Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros
s section will be maintained during deformation. A cold-worked tensile strength in excess of 840 MPa (122,000 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 10 mm (0.40 in.). Explain how this may be accomplished. (10 points)
1 answer:
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Answer:
Head loss is 1.64
Explanation:
Given data:
Length (L) = 200 m
Discharge (Q) = 0.16 m3/s
According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m
We know,
where, f = Darcy friction factor
V = flow velocity
g = acceleration due to gravity
We know, flow rate Q = A x V
solving for V
obtained Darcy friction factor
calculate Reynold number (Re) ,
where, = density of water
= Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2
so reynold number is
= 6.4 x 10^5
For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm
Relative roughness (e/D) = 0.0015 / 318 = 0.00005
from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126
Therefore head loss is
HL = 1.64 m