Skip to main content

Unconventional Machining Process - Syllabus


              Unconventional Machining Process               


UNIT I INTRODUCTION AND MECHANICAL ENERGY BASED PROCESSES  

Unconventional machining Process – Need – classification – merits, demerits and applications. Abrasive Jet Machining – Water Jet Machining – Abrasive Water Jet Machining - Ultrasonic Machining. (AJM, WJM, AWJM and USM). Working Principles – equipment used – Process parameters – MRR- Applications.

 

UNIT II THERMAL AND ELECTRICAL ENERGY BASED PROCESSES

Electric Discharge Machining (EDM) – Wire cut EDM – Working Principle-equipments-Process Parameters-Surface Finish and MRR- electrode / Tool – Power and control Circuits-Tool Wear – Dielectric – Flushing –– Applications. Laser Beam machining and drilling, (LBM), plasma, Arc machining (PAM) and Electron Beam Machining (EBM). Principles – Equipment –Types - Beam control techniques – Applications.

 

UNIT III CHEMICAL AND ELECTRO-CHEMICAL ENERGY BASED PROCESSES

Chemical machining and Electro-Chemical machining (CHM and ECM)- Etchants – Maskant - techniques of applying maskants - Process Parameters – Surface finish and MRR-Applications. Principles of ECM- equipments-Surface Roughness and MRR Electrical circuit-Process Parameters- ECG and ECH - Applications.

 

UNIT IV ADVANCED NANO FINISHING PROCESSES

Abrasive flow machining, chemo-mechanical polishing, magnetic abrasive finishing, magneto rheological finishing, magneto rheological abrasive flow finishing their working principles, equipments, effect of process parameters, applications, advantages and limitations.

 

UNIT V RECENT TRENDS IN NON-TRADITIONAL MACHINING PROCESSES

Recent developments in non-traditional machining processes, their working principles, equipments, effect of process parameters, applications, advantages and limitations. Comparison of non-traditional machining processes.

Labels:

Comments

Post a Comment

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...
Post a Comment
Read more

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      
Post a Comment
Read more

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...
Post a Comment
Read more