For calculating the -per person- part, we are going to be using the population of the USA, i.e, $3\times10^8$.
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What have you done so far? – an4s Mar 08 '18 at 06:45
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@an4s 1 quad = 2.9329 x 10^11 kWhr... then I divided both sides by 365 days to get to -> 8.0353 x 10^8 kWhr / day.... then I divide it by the number of people. I don't know though if what I did was right? – user3489850 Mar 08 '18 at 07:12
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Sounds about right. – an4s Mar 08 '18 at 07:20
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Well, first of all we know that:
$$x\space\space\text{quad}=1055055852620000000\cdot x\space\space\text{joule}=\frac{2637639631550\cdot x}{9}\space\space\text{kWh}\tag1$$
So, when we have $x$ quad per year we know that one year has the following number of days:
$$\eta\approx365.242199\tag2$$
And the number of people in a country is:
$$\mathcal{P}\tag3$$
So, when we want to find kWh/day/person we need to find:
$$\text{E}\approx\frac{\left(\frac{\frac{2637639631550\cdot x}{9}}{365.242199}\right)}{\mathcal{P}}=\frac{376805661650000000}{469597113}\cdot\frac{x}{\mathcal{P}}\approx8.02402\cdot10^8\cdot\frac{x}{\mathcal{P}}\tag4$$
Jan Eerland
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