Thermodynamic - Lecture

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.

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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...

ME8691 COMPUTER AIDED DESIGN AND MANUFACTURING

ME8691 COMPUTER AIDED DESIGN AND MANUFACTURING UNIT I INTRODUCTION Text Book Link: https://drive.google.com/file/d/1jcXUoP-axkNTd2S-AbiWhbUMCyqB643D/view?usp=sharing UNIT II GEOMETRIC MODELING Text Book Link: https://drive.google.com/file/d/1FMm2whgkPH0YdIz6z84csG9DonJJxJSR/view?usp=sharing UNIT III CAD STANDARDS Text Book Link: https://drive.google.com/file/d/1KBsbpx-PMudXsLasu2G6624ctD1ZgoMk/view?usp=sharing UNIT IV FUNDAMENTAL OF CNC AND PART PROGRAMING Text Book Link: https://drive.google.com/file/d/1LSSaf9Hr0olLyiyqE2nJNLzxs-lvjByo/view?usp=sharing UNIT V CELLULAR MANUFACTURING AND FLEXIBLE MANUFACTURING SYSTEM (FMS) Text Book Link: https://drive.google.com/file/d/1PxFAEx8YoFha_G_ZrDM4xOZpT91KPcbe/view?usp=sharing CAD&M Notes Link: https://drive.google.com/file/d/1dra3jcfkrF02HxXPLcgVyIcA234yPLQ2/view?usp=sharing CAD&M QB Link: https://drive.google.com/file/d/18fIwlt-Kbn_jVdJcJL97fSysuTfLINp1/view?usp=sharing

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...

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