Calculating Maximum Power Cable length

 

Calculating Maximum Power Cable length

We calculate maximum power cable length on single phase AC supply:-

 

Lmax = (1000 X V X VD) / (2 X I X (Rcos ᶿ + Xsinᶿ ))

 

Where

 

Lmax = Maximum power cable length (m)

V = Maximum Permissible Voltage Supply (V)

VD = Voltage drop (%)

I = Starting Current (A)

R = Power Cable Resistance (Ω/m)

Cos ᶿ = Power Factor

X = Conductor inductive reactance (Ω/m)

ᶿ = Phase angle of load (°)

We already learned how to calculate Power Cable Resistance in article Power Cable Size &  Calculation of Voltage Drop.

  

Let we assume:-

 

Power factor (Cos ᶿ) = 1

Phase angle of load (ᶿ) = 90°

 

Lmax = (1000 X V X VD) / (2 X I X (R X 1 + Xsin90°))

 

We know that sin90° = 0, i.e.

 

Lmax = (1000 X V X VD) / (2 X I X (R + 0))

 

And we get final result-

 

Lmax = (1000 X V X VD) / (2 X I X R)

 

 

We calculate maximum power cable length on Three phase AC supply:-

 

Lmax = (1000 X V X VD) / (√3 X I X (Rcos ᶿ + Xsinᶿ))

 

Where

 

Lmax = Maximum power cable length (m)

V = Maximum Permissible Voltage Supply (V)

VD = Voltage drop (%)

I = Starting Current (A)

R = Power Cable Resistance (Ω/km)

Cos ᶿ = Power Factor

X = Conductor inductive reactance (Ω/km)

ᶿ = Phase angle of load (°)

 

Let we assume:-

 

Power factor (Cos ᶿ) = 1

Phase angle of load (ᶿ) = 90°

 

Same theory applicable here then we get result:-

 

Lmax = (1000 X V X VD) / (√3 X I X R)

 

 

We calculate maximum power cable length on DC supply:-

 

Lmax = (1000 X Vdc X VD) / (2 X I X R)

 

Lmax = Maximum power cable length (m)

Vdc = Voltage Supply (V)

VD = Voltage drop (%)

I = Starting Current (A)

R = Power Cable Resistance (Ω/km)

 

Example1:-

 

V = 230V

I = 25A

R = 5.9

Cos ᶿ = 1

ᶿ = 90°

Phase = 1ph

VD = 5%

Lmax = ?

 

Solution:-

 

We use formula:-

 

Lmax = (1000 X V X VD) / (2 X I X R)

 

Putting all value in formula:-

 

Lmax = (230 X 1000 X 5%) / (2 X 25 X 5.9)

 

Lmax = 11500 / 295

 

Lmax = 38.98m

 

We can use upto 38.98m of cable length in this parameter.

 

Example2:-

 

V = 440V

I = 120A

R = 0.495

Cos ᶿ = 1

ᶿ = 90°

Phase = 3ph

VD = 5%

Lmax = ?

 

Solution:-

 

We use formula:-

 

Lmax = (1000 X V X VD) / (√3 X I X R)

 

Putting all value in formula:-

 

Lmax = (1000 X 440 X 5%) / (√3 X 120 X 0.495)

 

Lmax = 22000 / 102.88

 

Lmax = 213.84m

 

We can use upto 213.84m of cable length in this parameter.