Question

Explain Carnot Engine!?

Answer 1

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IT IS THEORETICAL THERMODYNAMIC CYCLE..

• IT WAS GIVEN LEONARD CARNOT.
• IT GIVES THE MAXIMUM POSSIBLE EFFICIENCY HEAT ENERGY.

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Answer 2

EXPLANATION.

Carnot cycle :

(1) = It is a cyclic process which involves four steps,

(2) = The engine based on Carnot cycle is called Carnot engine Or four steps engine.

(3) = The efficiency of Carnot engine is maximum but not 100%.

The four steps involved in the Carnot cycle are as follows :

Stoke = 1.

Isothermal reversible expansion of ideal gases.

Stoke = 2.

Adiabatic reversible expansion of ideal gases.

Stoke = 3.

Isothermal reversible compression of ideal gases.

Stoke = 4.

Adiabatic reversible compression of ideal gases.

T₁ = Lower temperature = Temperature of sink.

T₂ = Higher temperature = temperature of source.

T₃ = Heat rejected by a cycle at T₁.

T₄ = Heat absorbed by cycle at T₂.

The efficiency of Carnot engine can be 100%.

If temperature of sink becomes to 0 kelvin or temperature of source becomes to ∞. but both are not possible so efficiency can never be 100%.

The efficiency can be increased by decreasing temperature of sink or by increasing temperature of source or both.

STATEMENT.

(1) = If temperature of sink decreases by ΔT or temperature of source increases by same ΔT then decreasing of temperature of sink is more efficient.

(2) = Any heat engine operating in reversible cycle between two temperature [ T₁ & T₂ will have the same efficiency ].

(3) = The efficiency of heat engine operating on reversible cycle is greater than that operating on irreversible cycle between the same two temperatures.

(4) = The efficiency of any Carnot cycle is independent from nature of fluid used.

Equations :

Isothermal expansion [ A → B ].

⇒ W₁ = Q₁ = μRT₁㏑(v₂/v₁).

Adiabatic expansion [ B → C].

⇒ W₂ = μR/(γ - 1) (T₁ - T₂).

Isothermal compression [ C → D ].

⇒ Q₂ = W₃ = μRT₂㏑(v₄/v₃) = Q₂ = μRT₂㏑(v₄/v₃).

Adiabatic compression [ D → A ].

⇒ W₄ = μR/(γ - 1) (T₂ - T₁).

The cycle of operations is called Carnot cycle.

In first two steps work is done by engine W₁ and W₂ are positive.

In last two steps work is done on gas.

W₃ and W₄ are negative.

The work done in complete cycle W = The area of the closed part of the p-v cycle.

⇒ W = W₁ + W₂ + W₃ + W₄.

$\sf \implies W = \mu RT_1 ln\dfrac{v_2}{v_1} + \dfrac{\mu R}{\gamma - 1} (T_1 - T_2) + \mu RT_2ln\dfrac{v_4}{v_3} + \dfrac{\mu R}{\gamma - 1} (T_2 - T_1)$

$\sf \implies W = \mu RT_1 ln\dfrac{v_2}{v_1} + \mu RT_2 ln\dfrac{v_4}{v_3}$

Efficiency of Carnot engine,

⇒ η = W/Q₁ = μRT₁㏑(v₂/v₁) + μRT₂㏑(v₄/v₃) / μRT₁㏑(v₂/v₁).

B to C and D to A are adiabatic paths.

So,

$\sf \implies T_1 V_2^{(\gamma - 1)} = T_2V_3^{(\gamma - 1)}$

$\sf \implies T_1V_1^{(\gamma - 1)} = T_2V_4^{(\gamma - 1)}$

⇒ v₂/v₁ = v₃/v₄.

⇒ η = T₁ - T₂/T₁ = Q₁ - Q₂/Q₁.

⇒ η = 1 - Q₂/Q₁ = 1 - T₂/T₁.

⇒ Q₁/T₁ = Q₂/T₂

⇒ η = T₁ - T₂/T₁ x 100%.

⇒ η = Q₁ - Q₂/Q₁ x 100%.

CARNOT THEOREAM.

No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) working between same hot and cold reservoirs.

η(r) = η(i).

⇒ 1 - T₂/T₁ > 1 - Q₂/Q₁.