Power output of a coal burning power plant

AI Thread Summary
A coal power plant operates at 30% efficiency, meaning it converts only 30% of the coal's energy into useful power, while 70% is discarded as waste energy. The discarded energy is indeed the energy that is not converted into useful energy. To calculate the rate at which water must flow away from the plant, one can use the formula for energy transfer, E=mcΔθ, where the rate of energy transfer to the water equals the rate of discarded energy. The solution involves determining the power output and the discarded energy, then applying the energy transfer equation to find the water flow rate. Understanding these concepts clarifies the calculations and the overall energy dynamics of the coal power plant.
SeanWade
Messages
2
Reaction score
0

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)
 

Attachments

  • Screen Shot 2012-08-31 at 12.39.11 PM.png
    Screen Shot 2012-08-31 at 12.39.11 PM.png
    34.5 KB · Views: 1,127
Physics news on Phys.org
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top