TITLE.PM5

178 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 Water out h = 192 kJ/kg 2 h = 240 kJ/kg Water in 1 2 1 20 m (Z – Z ) 21 ...

FIRST LAW OF THERMODYNAMICS 179 dharm /M-therm/Th4-5.pm5 C 22 2 = (h^1 – h^2 ) + C 12 2 = (915 – 800) × 1000 + 300 2 2 ∴ C 22 = ...

180 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 Capacity of electric motor : Steady flow energy equation is given by mu ...

FIRST LAW OF THERMODYNAMICS 181 dharm /M-therm/Th4-5.pm5 Fuel air ratio = 0.0180 If, mass of air, kg, then mass of fuel kg and m ...

182 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 (iii)The velocity at exit from the nozzle, assuming no heat loss. Take t ...

FIRST LAW OF THERMODYNAMICS 183 dharm /M-therm/Th4-5.pm5 (ii)Turbine : Power output of turbine : Energy equation for turbine giv ...

184 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 T 2 Ta T 1 pa p 1 p 2 2 p v 12 = v v a 1 (a) vv 12 = p 2 a p 1 s 1 s 2 s ...

FIRST LAW OF THERMODYNAMICS 185 dharm /M-therm/Th4-5.pm5 If after cooling the condition of steam remains wet, then the mass frac ...

186 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 Steam at 80 bar and 350°C is in a superheated state, and the specific vo ...

FIRST LAW OF THERMODYNAMICS 187 dharm /M-therm/Th4-5.pm5 Constant pressure Heating or Cooling. Fig. 4.59 (a), (b) and (c) shows ...

188 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 x 1 1 a T 1 T 2 p= p 12 2 h s (c) Fig. 4.59. Constant pressure process. ...

FIRST LAW OF THERMODYNAMICS 189 dharm /M-therm/Th4-5.pm5 2 × 10^512 p (N/m )^2 v = 0.885 1 v = 1.316 2 v (m /kg)^3 Fig. 4.60 (i) ...

190 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 6.55 = 2.0457 + x 1 × 4.6139 ∴ x 1 6 0457 4 = .55 2.− .6139 = 0.976 Fig. ...

FIRST LAW OF THERMODYNAMICS 191 dharm /M-therm/Th4-5.pm5 This process is limited to wet region only. Hyperbolic process (pv = co ...

192 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 (c) Fig. 4.62. Constant temperature or isothermal expansion. WpdvC v dv ...

FIRST LAW OF THERMODYNAMICS 193 dharm /M-therm/Th4-5.pm5 1 2 165° C 1.5 × 10^5 7 × 10^5 p (N/m )^2 v 1 v 2 v (m /kg) 3 Fig. 4.63 ...

194 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 (iii)Work done : From non-flow energy equation, Q = (u 2 – u 1 ) + W ∴ W ...

FIRST LAW OF THERMODYNAMICS 195 dharm /M-therm/Th4-5.pm5 Fig. 4.64 The constant is either p 1 v 1 or p 2 v 2 i.e., W = 5.5 × 10^ ...

196 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-5.pm5 Fig. 4.65 (i)Heat supplied : Now, heat supplied, Q = shaded area = T(s 2 ...

FIRST LAW OF THERMODYNAMICS 197 dharm /M-therm/Th4-6.pm5 In case the process is steady flow reversible adiabatic, then first law ...