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I'm currently studying the textbook Fundamentals of Electric Circuits, 7th edition, by Charles Alexander and Matthew Sadiku. Chapter 1.5 Power and Energy gives the following practice problem:

Practice Problem 1.6
A home electric heater draws 12 A when connected to a 115 V outlet. How much energy is consumed by the heater over a period of 24 hours?

The answer is said to be 33.12 k watt-hours.

The chapter says that the energy absorbed or supplied by an element from time \$t_0\$ to time \$t\$ is $$w = \int_{t_0}^t p \ dt = \int_{t_0}^t \nu i \ dt, \tag{1.9}$$

where \$ \nu \$ is the voltage, \$ i \$ is the current, and \$w\$ is energy in Joules. It then says that 1 Wh = 3,600 J.

So, by my calculation, we have

$$\int_0^{24} 12 \times 115 \ dt = \left[ 1380t \right]_0^{24} = 33120 \text{ J}.$$

But, converted to watt-hours, as in the textbook's answer, this would be \$ \dfrac{33120}{3600} = 9.2 \text{ Wh}\$. Am I misunderstanding something? Are my units incorrect? Or is this, perhaps, a textbook error?

The Pointer
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3 Answers3

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Let's do some back-of-the-envelope stuff before we start losing decimal places.

The heater uses 12 A at 115 V, that's 1380 W, or just over 1 kW. So in one hour, it uses just over 1 kWh. In 24 hours, it must use a bit more than 24 kWh. To do it exactly, in 24 hours, it uses 1.38 * 24 = 33.12 kWh, as in the book.

So where's your mistake? A watt is a Joule per second, not hour. Multiply equation (1.9) by 3600 to get from hours to seconds, and you'll get the right answer.

Neil_UK
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  • I don't understand. If the textbook says that 1 Wh = 3,600 J, and we have that $w = 33120 \text{ J}$, then don't we just divide 33120 by 3600 to get the amount in Wh? – The Pointer Nov 15 '21 at 18:15
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    You’re integrating over 24 seconds not 24 hours, so your joule figure is 3600 times too small – Frog Nov 15 '21 at 18:20
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    @Frog Ohh, yes, I see what you mean. The limits of integration are indeed in units of seconds – not hours – so computing the upper limit should be $1380 \times 24 \times 3600 $, so we then get $\dfrac{1380 \times 24 \times 3600}{3600} = 33120 \ \text{Wh}$, as required. Thanks for the clarification. – The Pointer Nov 15 '21 at 18:25
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    It goes without saying that unit consistency is very important. Keep in mind the the Mars Climate Observer was loss back in 1993 because of a navigation error caused by a failure to translate English units to metric. – SteveSh Nov 15 '21 at 18:29
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Am I misunderstanding something?

Yes. The result of the integration from 0 to 24 hours is the total energy in watt-hours, not in joules. You show 33120 J, where it actually is 33120 Wh. Multiply this by 3600 to get 119.232 MWh, or 119.232 MJ.

If you want the final result in joules rather than kWh, a more direct method is to integrate in seconds rather than hours. If you integrate from 0 to 86400, the result in in joules (or watt-seconds).

AnalogKid
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Physical units means a lot , always keep them in mind to avoid confusion. I’ll give you another solution.

P = VI
E = Pt ==> E = VQ
Q = It = 12 coulombs (amperes × s)
through 24 hours is 1 × 60 × 60 × 24 = 86400 s.

So Q through the day is 12 × 86400= 1,036,800 coulomb (C)
so E = 115 × 1,036,800 = 119,232,000 joules (J).

So as you can see, when confused stick to basics.

Transistor
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Abd Alhaleem Bakkor
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