond to lengths of from 1 to 100 yards, and areas from .001 to .1 sq. inches, with the left-hand index marked 1 yd. and .001 sq. ins.

The voltage drop is obtained by the following formula:-

The Voltage Drop scale is so placed with reference to the other scales that the multiplication by the resistance of copper is automatically performed in reading the answer on the Voltage Drop scale.

EXAMPLE: What is the voltage drop when 30 amps flow along a conductor 50 yards in length having a cross sectional area of .0045 sq. ins.

Set the hair line to 3 on the amps scale, i.e., 30 amps, and bring the 4.5, i.e., .0045 sq. ins on the slide to correspond with it. Move the hair line along to 50 yards and read off the corresponding value on the Voltage Drop scale of 8.1 volts.

EXAMPLE: What is the voltage drop when 20 amps flow along a conductor 1200 yards in length, having a cross-sectional area of .25 sq. ins.?

From a knowledge of the rule it is obvious that 1200 yards cannot be set on the rule, nor can .25 sq. inches; however, as the resistance of the conductor and also the voltage drop vary directly with the length, and inversely with the cross-sectional area, a length of 12 yards and an area of .0025 sq. inches will give the same result.

Set the hair line to 20 amps and bring .0025 sq. ins. under it. Move the hair line along to 12 yards and read the result under it as 2.32 volts.

EXAMPLE: What is the voltage drop in a conductor 900 yards long with an area of .0145 sq. ins. when 2.5 amps flow along it?

The current of 2.5 amps cannot be directly read on the rule, but as the voltage drop varies directly as the length and current flowing in the conductor, 2.5 amps and 900 yards would have the same effect as 25 amps along 90 yards. The problem then is to find the voltage along the shorter length with the greater current and the same cross section.

Set the hair line to 25 amps on the amps scale and bring .0145 sq. ins. on the slide to correspond with it. Move the hair line along to 90 yards and read the result under it as 3.75 volts.

These scales can also be used to find the cross sectional area for a fixed voltage drop, the formula being:

In this case also, the slide automatically multiplies by the resistance of copper.

EXAMPLE: What is the area required to carry 120 amps 50 yards with a drop of 2.5 volts?

Set the hair line to 2.5 volts and bring 50 yards on the length scale to correspond with it, then corresponding with 120 amps will be found the answer .0582 sq. inches.

EXAMPLE: What is the area required to carry 6.2 amps 600 yards for a voltage drop of 1 volt?

As 6.2 amps cannot be found directly on the rule, the problem can be re-stated as the area to carry 62 amps 60 yards for 1 volt drop, because the drop varies directly as the current and the length of cable.

Set the hair line to 1 volt and bring 60 yards to correspond with it, and read off the area .090 sq. inches opposite 62 amps.

All of the above examples have been for the standard resistance of 60° F. Voltage drops for the other temperatures can readily be found by the use of the Temperature scale.

EXAMPLE: Find the area required to carry 6.2 amps 600 yards for a voltage drop of 1 volt at a temperature of 90° F.

As the resistance and voltage drop increase with temperature, the permissible voltage drop at 90° F. will be reduced for the standard temperature of 60° F.

Set the hair line to 1 volt and bring 90° on the Temperature scale to correspond with it. Then move the hair line to 60° and bring 60 yards to correspond with it and read off .096 sq. inches opposite 62 amps.

EXERCISES ON ELECTRICAL PROBLEMS

62. What is the output of a dynamo of 90% efficiency with an input of 25 horse-power?

63. What is the efficiency of a dynamo giving 15 kilowatts with an input of 27 horse-power?

64. What is the efficiency of a 25 horse-power motor having an input of 20 kilowatts?

65. What is the input of a 30 horse-power motor which has an efficiency of 85%?

66. What is the voltage drop when 20 amps flow along a conductor 45 yards in length having a cross-sectional area of .0030 sq. ins.?

67. What is the voltage drop when 25 amps flow along a conductor 1500 yards in length, having a cross-sectional area of .35 sq. ins.?

ANSWERS TO EXERCISES

62. 16.8 K.W.

63. 74.5%

64. 93.3%

65. 26.3 K.W.

66. 7.3 Volts

67. 2.6 Volts