Skip to main content

Engineering Thermodynamics - Study Material

 UNIT – I        BASIC CONCEPTS AND FIRST LAW


PPT :-  1. Basics : https://www.file-upload.com/v037x5fcdv1z

PPT :- 2. Problems : https://www.file-upload.com/410vhxr7fzzy

HANDWRITTEN NOTES :- 

1. Closed System : https://www.file-upload.com/77blx0sdty25

2. Open System : https://www.file-upload.com/fb1x04zxve86

BOOK LINK :- https://www.file-upload.com/2nva9e1fnk73


UNIT - II    SECOND LAW AND AVAILABILITY ANALYSIS


PPT :-  https://www.file-upload.com/xfux1670ath1

HANDWRITTEN NOTES :- https://www.file-upload.com/aq0vlmfyihur

BOOK LINK :- https://www.file-upload.com/1wol83ttmyh7


UNIT - III  PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE


PPT :-  https://www.file-upload.com/3m3ss76y8702

HANDWRITTEN NOTES :- https://www.file-upload.com/r4auu9gqp9af

BOOK LINK :- https://www.file-upload.com/w7lrrkuzcr55


UNIT - IV    IDEAL AND REAL GASES, THERMODYNAMIC RELATIONS


PPT :- https://www.file-upload.com/qfwp8o70dp12

HANDWRITTEN NOTES :- https://www.file-upload.com/vlsfy0zpd17u

BOOK LINK :- https://www.file-upload.com/bu2siqcwgiti


UNIT – V   GAS MIXTURES AND PSYCHROMETRY


PPT :- https://www.file-upload.com/ovgbrhj1rq24

HANDWRITTEN NOTES :-  https://www.file-upload.com/i3frzq4xis3r

BOOK LINK :- https://www.file-upload.com/ymdxjn99qypv



Comments

Popular posts from this blog

Thermodynamics - Lecture - 11 - Unit - 3 - Steam Properties and Problems

Problem: 1   A vessel of volume 0.04 m 3 contains a mixture of saturated water and steam at a temperature of 250°C The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and internal energy. Given data: V=0.04m 3 T= 250°C m 1 = 9 kg To find: p, m, v, h, S, and U   Solution: From Steam Tables corresponding to 250°C, v f =v 1 = 0.001251 m/kg V g = V s = 0.050037 m/kg p= 39.776 bar   Total volume occupied by the liquid, V 1 = m 1 v 1 = 9 x 0.001251 = 0.0113 m 3 Total volume of the vessel, V = Volume of liquid + Volume of steam V 1 +V s V s =0.0287 m 3   Mass of steam, m s = V s / V s   =0.0287 / 0.050037 =0.574 kg Mass of mixture of liquid and steam, m=m 1 +m s = 9+ 0.574 = 9.574 kg   Total specific volume of the mixture, v = V / m   = 0.04 /9.574 = 0.00418 m 3 /kg   We know that, v=v f +x V fg V fg = v g - v f 0.00418 = 0.001251 + x (0.05003...

Engg Thermodynamics - Lecture - 13 - Unit - 4 - Ideal Gas and Real Gas & TD Relations

 Ideal Gas Imaginary Substance Obeys the law of PV=RT At low Pressure and High Temperature – density of gas Decreases Real Gas At High Pressure – Gas start to Deviate from Ideal Gas Measuring of Deviation – Compressibility Factor PV=ZRT Z=PV / RT Z = Vactual / Videal For Ideal Gas Z = 1 For Real Gas Z > 1 Important Laws (Ideal Gas) Boyle's Law (constant temperature) P = constant / V Charles Law (constant pressure) V = constant x T Gay-Lussac’s Law (constant volume) P = constant x T THE ENTHALPY OF ANY SUBSTANCE h=u+pv for an ideal gas h=u+RT h=f(T) dh=du+RdT since R is constant ∆h=∆u+R∆T =Cv∆T+R∆T = (Cv+R)∆T Since h is a function of T only, Cp=(∂h/∂T)p Entropy change of an ideal gas: (Eqn – 1) From the general property relations Q = W+ U Tds=du+pdv And for an ideal gas, du=CvdT, dh=CpdT, and pv=RT, the entropy change between any two states 1 and 2 may be computed as given below ds=du/T+p/Tdv =CvdT/T+Rdv/v S2-s1=Cv ln T2/T1+R ln v2/v1 . Entropy change of an ideal gas: (Eqn -2) Fr...

Thermodynamic - Lecture - 2

  Thermodynamic Process: Process which contains transformation of heat into work or work into heat  System:  If thermodynamics  Process carried in any space is known as System Types:  Closed System, Open system and Isolated system Surroundings:  Every thing apart from the system Boundary :  Separation of System and Surrounding Universe:  Everything outside of Surroundings  Types:  Closed System, Open system and Isolated system Closed System:  can exchange only energy with its surroundings, not matter.  Open system:  can exchange both energy and matter with its surroundings.   Isolated system:  cannot exchange either matter or energy with its surroundings. Closed System, can exchange only energy with its surroundings, not matter (mass). Open system can exchange both energy and matter (mass) with its surroundings.  Isolated system: cannot exchange either matter or energy with its surroundings.