I M: Is Power Liner Function of Current

For the power measurement, not only the current is used, but the voltage is also used as input. The Watt meter (power meter) uses both current and voltage and it shows the value of √3 x V x I x P.F from the voltage and current input values. If the transducer is used, it generates the mA corresponding to the value of √3 x V x I x P.F. If the power meter is used for measuring the motor load, it measures the motor input power, so efficiency is not a factor here.

Your concept about measuring the power ("what I thought about measurement system is that only current is measured") is not correct.

See this link for the single phase and three phase wattmeter wiring connection.

http://www.panasonic-electric-works.com/peweu/en/downloads/sf_x630_en_eco_powermeter_kw4m.pdf (see the wiring diagram at the last page)

- MS

Thank U very much for Reply.

Now I have a transducer "CELSA Power Transducer Type CPNV Class 0.5 4 KV" it mentions the measurment principle as follows

1. Measure the Voltage 2. Measure the Current 3. Multiply both current and Voltage 4. perform Integration on their product 5. Generate a corrosponding mA signal

Now Voltage is constant, at Low Loads (Below 50 %) motor Efficiency and P.F changes (comes down ),Now We have three varibales, current, Power factor, Efficiency. This integration could only handle variation current but not the other two

Now look at the Motor characteristics Efficiency and P.F with respect to loads. Read the first two pages

http://www.p2pays.org/ref/40/39569.pdf

Now consider the voltage is (constant) 400 V when the motor is 37.5 % load (i-e 18.26 A, 10 mA, 9.75 KW) at this stage consider motor efficiency is 75 % and P.F is 0.65 (Approximate values are taken from the graphs)

now put in the motor power formula at 18.26 A

P = 1.732*400*18.26*.65*.75 = 6.167 KW while above method will show power of 9.75 Kw.

Waiting for Ur reply

Thanks

All these steps are done within the transducer using the instantaneous power eqauation.

The average electrical power equation:

Where v(t) and i(t) are, respectively, the instantaneous voltage and current as functions of time.

T is generally considered the 1/2 cycle time period.

The transducer generates the mA corresponding to the value of P.

- MS

What U r mentioning is the measurement of Electrical power consumed by the motor.

What I want to say about the measurment of Mechanical Power (Name Plate Mentioned power) generated by the motor.

Electrical input power is the function of Voltage, current and P.F

Output power Power of a motor (Mechanical Power) depend on another factor called Efficiency.

Simens Simocode (In LV Switch gears) measure the current of each phase and according to the current drawn they give the output power (Mechanical) of the motor.

What I think that if motor is operating below 60% load, values of Mechanical power given by Simocode have error to the actual values based on the graph in the report I mention in previous post

The currents given in your first post are actual currents as measured?

The linear relationships that case are not correct. Since the pf, which is highly load dependant is not accounted for.

However the usual power transducers (active/ or reactve- both are there) work no vector multiplication of the V & I and not multiplication per se

So the output current will be proportional to power directly (electrical) - however since it is on electrical side, it can not account for the efficiency, which will be highly architectural dependant.

We are measuring the Output Power ( Mechanical) of a Motor. We are not measuring the power of a utility. The meter mentioned measures the input power (Electrical) of a motor very well but output power have a factor known as efficiency and this factor depend on the load

I don't really understand your question but perhaps this will help.

Electrical power is (in principle) easy to measure. It's instantaneous voltage * instantaneous current. This is usually done using RMS volt * RMS current * power factor (to account for the phase difference between V & I). There are other formulas that assume constant voltage etc, no harmonics etc.

Mechanical power is a bit more complicated. It's possible to ESTIMATE the power using the electrical input, allowing for resistive, magnetic & friction losses. this usually takes the form of an "efficiency factor". Mech power = elec power * efficiency

Naturally, this is only approximate. If you really want mech power you'll need a rotary torque sensor and a tacho, then use P = 2∏RPM/60*Τ

Ask an engineer, he/she will be able to help.

Why not measuring the difference in speed? (no load speed against actual speed ).

The relation speed difference against mechanical output power is much more lineair than the relation current against mechanical output power.

We have Simocodes (Siemens) in case of LV switch gears. In simocode there are three C.T one for each phase. There is no P.T.From all these discussion I came to the conclusion that power is not shown in the control room but the percentage of current.

If the motor take 75 % of the rated current, operator sitting there often tell us the it is operating at 74 % load which we consider that the output power of the motor is 74 % of the rated Power mentioned on the nameplate.

Thanks for Ur participation

Power is not going to be linear with current. From Ohm's Law, we have that V=I*Z, Z being the impedance of the motor, and the point at which the power factor enters the equation. Power is given as P= V*I, therefore P= Z* I². Power is a function of the square of the current, not linear with current.

What you are saying is correct, but is has not much to do with electrical motors.

These formule is used when calculating the electrical power of one phase systems.