מעגל קרנו - בעיות ופתרונות

מעגל קרנו - בעיות ופתרונות

1. אם חוֹם absorbed by the engine (Q₁) = 10,000 Joule, what is the work done by the מנוע קרנו?

ידוע:

טמפרטורה נמוכה (טמפרטורת₂) = 400 אלף

טמפרטורה גבוהה (טמפרטורת₁) = 800 אלף

קלט חום (Q₁) = 10,000 ג'אול

מבוקש: Work done by Carnot engine (W)

פתרון:

יעילות מנוע קרנו:

תיק עבודות was done by Carnot engine :

W = e Q₁

W = (1/2)(10,000) = 5000 ג'אול

2.

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

ידוע:

טמפרטורה נמוכה (טמפרטורת_L) = 400 אלף

טמפרטורה גבוהה (טמפרטורת_H) = 600 אלף

קלט חום (Q₁) = 600 ג'אול

מבוקש: Work was done by Carnot engine (W)

פתרון:

יעילות מנוע קרנו:

Work done by Carnot engine :

W = e Q₁

W = (1/3)(600) = 200 ג'אול

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

ידוע:

טמפרטורה נמוכה (טמפרטורת_L) = 350 אלף

טמפרטורה גבוהה (טמפרטורת_H) = 500 אלף

מבוקש: Efficiency of Carnot engine (e)

פתרון:

Efficiency of Carnot engine :

ה = (T_H ת_L) / ת_H

ה = (500 – 350) / 500

ה = 150 / 500

e = 0.3

ה = 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.

ידוע:

טמפרטורה נמוכה (טמפרטורת_L) = 400 אלף

טמפרטורה גבוהה (טמפרטורת_H) = 600 אלף

מבוקש: heat output (Q₂)

פתרון:

Efficiency of Carnot engine :

ה = (T_H ת_L) / ת_H

ה = (600 – 400) / 600

ה = 200 / 600

ה = 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 - וו

Q₂ = 3W – W

Q₂ = 2W

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

ידוע:

טמפרטורה נמוכה (טמפרטורת_L) = 500 אלף

טמפרטורה גבוהה (טמפרטורת_H) = 800 אלף

תפוקת חום (Q₂) = 3000 ג'אול

מבוקש: קלט חום (Q₁)

פתרון:

Efficiency of Carnot engine :

ה = (T_H ת_L) / ת_H

ה = (800 – 500) /8600

ה = 300 / 800

ה = 3/8

Work done by Carnot engine :

W = e Q₁

W = (3/8)(Q₁)

8W/3 = Q₁

Q₂ = Q₁ - וו

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 ג'אול

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.

ידוע:

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

מבוקש: טמפרטורה גבוהה (טמפרטורת_H) if efficiency (e) = 80% = 0.8

פתרון:

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?

ידוע:

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

מבוקש: טמפרטורה גבוהה (טמפרטורת_H) if efficiency (e) = 75% = 0.75

פתרון:

טמפרטורה גבוהה (טמפרטורת_H) if efficiency (e) = 75 % ?

High temperature = 1440 Kelvin

1. What is the Carnot cycle? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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? תשובה: 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.