Efficiency of engine taking diatomic ideal gas through cycles

In summary, the conversation is discussing the thermal efficiency of an engine operating by taking n moles of diatomic ideal gas through a specific cycle. The efficiency formula is mentioned (e = W/Qh) and it is stated that the answer is 10.5. The individual is asking for help understanding how this answer is generated and how to find temperatures at different points in the cycle. The response explains that there are two isochoric and two isobaric processes in the cycle and suggests finding temperatures at each point and calculating the heat absorbed and net work done. The efficiency is also mentioned to always be less than one and that it is equal to 10.5%. The conversation ends with the individual still needing clarification and
  • #1
jensgt
7
0

Homework Statement



What is the thermal efficiency of an engine that operates by taking n moles of diatomic ideal gas through the cycle 1 -------> 2 -------> 3 -------> 4 ------> 1 shown in the figure ?

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Homework Equations



e = W/Qh



The Attempt at a Solution




I know the answer is 10.5

I am trying to prep for a test today and this is the last thing from my homework that I don't understand...can someone explain to me how the answer is generated? I am sort of lost on it and I figure it has to be easier than it seems.
 
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  • #2
There are two isochoric and two isobaric processes in the cycle. Find temperatures at all four points and calculate the heat absorbed from the source and the net work done.
The efficiency is always less than one thus it must be 10.5%.
 
  • #3
How to find temp with no p or v numbers?
 
  • #4
It is P0V0/nR at 1.
 
  • #5
I'm still confused...po is not a numerical value.
 
  • #6
say P0V0/nR = T0 then what are temperatures at 2, 3 and 4 ?
 
  • #7
these are 2T0, 4T0 and 2T0 respectively.
Now find the heat absorbed in isochoric and isobaric processes using Q = n Cv ΔT and Q = n Cp ΔT
 

Related to Efficiency of engine taking diatomic ideal gas through cycles

1. What is the efficiency of an engine taking diatomic ideal gas through cycles?

The efficiency of an engine taking diatomic ideal gas through cycles is determined by the Carnot efficiency formula, which states that it is equal to the difference between the hot and cold temperatures divided by the hot temperature. It can also be expressed as the ratio of the work output to the heat input.

2. How does the type of gas affect the efficiency of an engine?

The type of gas used in an engine can affect its efficiency because different gases have different properties, such as specific heat capacities and molecular weights, that can impact the amount of work that can be extracted from the engine. Diatomic ideal gases, such as hydrogen, nitrogen, and oxygen, are commonly used in engines due to their high efficiency.

3. What are the key factors that influence the efficiency of an engine taking diatomic ideal gas through cycles?

The key factors that influence the efficiency of an engine taking diatomic ideal gas through cycles include the temperature of the hot and cold reservoirs, the type of gas used, the compression ratio, and the heat transfer processes involved in the cycle. Improving these factors can lead to a higher efficiency for the engine.

4. Can the efficiency of an engine taking diatomic ideal gas through cycles be improved?

Yes, the efficiency of an engine taking diatomic ideal gas through cycles can be improved through various methods, such as increasing the temperature difference between the hot and cold reservoirs, using a more efficient gas, and optimizing the compression ratio and heat transfer processes. However, the Carnot efficiency limit states that it is impossible to achieve 100% efficiency.

5. How does the efficiency of an engine taking diatomic ideal gas through cycles compare to other types of engines?

The efficiency of an engine taking diatomic ideal gas through cycles can be higher compared to other types of engines, such as internal combustion engines, which typically have efficiencies around 30%. However, it is still limited by the Carnot efficiency and may not be as efficient as other types of engines, such as fuel cells, which can have efficiencies over 60%.

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