TITLE.PM5

278 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 For constant volume process = 20 × 0.7432 loge 2 = 10.3 kJ/K. (Ans.) Exa ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 279 dharm /M-therm/th5-4.pm5 Example 5.32. A rigid cylinder containing 0.004 m^3 of nit ...

280 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 Example 5.34. 1 kg of air initially at 8 bar pressure and 380 K expands ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 281 dharm /M-therm/th5-4.pm5 2 T S pV = Cn T 1 1 4 T 2 dS 3 = 0 287 14 1 290 6 380 0 28 ...

282 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 Mean absolute temperature, Tmean = TT^12 2 290 666 2 2 + = +. = 478.1 K ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 283 dharm /M-therm/th5-4.pm5 ∴ Overall change of entropy = (S 2 – S 1 ) + (S 3 – S 2 ) ...

284 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 = 0.0209 × (0.718 + 0.287) × loge 300 1500 F HG I KJ = – 0.0338 kJ/K. (A ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 285 dharm /M-therm/th5-4.pm5 Expression for entropy change in terms of pressure, volum ...

286 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 Entropy variation for a constant volume process is given by : dS = mcv d ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 287 dharm /M-therm/th5-4.pm5 Solution. Given : VA = 0.25 m^3 ; pAi = 1.4 bar ; TAi = 29 ...

288 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 Temperature fall when expanded reversibly and adiabatically = 165 K Temp ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 289 dharm /M-therm/th5-4.pm5 Heat pump Source T 1 Sink T 2 W Q 1 Q 2 Solution. The arra ...

290 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 W Q 3 Q 1 Source-1 500 K Source-2 400 K Source-3 300 K Q 2 Heat engine F ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 291 dharm /M-therm/th5-4.pm5 W Q 1 Q– W 1 System T = 250 K 1 HE Reservoir T = 125 K 2 ( ...

292 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 (∆ S)working fluid in (^) HE = 0 ∴ (∆ S)universe = (∆ S)system + (∆ S)re ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 293 dharm /M-therm/th5-4.pm5 Solution. Fig. 5.49 shows the isolated system before mixin ...

294 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 Solution. Mean temperature of the mixture = GHFTT^12 + 2 IKJ. Thus chang ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 295 dharm /M-therm/th5-4.pm5 System (0 + 273 = 273 K) Q Reservoir (90 + 273 = 363 K) (i ...

296 ENGINEERING THERMODYNAMICS dharm /M-therm/th5-4.pm5 (ii)What is the minimum amount of work necessary to convert the water ba ...

SECOND LAW OF THERMODYNAMICS AND ENTROPY 297 dharm /M-therm/th5-4.pm5 Fig. 5.52 The entropy change of the working fluid in the r ...