F
fjr1300
Hi All
I’ve got to install electricity in 7 adjoining barns (10M*17M each)
Each barn will have 2*(Double waterproof Fluorescent lights) and 5 (58W) heated water troughs making a load of 438W per barn.
The loads are single phase 230V and will come from a 3 phase supply – load balancing 2 barns per phase.
The supply is around 250M away and will be underground so Method D – 4 core cable.
Now I'm not sure of which column to use on tables 4D4A and 4D4B
(Pages 280-281 17tg regs) because of the single loads but using 4 core 3 phase cables.
I'm doing the installation in France where they use non-armoured cable (less weight thank god ) (60-80cm deep in ground) but assume the armoured tables would be ok.
I obviously would not use 1.5mm cable because the volt drop is too high and I think I'd allow for 16Amps load per phase (3680W) so that power sockets could be fitted but I'd like to see if I've got the correction factors and tables right.
I don't want to oversize the cable since the weight of 250M-300M, 4 core, is going to be fun installing .
Thanks
Leigh
Calculations:-
Voltage = 230V
Max Voltage drop in is 4% = 9.2V
Length of Cable 300 (250M + approx middle of barns)
Required Power per barn 434 watts
Since there are 7 Barns, some of then will share the same phase
Max power, IE 2 Barns 868 Watts
Design Current 3.77 Ib Amps
Selected Breaker 5 In Amps
(Manual Selection)
Factors
1 Ca Equals 1 in 30Deg C Temp
0.65 Cg 4 Circuits (L1/L2/L3/nN)
1 Ci Equals 1 since in Ground and not Insulation involved
0.9 Cc Equals 0.9 since in ground
Cable Current Capacity 8.55 Amps It
It ≥ In / (Ca * Cg * Ci * Cc)
Select Cable size that has current capactiy ≥ to It
1.5mm
(From Table 4D4A Page 280 17th Regs)
From Table 4D4B (4) the mV/A/m) = 29 (Single phase Col 3)
Single Phase Loads
Therefore
Voltage Drop = (mV/A/m)*Ib*L / 1000 where IB=3.77 and L = 300
32.83 Vdrop
I’ve got to install electricity in 7 adjoining barns (10M*17M each)
Each barn will have 2*(Double waterproof Fluorescent lights) and 5 (58W) heated water troughs making a load of 438W per barn.
The loads are single phase 230V and will come from a 3 phase supply – load balancing 2 barns per phase.
The supply is around 250M away and will be underground so Method D – 4 core cable.
Now I'm not sure of which column to use on tables 4D4A and 4D4B
(Pages 280-281 17tg regs) because of the single loads but using 4 core 3 phase cables.
I'm doing the installation in France where they use non-armoured cable (less weight thank god ) (60-80cm deep in ground) but assume the armoured tables would be ok.
I obviously would not use 1.5mm cable because the volt drop is too high and I think I'd allow for 16Amps load per phase (3680W) so that power sockets could be fitted but I'd like to see if I've got the correction factors and tables right.
I don't want to oversize the cable since the weight of 250M-300M, 4 core, is going to be fun installing .
Thanks
Leigh
Calculations:-
Voltage = 230V
Max Voltage drop in is 4% = 9.2V
Length of Cable 300 (250M + approx middle of barns)
Required Power per barn 434 watts
Since there are 7 Barns, some of then will share the same phase
Max power, IE 2 Barns 868 Watts
Design Current 3.77 Ib Amps
Selected Breaker 5 In Amps
(Manual Selection)
Factors
1 Ca Equals 1 in 30Deg C Temp
0.65 Cg 4 Circuits (L1/L2/L3/nN)
1 Ci Equals 1 since in Ground and not Insulation involved
0.9 Cc Equals 0.9 since in ground
Cable Current Capacity 8.55 Amps It
It ≥ In / (Ca * Cg * Ci * Cc)
Select Cable size that has current capactiy ≥ to It
1.5mm
(From Table 4D4A Page 280 17th Regs)
From Table 4D4B (4) the mV/A/m) = 29 (Single phase Col 3)
Single Phase Loads
Therefore
Voltage Drop = (mV/A/m)*Ib*L / 1000 where IB=3.77 and L = 300
32.83 Vdrop