Regarding power supply for the PLC control cabinets in the workshop
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The workshop requires a PLC power cable with a power rating of 4KW at 220V. It is necessary to supply power from the power cabinet in the control room’s cabinet area (via the UPS circuit). The length of the cable needed is around 210 meters; it should be installed in galvanized cable trays for instrumentation, and separated from signal cables by partitions. My question is: what size cable should I choose? Is the specification ZR-YJV-0.6/1KV 3×6 acceptable? How is the intermediate pressure drop loss calculated? It’s quite difficult to have the meter technician calculate these. Is there a manual available?
At an operating temperature of 30°C, the current-carrying capacity under a continuous 90% load for a long period is as follows: 1.5 square millimeters – 18A； 2.5 square millimeters – 26A ； 4 square millimeters – 26A ； 6 square millimeters – 47A ； 10 square millimeters – 66A 16 square millimeters – 92A ； 25 square millimeters – 120A ； 35 square millimeters – 150A. Power P = Voltage U × Current I = 220 volts × 18 amps = 3960 watts. Values for the current load that wires can handle, as specified in standard GB4706.1-1992/1998 (partial list): For copper-wired cables: Cross-sectional area of the copper wire… Allowable continuous current: 2.5 square millimeters (16A–25A); 4 square millimeters (25A–32A); 6 square millimeters (32A–40A). For aluminum-wired cables: Cross-sectional area of the aluminum wire… Allowable continuous current: 2.5 square millimeters (13A–20A); 4 square millimeters (20A–25A); 6 square millimeters (25A–32A). The formula for calculating voltage drop in cables is ΔU = (P × L) / (A × S), where P represents the load on the circuit ； L is the length of the line ； A is the conductor material coefficient (around 77 for copper, around 46 for aluminum) ； S is the cross-sectional area of the cable.