Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
INVERTER TYPE POWER SOURCESFor site welding and repair applications, a need was felt for long time to have light weight arc welding power source. Even though, a petrol driven rotary converter is portable, the weight is considerable and it produces lot of noise and smoke. Where power connection is available a simple rectifier type may do the job but not always. For example, for remote and elevated level welding, it is highly inconvenient to drag heavy welding cables and the cost of long welding cables is also high. Therefore to avoid these problems a new type of power source has been developed. This is called inverter type. A schematic is shown in Fig. The weight and size reduction is achieved by using a HF step down transformer, to produce necessary arc voltage. Transformer size and weight are inversely related to the frequency of operation. A Normal 50 Hz step down welding transformer is bulky and heavy because the frequency is low. If the frequency is increased to say 5 to 10 kHZ, then the transformer (made of ferrite cores) size and weight are drastically reduced. This is the principle behind inverter type power sources. Input voltage is directly rectified and is fed to a inverter circuit. The inverter circuit may be made of either thyristors or transistors. The inverter circuit generates HF AC (5 to 10 kHz) and the voltage is step down by a HF transformer. The transformer output is rect ified to give DC output . The inverter circuit can be easily controlled in a closed loop manner to give the desired output characteristics.
Find modulus of elasticity of the composite under isostrain: A composite is made of alternate layers of 60% E-glass and 40% epoxy resin. If modulii of elasticity of E-glass an
Reaction Bonding Such bonding is attained by arranging for two or more components of a compacted body to respond to form both a desired phase and a bond among particles. The r
Calculate the Cutting Speed in Drilling Operation Calculate the cutting speed if a drill of 0.04 m dia is operated at a speed of 250 rpm. Solution D = 0.04 m N = 25
Calculate Machining Time and Material Removal Rate A hole of 40 mm diameter and 50 mm depth is to be drilled in a mild steel component. The cutting speed can be taken as 65 m/
derivation for varing shear stress for thin circular tubes.
What are objectives of the footings? Objectives of the footings: After studying this unit, you should be able to: a. Know different types of footings and the situation wh
In MOTORCYCLE REPAIRING we have gone through various complaints related to engine starting, problem related to gear, clutch, etc. and their possible reasons of trouble making. The
what is the application of the rank order clustering
whats the example of opportunity cost?
Q. Show the Coating Application? Immediately following surface preparation, the cleaned pipe shall be uniformly preheated by a non-contaminating method to the application tempe
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +1-415-670-9521
Phone: +1-415-670-9521
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd