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[Aokiji_Master]

A 50 km long underground transmission line supplies a load of 5 MVA at 0.8 p.f. Lagging at 33 kV. The efficiency of transmission is 90%. Calculate the volume of aluminium conductor required for the line when: (i) Single phase, 2-wire system is used (ii) 3-phase, 3-wire system is used (star connection). The specific resistance of aluminium is 2.85 x 10-8 m

 

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Firstly, let's break down the problem.

The data given is:

• Length of transmission line (L) = 50 km = 50000 m
• Load (P) = 5 MVA = 5 * 10^6 VA
• Power factor (pf) = 0.8 lagging
• Voltage (V) = 33 kV = 33 * 10^3 V
• Efficiency (η) = 90% = 0.90
• Specific resistance (ρ) of aluminium = 2.85 x 10^-8 Ω.m

We also know that the volume of a conductor (V) can be calculated using the formula V = A*L, where A is the cross-sectional area.

(i) For a single-phase, 2-wire system, the current (I) can be calculated using the formula P = √3 * V * I * pf.

However, since this is a single-phase system, we simplify it to P = V * I * pf.

Rearranging for I, we get I = P / (V * pf) = 5 * 10^6 / (33 * 10^3 * 0.8) = 189.39 A.

We also know that the power loss in the line, which is I²R, should be equal to 10% of the power transmitted, as the efficiency is 90%.

So, 0.10 * P = I² * R.

Rearranging for R, we get R = 0.10 * P / I² = 0.10 * 5 * 10^6 / (189.39)² = 13.97 Ω.

The resistance (R) of the conductor is also given by the formula R = ρ * (L/A).

Rearranging for A, we get A = ρ * L / R = 2.85 x 10^-8 * 50000 / 13.97 = 0.1018 m².

The volume of the conductor (V) is then V = A * L = 0.1018 * 50000 = 5090 m³.

(ii) For a 3-phase, 3-wire system (star connection), the current (I) is given by the formula P = √3 * V * I * pf.

Rearranging for I, we get I = P / (√3 * V * pf) = 5 * 10^6 / (√3 * 33 * 10^3 * 0.8) = 109.36 A.

Repeating the steps as above for resistance and cross-sectional area, we will get R = 0.10 * P / I² = 0.10 * 5 * 10^6 / (109.36)² = 41.82 Ω and A = ρ * L / R = 2.85 x 10^-8 * 50000 / 41.82 = 0.034 m².

The volume of the conductor for 3-phase system is therefore V = 3 * A * L = 3 * 0.034 * 50000 = 5100 m³.

Please note that these calculations are simplified, and may not take into account all factors that could influence the actual volumes of conductor required in real-world applications.