Page 83 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)

80 Chapter 1 Fundamentals oF Vibration
But the shear stress is given by
dv
t = -m (E.2)
dy
where the negative sign is consistent with a decreasing velocity gradient [1.33]. Using Eq. (E.2) in
Eq. (E.1), we obtain
2
d v
F = -pDl dy m (E.3)
dy 2
The force on the piston will cause a pressure difference on the ends of the element, given by

P 4P
p = = (E.4)
pD 2 pD 2
¢ ≤
4
Thus the pressure force on the end of the element is
4P
p1pD dy2 = dy (E.5)
D
where 1pD dy2 denotes the annular area between y and 1y + dy2. If we assume uniform mean veloc-
ity in the direction of motion of the fluid, the forces given in Eqs. (E.3) and (E.5) must be equal.
Thus we get
2
4P dy = -pDl dy m d v
D dy 2
or

2
d v 4P
= - (E.6)
2
dy 2 pD lm
Integrating this equation twice and using the boundary conditions v = -v 0 at y = 0 and v = 0 at
y = d, we obtain
2P y
2
v = 1yd - y 2 - v 0 ¢1 - ≤ (E.7)
2
pD lm d
The rate of flow through the clearance space can be obtained by integrating the rate of flow through
an element between the limits y = 0 and y = d:
d 2Pd 3 1
Q = vpD dy = pDJ - v 0 dR (E.8)
2
L 0 6pD lm 2
The volume of the liquid flowing through the clearance space per second must be equal to the vol-
ume per second displaced by the piston. Hence the velocity of the piston will be equal to this rate of
flow divided by the piston area. This gives
Q
v 0 = (E.9)
p 2
¢ D ≤
4