2420 / 10750 = .225 or a Coefficient of Performance (COP) = 22.5
25 percent
The machine efficiency is 35 percent (35/100).
200,000-150,000= 50,000 25 %
If the input energy is 210 joules and the efficiency of the system is 30%,then the output energy is30% of 210 = (0.3 x 210) = 63 joules.
If the waste heat is 15 compared to the total heat input of 20 (ignoring the thousands) this means the heat doing useful work is 5, which represents 25 percent of the total input heat, so this is the efficiency. Waste heat in a car is the sum of the engine cooling losses and the exhaust losses, the rest is converted to mechanical energy which is useful. 25%
25 percent
25%
25 percent
25 percent
The machine efficiency is 35 percent (35/100).
200,000-150,000= 50,000 25 %
If the input energy is 210 joules and the efficiency of the system is 30%,then the output energy is30% of 210 = (0.3 x 210) = 63 joules.
If the waste heat is 15 compared to the total heat input of 20 (ignoring the thousands) this means the heat doing useful work is 5, which represents 25 percent of the total input heat, so this is the efficiency. Waste heat in a car is the sum of the engine cooling losses and the exhaust losses, the rest is converted to mechanical energy which is useful. 25%
No. It is not reasonable. If that ever happened, we would have to ask where the additional 40 joules came from, since energy cannot be created or destroyed. If there were something inside the box that added 40 joules to the 110 passing through, then that 40 would need to be added to the "input" work.
Output energy divided by %eff.= input energy %eff. is energy out divided by energy in
The idea here is to divide the useful work by the input energy. (Note: Usually "power" would be used instead of "energy"; power is measured in joules/second.)
.33