TITLE.PM5

648 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-3.pm5

∴ pm =^1

113 1

45 21

()

()

rV

pVVpV p rV pV prV

c

− cc−+ cccc

−×

−

− −

−

L

N

M

O

Q

ρ P

ρ

γγ

r = 9, ρ = 1.15, γ = 1.4

p 1 = 1 bar, p 2 = 21.67 bar, p 3 = p 4 = 68 bar, p 5 = 3.81 bar

Substituting the above values in the above equation, we get

pm =

1

91

68 15 1^6815 81 9

41

21.67 9

() 41

()

−

−+×− ×

−

− −

−

L

N

M

O

Q

1. P
   
   
   1. 
      
      
      
      3. 
         
      
      
      4. 
         
         
         
         1. =^18 (10.2 + 109.77 – 31.67) = 11.04 bar
            Hence, mean effective pressure = 11.04 bar. (Ans.)
            Example 13.27. An I.C. engine operating on the dual cycle (limited pressure cycle) the
            temperature of the working fluid (air) at the beginning of compression is 27°C. The ratio of the
            maximum and minimum pressures of the cycle is 70 and compression ratio is 15. The amounts of
            heat added at constant volume and at constant pressure are equal. Compute the air standard
            thermal efficiency of the cycle. State three main reasons why the actual thermal efficiency is
            different from the theoretical value. (U.P.S.C. 1997)
            Take γ for air = 1.4.
         
         
         
         
      
      
      
      
   
   
   
   

Solution. Refer Fig. 13.23. Given : T 1 = 27 + 273 = 300 K ; p

p

3

1

= 70, v

v

v

v

1

2

1

3

= = 15

p

v

3 4

2

5

1

Fig. 13.23. Dual cycle.

Air standard efficiency, ηair-standard :

Consider 1 kg of air.

Adiabatic compression process 1-2 :

T

T

v

v

2

1

1

2

1

=F 15 14 1

HG

I

KJ

=

−

−

γ

(). = 2.954