Find the force, in N, required to pull a long tube 2.00 cm in diameter through a horizontal cylindrical hole 2.06 cm in diameter and 50.0 cm long, at a constant speed 3cm/s. assume that the tube is always centered in the hole and that the gap between the tube and the hole is filled with SAE 30 oil at 15.6 c?
Solution
Given,
Length of the long tube = l = 50 cm = 0.5 m
Hole radius = R2 = 1.03 cm = 0.0103 m
Radius of the long tube = R1 = 1.00 cm = 0.01 m
Speed of the tube = u o = 3 cm/s = 0.03 m/s
Viscosity of SAE 30 oil = 0.310 Pa.s
We know that,
Velocity of the fluid is u =Â Â u o * ln(R2/R)/ln(R2/R1)
Shear Stress around the tube = -(Viscosity of SAE 30 oil)*u 0 /(R1*ln(R2/R1))
= - 0.310 * 0.03/(0.01*ln(0.0103/0.01))
= - 31.4627 Pa
Total Shear force on the tube = Shear stress on tube * (2*pi*R1) * (Length of the tube)
==> F = -31.4627*2*3.14*0.01*0.5
==> F = - 0.987928 N
Therefore, the required force to pull the tube = - Shear Force = - F = 0.987928 N
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Find the force- in N- required to pull a long tube 2-00 cm in diameter.docx
1. Find the force, in N, required to pull a long tube 2.00 cm in diameter through a horizontal
cylindrical hole 2.06 cm in diameter and 50.0 cm long, at a constant speed 3cm/s. assume that
the tube is always centered in the hole and that the gap between the tube and the hole is filled
with SAE 30 oil at 15.6 c?
Solution
Given,
Length of the long tube = l = 50 cm = 0.5 m
Hole radius = R2 = 1.03 cm = 0.0103 m
Radius of the long tube = R1 = 1.00 cm = 0.01 m
Speed of the tube = u o = 3 cm/s = 0.03 m/s
Viscosity of SAE 30 oil = 0.310 Pa.s
We know that,
Velocity of the fluid is u =Â Â u o * ln(R2/R)/ln(R2/R1)
Shear Stress around the tube = -(Viscosity of SAE 30 oil)*u 0 /(R1*ln(R2/R1))
= - 0.310 * 0.03/(0.01*ln(0.0103/0.01))
= - 31.4627 Pa
Total Shear force on the tube = Shear stress on tube * (2*pi*R1) * (Length of the tube)
==> F = -31.4627*2*3.14*0.01*0.5
==> F = - 0.987928 N
Therefore, the required force to pull the tube = - Shear Force = - F = 0.987928 N
