Carnotov kru ni proces
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p. V. Carnotov kružni proces. W = W AB + W BC + W CD + W DA. = Q 1 + (-  U) + (- Q 2 ) +  U. A. W = Q 1 – Q 2. T 2. B. T 1. D. C. Q 1. Q 2. W AB = Q 1. W CD = - Q 2. W BC = -  U. W DA =  U. Primjer: Temperatura grijača idealnog toplinskog stroja je

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Carnotov kružni proces

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Carnotov kru ni proces

p

V

Carnotov kružni proces

W = WAB + WBC + WCD + WDA

= Q1 + (- U) + (- Q2) + U

A

W = Q1 – Q2

T2

B

T1

D

C

Q1

Q2

WAB = Q1

WCD = - Q2

WBC = - U

WDA = U


Carnotov kru ni proces

Primjer: Temperatura grijača idealnog toplinskog stroja je

200 oC, a temperatura hladnijeg spremnika 20 oC.

a) Kolika je korisnost?

b) Odredite rad koji može dati stroj, ako od grijača primi količinu

topline 300 kJ.

Rješenje:

t1 = 200 oC

, T1 = 473 K

, T2 = 293 K

t2 = 20 oC

b) W = ?

Q1 = 300 kJ

a)  = ?

= 0,38

W = Q1

= 0,38300kJ

W = 114kJ

 = 38 %


Carnotov kru ni proces

Zadatak: U Carnotovu procesu plinu je pri temperaturi 400 K

dovedeno 8,37 kJ topline. Dobiveni rad u kružnom procesu je

2093 J. Kolika je korisnost, toplina predana hladnijem

spremniku i temperatura hladnijeg spremnika?

Rješenje:

T1 = 400 K

Q1 = 8,37 kJ

= 8370 J

W = 2093 J

W = Q1 – Q2

T2 = T1(1-)

Q2 = Q1 – W

 = 25 %

= 400 K·(1- 0,25)

= 8370 J – 2093 J

T2 = 300 K

Q2 = 6277 J


Motori s unutarnjim izgaranjem

p

p

V

V

Motori s unutarnjim izgaranjem

Ottov kružni proces

Dieselov kružni proces

3

2

3

2

4

4

1

1


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