Thermodynamics engine question

In summary, the conversation discusses an inventor's claim of developing an engine that takes in 10^8 J at a temperature of 400 K and rejects 4x10^7 J to a reservoir at 200 K, while delivering 15 kilowatt hours of mechanical work. The efficiency of this hypothetical engine is calculated to be 54%, which is higher than the maximum efficiency of a Carnot engine (50%). However, the feasibility of this invention is questioned due to the difficulty of finding a reservoir to output at -73C. Ultimately, the conversation concludes that investing in this engine may not be advisable.
  • #1
ChronicQuantumAddict
39
0
Ok, my question is as follows:

An inventor claims to have developed an engine that takes in 10^8 J (Q_in) at a temperature of 400 K (T_2), and rejects 4x10^7 J (Q_out) to a reservoir of Temperatue of 200 K (T_1). The engine delivers 15 kilowatt hours of mechanical work (which = 3600 sec/hour *15 * 10^3 watts = 5.4x10^6 Joules). Would you advise investing money to put this engine on the market?

the way i approached it was to calculate the max efficiency that a carnot engine would have, which is:

efficiency = 1 - T_1/T_2 = 1 - 200/400 = 0.5 or 50%​

Now, using the expression,
Efficiency = Work output/Heat input​
for the hypothetical engine gives. This gave me an efficiency of roughly 54%, and by Carnot's theorem, no engine can be more efficient than a carnot engine, or:

Efficiency(carnot) > Efficiency(hypothetical)​
and in this case, it doesn't hold, i.e.:
50 % > 54 % is not true.​

Is this the correct way to do this problem?
Thanks
 
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  • #2
I didn't check your number crunching, but you've approached the problem correctly.
 
  • #3
Thanks very much
 
  • #4
ChronicQuantumAddict said:
Ok, my question is as follows:
An inventor claims to have developed an engine that takes in 10^8 J (Q_in) at a temperature of 400 K (T_2), and rejects 4x10^7 J (Q_out) to a reservoir of Temperatue of 200 K (T_1). The engine delivers 15 kilowatt hours of mechanical work (which = 3600 sec/hour *15 * 10^3 watts = 5.4x10^6 Joules). Would you advise investing money to put this engine on the market?
So what is your answer and why?

My reason for not investing (besides the fact that the claim cannot be true as you have shown) would be: where are you going to find a reservoir to output at -73C?

AM
 

Related to Thermodynamics engine question

1. What is a thermodynamics engine?

A thermodynamics engine is a device that converts thermal energy into mechanical work, or vice versa. It operates based on the principles of thermodynamics, specifically the laws of conservation of energy and entropy.

2. How does a thermodynamics engine work?

A thermodynamics engine works by taking in heat energy from a hot source, such as burning fuel, and converting it into mechanical work. This is achieved through a series of processes, including compression, expansion, and heat transfer, that utilize the working fluid within the engine.

3. What are the types of thermodynamics engines?

The two main types of thermodynamics engines are heat engines and refrigeration/heat pump engines. Heat engines, such as steam engines and gas turbines, convert heat energy into mechanical work. Refrigeration/heat pump engines, like air conditioners and refrigerators, transfer heat from a cold source to a hot source using mechanical work.

4. What is the efficiency of a thermodynamics engine?

The efficiency of a thermodynamics engine is the ratio of the work output to the heat input. This is known as the Carnot efficiency and is limited by the laws of thermodynamics. It can never be 100%, as some heat energy will always be lost in the conversion process.

5. What are some real-world applications of thermodynamics engines?

Thermodynamics engines have a wide range of applications, including power generation, transportation, and refrigeration. They are used in power plants to generate electricity, in vehicles to power engines, and in air conditioning units to cool buildings. They are also used in industrial processes, such as food production and chemical manufacturing.

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