Transformer "design" questions- help! (1 Viewer)

[DementedDonkey]

We have an assessment in which one section requires us to answer the following three sections.. which i am totally confused by.. any help asap would be soooooooo appreciated!

The Question.

A power station generates electric power at 110kW with a voltage of 415V AC. It sends this power to a farming town 25km away through transmission lines that have a resistance of 0.05 ohms per km.

i) describe the details of a transformer needed at the power station so that the power loss in the tranmission lines is only 10% of generated power. This transformer can be assumed to have 1000 turns in its primary coil... 10 marks

ii) What is the nature of the electricity that arrives at the end of the end of the transmission lines ? (power, voltage, current).. 3 marks

iii) Another transformer is to be designed so that 240V AC is supplied to each home in the town. Describe this transformer if its size is limited to have 100 turns in its secondary coil....6 marks

 

[wogboy]

i) The current in the generator is 110000/415 A = 265.06 A (using P = V*I)

For the transmission line loss to be 10% of the total power, the current in the transmission line (I) is found by:
I^2 * R = 0.1 * 110000
I^2 * 0.05 * 50 = 0.1 * 110000 (total length of wire = 50 km)
I = 66.3 A

So the turns ratio is 265.05/66.3 ~ 4. So the number of turns on the secondary coil is 4000 (a step up transformer, stepping up the voltage by a factor of 4).

ii) The voltage at the secondary coil terminal is 415*4 V = 1.66 kV. The voltage drop across the resistive transmission lines is I*R = 66.3 * 0.05 * 50 V =165.75 V. So the load voltage (at the end of the transmission line) is the (secondary coil terminal voltage) - (voltage drop) = 1660 - 165.75 V = 1494.25 V. The load current (at the end of the transmission line) is 66.3 A (the same as the current through the secondary coil of the transformer). The power dissipated in the load is given by P = V*I = 1494.25 * 66.3 W = 99 kW (which, not surprisingly, is ~ 90% of the power generated)

iii) This transformer needs to step down 1494.25 V to 240V. The turns ratio needed is 1494.25/240 ~ 6.23 . So if the secondary coil has 100 turns, the primary coil must have ~ 623 turns.

NB: In using the formula P = V*I, you assume the power factor of this circuit is 1 (i.e. the load is purely resistive, no capacitances nor inductances are present). Since the power factor concept is completely outside of the HSC physics syllabus, P = V*I is a valid formula to use. In reality though, capacitances and inductances exist on the transmission lines.

 

[DementedDonkey]

Thanks heaps.. pardon me for being skeptical but are you sure that's it? its just part i) doesnt seem enough work for a ten marker?

 

[xiao1985]

maybe discuss the strategies to deal with power loss... say laminatin of iron core, use ferrites (iron oxide, which is a good B conductor, poor E conductor) and hence acheive the 10% power loss...

 

[DementedDonkey]

well the marking criteria for that qu. says "understands the design procedure, correct formulae and calculations".. dunno if that helps