Comparing Energy Prices

Source

# from python
from fractions import Fraction

def print_energy(megajoules_per_dollar: float, source: str) -> None:
    """Print the energy provided by the source.

    Args:
     megajoules_per_dollar: how many megajoules provided by the source for each dollar
     source: where the energy is coming from
    """
    print(f"{source} provides {megajoules_per_dollar:0.2f} megajoules per dollar.")
    print(f"{source} costs ${1/megajoules_per_dollar:0.4f} per megajoule.")
    return

COSTS = []

How Many Megajoules For a Dollar of Gas?

  • A gallon of gasoline carries with it \(\approx 1.3E8\) joules of energy.
  • You pay \(5 \frac{\text{dollars}}{\text{gallon}}\).
  • How many megajoules (\(10^6\) joules) can you get for a dollar?

Chain-Link Conversion

\begin{align} \require{cancel} \left(\frac{1 \cancel{gallon}}{5\textit{ dollars}}\right)\left(\frac{1.3 \times 10^8\textit{ joules}}{1 \cancel{gallon}}\right) &= \left(\frac{1.3 \times 10^8 \cancel{joules}}{5\textit{ dollars}}\right)\left(\frac{1 megajoule}{10^6 \cancel{joule}}\right) \\ &=\frac{1.3\times 10^8 megajoules}{(5 \textit{ dollars})\left(10^6\right)}\\ &= 26 \frac{megajoules}{dollar} \end{align}

In Python

gallons_per_dollar = Fraction(1, 5)
joules_per_gallon = 13 * 10**7
joules_per_dollar = joules_per_gallon * gallons_per_dollar
MEGAJOULES_PER_JOULE = Fraction(1, 10**6)
gas_megajoules_per_dollar = joules_per_dollar * megajoules_per_joule
gasoline = 1/gas_megajoules_per_dollar

print_energy(float(gas_megajoules_per_dollar), "Gasoline")
COSTS.append((gasoline, "Gasoline"))

Gasoline provides 26.00 megajoules per dollar.
Gasoline costs $0.0385 per megajoule.

How Many MegaJoules For a Dollar of Electricity?

  • Electricity is $0.05 per kilowatt-hour.
  • \(1\text{ watt} = 1 \frac{joule}{second}\)
  • \(1\text{ kilowatt-hour} = 1,000\text{ watts} \times 3,600\text{ seconds}\)

Kilowatt-Hour To Joules

\begin{align} 1 kilowatt-hour &= (1,000\text{ watts})(3,600\text{ seconds})\\ &= \left(\frac{1,000\text{ joules}}{\cancel{second}}\right)\left(3,600 \cancel{seconds}\right)\\ &= 36 \times 10^5 \text{ joules} \end{align}

A Dollar's Worth Of Megajoules

\begin{align} \frac{1 \text{ kilowatt-hour}}{0.05\text{ dollars}} &= \left(\frac{3.6 \times \cancel{10^6 joules}}{0.05\text{ dollars}}\right)\left(\frac{1 \text{ megajoule}}{\cancel{10^6 joules}}\right)\\ &= \left(\frac{3.6\ megajoules}{0.05\text{ dollars}}\right)\left(\frac{20}{20}\right) \\ &= 72 \ \frac{megajoules}{dollar} \end{align}

Check The Math

kilowatt_hours_per_dollar = 20 # 1/0.05 = 20
megajoules = 3.6
electricity_megajoules_per_dollar = kilowatt_hours_per_dollar * megajoules
electricity = 1/electricity_megajoules_per_dollar

print_energy(electricity_megajoules_per_dollar, "Electricity")
COSTS.append((electricity, "Electricity"))

Electricity provides 72.00 megajoules per dollar.
Electricity costs $0.0139 per megajoule.

How Many Megajoules For a Dollar Of Natural Gas?

  • A standard cubic foot of natural gas has about \(1.1 \times 10^6\) joules of energy.
  • You can get about \(5\times 10^5\) BTUs of gas for a dollar.
  • There are about 1,030 BTUs in a cubic foot.
\begin{align} \left(1.1 \times 10^6 \frac{joules}{\cancel{cubic foot}}\right)\left(\frac{1\ \cancel{cubic foot}}{1.03 \times 10^3\ BTU}\right) &= \left(\frac{1.1 \times 10^3\ joules}{1.03\ \cancel{BTU}}\right)\left(\frac{5 \times 10^5\ \cancel{BTU}}{dollar}\right)\\ &= \left(\frac{5.5 \times 10^8\ \cancel{joules}}{1.03\ dollars}\right)\left(\frac{1\ Megajoule}{10^6 \cancel{joules}}\right)\\ &= \left(\frac{5.5 \times 10^2\ Megajoules}{1.03 dollars}\right)\\ &\approx 533.98\ \frac{Megajoules}{dollar}\\ \end{align}

Check The Math

joules_per_cubic_foot = 1.1E6
cubic_feet_per_btu = 1/1030
btus_per_dollar = 5e5

joules_per_btu = joules_per_cubic_foot * cubic_feet_per_btu

joules_per_dollar = joules_per_btu * btus_per_dollar
natural_gas_megajoules_per_dollar = joules_per_dollar * MEGAJOULES_PER_JOULE
natural_gas = 1/natural_gas_megajoules_per_dollar
print_energy(natural_gas_megajoules_per_dollar, "Natural Gas")
COSTS.append((natural_gas, "Natural Gas"))

Natural Gas provides 533.98 megajoules per dollar.
Natural Gas costs $0.0019 per megajoule.

How Many Megajoules For a Dollar of Coal?

  • A Ton of coal holds about \(3.2 × 10^{10}\) joules of energy.
  • A Ton of coal costs $40.

How many Megajoules of energy in the form of coal can you get for a dollar?

\begin{align} \left(\frac{3.2 \times 10^{10}\ joules}{40\ dollars}\right) &= \left(\frac{8 \times 10^{8} joules}{dollar}\right)\\ &= \left(\frac{8 \times 10^{8}\ \cancel{joules}}{dollar}\right)\left(\frac{1\ Megajoule}{1 \times 10^6\ \cancel{joules}}\right)\\ &=8 \times 10^2 \frac{Megajoules}{dollar} \end{align}

Checking the Math

coal_megajoules_per_dollar = MEGAJOULES_PER_JOULE * 3.2e10/40
coal = 1/coal_megajoules_per_dollar
print_energy(coal_megajoules_per_dollar, "Coal")
COSTS.append((coal, "Coal"))

Coal provides 800.00 megajoules per dollar.
Coal costs $0.0013 per megajoule.

How Many Megajoules Per Dollar In Corn (Biodiesel)?

  • Corn oil costs about $0.10 per fluid ounce (wholesale).
  • A fluid ounce carries about 240 dietary calories (kilo-calories).
  • A calorie is about 4.2 joules.

How many Megajoules of energy in the form of corn oil can you get for a dollar?

\begin{align} \left(\frac{1\cancel{fluid-ounce}}{0.10\ dollar}\right)\left(\frac{240\ kilocalories}{\cancel{fluid- ounce}}\right) &= \left(\frac{240\ \cancel{kilocalories}}{0.10\ dollar}\right)\left(\frac{10^3\ calories}{1\ \cancel{kilocalorie}}\right)\\ &=\left(\frac{24\times 10^4\ \cancel{calories}}{0.10\ dollar}\right)\left(\frac{4.2\ joules}{1 \cancel{calorie}}\right)\\ &=\left(\frac{100.8\times 10^4\ joules}{1 \times 10^{-1} dollar}\right)\\ &=\left(\frac{1.008 \times 10^7 joules}{dollar}\right)\left(\frac{1\ megajoule}{10^6\ joules}\right)\\ &= 1.008 \times 10^{1}\ \frac{megajoules}{dollar}\\ &= 10.008 \end{align}

Checking the Math

ounces_per_dollar = 1/0.10
kilocalories_per_ounce = 240
joules_per_calorie = 4.2
calories_per_kilocalorie = 10**3

corn_megajoules_per_dollar = (ounces_per_dollar
                              * kilocalories_per_ounce
                              * calories_per_kilocalorie
                              * joules_per_calorie
                              * MEGAJOULES_PER_JOULE)
corn = 1/corn_megajoules_per_dollar
print_energy(corn_megajoules_per_dollar, "Corn")
COSTS.append((corn, "Corn"))

Corn provides 10.08 megajoules per dollar.
Corn costs $0.0992 per megajoule.

What Is the Ratio of the Cost of the Most Expensive to the Cheapest?

most_expensive, cheapest, cost, name = max(COSTS), min(COSTS), 0, 1
print(f"The most expensive form of energy is {most_expensive[name]} "
      f"at ${most_expensive[cost]:0.2f} per megajoule.")
print(f"The cheapest form of energy is {cheapest[name]} at "
      f"${cheapest[cost]:0.4f} per megajoule.")

The most expensive form of energy is Corn at $0.10 per megajoule.
The cheapest form of energy is Coal at $0.0013 per megajoule.

print(f"{most_expensive[name]} is {most_expensive[cost]/cheapest[cost]:0.2f} "
      f"times more expensive than {cheapest[name]}.")

Corn is 79.37 times more expensive than Coal.