Why Are Carnot Engines Not Built in Real Life?

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SUMMARY

The discussion centers on the theoretical nature of the Carnot engine, emphasizing that it cannot be constructed in reality due to practical limitations. Key reasons include the Carnot engine's zero power output and the excessive thermal pollution it would generate. Additionally, while the Carnot engine is an idealized model with maximum efficiency, real engines operate below this efficiency and are subject to the second law of thermodynamics. The conversation also touches on the calculation of maximum work output (Wmax) using the equation Wmax = Qc * (Th/Tc - 1).

PREREQUISITES
  • Understanding of thermodynamic cycles, particularly the Carnot cycle.
  • Familiarity with the second law of thermodynamics.
  • Knowledge of heat transfer principles and efficiency calculations.
  • Basic mathematical skills for solving equations related to work and heat.
NEXT STEPS
  • Research the principles of thermodynamic efficiency and its implications in real-world engines.
  • Study the second law of thermodynamics and its applications in heat engine design.
  • Explore real-world examples of heat engines and their efficiencies compared to the Carnot engine.
  • Learn about the mathematical derivation of work output in various thermodynamic cycles.
USEFUL FOR

Students of thermodynamics, mechanical engineers, and anyone interested in the theoretical and practical aspects of heat engines and their efficiencies.

xinlan
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Homework Statement


We have stressed that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot engine. However, no attempts to build a Carnot engine are being made. Why is that?
a. A Carnot engine would generate too much thermal pollution.
b. Building the Carnot engine is possible but is too expensive.
c. The Carnot engine has zero power.
d. The Carnot engine has too low an efficiency.

and also,
A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold reservoir. During that time, what is the maximum amount of work (Wmax) that the engine might have performed?

Homework Equations



for question 2.
Wmax = eff*Qc

The Attempt at a Solution


for question 1.
I think (d) and (c) are not correct, because carnot engine was made because its efficiency can reach up to 100%. That's why it is an ideal engine. so, it also can't have 0 power.
then I don't know the answer is (a) or (b) ..

for question 2.
I calculated but I got wrong
here is the calculation:
Wmax = ((Th-Tc)/Th)*Qc

Please help me..
thank you
 
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You are right, C and D are wrong. As for A and B, ask yourself this, does a more efficient engine produce more or less pollution? A real life example will help here--think about SUVs compared with hybrid cars.

As for your second question, if the real engine is performing the maximum amount of work possible it will behave according to the Carnot cycle. Use conservation of energy (heat in = work + heat out and be careful with signs), and Qc*Th=Qh*Tc (this follows from the Carnot cycle).
 
for question I. is the answer B? since efficient engine produce less pollution.
 
for question 2.
work = heat in (Qh) - Heat out (Qc)
then use Qc*Th = Qh*Tc --> Qh = (Qc*Th)/Tc
then work = (Qc*Th)/Tc - Qc

is that the answer?

thanks for helping me..
 
I got it..
thanks for helping me..
 
Hi

I have a question on Why the Carnot cycle is not used with real engines? I've got to refer to real cycles in my explanation.
 
dwintz02 said:
You are right, C and D are wrong. As for A and B, ask yourself this, does a more efficient engine produce more or less pollution? A real life example will help here--think about SUVs compared with hybrid cars.

As for your second question, if the real engine is performing the maximum amount of work possible it will behave according to the Carnot cycle. Use conservation of energy (heat in = work + heat out and be careful with signs), and Qc*Th=Qh*Tc (this follows from the Carnot cycle).
D is wrong but not C. C is the right answer.

Think about the Carnot engine. It uses a reversible cycle, which means that the engine is always in a state of thermodynamic equilibrium. The heat is transferred from the hot reservoir and dispersed to the cold reservoir reversibily - infinitessimal temperature differences mean it occurs over time scales approaching infinity. The work done divided by time gives the power, so the power of a Carnot engine is arbitrarily close to 0.

AM
 
xinlan said:
for question I. is the answer B? since efficient engine produce less pollution.
Efficiency is thermodynamic efficiency which has nothing to do with the chemical efficiency and completeness of combustion. It may not use any chemical combustion at all. You could run a near-Carnot engine using electricity to produce the heat and it would have no pollution whatsoever.

AM
 
xinlan said:
I think (d) and (c) are not correct, because carnot engine was made because its efficiency can reach up to 100%. That's why it is an ideal engine. so, it also can't have 0 power.
A Carnot engine may be an ideal engine, but that ideal is not 100% efficient.
A Carnot engine can only have 100% efficiency if the cold reservoir is at absolute 0.

AM
 
  • #10
xinlan said:

Homework Statement


We have stressed that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot engine. However, no attempts to build a Carnot engine are being made. Why is that?
a. A Carnot engine would generate too much thermal pollution.
b. Building the Carnot engine is possible but is too expensive.
c. The Carnot engine has zero power.
d. The Carnot engine has too low an efficiency.

and also,
A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold reservoir. During that time, what is the maximum amount of work (Wmax) that the engine might have performed?

Homework Equations



for question 2.
Wmax = eff*Qc

The Attempt at a Solution


for question 1.
I think (d) and (c) are not correct, because carnot engine was made because its efficiency can reach up to 100%. That's why it is an ideal engine. so, it also can't have 0 power.
then I don't know the answer is (a) or (b) ..

for question 2.
I calculated but I got wrong
here is the calculation:
Wmax = ((Th-Tc)/Th)*Qc

Please help me..
thank you

see as the engine is real we can no use carnot efficiency to calculate the work. we shud use 2nd law of thermodynamics.solution is as
let the temp of cold and hot reservoir be T2 and T1 resp.
then to have a feasible process dS>= 0
this implies dat (QC/T2)-[(W+Qc)/T1]>=0
here change in entropy of source and sink are taken into account. the working fluid doesn't has any chnge of entropy as it undergoes a cycle.
for maximum work done possible we equate to zero the above equation(to have a reversible process)
this gives W max = (Qc/T2)-(Qc/T1)*T1
=Qc[T1/T2 -1]
 

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