A single phase overhead transmission line delivers a power of 5500 kW ...
Degrees. The resistance and inductive reactance of the line are 0.5 ohm and 1 ohm respectively.
To calculate the sending end voltage and the power loss in the transmission line, we can use the following formulas:
Sending end voltage:
V_s = V_r + I * (R + jX)
Power loss in the transmission line:
P_loss = 3 * I^2 * R
Given:
Power, P = 5500 kW
Load voltage, V_r = 11 kV
Phase angle, δ = 45 degrees
Resistance, R = 0.5 ohm
Inductive reactance, X = 1 ohm
First, we need to convert the power from kW to W and the load voltage from kV to V:
P = 5500 * 1000 = 5,500,000 W
V_r = 11 * 1000 = 11,000 V
We can calculate the current using the power and the load voltage:
P = 3 * V_r * I * cos(δ)
5,500,000 = 3 * 11,000 * I * cos(45)
I * cos(45) = 5,500,000 / (3 * 11,000)
I * cos(45) = 55/3
I = (55/3) / cos(45)
I ≈ 12,903.3 A
Next, we can calculate the imaginary part of the current using the power factor (sin(δ)):
I * sin(δ) = P / (3 * V_r * sin(δ))
I * sin(45) = 5,500,000 / (3 * 11,000 * sin(45))
I * sin(45) = 55/3
I = (55/3) / sin(45)
I ≈ 12,903.3 A
Now, we can calculate the sending end voltage using the load voltage, current, resistance, and reactance:
V_s = V_r + I * (R + jX)
V_s = 11,000 + 12,903.3 * (0.5 + j * 1)
V_s = 11,000 + 12,903.3 * (0.5 + j)
V_s = 11,000 + 6,451.65 + j * 12,903.3
V_s ≈ 17,451.65 + j * 12,903.3 V
Finally, we can calculate the power loss in the transmission line using the current and resistance:
P_loss = 3 * I^2 * R
P_loss = 3 * (12,903.3)^2 * 0.5
P_loss = 3 * 166,811,810.89 * 0.5
P_loss ≈ 249,217,716.335 W
Therefore, the sending end voltage is approximately 17,451.65 + j * 12,903.3 V and the power loss in the transmission line is approximately 249,217,716.335 W.