วัฏจักรคาร์โนต์ – ปัญหาและวิธีแก้ไข

วัฏจักรคาร์โนต์ – ปัญหาและวิธีแก้ไข

1 ถ้า ความร้อน ดูดซับโดยเครื่องยนต์ (Q)₁ถ้า ) = 10,000 จูล งานที่ทำโดย คือเท่าใด เครื่องยนต์คาร์โนต์?

เป็นที่รู้จัก:

อุณหภูมิต่ำ (T)₂) = 400 K

อุณหภูมิสูง (T)₁) = 800 K

การป้อนความร้อน (Q)₁) = 10,000 จูล

ต้องการ: งานที่ทำโดยเครื่องยนต์คาร์โนต์ (W)

วิธีการแก้:

The efficiency of the Carnot engine :

งาน was done by Carnot engine :

W = e Q₁

W = (1/2)(10,000) = 5000 จูล

2.

Based on graph above, what is the work done by engine in a cycle?

เป็นที่รู้จัก :

อุณหภูมิต่ำ (T)_L) = 400 K

อุณหภูมิสูง (T)_H) = 600 K

การป้อนความร้อน (Q)₁) = 600 จูล

ต้องการ: Work was done by Carnot engine (W)

วิธีการแก้ปัญหา:

The efficiency of the Carnot engine :

Work done by Carnot engine :

W = e Q₁

W = (1/3)(600) = 200 จูล

3. Based on the graph below, what is the efficiency of the Carnot engine?

เป็นที่รู้จัก :

อุณหภูมิต่ำ (T)_L) = 350 K

อุณหภูมิสูง (T)_H) = 500 K

เป็นที่ต้องการ : Efficiency of Carnot engine (e)

วิธีการแก้ปัญหา:

Efficiency of Carnot engine :

e = (T_H - ท_L) / T_H

e = (500 – 350) / 500

e = 150 / 500

อี = 0.3

e = 30/100 = 30%

4. Based on graph below, the heat engine’s high temperature is 600 K and low temperature is 400 K. If the work done by engine is W, what is the heat output.

เป็นที่รู้จัก :

อุณหภูมิต่ำ (T)_L) = 400 K

อุณหภูมิสูง (T)_H) = 600 K

เป็นที่ต้องการ : heat output (Q₂)

วิธีการแก้ปัญหา:

Efficiency of Carnot engine :

e = (T_H - ท_L) / T_H

e = (600 – 400) / 600

e = 200 / 600

e = 1/3

Work done by Carnot engine :

W = e Q₁

W = work done by engine, e = efficiency, Q1 = heat input

W = (1/3)(Q₁)

3W = Q₁

Heat output :

Q₂ = Q₁ - ใน

Q₂ = 3W – W

Q₂ = 2W

5. Based on graph below, if the heat output is 3000 Joule, what is the heat input.

เป็นที่รู้จัก :

อุณหภูมิต่ำ (T)_L) = 500 K

อุณหภูมิสูง (T)_H) = 800 K

Heat output (Q₂) = 3000 จูล

เป็นที่ต้องการ : การป้อนความร้อน (Q)₁)

วิธีการแก้ปัญหา:

Efficiency of Carnot engine :

e = (T_H - ท_L) / T_H

e = (800 – 500) /8600

e = 300 / 800

e = 3/8

Work done by Carnot engine :

W = e Q₁

W = (3/8)(Q₁)

8W/3 = Q₁

Q₂ = Q₁ - ใน

Q₂ = 8W/3 – 3W/3

Q₂ = 5W/3

3Q₂ = 5W

W = 3Q₂/5 = 3(3000)/5 = 9000/5 = 1800

Heat absorbed by engine :

Q₁ = W + Q₂ = 1800 + 3000 = 4800 จูล

6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant.

เป็นที่รู้จัก :

If high temperature (T_H) = 800 K , efficiency (e) = 50% = 0.5

เป็นที่ต้องการ : อุณหภูมิสูง (T)_H) if efficiency (e) = 80% = 0.8

วิธีการแก้ปัญหา:

Low temperature = 400 Kelvin

What is the high temperature (T_H) if efficiency (e) = 80 % ?

High temperature = 2000 Kelvin

7. A Carnot engine works at high temperature 600 Kelvin with the efficiency of 40%. If the efficiency of the engine is 75% and the low temperature kept constant, what is the high temperature?

เป็นที่รู้จัก :

If high temperature (T_H) = 600 K , efficiency (e) = 40% = 0.4

เป็นที่ต้องการ : อุณหภูมิสูง (T)_H) if efficiency (e) = 75% = 0.75

วิธีการแก้ปัญหา:

อุณหภูมิสูง (T)_H) if efficiency (e) = 75 % ?

High temperature = 1440 Kelvin

1. What is the Carnot cycle? คำตอบ: The Carnot cycle is a theoretical thermodynamic cycle that represents the most efficient reversible heat engine cycle possible. It consists of two isothermal processes and two adiabatic processes.
2. Why is the Carnot cycle considered an ideal cycle? คำตอบ: The Carnot cycle is considered ideal because it represents the upper limit of efficiency for any heat engine. No real engine can be more efficient than a Carnot engine operating between the same two temperature reservoirs.
3. What are the four processes in a Carnot cycle? คำตอบ: The four processes in a Carnot cycle are:
   1. Isothermal expansion at the high temperature T_H†<.
   2. Adiabatic expansion (where the system is thermally insulated and cools down).
   3. Isothermal compression at the low temperature T_C†<.
   4. Adiabatic compression (where the system is thermally insulated and heats up).
4. Why is there no actual heat engine that operates on the Carnot cycle? คำตอบ: Real engines have irreversible losses, such as friction, and cannot maintain perfect insulation during the adiabatic processes. Furthermore, it would be impractical to achieve the infinitely slow isothermal processes required by the Carnot cycle.
5. What is the efficiency of a Carnot engine? คำตอบ: The efficiency $\eta$ of a Carnot engine operating between two temperature reservoirs T_H†< (hot) and T_C†< (cold) is given by:
   η=1-T_C/T_H†<†<†<

   where temperatures are in Kelvin.

6. Why can’t a Carnot engine have 100% efficiency? คำตอบ: A Carnot engine’s efficiency is dependent on the temperature difference between the hot and cold reservoirs. To achieve 100% efficiency, the cold reservoir’s temperature would need to be absolute zero (0 Kelvin), which is unattainable in practice.
7. What is the significance of reversibility in the Carnot cycle? คำตอบ: Reversibility ensures that there are no entropy-generating processes, which means the cycle can operate at maximum efficiency. Any irreversible process would decrease the cycle’s efficiency.
8. How is the Carnot cycle related to the second law of thermodynamics? คำตอบ: The Carnot cycle underpins the Second Law by establishing an upper limit on the efficiency of heat engines. The Second Law asserts that no engine can be more efficient than a Carnot engine operating between the same two temperatures.
9. Why is it impossible to have isothermal processes in real-world applications exactly as they appear in the Carnot cycle? คำตอบ: An isothermal process, as depicted in the Carnot cycle, requires an infinite amount of time, which is impractical in real-world applications. This is because to maintain the isothermal condition, heat transfer should take place infinitesimally slowly.

10. How does the Carnot cycle help engineers and scientists? คำตอบ: The Carnot cycle provides a theoretical benchmark for the maximum possible efficiency of heat engines. By comparing real engines to the Carnot cycle, engineers and scientists can identify areas for improvement and understand the fundamental limits of their designs.