Please help me to solve this problem?

Question

I hope Anyone who has sound knowledge in Electrical engg. Can only make a suggestion for the following question! I also need simple illustration. I must generate 200 watts of Electricity. But i have only these details.
1. Crank shaft pulley dia 8 cm.
2. Crank shaft speed 700 rpm.
3. 2 kg of force acting on the crank shaft.
4. Crank shaft length - 15 cm.
If I connect the crank shaft to an ordinary car alternator, will it be able to produce 200 watts?
I know nothing about Elec. Engg. So please make a good suggestion how to produce 200 watts.

Answer 1

The given data is not too clear, is "crank shart length" the offset of the crank? in other words, is the 2kgf (19.6N) acting on a 15cm radius? or is the force acting on a 4cm radius (i.e. tangent to the 8cm dia. pulley)?

since 700PRM = 73.3 radians per second, in the first case you have 19.6N * 0.15m = 2.94Nm(torque), 2.94Nm * 73.3rad/s = 215.5W, if you alternator is very efficient (and at 700 RPM it probably won't be) you might get 200W of electricity, but in reality you will probably fall a little short. An auto alternator will work better at a higher speed, which you could achieve with a belt drive using a smaller pulley on the alternator, but then you will have some losses in the drive.
In the second case where the 19.6N is acting on a 4cm radius, you only have 19.6 * 0.04 * 73.3 = 57.5W of mechanical power, no way can you get 200W of electrical.

[EDIT] you may notice a discrepancy between my answer and the first, but 2π * 9.15W = 57.5W and the 2π was missing from the first answer (see http://en.wikipedia.org/wiki/Torque#Relationship_between_torque.2C_power_and_energy )

Answer 2

2 kg ==> 19.6 N
T = (19.6 n)(0.04 m) ≈ 0.785 N-m
P = (0.785 N-m)(700 rpm)(1 min/60 s) ≈ 9.15 W
You need 22 times the speed or torque to generate 200 W.