2.4 A 2650-lb car is traveling at sea level at a constant speed. Its engine is running at 4500 rev/min and is
producing 175 ft-lb of torque. It has a drivetrain efficiency of 90%, a drive axle slippage of 2%, 15-inch–
radius wheels, and an overall gear reduction ratio of 3 to 1. If the car’s frontal area is 21.5 ft2, what is its drag
coefficient?
Solution
1- Calculate the speed of the vehicle (𝑉) using the formula:
2𝜋𝑟𝑛𝑒 (1 − 𝑖)
𝜀𝑜
Here, r is the radius of wheel, 𝑛𝑒 is speed of the engine, i is drive axle slippage, and 𝜀𝑜 is overall gear
resection ratio.
v=
Substitute 15 in. for r, 4500 rev/min for 𝑛𝑒 , 2 % is for i, and 3 for 𝜀𝑜 :
v=
2𝜋(15 in.×
v = 192.42 ft/s
1 ft
rev
1 min
2
)(4500
×
)(1 −
)
12 in
min
60 s
100
3
Thus, the speed of the vehicle is 192.42 ft/s
2- Calculate the engine-generated tractive effort (Fe) using the formula:
𝑀𝑒 𝜀𝑜 𝜂𝑑
𝑟
Here, Me is the engine torque and 𝜂𝑑 is the drivetrain efficiency.
𝐹𝑒 =
Substitute 175 ft-Ib for 𝑀𝑒 , 3 for 𝜀𝑜 , 0.9 for 𝜂𝑑 , and 15 in. for r.
175 × 3 × 0.9
15/12
𝐹𝑒 = 378 Ib
𝐹𝑒 =
Thus, the engine-generated tractive effort is 378 Ib.
Consider that the available tractive effort (F) is the engine-generated tractive effort (Fe), 𝐹 = 𝐹𝑒 .
Hence, the available tractive effort is 378 Ib.
3- Calculate the coefficient of rolling resistance (𝑓𝑟𝑙 ) using the formula:
𝑓𝑟𝑙 = 0.01 (1 +
V
192.42
) = 0.01 (1 +
) = 0.0231
147
147
Thus, the coefficient of rolling resistance is 0.0231
4- Calculate the coefficient of drag (𝐶𝐷 ) from the following relation:
𝜌
𝜌
𝐹 = 𝑚𝑎 + 𝑅𝑎 + 𝑅𝑟𝑙 + 𝑅𝑔 = 𝑚𝑎 + + 𝐶𝐷 𝐴𝑓 𝑉 2 + 𝑅𝑔 = 𝑚𝑎 + 𝐶𝐷 𝐴𝑓 𝑉 2 + 𝑓𝑟𝑙 𝑊 + 𝑊𝐺
2
2
Here, 𝐹 is the available tractive effort, 𝑚 is mass of the vehicle, 𝑎 is acceleration, 𝑅𝑎 is the aerodynamic
resistance, 𝑅𝑟𝑙 is the rolling resistance, 𝑅𝑔 is the grade resistance, 𝜌 is the density of air, 𝐶𝐷 is the
coefficient of aerodynamic drag, 𝐴𝑓 is the frontal area of the vehicle, 𝑊 is weight of the vehicle.
5- Consider, maintaining the speed of the vehicle, the tractive effort is equal to the summation of resistances.
Hence, no tractive effort will remain for vehicle acceleration, so 𝑚𝑎 is equal to zero, and for level road, 𝐺
is equal to zero.
𝜌
𝐹 = 0 + 𝐶𝐷 𝐴𝑓 𝑉 2 + 𝑓𝑟𝑙 𝑊 + 𝑊(0)
2
𝜌
= 𝐶𝐷 𝐴𝑓 𝑉 2 + 𝑓𝑟𝑙 𝑊
2
Substitute 378 Ib. for F, 21.5 ft2 for 𝐴𝑓 , 192.42 ft/s for 𝑉, 0.0231 for 𝑓𝑟𝑙 , and 2650 Ib for 𝑊, 0.002378
slugs/ft3 for 𝜌.
0.002378
𝐶𝐷 × 21.5 × (192.42)2 + 0.0231 × 2.650
2
𝐶𝐷 = 0.167
378 =
Thus, the drag coefficient is 0.167