TITLE.PM5

198 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 (^1) x 1 a 2 p 2 p 1 h s (c) Fig. 4.66. Reversible adiabatic or isentrop ...

FIRST LAW OF THERMODYNAMICS 199 dharm /M-therm/Th4-6.pm5 Final pressure of steam, p 2 = 38 bar From superheat tables, at 120 bar ...

200 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 (b) (c) Fig. 4.68. Polytropic process. = (h 2 – h 1 ) + (p 1 v 1 – p 2 v ...

FIRST LAW OF THERMODYNAMICS 201 dharm /M-therm/Th4-6.pm5 Q = 0 ∴ 0 = ∆u + W ∴ W = – ∆u = – (u 2 – u 1 ) = (u 1 – u 2 ) i.e., W = ...

202 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 Fig. 4.69 =^10 0 5 .1 (1.869 – 1.419) = 10^5 × 4.5 N-m/kg i.e., Work don ...

FIRST LAW OF THERMODYNAMICS 203 dharm /M-therm/Th4-6.pm5 Example 4.59. Steam enters a steam turbine at a pressure of 15 bar and ...

204 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 Process of expansion : Reversible adiabatic As the process is reversible ...

FIRST LAW OF THERMODYNAMICS 205 dharm /M-therm/Th4-6.pm5 Fig. 4.71 As the enthalpy and pressure of steam at the exit of the nozz ...

206 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 (^11) 2 2 p 1 p 2 T s s h 2 (wet) 2 (superheated) p 2 p 1 1 1 Fig. 4.73. ...

FIRST LAW OF THERMODYNAMICS 207 dharm /M-therm/Th4-6.pm5 At 18 bar : hf = 884.6 kJ/kg, hfg = 1910.3 kJ/kg ∴ 2776.4 = 884.6 + x 1 ...

208 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 Locate the point ‘1’ at an intersection of the 10 bar pressure line and ...

FIRST LAW OF THERMODYNAMICS 209 dharm /M-therm/Th4-6.pm5 (ii) Change in enthalpy = h 2 – h 1 = 2559.29 – 2692.64 = – 133.35 kJ/k ...

210 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-6.pm5 (b) Using Mollier chart. Refer to Fig. 4.77. Isentropic 2 1 15 ba r 7. 5 ...

FIRST LAW OF THERMODYNAMICS 211 dharm /M-therm/Th4-7.pm5 Filling a tank : Let m 1 = Initial mass of fluid, p 1 = Initial pressu ...

212 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-7.pm5 When the tank is fully insulated and thus no heat transfer takes place, ...

FIRST LAW OF THERMODYNAMICS 213 dharm /M-therm/Th4-7.pm5 ∴ Mass of air which left the receiver, m = m 1 – m 2 = 97.34 – 79.3 = 1 ...

214 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-7.pm5 and – 12°C on the other side. The cylinder walls and piston may be regar ...

FIRST LAW OF THERMODYNAMICS 215 dharm /M-therm/Th4-7.pm5 (ii)The piston will be in virtual equilibrium and hence zero work is ef ...

216 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-7.pm5 (v)Polytropic reversible process (pvn = constant) ∆u = cv (T 2 – T 1 ) ; ...

FIRST LAW OF THERMODYNAMICS 217 dharm /M-therm/Th4-7.pm5 The net work done per kg of gas in a polytropic process is equal to (a ...