Page 99 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

3.3 Data: p1=1.02 bar; T1=22 + 273 = 295 K;

3
v1= 0.015 m ; p2= 6.8 bar

From equation 3.8
−
 1
T  p  
2 =  2 
T 1  p 1 

 8 . 6  4 . 1 ( − 4 . 1 / ) 1
T2 = 295 x  
 . 1 02 

= 507.5 K

(where  for air = 1.4)

o
i.e. Final temperature = 507.5 – 273 = 234.5 C

From equation 3.6

p V   v  p   / 1
2 =  1  or 1 =  2 
V
p 1  2  v 2  p 1 


 . 0 015 =  8 . 6   4 . 1 / 1
v 2  . 1 02 

i.e. Final volume
3
v2 = 0.0038 m

For an adiabatic process,
W = u1 – u2

and for a perfect gas,

W = cv(T1- T2)
= 0.718(295-507.5)

= - 152.8 kJ/kg

i.e. work input per kg = 152.8 kJ

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