Topic 5 of 5
Reducing transmission losses
Transmission cables have resistance and warm when current flows. For the same delivered power, using a higher transmission voltage reduces current and therefore reduces cable heating.
Electrical power is P = VI for the stated operating conditions. For a resistance R carrying current I, the voltage drop across that resistance is V = IR.
Use the voltage across the cable
Let R be the total resistance of the outgoing and return cables. Their combined voltage drop is Vdrop = IR.
The receiving-end voltage is across the equipment receiving the useful power. It is not the same as the drop along the cables. Multiplying cable current by the full receiving-end voltage gives the delivered power in this model, not the cable loss.
A grid overview and a separate cable power model
Overview: change voltage before and after transmission
This overview has no numerical voltages. The outgoing and return conductors are both shown. Brown arrows show one instant of alternating current.
The receiving-end transmission voltage is across the step-down transformer's input. The final load receives the lower output voltage.
Direct cable model: 1000 V received
This is a direct cable-and-load model, separate from the overview. Its receiving voltage is across this load. The sending supply must provide the useful 10.0 kW plus the cable heating.
Direct cable model: 10000 V received
This is a direct cable-and-load model, separate from the overview. Its receiving voltage is across this load. The sending supply must provide the useful 10.0 kW plus the cable heating.
Compare two systems delivering 10.0 kW
Use an idealised power account with receiving-end power 10.0 kW = 10 000 W and total cable resistance 2.0 ohm. The receiving equipment is arranged to take that same useful power at either stated voltage; this is not a comparison of different voltages applied to an unchanged resistor.
Case A
1000 V at the receiving end
Cable current I = Preceived/Vreceived = 10 000/1000 = 10.0 A.
Cable drop Vdrop = IR = 10.0 x 2.0 = 20.0 V.
Cable heating Ploss = I2R = 10.02 x 2.0 = 200 W.
The sending-end voltage must be 1000 + 20.0 = 1020 V. Its input power is 1020 x 10.0 = 10 200 W: 10 000 W delivered plus 200 W lost in the cables.
Case B
10 000 V at the receiving end
Cable current I = 10 000/10 000 = 1.00 A.
Cable drop Vdrop = 1.00 x 2.0 = 2.00 V.
Cable heating Ploss = 1.002 x 2.0 = 2.00 W.
The sending-end voltage must be 10 000 + 2.00 = 10 002 V. Its input power is 10 002 x 1.00 = 10 002 W: 10 000 W delivered plus 2.00 W lost in the cables.
The receiving-end voltage is ten times larger in B, so the current is one tenth as large. Because heating depends on current squared, the cable loss is one hundredth as large: 2 W instead of 200 W.
Keep the full power account
A step-up transformer can raise the voltage before a long transmission cable. A step-down transformer reduces it afterwards to the voltage required by the receiving equipment. The transformers change the voltage/current combination; the cable resistance still causes a finite loss.
The advantage of higher voltage follows from keeping delivered power and cable resistance fixed. Increasing voltage does not universally reduce heating: across the same unchanged resistance, a higher voltage would increase current and heating. Name the conditions before making the comparison.
Real transformers and other equipment also have losses. Include them if supplied; the calculated 200 W and 2 W are the cable losses under these stated conditions.