TITLE.PM5

418 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-1.pm5 Again, m M = n and Mcv = Cv ∴ U = nCvT ...(9.23) Similarly, H = nCpT ... ...

GASES AND VAPOUR MIXTURES 419 dharm \M-therm\Th9-1.pm5 Fig. 9.4 Applying steady-flow energy equation to the mixing section (negl ...

420 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-1.pm5 In Fig. 9.5 (ii) a small quantity of water is introduced into the vessel ...

GASES AND VAPOUR MIXTURES 421 dharm \M-therm\Th9-1.pm5 Also, the characteristic gas equation is given by pV = mRT ...(ii) ∴ pV = ...

422 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-1.pm5 (i)Gas constant for air : Now using the equation, R = m m i ∑ Ri, we hav ...

GASES AND VAPOUR MIXTURES 423 dharm \M-therm\Th9-1.pm5 +Example 9.4. A vessel contains at 1 bar and 20°C a mixture of 1 mole of ...

424 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-1.pm5 i.e., Gas constant for the mixture = 0.2608 kJ/kg K. (Ans.) (iv) To find ...

GASES AND VAPOUR MIXTURES 425 dharm \M-therm\Th9-2.pm5 Solution. Composition of mixture by volume : H 2 = 78%, CO = 22% Final co ...

426 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-2.pm5 Consider 1 mole of the mixture. Constituent ni Mi mi = niMi mmi = Fracti ...

GASES AND VAPOUR MIXTURES 427 dharm \M-therm\Th9-2.pm5 T s v (^1) v 2 1 A 2 R loge clogve T T 1 2 v v 2 1 For isothermal process ...

428 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-2.pm5 M = 1 2 2 2 2 2 2 () () ()O () O N N CO CO CO CO m M m M m M m M f ++ +f ...

GASES AND VAPOUR MIXTURES 429 dharm \M-therm\Th9-2.pm5 xCO 2 = 6 44 4 28 6 44 + = 01364 01428 01364 . ..+ = 0.488. (Ans.) (ii)Th ...

430 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-2.pm5 Now cvCO 2 = RCO 2 γ− 1 = 8 314 44 1 286 1 . (. − ) [forCO.]Qγ^2 =1 286 ...

GASES AND VAPOUR MIXTURES 431 dharm \M-therm\Th9-2.pm5 Solution. Using the relation, n = pV RT 0 ∴ nO 2 = 810 18 8 314 10 323 5 ...

432 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-2.pm5 For an isothermal process 1-A : SA – S 1 = mR loge V V A 1 or SA – S 1 = ...

GASES AND VAPOUR MIXTURES 433 dharm \M-therm\Th9-2.pm5 16 × 10^5 × VA = 0.6 × 8.314 × 10^3 × (55 + 273) ∴ VA = 1.022 m^3 The mas ...

434 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-3.pm5 = 16 8 328 3 0 298 19 8 .. . ×+× = 323.5 K i.e., t = 323.5 – 273 = 50.5° ...

GASES AND VAPOUR MIXTURES 435 dharm \M-therm\Th9-3.pm5 Now pN 2 = nRT V N (^20) ...(i) and pO 2 = nRT V O (^20) ...(ii) Dividing ...

436 ENGINEERING THERMODYNAMICS dharm \M-therm\Th9-3.pm5 Again, xN 2 = n n N (^2) = p p N 2 ∴ pN 2 = n n N (^2) × p =^376 1376 . ...

GASES AND VAPOUR MIXTURES 437 dharm \M-therm\Th9-3.pm5 or pCO 2 + 12 = 18 pCO 2 = 6 bar Now pCO 2 = mRT V CO 222 CO CO ∴ mCO 2 = ...