TITLE.PM5

628 ENGINEERING THERMODYNAMICS

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\M-therm\Th13-2.pm5

∴ T 2 = T 4 = TT 13. Proved.

(b) Power developed, P :

T

T

m

1

3

310

1450

038

=

=

=

U

V

|

W

|

K

K

.kg

....(given)

Work done W = cv [(T 3 – T 2 ) – (T 4 – T 1 )]
T 2 = T 4 = TT 13 =×310 1450 = 670.4 K
∴ W = 0.71 [(1450 – 670.4) – (670.4 – 310)]
= 0.71 (779.6 – 360.4) = 297.6 kJ/kg
Work done per second = 297.6 × (0.38/60) = 1.88 kJ/s
Hence power developed, P = 1.88 kW. (Ans.)
Example 13.16. For the same compression ratio, show that the efficiency of Otto cycle is
greater than that of Diesel cycle.
Solution. Refer Fig. 13.14.

p

4

1

3

2

Otto cycle

V V

p

2 ′ 3 ′

4 ′

1 ′

Diesel cycle

Fig. 13.14

We know that

ηOtto = 1 –

1

()rγ−^1

and ηDiesel = 1 – 11 1

()rγ^11

γ

γ

ρ

− ρ

× −

−

R

S

|

T|

U

V

|

W|

As the compression ratio is same,

V

V

1

2

=

V

V

1

2

′

′ = r

If

V

V

4

3

′

′= r^1 , then cut-off ratio, ρ =

V

V

r

r

3

21

′

′

=