Academic help online
Water is to flow from reservoir A to reservoir B through the piping system shown in Figure 11.21. The flow rate when the valve is completely open is to
be 0.003 m3/s. The generalized head loss equation becomes
lLV
2
5
V
2
zA – zB =
+ a KLm
2gD
2g
m = 1
where zA – zB is the change in height between the free surfaces of the water in the two reservoirs, KLm are the minor loss coefficients at the locations shown in
Figure 11.21, l is the pipe friction coefficient, and V = 4Q/pD2 is the average velocity in the pipe. If
nvis = 1.3 * 10 – 6 m2/s and r = 1,000 kg/m3, then determine the pipe diameter (Answer: D = 0.04698 m).
The oscillations caused by a suddenly released fluid from a height Z that separates the fluid levels in two rectangular prismatic reservoirs connected
by a long pipeline of length L shown in Figure 11.22 can be determined from9
d2Z
+ signum a
dZ
b p a
dZ
b
2
+ qZ = 0
dt2
dt
dt
where
p =
fA1A2Le
q =
ga(A1 + A2)
2DaL(A1 + A2)
A1A2L
Moreover, it has been assumed that the motion of the liquid is mostly turbulent so that the head loss is proportional to the square of the velocity. The quantity Le is
the equiv-alent length of the pipe incorporating minor losses, g is the gravitational constant, f is the friction coefficient in the pipe, A1 and A2 are the surface
areas of the two reservoirs, a is the area of the pipeline, and D is its diameter.
A
KL1 = 0.5 (Entrance)
KL2 = 0.3 (Elbow)
3 m
B
KL5 = 1.0 (Exit)
Total length = 25 m k = 0.26 mm
= pipe roughness
KL3 = 0.3 (Elbow) KL4 = 0.15 (Valve fully opened)
Figure 11.21 Piping system between two reservoirs.
9 D. N. Roy, Applied Fluid Mechanics, Ellis Horwood Limited, Chichester, England, 1988, pp. 290–293.
Z
A1
A2
L
D, a, f
Figure 11.22 Interconnected reservoirs.
If p = 0.375 m – 1 and q = 7.4 * 10 – 4 s – 2 and the initial conditions are Z(0) = Zn m and dZ(0)/dt = 0 m/s, then determine the value of the first occurrence of tn
for which Z(tn) = 0 when Zn = 5, 10, Á , 50. Plot the results, which should look like those shown in Figure 11.23. Use interp1 to determine tn.
Consider a belt whose surface is moving vertically upward with velocity Vb = 1.5 m/s. There is a thin layer of oil on the surface whose thickness is held
constant. Gravity acts to pull the oil downward, while viscous forces drag the oil upward. This flow is governed by the equation
0w
= n
02w
– g
0t
0x2
280
260
240
(s)
220
crossing
200
first zero
180
160
to
Time
140
120
100
80
10
15
20
25
30
35
40
45
50
5
Initial height [Z(0)]
Figure 11.23 Values of the first occurrence of tn for which Z(tn) = 0 as a function of Z(0) = Zn.
1.6
1.4
1.2
(m/s)
1
t = 0.4 s
Velocity
0.8
0.6
Vertical
0.4
0.2
0
−0.2
0.002
0.004
0.006
0.008
0.01
0
y (m)
Figure 11.24 Velocity distribution in an oil layer on a belt surface at x = 0 moving vertically at speed 1.5 m/s. The velocity distributions are shown at equally
spaced inter-vals in time up to t = 0.4 s. The dotted line is the theoretical steady-state solution to the problem. It is seen that this curve is virtually coincident
with the curve at t = 0.4 s.
where w(x, t) is the velocity distribution through the thickness of the oil layer, n is the kinematic viscosity, and g = 9.81 m/s2 is the acceleration of gravity. At
the surface of the belt, the no-slip condition is enforced, w(0, t) = Ub, while at the free surface of the oil, the stress is zero,
rnc
0w
dx = h
= 0
0x
Let h = 0.01 m, r = 900 kg/ m3, and n = 0.0005 m2/s. Calculate the velocity profiles for times up to the point where the velocity profile reaches a steady state, which
occurs at approximately t = h2/n = 0.2 s. The results are shown in Figure 11.24 along with the theoretical steady-state solution
w =
rg x2
–
rgh x + Vb
2m
m
11.4 Consider a horizontal layer of fluid with r = 800 kg/m3, m = 0.02 Ns/m2, and h = 0.02 m that is initially at rest. There is a free surface y = h and at y = 0
the fluid is subjected to a constant stress (t = 0.01 N/m2) that is turned on at t = 0. Compute the velocity pro-files in the layer for times from t = 0 to t = 1.0 s.
The results are shown in Figure 11.25.
