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I have an experiment running which generates variable current due to a chemical reaction over a period of 8 hours. I sampled the current at each hour with a constant voltage supply of 12V. When i use P= I*V i get the power at that particular reading.

My assumption to calculate a the power consumed during the complete experiment was to plot a graph between the power at each hour and experiment duration. Calculating the area under the plot graph will be power calculated per experiment but not in kwh.

Is my assumption true or is there a mathematical formula that i have to use to calculate the power in kwh.

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    \$\begingroup\$ If you plot kW against time in hours why won’t you get kWh ? \$\endgroup\$
    – Solar Mike
    Nov 19 '17 at 22:32
  • \$\begingroup\$ No such thing as "power in kwh". Please learn the difference between power and energy or you will have a hard time with any electrical or electronic engineering. \$\endgroup\$
    – winny
    Nov 19 '17 at 22:54
  • \$\begingroup\$ The plot of kW against time in hours will yield the energy used in kWh when the area under the plot is calculated. There are no assumptions. \$\endgroup\$
    – Barry
    Nov 19 '17 at 23:35
  • \$\begingroup\$ The only assumption you are making is that your sample rate is adequate. Energy can be measured in Watt seconds or Watt minutes or Watt hours or Kilowatt hours (and various other units). It is just a simple unit conversion. Since you want kilowatt hours, draw your graph with kilowatts as the vertical unit and hours as the horizontal unit. The area under the curve is energy. \$\endgroup\$
    – mkeith
    Nov 20 '17 at 1:38
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  • Power is measured in watts (W).
  • Electrical energy is measured in watt-hours (Wh).

Your understanding of P = VI is correct and will be in watts. That gives you the power. Since your interval of measurement is one hour the watt-hours will be the same numerical value.

Assuming the power rises or falls continuously during the experiment (rather than randomly increasing and decreasing) I would use the average power for the interval: \$ P_{n} = \frac {P_{n-1} + P_n}{2} \$ where \$n\$ is the measurement number.

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Have you tried Calorimetry. Its quite a popular way of measuring the energy output in a reaction. Simply short your experiment in a container of water in the shape or from of a heater element and measure the temperature difference from start to end. Refer to

q= McT where 'T' is the change in temperature. once you find 'q' plug that into P=E/T where 'E' can be swapped for q.

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  • \$\begingroup\$ (1) Is "Calorimetry" a brand name (or why has it got a capital C)? (2) How do you "short your experiment"? (3) "... a container of water in the shape or from of a heater element ...". What? \$\endgroup\$
    – Transistor
    Nov 20 '17 at 0:20
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    \$\begingroup\$ If electrochemical work is being performed, the usual assumption that all energy is converted to heat will not be valid and it will not be a good application of calorimetry. Think of a battery, for example. Most of the energy added to a battery during charging does not get converted to heat. \$\endgroup\$
    – mkeith
    Nov 20 '17 at 1:41
  • \$\begingroup\$ @Transistor -- to 1) dunno why he capitalized it save for emphasis as it's not any sort of trade name \$\endgroup\$ Nov 20 '17 at 2:28
  • \$\begingroup\$ Sorry for the misunderstanding. @Transistor: 1) Just a punctuation error. 2) I mean short as in create a circuit load through heating alone, that way almost all of the electrical energy will be transformed in to heat energy. Only small almost immeasurable amounts of energy become another energy form such as and including: EM, sound, light, and Ionization (from being in water). Hope this helps without creating more questions. \$\endgroup\$ Nov 22 '17 at 4:15
  • \$\begingroup\$ No problem. You can edit your answer to correct / clarify and it may prove useful to someone. It's worth proof reading before hitting submit! Welcome aboard. \$\endgroup\$
    – Transistor
    Nov 22 '17 at 7:17

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