Power cable sizing for a motor!

[ieee]

Hello everybody,

I have a pump driven by a Heemaf motor 380VAC, 196 A current load, I wonder how big the cable i want to use to supply this motor from 450 meters far power supply.

Thank you

 

[Sahib]

Hello everybody,

I have a pump driven by a Heemaf motor 380VAC, 196 A current load, I wonder how big the cable i want to use to supply this motor from 450 meters far power supply.

Thank you

Are you following IEC guidelines? If yes, the cable size, in addition to having current capacity and allowance for voltage drop for motor current of 196A for a distance of 450 meters, should also be adequate for the fault level at motor terminals.

 

[JFletcher]

Hello everybody,

I have a pump driven by a Heemaf motor 380VAC, 196 A current load, I wonder how big the cable i want to use to supply this motor from 450 meters far power supply.

Thank you

Welcome. Assuming a 3ph motor, you'd need 4/0 copper to keep VD <10%, 600MCM to keep it below 5%, and 2 500MCM per phase for <3%. Here's a calculator:

http://www.southwire.com/support/voltage-drop-calculator.htm

At 450M (1476') with such a high loading, stepping up/down the voltage may result in a less expensive install overall tho I'd have to look at xfmr costs, conduit sizes, wire costs, labor, etc. Going 380 to 600V gets you to 1/0 CU at 8% VD vs 4/0.

eta: figuring on a 150kVA xfmr (x2), it's going to be cheaper just to use more wire and conduit. AL wire and RPVC are dirt cheap compared to copper and transformers. Getting the POCO to get their xmfr closer is worth looking into tho...

 

[Julius Right]

Let's say the cable is PVC insulated underground [30oC ambient, RHO 250].
According to IEC 60364-5-52 Table A.52-4 for 203 A 3*120 mm^2 copper conductor is recommended.
Voltage drop formula it is as following:
DV=1.73*Irat*[r*cos(fi)+x*sin(fi)]*length/1000
Let's use BS7671 Table 4D2B for 70oC [PVC] insulated copper conductor cable.
This standard shows the resistance and the reactance multiplied already by 1.73
So for 3*120 mm^2 we have r=0.33 and x=0.135 mV/A/m=ohm/km
An induction motor of 110 kW presents pf[cos(fi)]=0.85 then sin(fi)=sqrt(1-0.85^2)=0.527
DV=196*(0.33*0.85+0.135*0.527)*450/1000=30.9 V or 30.9/380*100=8.13%
The maximum permissible voltage drop it is 5%.Then you have to choose a bigger
conductor.
If you’ll take 3*240 mm^2 then r=0.165 x=0.13 then DV=18.35 V 18.35/380*100=4.83%
Now, the problem will be at start. If your motor is provided with VFD drive then this cable is good. If you have to start it D.O.L.[direct-on-line] then the pf=0.25 but the current will be 7*196 A so you need 4 parallel cables of 3*240 mm^2 copper[15% voltage drop].

 

[Sahib]

Now, the problem will be at start. If your motor is provided with VFD drive then this cable is good. If you have to start it D.O.L.[direct-on-line] then the pf=0.25 but the current will be 7*196 A so you need 4 parallel cables of 3*240 mm^2 copper[15% voltage drop].

I think DOL contactor will hold for voltage drop of up to 35℅. So such oversizing cable may not be necessary.

 

[Julius Right]

Thank you, Sahib. From our experience-power stations-the starting voltage has to be so high in order to produce at least 50% of the rated torque.
With respect to IEC 60034-12 Rotating_electrical machines Part 12: Starting performance of single-speed three-phase cage induction motors
Table 1 - Minimum values of torques for design N. the minimum voltage has to be:
For motor power equal or less than 16 kW 35%
For motor power equal or less than 160 kW 20%
That is recalculated on basis Tq=(Vstart/Vrated)^2*Ti [Ti= locked rotor torque-minimum]
However, it is not compulsory and it has to be verified for the specific case.

 

[topgone]

Thank you, Sahib. From our experience-power stations-the starting voltage has to be so high in order to produce at least 50% of the rated torque.
With respect to IEC 60034-12 Rotating_electrical machines Part 12: Starting performance of single-speed three-phase cage induction motors
Table 1 - Minimum values of torques for design N. the minimum voltage has to be:
For motor power equal or less than 16 kW 35%
For motor power equal or less than 160 kW 20%
That is recalculated on basis Tq=(Vstart/Vrated)^2*Ti [Ti= locked rotor torque-minimum]
However, it is not compulsory and it has to be verified for the specific case.

A 3 X 240mm2 will do. The OP can be assured all other requirements are complied with.

 

[Julius Right]

A 3 X 240mm2 will do. The OP can be assured all other requirements are complied with.

You are right, topgone. I am very sorry. I forgot to reduce the starting current since the voltage decreases
So for 7.5*Irat at 380 V at 295 V [ 22.5% voltage drop] is 5.81*Irat. Then a single cable will provide 22.5% voltage
[including 5% from 400 V to 380] drop and this it could work well. Thank you, topgone.:ashamed1:

 

[Sahib]

One last check remains: Checking for short circuit withstand ability of cable. The minimum size of cable for that purpose is given by
Area 'A'=(I/k)*Sq.Root(T), where I=short circuit current, 'k' a constant =113 for aluminium cable, T=time of operation of OCPD. Suppose I=10000A, T=1second. Then A=10000/113=88.50 Sq.mm. For copper cable k=83. Then A= 120 Sqmm. So 3 X 240mm2 size cable will do.

 

[Julius Right]

I agree with you, Sahib. However, only for my curiosity, from where you have extracted the k constant. From BS7671/2008 Table 43.1 k=115 for copper conductor less than 300 mm^2 and 76 for aluminum.
This meets also the NEC2014 Table 240.92(B) Tap Conductor Short-Circuit Current :
A=I*sqrt(t)/sqrt(o0.0297*log10((234+160)/(234+70)) [c.mils]
or IEEE 80/2013 formula 37.

 

[Sahib]

Julius Right, you must be right. I wrote from my memory. Let me confirm it. Thanks.