Energy and fuel consumption rates
Converting between L/100 km and km/L, comparing petrol, hybrid and electric cars by cost per kilometre and per year, payback time for a more efficient car, gas use in MJ, and solar power savings.
Fuel and energy are paid for by the litre, the kilowatt-hour or the megajoule, but used per kilometre or per hour. Comparing two cars, two heaters or two power plans means turning everything into the same rate first. This section is about making those comparisons fairly.
Revise: trip costs from L/100 km and EV charging are in Year 11 Running a car, and kWh = kW × h is in Year 11 Energy at home and in food.
L/100 km and km/L
Australia labels cars in litres per $100$ km: smaller is better, because the car uses less fuel. Some countries, and many drivers, think in kilometres per litre: there, bigger is better. The two are reciprocals, so you can always flip one into the other through $100$.
$\text{km/L} = \dfrac{100}{\text{L/100 km}}$ and $\text{L/100 km} = \dfrac{100}{\text{km/L}}$
A ute travels $14$ km on each litre. Write this in L/100 km. Then convert $8$ L/100 km to km/L.
Comparing vehicles
To compare a petrol car with an electric one, put both into the same money rate, such as dollars per $100$ km. Then scale to the distance you actually drive.
A small petrol car uses $6.5$ L/100 km with petrol at $195$c/L. An electric car uses $16$ kWh/100 km, charged at home at $30$c/kWh. Compare their running costs per $100$ km.
Set a $450$ km trip and check Example 2 against the cost per $100$ km for the small car and the electric car. Then raise petrol to $230$c/L: how much more does the SUV trip cost? Finally find the electricity price at which the electric car costs the same per km as the hybrid.
A family drives $15{,}000$ km a year. An SUV uses $9.8$ L/100 km and a hybrid $4.2$ L/100 km, with petrol at $\$1.95$/L. The hybrid costs $\$6000$ more to buy. How long until the fuel savings pay back the extra price?
Gas and solar
Gas is billed in megajoules (MJ). Gas heaters and cooktops are rated in MJ per hour, so the energy used is the rate times the time. Recall $1$ kWh $= 3.6$ MJ.
A gas heater is rated at $25$ MJ/h and runs for $4$ hours. Gas costs $4.2$c/MJ. Find the cost, and the energy in kWh.
On a sunny day, a home's solar panels generate $22$ kWh. The home uses $9$ kWh of it, which saves buying power at $33$c/kWh, and exports the other $13$ kWh for a feed-in tariff of $5$c/kWh. What is the day's benefit?
Your turn
A car is advertised at $5.6$ L/100 km. How many km/L is that, and how much fuel would a $380$ km trip use?
Show the working
$100 \div 5.6 \approx 17.9$ km/L. Fuel: $5.6 \times 380 \div 100 = 21.28$ L.
Convert $8$ L/100 km to km/L.
$100 \div 8 = 12.5$ km/L.
Convert $14$ km/L to L/100 km, to $2$ decimal places.
$100 \div 14 \approx 7.14$ L/100 km.
Which car is the most fuel efficient?
Convert to km/L: $9.8 \to 10.2$, $6.5 \to 15.4$. So $24$ km/L goes furthest on each litre.
An electric car uses $16$ kWh/100 km and power costs $30$c/kWh. What is the cost, in dollars, per $100$ km?
$16 \times 0.30 = 4.80$.
A car uses $9.8$ L/100 km. How many litres does it use in $15{,}000$ km?
$15{,}000 \div 100 = 150$ lots of $100$ km. $9.8 \times 150 = 1470$ L.
At $\$1.95$/L over $15{,}000$ km, how much does a $4.2$ L/100 km hybrid save, in dollars, compared with a $9.8$ L/100 km SUV?
Difference in fuel: $(9.8 - 4.2) \times 150 = 840$ L. $840 \times 1.95 = 1638$.
A heater rated at $25$ MJ/h runs for $4$ hours with gas at $4.2$c/MJ. What is the cost, in dollars?
$25 \times 4 = 100$ MJ. $100 \times 0.042 = 4.20$.
Solar power used at home is worth more than solar power exported because:
Each kWh used at home avoids buying one at, say, $33$c, while each kWh exported might earn only $5$c.
Exam style: Panels make $22$ kWh in a day. $9$ kWh is used at home (saving $33$c/kWh) and the rest is exported at $5$c/kWh. What is the total benefit, in dollars?
$9 \times 0.33 = 2.97$. $13 \times 0.05 = 0.65$. Total $= 2.97 + 0.65 = 3.62$.
Exam style: A hybrid costs $\$6000$ more than an SUV and saves $\$1638$ a year in fuel. How many years does it take to pay back the extra cost, to $1$ decimal place?
$6000 \div 1638 \approx 3.66$, which is $3.7$ years to $1$ decimal place.
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