What is th proper method of determining the required/calculated cable ampacity of an unbalanced 3Ph panel. Assume a 3Ph-3W delta configured panel that has 15kVA PhAB, 10kVA PhBC and 12kVA PhCA. I understand the method of calulating line currents of unbalanced systems, but what is the appropriate method for sizing the circuit protection and cable? I have been told typically that the CB or cable is based on the maximum of the three line currents multiplied by SQRT(3) not the RMS value of thre three line currents.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations MintJulep on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Required Cable/Protection of Unbalanced 3Ph Panel
- Thread starter 7JLAman4
- Start date
First of all, the current has to be measured in A .Second: the panel feed a transformer or a heater delta connected as the panel itself [I cannot imagine how could be] is not delta connected, does it?
Let's say the voltages are equilibrated and equally phased [120 degrees apart].
The total S=15+10+12=37 Kva [apparent power]
Let's say the voltage [L-L] is 400 V then the average current will be=1000*S/sqrt (3)/Voltage=53.4 A
The average Sav=37/3=12.33 kVA and the maximum 15/12.33=1.216
So the maximum may be 53.4*1.216=65 A
This is a conservative solution, but not exaggerated.
If you know the current phasor [vector] angles you may calculate exact the line currents.
According to relation Sum(I) =0 in one corner:
Ia=Iac-Iba ; Ib=Iba-Icb ; Ic=Icb-Iac
If the supplied device is a heater for instance the pf=1 [pure resistance], so the current angle with respect the voltage angle will be 0.
In order to measured the actual angle if pf<1 you need a single-phase wattmeter and the angle with respect the voltage will be = arccosine (P[watts]/S[VA]).
All it remains is plane geometry [in order to calculate one edge when you know the other two and the
angle between in a triangle].
Iab=23.4*1000/400=58.575A [11% less than above approximate calculation.]
Let's say the voltages are equilibrated and equally phased [120 degrees apart].
The total S=15+10+12=37 Kva [apparent power]
Let's say the voltage [L-L] is 400 V then the average current will be=1000*S/sqrt (3)/Voltage=53.4 A
The average Sav=37/3=12.33 kVA and the maximum 15/12.33=1.216
So the maximum may be 53.4*1.216=65 A
This is a conservative solution, but not exaggerated.
If you know the current phasor [vector] angles you may calculate exact the line currents.
According to relation Sum(I) =0 in one corner:
Ia=Iac-Iba ; Ib=Iba-Icb ; Ic=Icb-Iac
If the supplied device is a heater for instance the pf=1 [pure resistance], so the current angle with respect the voltage angle will be 0.
In order to measured the actual angle if pf<1 you need a single-phase wattmeter and the angle with respect the voltage will be = arccosine (P[watts]/S[VA]).
All it remains is plane geometry [in order to calculate one edge when you know the other two and the
angle between in a triangle].
Iab=23.4*1000/400=58.575A [11% less than above approximate calculation.]
- Thread starter
- #3
Well thank you for the graphics! I have a Delta 3Ph 3Wire secondary 240V (L-L) that directly supplies a panel with resistive loads. When I calculate the line currents of the loads described in the OP, I calculate ILA = 97.6A, ILB = 90.8A, ILC = 79.5A. Typically I am accustomed to adding another 25% of the largest load for determining the panel demand load. However that is when dealing with all 3Ph loads. My question is, when there are multiple single phase and three phase loads on a panel, how is the panel demand load calculated for sizing feeder cable size and circuit protection.
The feeder cable maximum current for protection setting depends upon:
1) the cable insulation type [PVC support 70 Co, XLPE 90 Co, EPR 90 Co- for instance ]
2) cable conductor cross section area
3) lay out way -exposed, in conduit, on cable tray, underground or else.
4) other power cables vicinity
5) ambient [air, ground] temperature.
The circuit protection has to avoid cable overheating within undefined time.
The cable manufacturer [or the local standard as NEC for US, for example] will indicate this maximum permissible current.
Usually low voltage cable has good resistance to short circuit currents.
If it is necessary, you have to check the thermal stability of the conductor for a certain short circuit current value and, for an existing cable, you have to set the breaker trip time to avoid short time overheating [in this case PVC support 130 Co and XLPE 250 Co].
If the cable conductor insulation would be hotter than permissible you have to change the cable for another bigger, of course.
Example:
From NEC APPENDIX B Table B-310-3 a 3*1 AWG [PVC insulated 60Co] the ampacity if the cable run in Free Air will be 107 A.
Now if the short circuit current will be 10 KA-for example- and using Clyde38 post method [see thread238-231532] and adjust the maximum temperature to 130 degrees C and the ambient to 40 :
33 * (I/A)^2 * S = log( (Tm - Ta) / (234 + Ta) + 1 )
I = current in Amperes 10 kA=10000 A
A = area of wire in circ. mils for 1 AWG is 83694
S = time the current flows in seconds
Tm = maximum permitted, 130 C
Ta = ambient temp, 40 C
S=*A^2/I^2*log( (Tm - Ta) / (234 + Ta) + 1 )/33
S=(83694/10000)^2*log( (130 - 40) / (234 + 40) + 1 )/33= 0.26 sec.
If you think this will limit the actual load, you have to take a bigger one and it will be 1/0 AWG with 124 A setting and 105531 circular mils.
S= (105531/10000) ^2*log ((130 - 40) / (234 + 40) + 1)/33= 0.42 sec.
1) the cable insulation type [PVC support 70 Co, XLPE 90 Co, EPR 90 Co- for instance ]
2) cable conductor cross section area
3) lay out way -exposed, in conduit, on cable tray, underground or else.
4) other power cables vicinity
5) ambient [air, ground] temperature.
The circuit protection has to avoid cable overheating within undefined time.
The cable manufacturer [or the local standard as NEC for US, for example] will indicate this maximum permissible current.
Usually low voltage cable has good resistance to short circuit currents.
If it is necessary, you have to check the thermal stability of the conductor for a certain short circuit current value and, for an existing cable, you have to set the breaker trip time to avoid short time overheating [in this case PVC support 130 Co and XLPE 250 Co].
If the cable conductor insulation would be hotter than permissible you have to change the cable for another bigger, of course.
Example:
From NEC APPENDIX B Table B-310-3 a 3*1 AWG [PVC insulated 60Co] the ampacity if the cable run in Free Air will be 107 A.
Now if the short circuit current will be 10 KA-for example- and using Clyde38 post method [see thread238-231532] and adjust the maximum temperature to 130 degrees C and the ambient to 40 :
33 * (I/A)^2 * S = log( (Tm - Ta) / (234 + Ta) + 1 )
I = current in Amperes 10 kA=10000 A
A = area of wire in circ. mils for 1 AWG is 83694
S = time the current flows in seconds
Tm = maximum permitted, 130 C
Ta = ambient temp, 40 C
S=*A^2/I^2*log( (Tm - Ta) / (234 + Ta) + 1 )/33
S=(83694/10000)^2*log( (130 - 40) / (234 + 40) + 1 )/33= 0.26 sec.
If you think this will limit the actual load, you have to take a bigger one and it will be 1/0 AWG with 124 A setting and 105531 circular mils.
S= (105531/10000) ^2*log ((130 - 40) / (234 + 40) + 1)/33= 0.42 sec.
- Status
Similar threads
- Question
- Question
- Question