Thermodynamics: Waste heat and total ice converted to steam

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SUMMARY

The discussion focuses on calculating the thermal energy exhausted by a coal-fired power plant generating 600 MW of electric power and the potential conversion of ice to steam using this exhaust heat. The thermal energy exhausted each day is determined to be 1.584 x 1014 J after accounting for the energy used in electricity generation. The participant seeks clarification on the correct application of thermodynamic equations to find the mass of ice that can be converted to steam, utilizing specific heat capacities and latent heat values for ice and water.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically heat transfer.
  • Familiarity with the equations Q=mL and Q=mcΔT.
  • Knowledge of specific heat capacities and latent heat values for water and ice.
  • Basic proficiency in unit conversions and energy calculations.
NEXT STEPS
  • Research the efficiency of coal-fired power plants and heat recovery systems.
  • Learn about the thermodynamic properties of water and ice, including specific heat and latent heat.
  • Explore advanced thermodynamic cycles, such as the Rankine cycle, for power generation.
  • Investigate methods for calculating energy losses in thermal systems.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, engineers involved in energy production, and anyone interested in the efficiency of coal-fired power plants and heat recovery applications.

jolierouge
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Homework Statement


A coal-fi red plant generates 600 MW of electric power. The plant uses
4:8106 kg of coal each day. The heat produced by the combustion of coal is 3:3107 J/kg.
The steam that drives the turbines is at a temperature of 300C, and the exhaust water is at
37C. (a) How much thermal energy is exhausted each day by this plant? (b) How much ice
at -10C can be turned into steam at 100C per year by the exhaust from this power plant?

Homework Equations


Q=mL
Q=mcΔT


The Attempt at a Solution


For question a) all I did was take the coal used each day and multiplied it by the heat produce by its combustion. I found the thermal energy exhausted each day to be 1.584 x 10^14J. (I think this is right)
Question B is where I am having troubles.
This is the equation I got:
Q=mCice(0-(-10))+mLice+mCwater(100-0)+mLvaporization
q/(Cice(10)+Lice+Cwater(100)+Lvice)=m
where:
Lice=334 J/g
Cice=2.1J/gK
Cwater=4.2 J/gk
Lvice=2.256J/g
Q=1.584 x 10^14
Is this right? I getting confused about what energy I should and shouldn't take when trying to find the mass.
 
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jolierouge said:
For question a) all I did was take the coal used each day and multiplied it by the heat produce by its combustion.
That's the energy released by burning the coal, but some went into producing electricity. The exhausted heat is whatever's left.
Question B is where I am having troubles.
This is the equation I got:
Q=mCice(0-(-10))+mLice+mCwater(100-0)+mLvaporization
q/(Cice(10)+Lice+Cwater(100)+Lvice)=m
where:
Lice=334 J/g
Cice=2.1J/gK
Cwater=4.2 J/gk
Lvice=2.256J/g
Q=1.584 x 10^14
Is this right? I getting confused about what energy I should and shouldn't take when trying to find the mass.
You switched Lvap to Lvice in there, and the value seems much too low.
 
So, for the first part I take the "Q" I calculated and subtract the energy converted to energy which is 600MW (which is 600*10^6J) and I should be left with the thermal energy exhausted? And I use this for the q in the second part?

Ah, I see where I switched Lvap. And I see why my Lvap is too low...conversion issues(my bad).

(By the way thanks!)
 

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