Mathematical Problem | Spinning

Design a Process sequence from draw frame to ring frame to produce 60 Ne yarn if the maximum draft of roving frame is 3 and ring frame is 40 [ sliver hank=0 .12]

Solution:

 

1st step: Draft = Yarn count / Roving count [where, yarn count = 60 Ne, Draft = 40] 
              So, Roving count= 60/40=1.5 Ne

2nd step: Required draft= roving hank / sliver hank, [where, roving hank=1.5, sliver hank= 0.12] 
                = 1.5/0.12= 12.5 

Here, required draft of roving frame is 12.5, but maximum draft of roving frame is 3. So, one m/c is not sufficient for production. That’s why; another m/c must be used. 



3rd step: For new roving frame: Roving frame hank= 0.12*3=0.36 [where, Draw sliver hank=0.12,           Roving draft= 3]

4th step: Required draft=1.5/0.36=4.16 [where, roving hank=1.5, sliver hank= 0.36]

Here, required draft of roving frame is 4.16, but maximum draft of roving frame is 3. So, two m/c s is not sufficient for production. That’s why; another m/c must be used.



 
5th step: Roving frame sliver hank (2) =3*0.36=1.08 [where, roving hank (1) =0.36, roving frame draft=3]

6th step: Required draft=1.5/1.08=1.38 [where, roving hank=1.5, sliver hank= 1.08]


Total Draft: 3*3*1.38= 12.42 [Ans]