Mechanical Engineering - Automobile Engineering - Discussion
Discussion Forum : Automobile Engineering - Section 6 (Q.No. 41)
41.
A petrol engine of a car develops 125 N-m torque at 2700 r.p.m. The car is driven in second gear having gear ratio of 1.75. The final drive ratio is 4.11. If the overall transmission efficiency is 90%, then the torque available at the driving wheels is
Discussion:
4 comments Page 1 of 1.
Sai venkata raja said:
6 years ago
Please explain clearly.
Aarthie Ravichandran said:
7 years ago
Torque =gear ratio * final drive ratio * overall transmission efficiency.
T = 1.75 * 4.11 * 90/100 * 125,
T = 809.1 Nm.
T = 1.75 * 4.11 * 90/100 * 125,
T = 809.1 Nm.
(1)
LALIT MLAV said:
1 decade ago
T = G*N*Te.
1.75*4.11*90%*125 = 809.1N-M.
1.75*4.11*90%*125 = 809.1N-M.
Ajith said:
1 decade ago
Engine torque * gear ratio * final drive ratio * overall transmission efficiency.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers