wire length versus energy consumption 2

[mwemag]

I'm looking for a way to calculate the additional energy consumption of an electrical device, if the same current used for the device has to run through a wire with a given length, diameter and resistance, before reaching the device.

As an example, take a lamp with an energy consumption of around 60W, 5 amps and 12V, when connected near the power source, then insert an additional wire between the lamp and the source, e.g. a copper wire with diameter=1mm and length=10m. The heat losses caused by the electrical resistance of the wire will lead to a current reduction in the device, so if I want tho have still 5 amps running through the lamp, more energy has to be consumed.

Is there a way to calculate the additional energy consumption?

Appreciate any help, tanks

 

[jghrist]

Energy = I²·r·D·T, where I is the current in the device, r is the resistance per unit length of the wire, D is the length of the wire, and T is the amount of time that the device is running.

r = rho/A where rho is the resistivity of the wire and A is the wire area.

rho = 0.01724 mm²·ohm/m for annealed copper at 20°C. (0.01724=1/58)

 

[mwemag]

Thanks. I assume the result will be in units of watts per second. Since I need to compare to the energy consumption of the original device in Watts (e.g. 60W), how to convert the watts per second to watts?

 

[Skogsgurra]

Watts are "energy per second". If you want to calculate energy, you just multiply watts with seconds to get watt-seconds, aka joules.

Watts per second is not a meaningful unit for everyday calculations.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[xnuke]

If you're looking for watts, you're looking for power loss, not energy. If 60 W is the power rating of your device, i.e., the rate at which energy is consumed (watts are joules consumed per second). The result of the formula jghrist gave is in joules. For watts, leave out the T term.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[xnuke]

Either you're fast or I'm slow, Gunnar.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[mwemag]

Just to be sure on the units:

Watts = I²·r·D

I=5 (amps)
r=0.01724 (mm²·ohm/m)
D=10 (meters)

=5²·0.01724·10
=4.31 Watts

Or do I have to adapt the effective diameter aerea (0.5mm·pi=0.785mm²) for the r value?

 

[xnuke]

Watts = I²·r·D

I=5 (amps)
r=0.01724 (mm²·ohm/m)/0.785mm²
D=10 (meters)

=5²·0.01724/.785·10
=1.1 Watts

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[xnuke]

Oops, forgot to square the 5.

Shoul;d be 5.5 W.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[mwemag]

Thanks again.

So, connecting a 60W-bulb to a wire descibed above will consume 65.5 Watts (theoretically). Correct?

 

[dpc]

Not exactly. If you are trying to be very precise, it is necessary to take into account that power consumed by the end device (e.g. 60W light) is often a function of the voltage at its terminals.

An incandescent lamp is basically a resistance, so as the voltage goes down, the current goes down, and therefore, the voltage drop in the wire goes down. And power consumption goes down. Other types of loads can behave differently as a function of voltage.

 

[mwemag]

dpc, let's assume that the lamp would be a low voltage halogen bulb, e.g. ratet at 60W and 12V, and that this rate corresponds to the voltage and current in the active state (including the losses by the increased resistance of the tungsten filament when heated).

I don't understand much of voltage drop, but it seems logic that the lamp manufacturers will rate the wattage of the active state rather than the initial energy consumption. The latter would be orders of magnitude higher because tungsten has a much lower resistance in the cold state than in the hot state.

 

[dpc]

Yes, but it may not be 60W if you don't have EXACTLY 12 V at the lamp terminal.

And yes, this is ignoring the initial warm-up phase - I'm talking steady-state conditions.

In power systems, all voltages are NOMINAL, and variations of +/- 5% are normal and +/- 10% not uncommon.

For an incandescent lamp, operating at 90% voltage reduces the power consumption to 81% of nominal.

Utilities often take advantage of this by slightly reducing voltage supplied to customers to reduce peak demand.

 

[mwemag]

Yes, fluctuations of the voltage are very common. Additionally, halogen bulbs are sometimes produced to run underdriven to increase their lifetime, but at the price of reduced brightness and color temperature.

Therefore exact values of power consumption will be easier to measure than to calculate. However, I guess the formula ist still usefull to estimate a relative increase in power consumption compared to the original device.

 

[jghrist]

The 65.5 watt total would be correct if the voltage were adjusted to get 5A through the lamp.

One caveat. The distance D has to include both the wires running to the lamp. So, if there are two 0.785 mm² wires, each 10 m long, then the total power is 71 watts.

 

[mwemag]

Very valuable answers, thanks to all!

 

[itsmoked]

I would also say you would normally NOT be able to tweak the voltage to keep the bulb at 12.0V That would add some serious expense to the system.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com