Problem on the second law of thermodynamics

In summary, the problem involves determining the rate of mass flow of the working fluid (steam) through a theoretically best efficiency steam engine as it climbs a hill with a given total train mass, temperature of steam in and out, and height. The solution involves calculating the efficiency of the Carnot cycle and finding the power and work per unit mass of water produced by the steam engine.
  • #1
mathlete
151
0
A locomotive steam engine opperating with a total train mass of 99.9x10^6 kg climbs a hill of height 1000m in 10 minutes. If the steam engine produces steam at a temperature of 390k and exhausts steam at 300k, and if the engine were at the theoretically best efficiency possible, what would be the rate, in mass per unit time, at which the same is exhausted be?

I figured out the efficiency, and the work done by the try... I get the heat in, then use that to get the heat exhausted (I get 4.22x10^11 J for reference).. but then I'm stuck. It's more not really understanding the question than a physics then.. I just don't really know what to do with this, almost like I can't figure out how I should manipulate the units :yuck:
 
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  • #2
I didn't understand what the problem's asking...You can find:heat consumed,work done and heat exhausted.Now,what do they mean by "the same is exhausted"...?And mass of what...?

Daniel.
 
  • #3
mathlete said:
A locomotive steam engine opperating with a total train mass of 99.9x10^6 kg climbs a hill of height 1000m in 10 minutes. If the steam engine produces steam at a temperature of 390k and exhausts steam at 300k, and if the engine were at the theoretically best efficiency possible, what would be the rate, in mass per unit time, at which the same is exhausted be?

I figured out the efficiency, and the work done by the try... I get the heat in, then use that to get the heat exhausted (I get 4.22x10^11 J for reference).. but then I'm stuck. It's more not really understanding the question than a physics then.. I just don't really know what to do with this, almost like I can't figure out how I should manipulate the units :yuck:

The problem requests you determine the Mass/Time ([tex]\breve{M}[/tex]) flow rate of the working fluid (steam) through the turbine (in at 390 degK and exhausted at 300 degK), which produces the work. If [tex]\breve{M}[/tex] is the problem "unknown", how did you calculate all the above mentioned quantities ("heat in", "heat exhausted", etc.) without [tex]\breve{M}[/tex]?? Or to phrase it differently, if you have indeed computed [tex]\breve{M}[/tex], then you've completed the problem, and you should be happy!

Best way to proceed here is to show your work, both symbolic equations and subsequent substitutions, and in particular, how you computed [tex]\breve{M}[/tex], which is the problem requirement.

(Hint: For the theoretical Rankine Cycle ("Steam Engine"), the Turbine segment is considered Isentropic and Adiabatic. {Heat Capacity of Steam}={2x10^3 Joules/Kg/degK} )


d - Didn't see your msg. Maybe we're asking the same thing!
~~
 
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  • #4
dextercioby said:
I didn't understand what the problem's asking...You can find:heat consumed,work done and heat exhausted.Now,what do they mean by "the same is exhausted"...?And mass of what...?

Daniel.
You see, that's is the same exact problem I'm having. Like I said, it's not as much as a problem with the physics as with the wording. I have no clue what the problem is asking.

xanthym said:
The problem requests you determine the Mass/Time ([tex]\breve{M}[/tex]) flow rate of the working fluid (steam) through the turbine (in at 390 degK and exhausted at 300 degK), which produces the work. If [tex]\breve{M}[/tex] is the problem "unknown", how did you calculate all the above mentioned quantities ("heat in", "heat exhausted", etc.) without [tex]\breve{M}[/tex]?? Or to phrase it differently, if you have indeed computed [tex]\breve{M}[/tex], then you've completed the problem, and you should be happy!

Best way to proceed here is to show your work, both symbolic equations and subsequent substitutions, and in particular, how you computed [tex]\breve{M}[/tex], which is the problem requirement.

(Hint: For the theoretical Rankine Cycle ("Steam Engine"), the Turbine segment is considered Isentropic and Adiabatic. {Heat Capacity of Steam}={2x10^3 Joules/Kg/degK} )


d - Didn't see your msg. Maybe we're asking the same thing!
~~
Interesting, the answer in the back of the book is "270 kg/s"... I have no idea where they got that from. The question I posted is word for word, so I'm afraid I can't really clarify further.
 
  • #5
mathlete said:
A locomotive steam engine opperating with a total train mass of 99.9x10^6 kg climbs a hill of height 1000m in 10 minutes. If the steam engine produces steam at a temperature of 390k and exhausts steam at 300k, and if the engine were at the theoretically best efficiency possible, what would be the rate, in mass per unit time, at which the same is exhausted be?

I figured out the efficiency, and the work done by the try... I get the heat in, then use that to get the heat exhausted (I get 4.22x10^11 J for reference).. but then I'm stuck. It's more not really understanding the question than a physics then.. I just don't really know what to do with this, almost like I can't figure out how I should manipulate the units :yuck:
I believe the maximum possible theoretical efficiency of a steam engine is the Carnot cycle:
[tex]\eta = \frac{T_H - T_C}{T_H}[/tex]

which, for these temperatures is 23.1%.

The problem first asks you to find the power (work/time) that the steam engine produces. Then you have to find the amount of work per unit mass of water that the steam engine produces (this is just the heat energy/kg of steam that is converted to work - which is 23.1% of the amount of heat needed to change the temperature of 1 kg of water from 300C to 390K) Once you have those two numbers you can answer the question.

AM
 
  • #6
Andrew Mason said:
I believe the maximum possible theoretical efficiency of a steam engine is the Carnot cycle:
[tex]\eta = \frac{T_H - T_C}{T_H}[/tex]

which, for these temperatures is 23.1%.

The problem first asks you to find the power (work/time) that the steam engine produces. Then you have to find the amount of work per unit mass of water that the steam engine produces (this is just the heat energy/kg of steam that is converted to work - which is 23.1% of the amount of heat needed to change the temperature of 1 kg of water from 300C to 390K) Once you have those two numbers you can answer the question.

AM
Can you clarify the part after finding the power? It seems like one big run on sentence. Is it the amount of (work per unit mass of water) that the engine produces? Or the amount of work per unit (mass of water that the engine produces)? And is which of those is the heat energy/kg of steam converted to work? And is it the (heat energy/kg of steam converted) that is 23.1% of the amount of heat needed to change the temperature 90K? Or is it the heat energy/(kg of steam that is converted to work which is 23.1% of the amount of heat needed to change the temperature 90K)? Thanks.
 
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  • #7
mathlete said:
Can you clarify the part after finding the power? It seems like one big run on sentence. Is it the amount of (work per unit mass of water)
that the engine produces?
Yes.
Or the amount of work per unit (mass of water that the engine produces)?
No. That makes absolutely no sense.

And is which of those is the heat energy/kg of steam converted to work? And is it the (heat energy/kg of steam converted) that is 23.1% of the amount of heat needed to change the temperature 90K? Or is it the heat energy/(kg of steam that is converted to work which is 23.1% of the amount of heat needed to change the temperature 90K)? Thanks.
The energy input is that required to heat the water from 300-390 K. The energy output is the work, which is 23.1% of the input energy.

Am
 

1. What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. This means that energy will always flow from a state of higher concentration to a state of lower concentration, resulting in a decrease in available energy for work.

2. How does the second law of thermodynamics relate to energy efficiency?

The second law of thermodynamics implies that it is impossible to convert all of the energy in a system into useful work. This means that there will always be some energy loss in any energy conversion process, leading to a decrease in overall energy efficiency.

3. Can the second law of thermodynamics be violated?

No, the second law of thermodynamics is a fundamental law of nature and has been proven through numerous experiments. It is a reflection of the fact that entropy, or disorder, always increases over time in a closed system.

4. How does the second law of thermodynamics apply to living organisms?

Living organisms are open systems that are constantly exchanging energy and matter with their surroundings. The second law of thermodynamics still applies to living organisms, as they also experience an increase in entropy over time. However, living organisms are able to maintain their internal order and decrease their entropy by constantly taking in energy from their environment.

5. What are some real-world applications of the second law of thermodynamics?

The second law of thermodynamics has many practical applications, including in the design of efficient engines and power plants, understanding chemical reactions, and predicting weather patterns. It is also essential in fields such as biology, ecology, and economics, where the concept of entropy is used to explain the behavior of complex systems.

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