618 ENGINEERING THERMODYNAMICS

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or 0.72 (T 3 – 689.1) = 1500

or T 3 =
1500
072.

• 689.1 = 2772.4 K. (Ans.)

Also, p

T

p

T

2

2

3

3

= ⇒ p 3 =

pT

T

23

2

= 18.379 2772.4

689.1

× = 73.94 bar. (Ans.)

Adiabatic Expansion process 3-4 :

T

T

V

V

(^3) r
4
4
3
1
=F^11418
HG
I
KJ

−
−−
γ
()γ ( ). = 2.297
∴ T 4 = T^3
297
2772.4

2. 
   
   
   
   2. 297
      = = 1206.9 K. (Ans.)
   
   
   
   

Also, pv 33 γγ=pv 44 ⇒ p 4 = p 3 ×

v

v

3

4

F

HG

I

KJ

γ

= 73.94 ×

1

8

F^14

HG

I

KJ

.

= 4.023 bar. (Ans.)

(ii)Specific work and thermal efficiency :

Specific work = Heat added – heat rejected

= cv (T 3 – T 2 ) – cv(T 4 – T 1 ) = cv [(T 3 – T 2 ) – (T 4 – T 1 )]

= 0.72 [(2772.4 – 689.1) – (1206.9 – 300)] = 847 kJ/kg. (Ans.)

Thermal efficiency, ηth = 1 –^11

()rγ−

= 1 –

1
() 8 14 1. −
= 0.5647 or 56.47%. (Ans.)
Example 13.10. An air standard Otto cycle has a volumetric compression ratio of 6, the
lowest cycle pressure of 0.1 MPa and operates between temperature limits of 27°C and 1569°C.
(i) Calculate the temperature and pressure after the isentropic expansion (ratio of specific
heats = 1.4).
(ii)Since it is observed that values in (i) are well above the lowest cycle operating condi-
tions, the expansion process was allowed to continue down to a pressure of 0.1 MPa. Which
process is required to complete the cycle? Name the cycle so obtained.
(iii)Determine by what percentage the cycle efficiency has been improved. (GATE, 1994)

Solution. Refer Fig. 13.8. Given : v

v

v

v

1

2

4

3

= = r = 6 ; p 1 = 0.1 MPa = 1 bar ; T 1 = 27 + 273

= 300 K ; T 3 = 1569 + 273 = 1842 K ; γ = 1.4.

(i)Temperature and pressure after the isentropic expansion, T 4 , p 4 :

Consider 1 kg of air :

For the compression process 1-2 :

pv p v p p v

11 2 2 (^21) v
1
2
γγ 16 123 14
γ
=⇒=×
F
HG
I
KJ
=×( ). =. bar
Also
T
T
v
v
2
1
1
2
1
=F 6 14 1
HG
I
KJ

−
−
γ
(). = 2.048
∴ T 2 = 300 × 2.048 = 614.4 K