00:01
Hi, here in this given problem here, maximum power produced by v12 engine, that is p1 is equal to 447 kilowatt or we can say this is 447 multiplied by 10 raised to the power 3 watt.
00:33
Engine is rotating at the rate of 6 ,500 revolutions per minute.
00:46
And if maximum torque is, let it be tau max.
01:01
If it is 700 newton into meter, then the rate of rotation of the engine, let it be f2, that is given as 1500 revolutions per minute.
01:16
In the first part of the problem, using an expression for the power, instantaneous power, consumed in an engine that is given as the product of torque acting on it and its angular velocity.
01:31
Angular velocity, this is given as 2 pi into f.
01:35
So torque in that case, in this first case when p1 and f1 are given in that case torque tau 1 will be given by p1 divided by 2 pi f1 means this is 447 triple 0 divided by 2 times of 3 .14 multiplied by f1 which is 6 ,500 revolutions per minute or divided by 60 revolutions per 7.
02:09
So this torque is calculated to be equal to 657 .0 newton into meter.
02:19
Answer for this given problem...