SIR IN 2ND NUMERICAL FOR FINDIND EXIT VELOCITY OF SLAB WE TOOK 40 FT/MIN AS OUR INITIAL VELOCITY BUT IN 2ND PASS WE AGAIN TOOK 40 FT/MIN.... WHY WE TOOK 40 FT/SEC. SOULD NOT WE TAKE FINAL VELOCITY OF 1ST ROLL AS AN INITIAL VELOCITY OF 2ND ROLL WHICH IS 51.78 FT/MIN?
Dear student, in our live session of this topic, i made a correction in the figure, it was discussed that we don't have series of 3 roll mills (As drawn in this video) instead we have 3 pass from a single roll mill. so that's we took the same velocity for each pass.
Hlo sir, for calculating fina wide , we will take 3% of each step then we are take 0.03 of initial wide but in u r solution you take it as 1.03 of intial wide how 1.03 will come....
Hi, it says the width of the slab is increasing by 3% upon each pass, hence after step 1 the final width will be = 10 + (10*0.03) =10.3. you can write it as = 10*1.03 = 10.3 as well. Hence for 2nd step it will be =10.3 + (10.3*0.03)= 10.609 or you can write it as = 10*1.03*1.03= 10.609 In similar way, for the 3rd pass 10*1.03*1.03*1.03
SIR IN 2ND NUMERICAL FOR FINDIND EXIT VELOCITY OF SLAB WE TOOK 40 FT/MIN AS OUR INITIAL VELOCITY BUT IN 2ND PASS WE AGAIN TOOK 40 FT/MIN.... WHY WE TOOK 40 FT/SEC. SOULD NOT WE TAKE FINAL VELOCITY OF 1ST ROLL AS AN INITIAL VELOCITY OF 2ND ROLL WHICH IS 51.78 FT/MIN?
Dear student, in our live session of this topic, i made a correction in the figure, it was discussed that we don't have series of 3 roll mills (As drawn in this video) instead we have 3 pass from a single roll mill. so that's we took the same velocity for each pass.
Same doubt I also had that in second pass ur taking initial velocity same as first pass 40 ??
Hlo sir, for calculating fina wide , we will take 3% of each step then we are take 0.03 of initial wide but in u r solution you take it as 1.03 of intial wide how 1.03 will come....
Hi, it says the width of the slab is increasing by 3% upon each pass, hence after step 1 the final width will be = 10 + (10*0.03) =10.3. you can write it as = 10*1.03 = 10.3 as well.
Hence for 2nd step it will be
=10.3 + (10.3*0.03)= 10.609 or you can write it as = 10*1.03*1.03= 10.609
In similar way, for the 3rd pass
10*1.03*1.03*1.03