A car moves on a flat road with a steady velocity of 80 km/h. It consumes gasoli
ID: 1816827 • Letter: A
Question
A car moves on a flat road with a steady velocity of 80 km/h. It consumes gasoline at a rate of 0.1 liter per km. Friction of the tires on the road and bearing losses are proportional to the velocity and, at 80 km/h, introduce a drag of 222 N. Aerodynamic drag is proportional to the square of the velocity with a coefficient of proportionality of 0.99 when the force is measured in N and the velocity in m/s.What is the efficiency of fuel utilization? Assuming that the efficiency is constant, what is the 'kilometrage" (i.e., the number of kilometers per liter of fuel) if the car is driven at 50 km/h?
The density of gasolone in 800 kg per cubic meter, and its heat of combustion is 49 MJ per kg.
Explanation / Answer
speed = 80e3 m/ h = 80e3 / 60*60 m/s = 22 .222 m / s Total drag = friction + the air drag = 222N + (22.2222)^2 * 0.99 = 222N + 488. 88888 N appoximately 711 N the gas used per km is 0.1 liters 1 liter = 10cm x 10cm x 10cm = (0.1)^3 m^3 = 0.001 m^3 800 kg ---> 1 cubic meter so 1 liter has 800g for 0.001 cubic meter so 0.1 liters is 80g of gas the combustion rate is 49 MJ per kg = 49 KJ per g so 0.1 liters gives you 80g * 49 KJ = 3.92 Million Joules now consider how far the car went with 0.1 liters. Force done by the gas pushing the car cancels out the friction/drag so F is what we found above. the work done by the gas pushing the car = Force * distance = 711 N * 1 km = 711, 000 Joules the fuel efficency when the car is traveling at 80 km per hr is then 711, 000 / 3,920,000 = 18 % efficiency . The same calculation can be done when the car is traveling at 50 km/h but you need to scale friction force, which is proportional to speed, and also scale the drag force, which is proportional to the velocity squared. good luck have fun
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