Question

5.1 A car is traveling along a highway at 65 miles per hour. The road is horizontal (0% slope). If the wind resistance and rolling resistance at the wheels creates a combined resistive force of 950 N, what is the power (kW) developed at the rear wheels?

5.2 How much power (kW) must the car engine in problem 5.1 develop if the overall mechanical efficiency of the transmission and drive train is 94%?

5.3 Assume the car engine in problem 5.1 is operated on gasoline with an energy content of 113,500 Btu's per gallon, and that the thermal efficiency of the spark ignition engine and drivetrain is 28.6%. What is the fuel efficiency of the car in miles per gallon?

5.4 The car in problem 5.1 approaches a mountain and begins ascending a grade of 6.0%. If the car maintains an uphill speed of 70 miles per hour, how much additional power (kW) must be developed by the engine to overcome the change in elevation? Assume the car has a mass of 1400 kg.

5.5 Assume the car above is operated on a blend of 85% ethanol and 15% gasoline (E85). If the energy content of ethanol is 80,460 Btu's per gallon, what is the energy content of the E85 fuel (Btu/gal)?

5.6 Assume the overall thermal efficiency of the engine and drivetrain of the car above drops to 24.5% when operated on E85 fuel. What is the estimated fuel economy (km/L) of the car above when subjected to the uphill operating conditions above (Problem 5.4)?

Answer 1

(5.1) it is given in the question,

speed of the car = 65miles/hour = 104.61km/h = v

resisting force = 950N = F

so, power developed by rear wheel = F. v = 950 * 104.61 = 99379.5 W = 99.3795 KW = 99.4KW

(5.2)it is given overall efficiency = 94%= 0.94

power by rear wheels(output power) = 99.4KW (see question 5.1)

overall efficiency of the car = power by rear wheels(output power)/ power by engine(input power)

so, 0.94 = 99.4/power by engine

now, power by engine = 99.4/0.94 = 105.74 KW

(5.3) energy of the gasoline is geven = 113500 Btu per gallon = 119742500 joul/ gallon

   28.6% consume in thermal efficiency , drive train etc,

(100-28.6)% = 71.4% consumes in kinetic energy of the car

then, 71.4Z% of 119742500 = 85496145 joul per gallon

   and from 5.1 , power of engine = F.V = 950*V = 85496145

   V = 85496145/950 = 89995.94 metre/ gallon

   V = 89.995KM per gallon

   V = 55miles and 16 yard per gallon

(5.4) additional power should be added to the car to overcome the gravitational potentional energy

it is given ascending grade = 6% = 0.06 = h/d

h = 0.06 d

potential energy of the car to cover the height h = mgh

where m is mass and g is acceleration due to gravity = 9.8m/s²

E = mg . 0.06 d

rate of energy is power = E/t = mg. 0.06d/t where d/t = speed of the car

p = 1400*9.8*0.06*112.65 70 mi= 112.65k/m

= 92733.48W = 92.733kw

this is the additional power needed to the car

5.5. consider the total fuel in the car including ethanol and gasoline = x

   85% of x = 80460

   so, x = 80460*100/85 = 94658.82

   now 15% of 94658.82 = 14198.82 Btu/ gal

5.6. efficiency of the car = out power / input power

proceed the equation go further due to time lag i droped this question