Determine the efficiency of a tractor that consumed 1.5 kg of fuel with a specific heat of combustion
August 31, 2021 | education|
Determine the efficiency of a tractor that consumed 1.5 kg of fuel with a specific heat of combustion of 42 MJ / kg to perform work of 18.9 MJ.
Useful = 18.9 MJ = 18.9 * 10 ^ 6 Joules – the work done by the tractor;
m = 1.5 kilograms is the mass of fuel consumed by the tractor;
q = 42 MJ / kg = 42 * 10 ^ 6 Joule / kilogram – specific heat of combustion of fuel.
It is required to determine N – tractor efficiency.
Let’s find the amount of energy that was released during the complete combustion of the fuel:
Q = q * m = 42 * 10 ^ 6 * 1.5 = 63 * 10 ^ 6 Joules.
Then the efficiency of the tractor will be equal to:
N = A useful / Q = 18.9 * 10 ^ 6 / (63 * 10 ^ 6) = 18.9 / 63 = 0.3 = 30%.
Answer: The efficiency of the tractor is 30%.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.