TITLE.PM5

182 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

(iii)The velocity at exit from the nozzle, assuming no heat loss.
Take the enthalpy of air as h = cpt, where cp is the specific heat equal to 1.005 kJ/kg°C and
t the temperature.
Solution. Refer Fig. 4.56.

(^12)
3 4
Q
Turbine
Nozzle
Heat
exchanger
W
Fig. 4.56
Temperature of air, t 1 = 20°C
Velocity of air, C 1 = 40 m/s.
Temperature of air after passing the heat exchanger, t 2 = 820°C
Velocity of air at entry to the turbine, C 2 = 40 m/s
Temperature of air after leaving the turbine, t 3 = 620°C
Velocity of air at entry to nozzle, C 3 = 55 m/s
Temperature of air after expansion through the nozzle, t 4 = 510°C
Air flow rate, m& = 2.5 kg/s.
(i)Heat exchanger :
Rate of heat transfer :
Energy equation is given as,
mh 1 C^1 Zg
2
++ 2 1
F
HG
I
KJ

• Q1–2 = mh 2 C^2 Zg
  2
  ++ 2 2
  F
  HG
  I
  KJ


• W1–2
  Here, Z 1 = Z 2 , C 1 , C 2 = 0, W1–2 = 0
  ∴ mh 1 + Q1–2 = mh 2
  or Q1–2 = m(h 2 – h 1 )
  = mcp (t 2 – t 1 ) = 2.5 × 1.005 (820 – 20) = 2010 kJ/s.
  Hence, rate of heat transfer = 2010 kJ/s. (Ans.)