Power loss in the transmission system with a transformer

AI Thread Summary
The discussion revolves around calculating power loss in a transmission system connected to a hydroelectric generator. Initially, the farmer's setup experiences a power loss of 50 kW with an efficiency of 70.6%. An engineer suggests using a transformer to increase the voltage to 2400 V, which reduces current and minimizes power loss in the cables. The calculations confirm that with the transformer, the power loss in the new system is 500 W, demonstrating the effectiveness of voltage transformation in reducing transmission losses. The conversation emphasizes understanding transformer principles and accurate calculations for power loss.
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Homework Statement


A farmer installs a private hydroelectric generator to provide power for equipment rated at 120 kW 240 V AC. The generator is connected to the equipment by two conductors which have a total resistance of 0.20 Ω. The system is shown schematically in Fig. 1.

b83acd9e4768.jpg


(a) The equipment is operating at its rated power. Calculate: (i) the power loss in the cables, (ii) the voltage which must be developed by the generator, (iii) the efficiency of the transmission system.

(b) An engineer suggests that the farmer uses a transformer to convert the generator output to give a PD of 2400 V at the end of the transmission line, as shown in Fig. 2. A second transformer is to be used to step down this PD to 240 V.

e9a7984d235d.jpg


(i) Explain briefly how a transformer makes use of electromagnetic induction to produce an output voltage several times bigger than the input voltage.
(ii) The transformers are 100 % efficient. Calculate the power loss in the new transmission system.

Answers: (a) (i) 50 kW, (ii) 340 V, (iii) 70.6 %, (b) (ii) 500 W.

2. The attempt at a solution
(a) (i) P = V I → I = P / V = 120 000 / 240 = 500 A. P = I2 R = 5002 * 0.2 = 50 kW.

(a) (ii) V = P / I = 50 000 / 500 = 100 V in the cables plus 240 V in the equipment so 340 V in total.

(a) (iii) Efficiency = 120 000 / (120 000 + 50 000) = 0.706 or 70.6 %.

(b) (ii) This one I'm not sure. I found the power: P = V2 / R = 24002 / 0.2 = 28 800 000 W and current I = V / R = 2400 / 0.2 = 12 000 A. And also current in the equpment: I = P / V = 120 000 / 240 = 500 A. But I'm not sure whether I'm going in the right direction. The numbers are so largely different.
 
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moenste said:
(b) (ii) This one I'm not sure. I found the power: P = V2 / R = 24002 / 0.2 = 28 800 000 W and current I = V / R = 2400 / 0.2 = 12 000 A. And also current in the equpment: I = P / V = 120 000 / 240 = 500 A. But I'm not sure whether I'm going in the right direction. The numbers are so largely different.

The 2400 V is NOT developed across the cable resistance! The 2400 V is developed across the primary of the step-down load transformer. So your power calculation for the cable loss is not correct.

What you want to do is determine the current in the cables and use that to find the power lost in the cable resistance.
 
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moenste said:
And also current in the equpment: I = P / V = 120 000 / 240 = 500 A.
Right.
moenste said:
. I found the power: P = V2 / R = 24002 / 0.2 = 28 800 000 W and current I = V / R = 2400 / 0.2 = 12 000 A.
No. Have you studied how a transformer works? Power input to the primary=power output at the secondary(+transformer losses(which are zero here)) is the principle here. What is the input power to the primary?
 
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gneill said:
The 2400 V is NOT developed across the cable resistance! The 2400 V is developed across the primary of the step-down load transformer. So your power calculation for the cable loss is not correct.

What you want to do is determine the current in the cables and use that to find the power lost in the cable resistance.
cnh1995 said:
Right.

No. Have you studied how a transformer works? Power input to the primary=power output at the secondary(+transformer losses(which are zero here)) is the principle here. What is the input power to the primary?
I E = I E
2400 I = 240 * 500
I = 50 A.

P = I2 R = 502 * 0.2 = 500 W.

Should be correct.
 
moenste said:
I E = I E
2400 I = 240 * 500
I = 50 A.

P = I2 R = 502 * 0.2 = 500 W.

Should be correct.
Good job!
 
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