
A small town with a demand of $ 800kW $ of electric power at $ 220V $ is situated $ 15km $ away from an electric plant generating power at $ 440V $ . The resistance of the two wire lines carrying power is $ 0.5\;\Omega {\text{ }}per{\text{ }}km $ . The town gets power from the line through a $ 4000 - 220V $ step-down transformer at a substation in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.
Answer
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Hint :To solve this question, we first need to find the rms value of the current. After that the power loss can be calculated by using the relationship of power with current and resistance. Then, the supply power can be found by adding power demand to the power loss. Finally, by determining primary and secondary voltage, we will obtain the transformer rating.
Complete Step By Step Answer:
We are given that the town gets power from the line through a $ 4000 - 220V $ step-down transformer at a substation in the town. Let us take $ V = 4000V $ to find the rms current in the coil and the demand of power $ {P_d} = 800kW $ .
$ I = \dfrac{{{P_d}}}{V} = \dfrac{{800 \times {{10}^3}}}{{4000}} = 200A $
Answer (a): The power loss in the form of heat is given by the formula: $ {P_l} = {I^2}R $
It is given that the resistance of the two wire line carrying power is $ 0.5\;\Omega {\text{ }}per{\text{ }}km $ and the town is located $ 15km $ away from the plant.
$ \Rightarrow R = \left( {2 \times 0.5} \right) \times 15 = 15\Omega $
$ \Rightarrow {P_l} = {I^2}R = {200^2} \times 15 = 600kW $
Answer (b): The plant should supply the total power which is equal to the sum of demand power and power loss.
$ P = {P_d} + {P_l} = 800 + 600 = 1400kW $
Answer (c):
Let us first consider the voltage drop $ V = IR = 200 \times 15 = 3000V $
Therefore, the secondary voltage will be $ 4000 + 3000 = 7000V $
The power generation takes place at the primary voltage which is $ 440V $ .
Thus, we can say that the range of transformers is $ 440 - 7000V $ .
Note :
In this question, we need to be careful while putting the value of the resistance. This is because the given value is for only one wire and it is said that there are two wires. Moreover, the value of resistance is given per kilometer, so we need to multiply it by the total distance between the town and the plant.
Complete Step By Step Answer:
We are given that the town gets power from the line through a $ 4000 - 220V $ step-down transformer at a substation in the town. Let us take $ V = 4000V $ to find the rms current in the coil and the demand of power $ {P_d} = 800kW $ .
$ I = \dfrac{{{P_d}}}{V} = \dfrac{{800 \times {{10}^3}}}{{4000}} = 200A $
Answer (a): The power loss in the form of heat is given by the formula: $ {P_l} = {I^2}R $
It is given that the resistance of the two wire line carrying power is $ 0.5\;\Omega {\text{ }}per{\text{ }}km $ and the town is located $ 15km $ away from the plant.
$ \Rightarrow R = \left( {2 \times 0.5} \right) \times 15 = 15\Omega $
$ \Rightarrow {P_l} = {I^2}R = {200^2} \times 15 = 600kW $
Answer (b): The plant should supply the total power which is equal to the sum of demand power and power loss.
$ P = {P_d} + {P_l} = 800 + 600 = 1400kW $
Answer (c):
Let us first consider the voltage drop $ V = IR = 200 \times 15 = 3000V $
Therefore, the secondary voltage will be $ 4000 + 3000 = 7000V $
The power generation takes place at the primary voltage which is $ 440V $ .
Thus, we can say that the range of transformers is $ 440 - 7000V $ .
Note :
In this question, we need to be careful while putting the value of the resistance. This is because the given value is for only one wire and it is said that there are two wires. Moreover, the value of resistance is given per kilometer, so we need to multiply it by the total distance between the town and the plant.
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