A tube has inner diamtere 50 mm and its thickness 5 mm is ex

A tube has inner diamtere 50 mm and its thickness 5 mm is extruded which its inner diameter becomes 40 mm and its thickness 4 mm. Find the percentage of reduction and the ratio between enrty and exit velocites.

Solution

Initial Diameter of tube = 50mm + (2*5)

= 60 mm

Final Diameter = 40 mm +(2*4)

= 48 mm

Percentage Reduction = [(Di -Df)/Di]*100

= 20 %

We know that V_i *Di² = V_f * Df²

   V_i / V_f   = (Df/Di)²

= (48/60)² = 0.64