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I have a heating system with which I heat up 10 liters of water by 24.2°C in 610 seconds.

My system consumed 0.5 KW of electric energy in this time.

Now I used this formula: Specific heat of water * volume * temperature difference / time

and I get 1,66 KW (created)

efficiency 332%

Is this correct? Or must I use some other formula to calculate efficiency?

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  • $\begingroup$ Energy conversion efficiency greater than 100% is never correct. $\endgroup$ – Chris Mueller Jun 20 '16 at 12:00
  • $\begingroup$ @ChrisMueller not entirely true - heat pumps can consume 0.5 kWh of electricity, and deliver 1.66 kWh of heat. They have typically efficiencies in the range 200-500%. Though normally it's referred to as a COP (coefficient of performance) of 2-5. $\endgroup$ – EnergyNumbers Jun 20 '16 at 14:20
  • $\begingroup$ @EnergyNumbers: That calculation completely ignores the heat extracted from the environment, which must be included in order to calculate the true efficiency. $\endgroup$ – Dave Tweed Jun 20 '16 at 16:00
  • $\begingroup$ @DaveTweed I know. And that doesn't change the fact that what I described is how we calculate the numbers for heat pumps. $\endgroup$ – EnergyNumbers Jun 20 '16 at 16:05
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You are confusing units, which is making your question difficult to understand.

Electrical energy is measured in kW-hours (or Joules), not kW. If your heater consumed 500 W of power for 610 s, then you used 305 kJ of energy to heat your water.

Similarly, if 10 l (10 kg) of water was raised by 24.2 °C, then it absorbed

$$4.186 \frac{J}{g ^\circ C} \cdot 10 kg \cdot 24.2 ^\circ C = 1013 kJ$$

of energy. Clearly, something is wrong with your description of what actually happened.

If the heater actually used 0.5 kWh of energy (i.e., 2.95 kW for 610 seconds), this corresponds to 1800 kJ, which would indicate that the process is only about 1013/1800 = 56.3% efficient. You lost quite a lot of the incoming energy in the form of heat that ended up somewhere other than in the water.

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  • $\begingroup$ I use 2.2Kw electric motor and I installed kwh-meter on it and i get this data $\endgroup$ – RoX Jun 20 '16 at 6:10
  • $\begingroup$ @Rox how are you using an electric motor to heat water? $\endgroup$ – EnergyNumbers Jun 20 '16 at 14:20
  • $\begingroup$ I am experimenting with copper and magnets. Where motor move magnets which heat copper and copper heat water. But now I have 2,2kw motor with 1400rpm. Now i will replace it with 1,1kw 2800rpm and transfer rpm to 1400 and I think I will get the same heating and half power consumed and i think there will be more than 100% efficiency. What do you think? $\endgroup$ – RoX Jun 20 '16 at 15:54
  • $\begingroup$ I think you're nuts. You can't get more than 100% efficiency. More likely, it will simply take the smaller motor a longer time to heat the water. In any case, what's the point of using the motor? Induction heating by applying the AC magnetic field directly to the copper plate is much more efficient. $\endgroup$ – Dave Tweed Jun 20 '16 at 15:58
  • $\begingroup$ @Rox No. You will not get over 100% efficiency like that. You'll get very close to 100% by using the copper as a resistance heater. And without a heat-pump, that's the best you can do. $\endgroup$ – EnergyNumbers Jun 20 '16 at 16:07
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The formula's you use seem correct. The value for efficiency however is impossible as you can never reach an efficiency of 100% let alone over 100%.

Efficiency = ( Usefull energy / Total energy ) * 100% or ( Usefull power / total power) * 100%

Energy = Cp (assume constant) * M * (Tf - Ti) in which M is equal to V * rho

Power = Energy / Time

So, either the time it took to heat up is invalid or the system you use has a lot more power than it says it does.

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