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I am a mechanical engineering student and i need to measure electrical output. I have some voltage and current measuring instruments. I used them to measure the voltage & current of a blower(resistance heater). The results were 3 ampere and 155 volts. But the appliance is rated at 2kW. So according to the formula, P=VI, the load of the appliance should be 155*3=465W. Why does this deficit exist between the rated power and the one i measured. Do i need to introduce power factory in the P=VI equation? Someone told me it involves the concepts of apparent power reactive power, real power. If so can someone please explain those terms to me in a simplified manner, my knowledge of electrical engineering and circuits is poor.

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    \$\begingroup\$ Does the heater have a temperature control? Are there high and low settings for the fan speed? All these will affect the total power drawn, as well as power factor which is related to the motor and not the heater element... \$\endgroup\$ – Solar Mike Jan 12 at 13:48
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    \$\begingroup\$ 155 V is non-standard. What voltage is the unit supposed to be powered from? 230 V? 400 V? If there is a rating plate on the unit then please add a photo to your question. \$\endgroup\$ – Transistor Jan 12 at 13:48
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    \$\begingroup\$ Was hot air coming out of the blower when you made this measurement? I am suspicious that the blower was running but the heater was not. \$\endgroup\$ – Elliot Alderson Jan 12 at 13:48
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    \$\begingroup\$ That 2kW rated power is similar to a car with 500bhp - you don’t always use all of them... \$\endgroup\$ – Solar Mike Jan 12 at 13:50
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    \$\begingroup\$ @Transistor its supposed to work on 230V. \$\endgroup\$ – Mohammad Nayef Jan 12 at 13:53
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From your readings you can calculate the effective resistance of the load at \$ R = \frac {V}{I} = \frac {155}{3} = 52 \ \Omega \$.

If a resistance of this value was connected to a 230 V supply then the power would be given by \$ P = \frac {V^2}{R} = \frac {230^2}{52} = 1 \ \text {kW} \$.

You would need to check:

  • Loss of one or more heater elements. My bathroom heater has 2 x 1 kW elements in parallel and one is switched off by thermostat once the air reaches a certain temperature.
  • Power selector switch.
  • Any kind of dimmer power control.

I have assumed that the fan motor is a small load - maybe 50 W - in comparison with the heaters.

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There are multiple types of "power ratings" which can produce very different results.

  1. Power on the label: This is determined by local legal procedures which may or may not represent actual use cases.
  2. Power in product documentation and advertising. This is typically determined by what marketing thinks "sells best" and doesn't involve outright lying (although sometimes it does)
  3. What the device actually does: which can be measured but may vary a lot with the specific use case and/or environmental conditions

For example let's look at https://www.crownaudio.com/en/products/ma-12000i

The name implies that it's a 12000W amplifier, the data sheet lists the highest use case as 9000W, the power on the label is 1950W and I'd expect it to draw maybe 1000W or less in a real use case.

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2 kW is the rating of the Blower for a standard voltage (say 220 V or 110 V for AC Systems). for 220 V max ampere is around 9A. for 110 V around 18 A.

This ampere is dependent on the load on the system (here the blower).

I am not sure of working of Blower, I think the blower has a resistance inside.

you can adjust this resistance to get max ampere of blower.

check for the standard voltage of blower and try adjusting the resistance(change the settings-hot/cold) and calculate results. You would probably get a value near the rated value.

Since you have mentioned 230 V is rated voltage, you can apply that and adjust the setting and find value closer to rated one.

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    \$\begingroup\$ 'V' for volt, 'kW' for kilowatt. Capitals matter. \$\endgroup\$ – Transistor Jan 12 at 15:02

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