Thermodynamics: Waste heat and total ice converted to steam

In summary, the coal-fired plant generates 600 MW of electric power using 4.8*10^6 kg of coal each day. The heat produced by combustion is 3.3*10^7 J/kg. The thermal energy exhausted each day is found to be 1.584*10^14 J. For question B, the correct equation is Q=mCice(0-(-10))+mLice+mCwater(100-0)+mLvaporization, where Lice=334 J/g, Cice=2.1 J/gK, Cwater=4.2 J/gK, and Lvap=2.256 J/g. The correct value for Lvap should be used in the
  • #1
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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  • #2
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.
 
  • #3
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!)
 

FAQ: Thermodynamics: Waste heat and total ice converted to steam

1. What is thermodynamics?

Thermodynamics is a branch of physics that studies the relationships between heat, energy, and work. It is concerned with how these factors interact and how they affect the behavior of matter.

2. What is waste heat?

Waste heat is the energy that is lost or dissipated during a thermodynamic process. It is the difference between the heat input to a system and the useful work that is produced.

3. How is waste heat converted to steam?

Waste heat can be converted to steam through a process called waste heat recovery. This involves using a heat exchanger to transfer the waste heat from one system to another, typically a boiler. The heat is then used to generate steam, which can be used for various purposes.

4. What is total ice?

Total ice refers to the sum of all ice present in a system, including both solid and liquid ice. In thermodynamics, total ice is used to calculate the energy required to melt or freeze a given amount of ice.

5. How does waste heat and total ice relate to steam?

In the process of converting waste heat to steam, the waste heat is used to melt the total ice present in the system. This melted ice then turns to steam as it is heated. Therefore, waste heat and total ice are both important factors in the production of steam through waste heat recovery.

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