M E 320 Professor John M. Cimbala Lecture 22
Today, we will:
•
Do more example problems – major and minor losses in pipe flows
•
Discuss diffusers and show a flow loop demonstration
Example Problem – Major and Minor Losses in a Piping System
Given: Water (
ρ
= 998.
kg/m
3
,
μ
= 1.00 × 10
-3
kg/m⋅s) flows by gravity
alone from one large tank
to another, as sketched.
The elevation difference
between the two surfaces is
H = 35.0 m. The pipe is 2.5
cm I.D. with an average
roughness of 0.010 cm.
The total pipe length is
20.0 m. The entrance and
exit are sharp. There are
two regular threaded 90-
degree elbows, and one
fully open threaded globe valve.
To do: Calculate the volume flow rate through this piping system.
Solution:
• First we draw a control volume, as shown by the dashed line. We cut through the surface of
both reservoirs (inlet 1 and outlet 2), where we know that the velocity is nearly zero and the
pressure is atmospheric. The rest of the control volume simply surrounds the piping system.
• We apply the head form of the energy equation from the inlet (1) to the outlet (2):
1
P
g
ρ
2
1
1
2
V
g
α
+
1pump,u
zh++
2
P
g
ρ
=
2
2
2
2
V
g
α
+
2 turbine,e
zh++
L
h
+
Therefore, the energy equation reduces to
12L
hzzH
=
−=
• Next, we add up all the irreversible head losses, both major and minor. Since the reference
velocity is the same for all the major and minor losses (the pipe diameter is constant
throughout), we may use the simplified version of the equation for h
L
, i.e., Eq. 8-59:
2
2
LL
L
V
hf K
Dg
⎛⎞
=+
⎜⎟
⎝⎠
∑
, &
Re DV
ρ
μ
=
2
4
D
V
π
=
V
0.010 cm 0.004
2.5 cmD
ε
==
• We also need either the Moody chart or one of the empirical equations that can be used in
place of the chart (e.g., the Colebrook equation).
The rest of this problem will be solved in class.
D H
2
1
Control volume
P
1
= P
2
= P
at
m
V
1
= V
2
≈ 0
