Carnot cycle – problems and solutions

Карно мөчлөг - асуудлууд ба шийдэл

1 дулаан absorbed by the engine (Q₁) = 10,000 Joule, what is the work done by the Карно хөдөлгүүр?

Мэдэгдэж байгаа: Carnot cycle – problems and solutions 1

Бага температур (T)₂) = 400 K

Өндөр температур (T₁) = 800 K

Дулааны оролт (Q)₁) = 10,000 Жоуль

Хүссэн: Work done by Carnot engine (W)

Шийдэл:

Карно хөдөлгүүрийн үр ашиг:

Carnot cycle – problems and solutions 2

Ажлын was done by Carnot engine :

W = e Q₁

W = (1/2)(10,000) = 5000 Жоул

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

Мэдэгдэж байгаа: Carnot cycle – problems and solutions 3

Бага температур (T)_L) = 400 K

Өндөр температур (T_H) = 600 K

Дулааны оролт (Q)₁) = 600 Жоуль

Хүссэн: Ажлыг Карно хөдөлгүүр (W) гүйцэтгэсэн.

Шийдэл:

Карно хөдөлгүүрийн үр ашиг:

Work done by Carnot engine : Carnot cycle – problems and solutions 4

W = e Q₁

W = (1/3)(600) = 200 Жоул

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

Мэдэгдэж байгаа: Carnot cycle – problems and solutions 5

Бага температур (T)_L) = 350 K

Өндөр температур (T_H) = 500 K

Хүссэн: Efficiency of Carnot engine (e)

Шийдэл:

Efficiency of Carnot engine :

e = (T_H - Т_L) / Т_H

e = (500 – 350) / 500

e = 150 / 500

e = 0.3 байна

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.

Мэдэгдэж байгаа: Carnot cycle – problems and solutions 6

Бага температур (T)_L) = 400 K

Өндөр температур (T_H) = 600 K

Хүссэн: heat output (Q₂)

Шийдэл:

Efficiency of Carnot engine :

e = (T_H - Т_L) / Т_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

мөн үзнэ үү Механик энергийн мөргөлдөөн ба хадгалалт - зонд ба уусмалууд

W = (1/3)(Q₁)

3W = Q₁

Heat output :

Q₂ = Q₁ - W

Q₂ = 3W – W

Q₂ = 2W байна

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

Мэдэгдэж байгаа:

Бага температур (T)_L) = 500 K Carnot cycle – problems and solutions 7

Өндөр температур (T_H) = 800 K

Дулааны гаралт (Q)₂) = 3000 Жоуль

Хүссэн: Дулааны оролт (Q)₁)

Шийдэл:

Efficiency of Carnot engine :

e = (T_H - Т_L) / Т_H

e = (800 – 500) /8600

e = 300 / 800

e = 3/8

Work done by Carnot engine :

W = e Q₁

W = (3/8)(Q₁)

8Вт/3 = Q₁

Q₂ = Q₁ - W

Q₂ = 8Вт/3 – 3Вт/3

Q₂ = 5Вт/3

3Q₂ = 5W байна

W = 3Q₂/5 = 3(3000)/5 = 9000/5 = 1800

Heat absorbed by engine :

Q₁ = W + Q₂ = 1800 + 3000 = 4800 Жоуль

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.

Мэдэгдэж байгаа:

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

Хүссэн: Өндөр температур (T_H) if efficiency (e) = 80% = 0.8

Шийдэл:

Carnot cycle – problems and solutions 8

Low temperature = 400 Kelvin

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

Carnot cycle – problems and solutions 9

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?

Мэдэгдэж байгаа:

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

Хүссэн: Өндөр температур (T_H) if efficiency (e) = 75% = 0.75

Шийдэл:

Carnot cycle – problems and solutions 10

Өндөр температур (T_H) if efficiency (e) = 75 % ? Carnot cycle – problems and solutions 11

High temperature = 1440 Kelvin

What is the Carnot cycle? хариулт: 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.
Why is the Carnot cycle considered an ideal cycle? хариулт: 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.

What are the four processes in a Carnot cycle? хариулт: 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).
Why is there no actual heat engine that operates on the Carnot cycle? хариулт: 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.
What is the efficiency of a Carnot engine? хариулт: The efficiency $η$ 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.
Why can’t a Carnot engine have 100% efficiency? хариулт: 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.

What is the significance of reversibility in the Carnot cycle? хариулт: 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.
How is the Carnot cycle related to the second law of thermodynamics? хариулт: 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.

Why is it impossible to have isothermal processes in real-world applications exactly as they appear in the Carnot cycle? хариулт: 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.

How does the Carnot cycle help engineers and scientists? хариулт: 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.