Calculating power measurement error from V and I error margins

Question

I am using a Yokogawa WT310 for measuring DC power. The manual (available here) says that both current and voltage have accuracy of +/-(0.1% of reading + 0.2% of range).

Assume this example setup:

Input current: 700mA
Current Range: 1A
Input voltage: 10v
Voltage Range: 15v

Based on that setup and the meter specifications, we can calculate the following:

P = V * I = 7w
V_error = +/-(0.001 * 10v + 0.002 * 15v) = +/- 0.04v
I_error = +/-(0.001 * 0.7A + 0.002 * 1A) = +/- 0.0027A

My goal is to calculate the power measurement's error margin. Here's what I've done:

P_error = V_error * I_error = 0.04v * 0.0027A = 0.000108w

Note that in this example, P_error is less than both V_error and I_error.

Now imagine I'm using a less accurate meter and measuring much higher voltage and current ranges. Assume I get these results for error margins:

V_error = 1.17v
I_error = 1.2A
P_error = V_error * I_error = 1.17v * 1.2A = 1.404w

Now, P_error is greater than both V_error and I_error.

This makes sense mathematically; that's just how multiplying numbers less than 1 and greater than 1 works. But it makes me feel like I'm missing something conceptually. Shouldn't P_error scale consistently relative to V_error and I_error? Am I just calculating P_error incorrectly?

Answer 1

Oh yes, you're definitely missing something. You can't focus on the errors without looking at the "non-errors". Let's use your second example, where you got a power error of 1.404 watts. What is the "non-error" component?

Just for grins, assume that the real voltage was 100 volts and the real current 20 amps. Then the real power was 2,000 W, but the inaccurate readings were 101.17 volts and 21.2 A, for an incorrect power calculation of 2144.8 watts, or a power error of 144.8 watts.

If you consider each measurement as a base plus an error, and the errors are normalized with respect to the base (such as you would do with percentage errors) then $$(1 + x)(1 + y) = 1 + x + y + xy$$ and the error term is the sum of the errors plus their product. For small errors the product term is negligible, and the error is simply the sum of the errors.

For non-normalized errors, as in your example, you need to compute $$P = (V + {\delta}v)(i+{\delta}i) = Vi + V{\delta}i + i{\delta}V +{\delta}V{\delta}i $$ and the power error E is simply $$E = V{\delta}i + i{\delta}V +{\delta}V{\delta}i $$

So, yes, your reservations were correct.

Figure 1. A graphical representation of the \$VI\$ term (green), \$V \delta i \$ term (red), \$ i \delta V \$, term (blue) and \$ \delta V \delta i \$ term (white). It can be seen that, in this case, the \$ \delta V \delta i \$ term contributes very little to the overall error and that about 80% of the error is due to the \$V \delta i \$ term.

Answer 2

I wonder why you are asking the question. For Yokogawa Power meters and analysers the accuracy for the power measurement is fully specified, including that for DC, so it is unnecessary to use the accuracy of the voltage and current measurements to derive the power accuracy. Using a formula is almost certain to produce a poorer specfication.

Please see the relevant specifications in the brochure for the WT300E at the top of page 12:

https://www.yokogawa.com/pdf/provide/E/GW/Bulletin/0000029826/0/BUWT300E-01EN.pdf

A power range is the multiple of the voltage and current ranges. In a Yokogawa power meter, these are rms ranges, not peak, so the actual power error due to the range error can be typically three times less than that for power meters that specify using peak ranges for voltage and current. This is of course is more relevant for AC waveforms than DC.

Answer 3

Consideration on difference between absolute and relative errors are quite correct.

But I belive what should be clearly pointed out about @skrrgwasme concerns is that is totally, absolutely, hopelessly meaningless to compare different quantities.

There is no chance to say wether 1 volt is more or less than 1 ampere, and so with watt or anything not equivalent to volt.