Calculating Decibels & Watts: Looking for Help

[rwooduk]

Homework Statement

An electronic device works at a power of 2000 W at depth of 6000 ft. It has 90% efficiency and the attenuation of the power signal through the wire is 0.1 dB/100m. Calculate the electrical power needed to power the device (at 2000 W) from the surface.

Relevant Equations

N/A

The above is just an example question to describe the situation. I am doing some simple calculations, but I think I am missing something. Is anyone here familiar with decibels?

It is a coax cable and I'm working at ~20 kHz where attenutation isn't listed on the data sheet - https://docs.rs-online.com/729b/0900766b815aad8e.pdf. But since attenuation is inversely proportional to frequency let's assumed a loss of around 0.1 db/100m.

Efficiency is just 90% of 2000 W. Then conversion of metres to feet. However, I get a loss of ~1.8 dB. I think this is only around 0.001 W, which cannot be correct. There's no way a signal would only loose 0.001 W traveling 6000 feet. I am expecting an electric power needed to power the device of around 10 kW at the surface.

Does anyone have any ideas?

Thanks for any help.

 

[berkeman]

Why are you driving lots of power through coax? Coax is usually for comm signals, and large gauge Romex-style twin lead is for transferring power. You list the powers, but not the voltages and currents. How much current are you trying to drive through that inner coax conductor...?

 

[rwooduk]

Why are you driving lots of power through coax? Coax is usually for comm signals, and large gauge Romex-style twin lead is for transferring power. You list the powers, but not the voltages and currents. How much current are you trying to drive through that inner coax conductor...?

Thanks for the reply. I just used coax as an example, at present using coax for only 50 W devices. Are the voltages / currents needed to determine attenuation of the signal? I thought it could be done using just the power used? My problem is the dB to Watts conversion which seems off for such long distance.

 

[berkeman]

Coax typically has a pretty small inner conductor, which is not suitable for high currents. And the breakdown voltage of typical coax cable is not real high, so to push lots of power you will end up with fairly high currents, which will give you attenuation in $I^2 R$ losses, in addition to the (small) dielectric AC losses.

If you want to do some 20kHz power application at the end of a log cable, it would seem better to run large gauge twin lead for power and low-loss/distortion coax for the signal. You would then put the transducer or transmitter or whatever at the end of the combined cable run.

Can you say anything about the application?

 

[rwooduk]

Coax typically has a pretty small inner conductor, which is not suitable for high currents. And the breakdown voltage of typical coax cable is not real high, so to push lots of power you will end up with fairly high currents, which will give you attenuation in losses, in addition to the (small) dielectric AC losses.

Agree, but other than SWR ratio how is it a function of distance? Especially for lower frequencies?

If you want to do some 20kHz power application at the end of a log cable, it would seem better to run large gauge twin lead for power and low-loss/distortion coax for the signal. You would then put the transducer or transmitter or whatever at the end of the combined cable run.

Can you say anything about the application?

This is actually really useful, thanks! At the end point the cable would be similar to used in oil / gas wells. Let's just call it a transmitter operating at depth. I also need to factor in temperature, but not got past the simplest part yet! For hydrophones you would usually convert voltage to decibels to pascals, then to W/m^2 which can be taken as a measure of power. But here I am converting dB/m directly to power, and this doesn't seem a correct conversion. Unless I've missed something?

 

[Steve4Physics]

However, I get a loss of ~1.8 dB. I think this is only around 0.001 W, which cannot be correct.

As noted by @berkeman, the RG213 coaxial cable may not be suitable.

However, you have made a basic maths error when using dB.

A power loss of 1.8dB means that:
$$10log \frac {P_{out}}{P_{in}} = -1.8$$or this can be written as$$10log \frac {P_{in}}{P_{out}} = 1.8$$This gives $$\frac {P_{in}}{P_{out}} = 10^{0.18} ≈ 1.5$$

So (roughly) for each kW transferred from the cable to the device, the power required into the cable at ground level needs to be 1.5kW.

It is not clear if the 1.8dB loss includes resistive (ohmic) losses which will be heavily dependent on the current.

 

[rwooduk]

As noted by @berkeman, the RG213 coaxial cable may not be suitable.

However, you have made a basic maths error when using dB.

A power loss of 1.8dB means that:
$$10log \frac {P_{out}}{P_{in}} = -1.8$$or this can be written as$$10log \frac {P_{in}}{P_{out}} = 1.8$$This gives $$\frac {P_{in}}{P_{out}} = 10^{0.18} ≈ 1.5$$

So (roughly) for each kW input to the device, the power required into the cable at ground level needs to be 1.5kW.

It is not clear if the 1.8dB loss includes resistive (ohmic) losses which will be heavily dependent on the current.

Ohh, I was going at this completely the wrong way, then went around in circles trying to convert the different units. Very helpful, thanks!