Total Work of 2 Carnot Engines | 890K & 670K

In summary, the two engines in this scenario have a hot reservoir with a temperature of 890K and a cold reservoir with a temperature of 670K. The first engine has a heat input of 4800J and the second engine uses the rejected heat from the first engine as input to extract additional work. The rejected heat from the second engine goes into a reservoir with a temperature of 420K. To find the total work delivered by the two engines, we can use the efficiency formula \eta = \frac{W_{output}}{Q_H} and the fact that \eta = 1 - \frac{T_C}{T_H}. Calculating the work output of the first engine and the heat transferred to the first cold reservoir
  • #1
KingTutATL
11
0
The hot reservoir for a Carnot engine has a temperature of 890K, while the cold reservoir has a temperature of 670K. The heat input for this engine is 4800J. The 670K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat of the first engine as input and extrats additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 420K. Find the total work delivered by the 2 engines?

I'm lost
 
Physics news on Phys.org
  • #2
KingTutATL said:
The hot reservoir for a Carnot engine has a temperature of 890K, while the cold reservoir has a temperature of 670K. The heat input for this engine is 4800J. The 670K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat of the first engine as input and extrats additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 420K. Find the total work delivered by the 2 engines?

I'm lost
What is the efficiency of a Carnot engine? Remember:

[tex]\eta = \frac{W_{output}}{Q_H}[/tex]

AM
 
  • #3
Also useful will be the fact that the efficiency is also

[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]

So, calculate the work output from the first engine, from that calculate the heat transferred to the first cold resevior... Wash, rinse, repeat to find the work output of the second engine, and you're pretty much in the clear.
 

1. What is the purpose of the Total Work of 2 Carnot Engines?

The Total Work of 2 Carnot Engines is a measure of the combined work output of two Carnot engines operating at different temperatures. It is used to determine the efficiency of the engines and can also be used to calculate the maximum possible work output of a heat engine.

2. How is the Total Work of 2 Carnot Engines calculated?

The Total Work of 2 Carnot Engines is calculated using the formula W = Qh - Ql, where Qh is the heat absorbed by the engine operating at the higher temperature (890K) and Ql is the heat released by the engine operating at the lower temperature (670K). This formula assumes that the engines are operating in a reversible cycle.

3. What is the efficiency of the Total Work of 2 Carnot Engines?

The efficiency of the Total Work of 2 Carnot Engines is determined by dividing the work output (W) by the heat input (Qh), resulting in an efficiency of W/Qh. This value represents the maximum possible efficiency of the engines and is known as the Carnot efficiency.

4. What factors affect the Total Work of 2 Carnot Engines?

The Total Work of 2 Carnot Engines is affected by the temperatures of the engines, as well as the heat transfer between them. In order to maximize the work output, there must be a large temperature difference between the two engines and efficient heat transfer must be maintained.

5. How is the Total Work of 2 Carnot Engines used in practical applications?

The Total Work of 2 Carnot Engines is primarily used as a theoretical concept in thermodynamics. It helps us understand the maximum possible efficiency of heat engines and can be used to compare the performance of real-world engines. However, it is not commonly used in practical applications as it assumes ideal conditions that are difficult to achieve in real-world scenarios.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
960
  • Introductory Physics Homework Help
Replies
18
Views
753
  • Introductory Physics Homework Help
Replies
1
Views
765
Replies
1
Views
764
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
774
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
8K
Back
Top