This question is relates vehicle dynamics . I Found this Question on a test paper.
A car mass of 1150 kg has road wheels of 0.7m effective diameter and reaches a speed of 110 km/h at an engine speed of 3000 rpm. Rolling resistance is 180 N and air resistance is 1000 N.What is the brake power developed developed by the engine if transmission efficiency is 90%, and the top gear ratio between engine and road wheels.
Please assist me to slove this problem.
There are several pieces of data missing. The first one that comes to mind would be the condition of the engine. If the engine was worn and could no longer keep compression, or could hold compression high, the results would be different.
If this is really a test, I hope you are not expecting someone here to do the work for you, after all your instructor (assuming you are a student) wants you to solve the problem and I doubt if he or she sees the the problem as you proving you can find someone else to do your work for you.
I think that the answer is “E” (all of the above).
Force on the edge of the tire = 1180 N (rolling + air resistance)
Actually 590 N per wheel with 2wd
Velocity of tire edge (and vehicle too) = 30.6 m/s
Brake power from the wheel = 1180 N * 30.6 m/s = 36100 N-m/s = 26600 ft-lb/s
Since 1 hp = 550 ft-lb/s then 26600 ft-lb/s = 48.4 hp
48.4 hp / 90% = 53.8 hp from the engine
Wheel torque = 1180 N * 0.7 m = 826 Nm
Wheel circumference = pi * 0.7 m = 2.2 m
Wheel rotational rate = 30.6 m/s / 2.2 m = 13.9 rps = 835 rpm
Drivetrain ratio = 3000 rpm / 835 rpm = 3.59:1
Engine torque = 826 N-m / 3.59 = 230 Nm = 170 lb-ft
(I recommend listening to Oscar Peterson when doing this kind of thing.)
There is a keyword missing from the O.P. “What is the brake power developed developed by the engine if transmission efficiency is 90%,”
How about “What is the brake horsepower developed…”