Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 2 (Q.No. 34)
34.
In a footstep bearing, if the speed of the shaft is doubled, then the torque required to overcome the viscous resistance will be
Discussion:
8 comments Page 1 of 1.
Sham Malik said:
6 years ago
Formula for torque of foot step bearing is; mu*π^2*N*R^4/60t.
N doubled then T doubled.
N doubled then T doubled.
Ajaygiri said:
6 years ago
T = ϲuNR^4/60t.
u - viscosity of the oil.
N - speed of the shaft.
R - radius of the shaft.
t - the thickness of the oil film.
And T proportional to N.
u - viscosity of the oil.
N - speed of the shaft.
R - radius of the shaft.
t - the thickness of the oil film.
And T proportional to N.
Rohit Koundal said:
7 years ago
Torque = π*N*u*R^4/60t.
Aamir said:
8 years ago
Apply formula of rotating cylinder viscometer, in which torque is directly pro to shaft speed.
Meet said:
8 years ago
Consider an annular strip of radius r and width dr.
F = Aμ (du/dy) = A μ(u/y)
u = 2πrN/60, y = h, A = 2πr dr y=h=clearance
Torque = Force * radius.
dT = 2πr dr μ(2πrN/60h)r.
dT proportional to N.
F = Aμ (du/dy) = A μ(u/y)
u = 2πrN/60, y = h, A = 2πr dr y=h=clearance
Torque = Force * radius.
dT = 2πr dr μ(2πrN/60h)r.
dT proportional to N.
Raja raha said:
9 years ago
Formula:
T = pie^2*N*R^4/60t.
T = pie^2*N*R^4/60t.
Shiva said:
1 decade ago
@Manoj.
No, Torque is directly proportional to speed both in journal and footstep bearing.
No, Torque is directly proportional to speed both in journal and footstep bearing.
Manoj said:
1 decade ago
T inversely proportional to N(speed of shaft).
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