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How we can find the suitable cross section of a cable for a load if we don't have tools and documents (Ex the load is 15kw 3Ph )?

Electrical Electrical Engineering Electrical Design
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Question ajoutée par AHMAD AL-KHATEEB , Sr. Electrical Eng. , Engicon
Date de publication: 2016/03/11
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Francisco Generosa
par Francisco Generosa , Electrical design Engineer/ consultant , Al-fadhli Engineering and Consulting

we cannot size the suitable cross-section of cable without Voltage.

Thamarai kannan Natesan
par Thamarai kannan Natesan , Electronics Engineer , Ceramic pipes company

A 10H.P (7.46kW) three phase squirrel cage induction motor of continuous rating using Star-Delta starting is connected through 400V supply by three single core PVC cables run in conduit from 250feet (76.2m) away from multi-way distribution fuse board. Its full load current is 19A. Average summer temperature in Electrical installation wiring is 35°C (95°F). Calculate the size of the cable for motor?   Solution:-
  • Motor load = 10H.P = 10 x 746 = 7460W     *(1H.P = 746W)
  • Supply Voltage = 400V (3-Phase)
  • Length of cable = 250feet (76.2m)
  • Motor full load Current = 19A
  • Temperature factor for 35°C (95°F) = 0.97 (From Table 3)
Now select the size of cable for full load motor current of 19A (from Table 4) which is 7/0.36” (23 Amperes) *(Remember that this is a 3-phase system i.e. 3-core cable) and the voltage drop is 5.3V for 100Feet. It means we can use 7/0.036 cable according Table (4).   Now check the selected (7/0.036) cable with temperature factor in table (3), so the temperature factor is 0.97 (in table 3) at 35°C (95°F) and current carrying capacity of (7/0.036”) is 23 Amperes, therefore, current carrying capacity of this cable at 40°C (104°F) would be: Current rating for 40°C (104°F) = 23 x 0.97 = 22.31 Amp.   Since the calculated value (22.31 Amp) at 35°C (95°F) is less than that of current carrying capacity of (7/0.036) cable which is 23A, therefore this size of cable (7/0.036) is also suitable with respect to temperature. Load factor = 19/23 = 0.826   Now find the voltage drop for 100feet for this (7/0.036) cable from table (4) which is 5.3V, But in our case, the length of cable is 250 feet.  Therefore, the voltage drop for 250 feet cable would be;   Actual Voltage drop for 250feet = (5.3 x 250/100) x 0.826 = 10.94V And maximum Allowable voltage drop = (2.5/100) x 400V= 10V   Here the actual Voltage drop (10.94V) is greater than that of maximum allowable voltage drop of 10V. Therefore, this is not suitable size of cable for that given load. So we will select the next size of selected cable (7/0.036) which is 7/0.044 and find the voltage drop again. According to Table (4) the current rating of 7/0.044 is 28Amperes and the volt drop in per 100feet is 4.1V (see Table 4). Therefore, the actual voltage drop for 250feet cable would be;   Actual Voltage drop for 250feet = = Volt drop per 100feet x length of cable x load factor (4.1/100) x 250 x 0.826 = 8.46V And Maximum Allowable voltage drop = (2.5/100) x 400V= 10V The actual voltage drop is less than that of maximum allowable voltage drop. So this is the most appropriate and suitable cable size for Electrical wiring Installation of given situation.  

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