Most Economical Size of Cable

In planning the layout of a factory distribution system, the four limitations stated on, page 55 relative to conductor size must be considered. The required mejchanical strength and the maximum allowable current-carrying capacity are both specified in the National Electric Code of the National Board of Fire Underwriters. The voltage drop allowable is fixed by the voltage variation which may be tolerated at the load. However, these three requirements may be satisfied by a conductor that is too small to meet the requirement of Lord Kelvin's Law, which is: For maximum economy the yearly cost of energy dissipated in the conductor must be equal to the yearly charges (depreciation, interest, taxes, and repairs) against the investment in the conductor.

Example 23-2. - In a large cement plant it is necessary to provide three-phase power at 500 amp to a 440-volt motor located 500 ft from the transformer substation. The cost of energy is $0.02 per kw-hr. The load is to be carried 12 hours every day of the year. Determine the proper size of conductor to be used. ;

Solution. - If rubber-covered cable is used on brackets in open air inside the building, the National Electric Code requires that a 700,000 cir mil cable be used to carry 500 amp. The resistance of 700,000 cir mil cable at 75 C is 0.0197 ohm per 1000 ft Resistance of 500 ft = 0.00985 ohm IR loss in each cable = 500 X 0.00985 = 4.92 volts

Voltage from line to neutral of a 440-volt motor is

Maximum tolerable voltage reduction at the motor terjminals is 10%, and the tolerable reduction in each conductor is

Phase displacement between line voltage drop and motor voltage may be neglected. Hence, an actual drop of 4.92 volts is well within the allowable limit. The mechanical strength is certainly sufficient. ;

Kelvin's Law should next be applied.

The cost of energy lost per conductor per year is I₂Rt times the cost per kilowatt-hour. Thus,

The cost of 700,000 cir mil cable is $527.80 per 1000 ft. For 500 ft, the cost is

This is the cost of one conductor. The fixed charges against such an investment with an unknown salvage value may be taken at 12%. The life expectancy is 20 years. The annual fixed charge is

Since the energy loss costs $215.71 per year, a larger cable will be more economical. The cable loss will be inversely proportional to the area of the cross-section. The cost of cable within a small range may be assumed to be directly proportional to the area of the cross-section.

Let X = the new cross-section. Then the new annual energy cost in terms of the new cross-section is

(a) Also, the annual fixed charge on the new cable is

(b) By Kelvin's Law, cost (a) must equal charge (b), Therefore,

and

The nearest commercial size is 1,800,000 cir mils. This, then, would be the most economical size of conductor.

The assumption of linear variation of cost of conductor with conductor cross-section cannot be justified exactly, but the error from such assumption is negligibly small. Because of difficulty in handling very large conductors it is sometimes advisable to substitute for each conductor two or more conductors in parallel.1

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Most Economical Size of Cable Author: E.E. Kimberly In planning the layout of a factory distribution system, the four limitations stated on, page 55 relative to conductor size must be considered. The required mejchanical strength and the maximum allowable current-carrying capacity are both specified in the National Electric Code of the National Board of Fire Underwriters. The voltage drop allowable is fixed by the voltage variation which may be tolerated at the load. However, these three requirements may be satisfied by a conductor that is too small to meet the requirement of Lord Kelvin's Law, which is: For maximum economy the yearly cost of energy dissipated in the conductor must be equal to the yearly charges (depreciation, interest, taxes, and repairs) against the investment in the conductor. Example 23-2. - In a large cement plant it is necessary to provide three-phase power at 500 amp to a 440-volt motor located 500 ft from the transformer substation. The cost of energy is $0.02 per kw-hr. The load is to be carried 12 hours every day of the year. Determine the proper size of conductor to be used. ; Solution. - If rubber-covered cable is used on brackets in open air inside the building, the National Electric Code requires that a 700,000 cir mil cable be used to carry 500 amp. The resistance of 700,000 cir mil cable at 75 C is 0.0197 ohm per 1000 ft Resistance of 500 ft = 0.00985 ohm IR loss in each cable = 500 X 0.00985 = 4.92 volts Voltage from line to neutral of a 440-volt motor is Maximum tolerable voltage reduction at the motor terjminals is 10%, and the tolerable reduction in each conductor is Phase displacement between line voltage drop and motor voltage may be neglected. Hence, an actual drop of 4.92 volts is well within the allowable limit. The mechanical strength is certainly sufficient. ; Kelvin's Law should next be applied. The cost of energy lost per conductor per year is I₂Rt times the cost per kilowatt-hour. Thus, The cost of 700,000 cir mil cable is $527.80 per 1000 ft. For 500 ft, the cost is This is the cost of one conductor. The fixed charges against such an investment with an unknown salvage value may be taken at 12%. The life expectancy is 20 years. The annual fixed charge is Since the energy loss costs $215.71 per year, a larger cable will be more economical. The cable loss will be inversely proportional to the area of the cross-section. The cost of cable within a small range may be assumed to be directly proportional to the area of the cross-section. Let X = the new cross-section. Then the new annual energy cost in terms of the new cross-section is (a) Also, the annual fixed charge on the new cable is (b) By Kelvin's Law, cost (a) must equal charge (b), Therefore, or and The nearest commercial size is 1,800,000 cir mils. This, then, would be the most economical size of conductor. The assumption of linear variation of cost of conductor with conductor cross-section cannot be justified exactly, but the error from such assumption is negligibly small. Because of difficulty in handling very large conductors it is sometimes advisable to substitute for each conductor two or more conductors in parallel.1
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