Download Course Outline-Thermodynamics-Lecture Slides and more Slides Thermodynamics in PDF only on Docsity! • Mixtures • Psychrometry • Combustion • Nozzles • Turbines • Compressors Course Outline docsity.com • Pure Substance: • A substance having a constant and uniform chemical composition. • A homogeneous mixture of gases can be treated as a pure substance provided no chemical reaction takes place. Example: Dry Air • The thermodynamic properties of a mixture of gases can be determined in the same way as for a single gas. • Two Sections: • Mixtures of perfect gas and perfect gas and vapour Combustion Calculation • Air and water vapour mixtures Psychrometry and air-conditioning Mixtures docsity.com Mixtures Volumetric Analysis of a Gas Mixture: Consider a volume V of a gaseous mixture at temperature T. Mixture constitutes of gases A, B and C. • Compress each of the constituent to total pressure, p, of the mixture. Temperature remains constant. • The partial volume occupied by each gas would be VA, VB and VC. m = mA + mB + mC = ∑ mi p = p A + p B + p C = ∑ p i n = nA + nB + nC = ∑ ni pV = mRT TR Vp m A A A VA P mA nA VB P mB nB VC P mC nC TR pV m A A A docsity.com Mixtures Volumetric Analysis of a Gas Mixture: AA pVVp TR pV TR Vp A A A A V p p V AA In general; i i i p p V V p p V From Dalton’s law p = ∑ pi VVi Amagat’s law / Leduc’s law: The volume of a mixture of gases is equal to the sum of the volumes of the individual constituents when each exists alone at the pressure and temp. of the mixture. docsity.com Mixtures Volumetric Analysis of a Gas Mixture: Amount of Substance, n (moles): It is that quantity which contains as many elementary entities as there are atoms in 0.012 kg of C-12; the elementary entities must be specified and may be atoms, molecules, ions, electrons or other particles or specific groups of such particles. Avogadro’s Hypothesis: The volume of 1 mol of any gas is the same as the volume of 1 mol of any other gas, when the gases are at the same temperature and pressure. VVi inn The total amount of substance in the vessel must equal the sum of the amounts of substance of the individual constituents. docsity.com Mixtures The molar mass and Specific Gas Constant: As mentioned before; The characteristic equation of state of a perfect gas is For a gas in a mixture, this can be written as Equating: Remember: Therefore: TRnpV ~ TRnVp ii ~ TR n pV ~ TR n Vp i i ~ n pV n Vp i i n n p p ii Vp p V ii V V n n p p iii The molar analysis is identical with the volumetric analysis and both are equal to the ratio of partial pressure to the total pressure docsity.com Mixtures The molar mass and Specific Gas Constant: Another method to determine Molar Mass: For individual constituent For mixture As done before As TRVpm iii TRpVm imm TR Vp RT pV i i i i R p R p mRR ~ ~ R mp R mp ii ~ ~ ~ ~ iimpmp ~~ n mn V mV p mp m iiiiii ~~~ ~ docsity.com Mixtures Specific Heat Capacities of a Gas Mixture: From Gibbs- Dalton Law: mu = ∑ mi ui For a perfect gas: u = CvT So, From Gibbs- Dalton Law: mh = ∑ mi hi For a perfect gas: h = CpT So, TCmTmC iviv iv i v C m m C TCmTmC ipip ip i p C m m C docsity.com Mixtures Adiabatic Mixing of Perfect Gasses (Non- Flow Process): VA pA mA nA TA| VB pB mB nB TB P, T m = mA + mB n = nA + nB V = VA + VB It is free expansion which is an irreversible process. Assume that the process is adiabatic, which means Entropy of the system will increase. In an adiabatic free expansion process, the internal energy remains constant. Refer FLOT. BvBAvA TCnTCnU BA ~~ 1 TCnCnU BA vBvA ~~ 2 ivi TCnU i ~ 1 ivi CnTU ~ 2 21 UU i i vi ivi Cn TCn T ~ ~ docsity.com Mixtures Adiabatic Mixing of Perfect Gasses (Steady Flow Process): mAhA1 mAhA1 mAhA1+ mBhB1 Apply steady-flow energy equation: 2211 BBAABBAA hmhmWQhmhm 0 0 TCh pAlso, TCmTCmTCmTCm pBpABpBApA ipiipi CmTTCm i ipi ipi Cm TCm T i ipi ipi Cn TCn T i~ ~ docsity.com