Need help with a question involving heat engines, the carnot cycle and entropy.

In summary, the conversation covered the topics of controlling temperature in a hot reservoir, entropy, and the efficiency of a Carnot cycle. The efficiency is greater when taking away temperature from the cold reservoir. The coefficient of performance is used to determine the minimum work needed to achieve a certain temperature, and there is another equation that relates the heat taken from a source to the change in temperature. Integrating this equation will give the total work needed.
  • #1
CharlieC89
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The Attempt at a Solution


For
a) I imagine it is easier to control the temperature of the hot reservoir because you could use a constant heat source e.g. bunsen burner.

c) melting/freezing water.

d) Entropy of the environment increases and the entropy of the universe always increases.

Any help with b) or e) would be massively appreciated
 
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  • #2
b)

[tex]\eta=\frac{W}{Q_{in}}=1-\frac{|Q_{out}|}{Q_{in}}=1-\frac{T_{min}}{T_{max}}[/tex]

If you raise the temp of hot reservoir by [tex]\Delta T[/tex] you'll have: [tex]T_{max}'=T_{max}+\Delta T[/tex], and if you lower the cold reservoir by the same amount you'll have:
[tex]T_{min}'=T_{min}-\Delta T[/tex].
So the new efficiencies will be:
[tex]\eta_1=1-\frac{T_{min}}{T_{max}'}=\frac{T_{max}-T_{min}+\Delta T}{T_{max}+\Delta T}[/tex]
and
[tex]\eta_2=1-\frac{T_{min}'}{T_{max}}=\frac{T_{max}-T_{min}+\Delta T}{T_{max}}[/tex]
If you rearrange those two:
[tex]\eta_1(T_{max}+\Delta T)=\eta_2T_{max}[/tex]

So it is obvious that [tex]\eta_1>\eta_2[/tex] the efficiency of Carnot cycle is greater when you take away the temp from cold reservoir. :)
 
  • #3
thats great, thanks

anybody have any idea for e), even just a prod in the right direction would be excellent.
 
  • #4
For e), the condition for minimum work to be achieved is the coefficient of performance [tex]\eta = \frac{dQ_{cool}}{dA} = \frac{T_{cool}}{T_{hot}-T_{cool}}[/tex]
Where:
[tex]dA[/tex] is the work needed to lower the temperature by [tex]dT[/tex]
[tex]dQ_{cool}[/tex] is the heat taken from the water/ice.
[tex]T_{hot}[/tex] is the temperature of the sink, which is the air in this problem.
[tex]T_{cool}[/tex] is the temperature of the source, which is the water/ice. Notice that [tex]T_{cool}[/tex] varies during the process.

There is another equation relating [tex]dQ_{cool}[/tex] with [tex]dT[/tex] (actually 2 equations, each corresponds to one process). Compute the integral, and you will obtain the total work needed.
 
  • #5
!

For b) The efficiency of a Carnot cycle is given by the equation e = 1- Tc/Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. Therefore, the efficiency of a Carnot cycle can be increased by increasing the temperature of the hot reservoir or decreasing the temperature of the cold reservoir. This can be achieved by using more efficient heat exchange methods and insulating the system to minimize heat loss.

For e) Entropy is a measure of the disorder or randomness in a system. In the Carnot cycle, entropy is conserved and remains constant throughout the cycle. However, in real-world systems, there is always some amount of energy lost as heat and the entropy of the system increases. This is due to the second law of thermodynamics, which states that in any energy conversion process, the total entropy of the universe will always increase. Therefore, it is important to minimize heat loss and increase efficiency in heat engines to reduce the overall increase in entropy in the universe.
 

1. How does a heat engine work?

A heat engine works by converting heat energy into mechanical work. This is achieved by using a source of heat, such as burning fuel, to produce high temperature and pressure. The high pressure steam or gas then drives a turbine or piston, which in turn generates mechanical energy.

2. What is the Carnot cycle and how does it relate to heat engines?

The Carnot cycle is a theoretical ideal cycle that describes the most efficient way to convert heat energy into mechanical work. It involves a reversible process of isothermal expansion and compression, as well as adiabatic expansion and compression. This cycle is used as a benchmark for evaluating the efficiency of real heat engines.

3. How does entropy play a role in heat engines?

Entropy is a measure of the disorder or randomness in a system. In heat engines, entropy increases as heat is converted into work, since some of the energy is lost as heat due to the second law of thermodynamics. This means that the efficiency of a heat engine is limited by the increase in entropy.

4. What factors affect the efficiency of a heat engine?

The efficiency of a heat engine depends on several factors, including the temperature difference between the hot and cold reservoirs, the type of working fluid used, and the design of the engine. The Carnot cycle shows that the efficiency increases as the temperature difference increases, but in real engines, other factors such as friction and heat loss also play a role.

5. Can heat engines be used for other purposes besides powering vehicles?

Yes, heat engines have a wide range of applications besides powering vehicles. They are used in power plants to generate electricity, in refrigerators and air conditioners to cool spaces, and in industrial processes to produce mechanical work. The principles of heat engines are also used in other fields, such as thermodynamics and engineering.

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