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The Carnot cycle sets the limit on thermal efficiency of a heat engine operating between two temperature limits. Show that ideal Carnot efficiency is nth= 1-(T1/T2). What is the thermal efficiency if T1 = 288 K and T2 = 2000 K?

Respuesta :

Answer:

η =0.856

Explanation:

Given that

T₁ = 288 K

T₂ = 2000 K

We know that ,Carnot cycle is having 4 process ,in two are constant temperature process and other two are constant entropy process.

Heat rejection

Qr=T₁ ΔS

Heat addition

Qa=T₂ ΔS

We know that efficiency given as

[tex]\eta =1-\dfrac{Q_r}{Q_a}[/tex]

Now by putting the values

[tex]\eta =1-\dfrac{T_1(S_a-S_b)}{T_2(S_a-S_b)}[/tex]

[tex]\eta =1-\dfrac{T_1}{T_2}[/tex]

[tex]\eta =1-\dfrac{288}{2000}[/tex]

η =0.856

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