Shibela Shinga
May 15
Calculate the minimum allowable cross section area of a pvc twin copper cable which Will Supply a 240v distribution fuse board 100m from supply point the total load is50A?
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ChatGPT
May 15
To calculate the minimum allowable cross section area of the PVC twin copper cable, we need to consider the voltage drop over the 100m distance and the total load of 50A.
First, we need to calculate the voltage drop using the formula:
Voltage drop = (2 x length x current x resistance) / cross sectional area
Assuming a resistivity of 0.0225 ohms per meter for copper, the resistance of the cable can be calculated as:
Resistance = resistivity x length / cross sectional area
Now, we can rearrange the formula for voltage drop to solve for the cross sectional area:
Cross sectional area = (2 x length x current x resistance) / voltage drop
Plugging in the values:
Length = 100m
Current = 50A
Voltage = 240V
Resistivity = 0.0225 ohms/m
Resistance = 0.0225 x 100 / cross sectional area
Voltage drop = (2 x 100 x 50 x (0.0225 x 100 / cross sectional area)) / cross sectional area
Solving for the cross sectional area, we get:
Cross sectional area = 0.0005625 m^2 or 562.5 mm^2
Therefore, the minimum allowable cross sectional area of the PVC twin copper cable should be 562.5 mm^2 to supply a 240V distribution fuse board 100m from the supply point with a total load of 50A.
First, we need to calculate the voltage drop using the formula:
Voltage drop = (2 x length x current x resistance) / cross sectional area
Assuming a resistivity of 0.0225 ohms per meter for copper, the resistance of the cable can be calculated as:
Resistance = resistivity x length / cross sectional area
Now, we can rearrange the formula for voltage drop to solve for the cross sectional area:
Cross sectional area = (2 x length x current x resistance) / voltage drop
Plugging in the values:
Length = 100m
Current = 50A
Voltage = 240V
Resistivity = 0.0225 ohms/m
Resistance = 0.0225 x 100 / cross sectional area
Voltage drop = (2 x 100 x 50 x (0.0225 x 100 / cross sectional area)) / cross sectional area
Solving for the cross sectional area, we get:
Cross sectional area = 0.0005625 m^2 or 562.5 mm^2
Therefore, the minimum allowable cross sectional area of the PVC twin copper cable should be 562.5 mm^2 to supply a 240V distribution fuse board 100m from the supply point with a total load of 50A.
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Google Gemini AI
May 15
Step 1: Calculate the voltage drop
The voltage drop (Vdrop) is the difference between the voltage at the supply point (Vsupply) and the voltage at the fuse board (Vload):
Vdrop = Vsupply - Vload
Assuming a voltage drop of 3% (a typical value for electrical installations), we have:
Vdrop = 0.03 - 240V = 7.2V
Step 2: Calculate the resistance of the cable
The resistance (R) of the cable is given by:
R = Vdrop / I
where I is the total load current (50A).
Substituting the values:
R = 7.2V / 50A = 0.144Ω
Step 3: Calculate the length of the cable
The length of the cable (l) is 100m (both ways).
Step 4: Calculate the cross-sectional area of the cable
The cross-sectional area (A) of the cable can be calculated using the equation:
A = (2 - l - R) / (ρ - V)
where:
- ρ is the resistivity of copper (0.017241 Ω⋅mm²/m)
- V is the voltage drop (7.2V)
Substituting the values:
A = (2 - 100m - 0.144Ω) / (0.017241 Ω⋅mm²/m - 7.2V) ≈ 20.04mm²
Therefore, the minimum allowable cross section area of the twin copper cable is 20.04mm².
The voltage drop (Vdrop) is the difference between the voltage at the supply point (Vsupply) and the voltage at the fuse board (Vload):
Vdrop = Vsupply - Vload
Assuming a voltage drop of 3% (a typical value for electrical installations), we have:
Vdrop = 0.03 - 240V = 7.2V
Step 2: Calculate the resistance of the cable
The resistance (R) of the cable is given by:
R = Vdrop / I
where I is the total load current (50A).
Substituting the values:
R = 7.2V / 50A = 0.144Ω
Step 3: Calculate the length of the cable
The length of the cable (l) is 100m (both ways).
Step 4: Calculate the cross-sectional area of the cable
The cross-sectional area (A) of the cable can be calculated using the equation:
A = (2 - l - R) / (ρ - V)
where:
- ρ is the resistivity of copper (0.017241 Ω⋅mm²/m)
- V is the voltage drop (7.2V)
Substituting the values:
A = (2 - 100m - 0.144Ω) / (0.017241 Ω⋅mm²/m - 7.2V) ≈ 20.04mm²
Therefore, the minimum allowable cross section area of the twin copper cable is 20.04mm².
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