Power output of a coal burning power plant

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SUMMARY

The discussion focuses on the power output of a coal-burning power plant operating at 30% efficiency, burning 10 kilograms of coal per day with a heat of combustion of 30 MJ/kg. The power output calculation reveals that 30% of the coal's energy is converted into useful energy, while 70% is discarded as thermal energy. The discarded energy rate is calculated using the formula P = (m/t)cΔθ, where Pdiscarded is the power of the discarded energy. The discussion clarifies misconceptions regarding energy conversion and provides a step-by-step explanation for calculating the water flow rate needed to maintain a temperature increase of no more than 5°C.

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  • Understanding of thermal efficiency in power plants
  • Knowledge of heat of combustion calculations
  • Familiarity with energy transfer equations, specifically E=mcΔθ
  • Basic principles of thermodynamics
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Homework Statement


A coal power plant with 30% efficiency burns 10 kilograms of coal per day. (Take the heat of combustion of coal to be 30 MJ/kg)
a) what is the power output of the plant?
b) At what rate is thermal energy being discarded by this plant?
c) If the thermal energy is carried away by water whose temperature is not allowed to increase by more than 5°C, calculate the rate at which water must flow away from the plant.

*note: I already have the solution, I just don't understand it.*
Specifically my questions would be
1) If the efficiency is 30%, then doesn't that mean that the plant only turns 30% of the coals energy into useful energy? The answer says that 70% is converted to useful energy.
2) Is discarded energy just the amount of energy that isn't converted into useful energy?
3) Lastly (and I realize that this is a very vague request), can someone explain from start to end the solution to part c? I can make guesses as to the other parts of the problem, but this one completely baffles me.

Homework Equations



I have no idea where the equations in the answer came from. We weren't provided any in the chapter that would be specifically relevant to this problem.

The Attempt at a Solution


I attached a picture of the solution (the text wouldn't copy properly)
 

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The calculation for part 'a' says 70% but the calculation uses 30%. So really you are correct and the actual calculation is correct, the wording is incorrect.

b) Yes, that is why they put 70% in the calculation i.e. 70% of the energy is not used.

c) From part b, you will get the rate of discarded energy in W (Pdiscarded).

The amount of energy required to raise the temperature of a substance is given by E=mcΔθ (see here for more information)

So the rate at which the energy will be transferred to the water is P=(m/t)cΔθ. Since you are transferring Pdiscarded, then P=Pdiscarded
 
Thanks, that helped a lot!
 

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