TITLE.PM5

158 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 Air out Centrifugal Electric motor compressor Q W Air in Fig. 4.35. Cent ...

FIRST LAW OF THERMODYNAMICS 159 dharm /M-therm/Th4-4.pm5 Applying energy equation to the system, we have : ∆PE = 0 and ∆KE = 0 s ...

160 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 Condensate out Steam in Water in Water out ejtw 2 ejtw 1 Fig. 4.38. Cond ...

FIRST LAW OF THERMODYNAMICS 161 dharm /M-therm/Th4-4.pm5 ∆PE = 0, ∆KE = 0 W = 0 [Q No work is absorbed or supplied] Applying the ...

162 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 4.13. Throttling Process and Joule-Thompson Porous Plug Experiment Throt ...

FIRST LAW OF THERMODYNAMICS 163 dharm /M-therm/Th4-4.pm5 Applying the energy equation to the system h 1 = h 2 This shows that en ...

164 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 If we carry out other series of experiments similar to described above s ...

FIRST LAW OF THERMODYNAMICS 165 dharm /M-therm/Th4-4.pm5 Solution. Flow of fluid = 10 kg/min Properties of fluid at the inlet : ...

166 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 Q = 55 kJ/s = 55 kJ / s 10 60 kg/s = 55 × 6 = 330 kJ/kg Now, ∆KE = CC^2 ...

FIRST LAW OF THERMODYNAMICS 167 dharm /M-therm/Th4-4.pm5 (i)Heat rejected, Q : Using the flow equation, h 1 + C^1 2 2 + Q = h^2 ...

168 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 Pressure of air at outlet to the compressor, p 2 = 7 bar Specific volume ...

FIRST LAW OF THERMODYNAMICS 169 dharm /M-therm/Th4-4.pm5 20 kJ/s, calculate the power developed by the turbine. Consider the boi ...

170 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 −−= −+ F HG I KJ − F HG I KJ × R S | | T | | U V | | W | | + −× L N M M ...

FIRST LAW OF THERMODYNAMICS 171 dharm /M-therm/Th4-4.pm5 Example 4.41. The working fluid, in a steady flow process flows at a ra ...

172 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 In the above equation : — the mass flow is in kg/s — velocity in m/s — i ...

FIRST LAW OF THERMODYNAMICS 173 dharm /M-therm/Th4-5.pm5 Conditions at ‘1’ : Pressure, p 1 = 7.5 bar = 7.5 × 10^5 N/m^2 , 750°C ...

174 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 Fig. 4.50 Specific volume, v 2 = 0.14 m^3 /kg Increase in enthalpy of ai ...

FIRST LAW OF THERMODYNAMICS 175 dharm /M-therm/Th4-5.pm5 (ii)Ratio of inlet to outlet pipe diameter, d d 1 2 : The mass of air p ...

176 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 Increase in enthalpy of circulating water, Qwater = – 92 kJ/kg Amount of ...

FIRST LAW OF THERMODYNAMICS 177 dharm /M-therm/Th4-5.pm5 Area, A 2 =? Mass flow rate, m& =? (i)Velocity at exit of the nozzl ...