A sport utility vehicle with a thermal efficiency etath of 2

A sport utility vehicle with a thermal efficiency (eta_th) of 25 % produces 250 hp of engine output while traveling at a velocity of 80 mph. Determine the rate of fuel consumption in kg/s if the heating value of the fuel is 43 MJ/kg. Express the rate of consumption in kilograms per second to five significant figures. If the density of the fuel is 800 kg/m^3, determine the fuel mileage of the vehicle in the unit of mi/gal. Express the fuel mileage in miles per gallon to two significant figures.

Solution

For the given SUV, we have the power output of 250 hp that the engine provides. Also, we know that the engine has an efficiency of 25 % which means that of the total energy produced by burning of the fuel, only 25% gets converted to the the power output.

We will use the relation for the efficiency to form a relation between the power output (Known) and power input (Unknown) and then resolve to determine the amount of fuel required per second.
It also needs to be noted that the for a fuel of heating value h MJ/kg, the amount of heat generated by burning m kg of fuel would be hm MJ

Part A.) Now, let us assume that the vehicle burns x kg of fuel per second.

Hence the heat produced would be 43x MJ/second

Now the power output we have is 250 hp or 186425 Watts or 186425 J/second

Also we know Power output / Power input = 0.25

That is, 186425 = 43x * 10^6 * 0.25

or, x = 17341.86 x 10^-6 Kg/s is the required rate of fuel consumption.

Part B.) We know that in one gallon we have 0.00378541 m^3

That is, mass of fuel in one gallon would be: 0.00378541 x 800 = 3.028328 Kg

As the rate of fuel consumption at the given speed is 17341.86 x 10^-6 Kg/s

The time for which the vehicle can continue to draw fuel at same rate with one gallon of fuel would be:

Time = 3.028328 / 1.734186 x 10^-2 = 174.6253 seconds.

Now the vehicle is travelling at a speed of 85 mph

or the distance travelled in 174.6253 seconds would be: D = 174.6253 x 85 / 3600 = 4.1231 Miles

Therefore in one gallon of fuel, the vehicle can travel a distance of 4.1231 Miles. That is the required mileage is 4.1231 Miles/gallon