Question

A car travels at a constant velocity of 90 km/h along an expressway under-ambient temperature and pressure conditions. The width of the car is 1.5 m while its height is 1.2 m. The wheelbase (length between the front and rear wheels) is 2.5 m. The drag coefficient is 0.22.
[Use $\rho$ air, ambient density of the air = 1.12 kg/m³ and, $\rho$ fuel, density of the fuel = 930 kg/m³. You may neglect any energy contained in the exhaust gases and any resistance to motion is only attributed to the drag force. The rolling resistance of the wheels is assumed to be small.]
(a) Recognise and write down the equation of the drag force acting on a car moving at a constant velocity, V. Identify all the variables you had used and state the unit of each of the variables.

The rate of fuel consumption of this car, ?̇ is 15 km/litre of fuel. The calorific value of the fuel is 44 MJ/kg.
(b) Apply the given data to show that the fuel consumption of the engine is 0.00155 kg/s.

(c) Interpret the engine power required and confirm that it is 4.725 kW.

(d) Realise that the efficiency of the engine is 6.93%.

(e) Explain why very little power is needed to propel this car at a constant velocity of 90 km/h while the typical engine power available in a medium size car is between 50 kW to 75 kW.

Answer 1

Given the car moves at const. velocity= 90kmph=90*5/18=25m/s

Also given, drag force is the only force that is overcome by car power.

Cd=coeff. of drag=0.22

$\rho$= density of air=1.12kg/m3

a.)hence, drag force is given by Fd=1/2*Cd*$\rho$*v2 *Frontal area = 0.5*0.22*1.12*25*25*1.5*1.2= 138.6N

b.) fuel consumption = fuel density*velocity/ rate of fuel consumption = 930*25m/s *10-3/ 15*1000 =0.00155kg/s

(following conversion of 1 litre=10-3 m3  is used in above calc)

c.) Since, rolling friction is negligible, so the traction force is just greater than equal to drag force to maintain the speed.

therefore, Traction force = Fd= 138.6N (as calculated at qery (a))

power required = force * velocity = 138.6N*25m/s = 3.46kW

But, I see as per the question, power required is 4.725kW, which is not matching. Need to verify the correctness of the value

d.) Power generated by burning fuel = mf*CV=0.00155kg/s*44*106 W = 68.2kW

efficeincy = power req/ power generated= 3.46/68.2= 0.05 or 5%

If I use the answer given query (c), where power req is 4.725kW, efficeincy will be 4.725/68.2= 0.0693 or 6.93%

e.) since, the resistance due to friction and other internal losses are not considered, the power required to move the vehicle is less. though the capcity of the vehicle is high, but the required force is less. Here the assumption considered is the total thermal energy is directly utilised in moving the vehicle, which is not practical, since, thermal energy being a low grade energy is not 100% convertible to high grade energy (mechanical energy).

Also efficiency of internal combustion engines is hardly 28-30%