TITLE.PM5

172 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-4.pm5

In the above equation :

— the mass flow is in kg/s

— velocity in m/s

— internal energy in J/kg

— pressure in N/m^2

— specific volume m^3 /kg

— the value of Q is in J/s

Then the unit of W will be J/s.

∴ W=−×+×−×+

F −

HG

I

KJ

L

N

M

M

O

Q

P

P

220

60

2000 1400 10 10 6 0 3

320 140

2

() (^35 .36 1.2 1.)^22 – 100 × 10^3

=^220

60

[]600 10×+×+ ×^3510 0.6 41.4 103 – 100 × 10^3

=^220

60

[600 × 10^3 + 60 × 10^3 + 41.4 × 10^3 ] – 100 × 10^3
= 10^3 × 2471.8 J/s [Q 1 kJ = 10^3 J]
= 2471.8 kJ/s or kW = 2.4718 MW
Hence power capacity of the system = 2.4718 MW. (Ans.)
+Example 4.42. A stream of gases at 7.5 bar, 750°C and 140 m/s is passed through a
turbine of a jet engine. The stream comes out of the turbine at 2.0 bar, 550°C and 280 m/s. The
process may be assumed adiabatic. The enthalpies of gas at the entry and exit of the turbine are
950 kJ/kg and 650 kJ/kg of gas respectively.
Determine the capacity of the turbine if the gas flow is 5 kg/s.
Solution. Refer Fig. 4.49.

Gas in

1

2

Gas turbine

Nozzle Gas out

W

Boundary

Fig. 4.49