1.3 Fluid Dynamics
23
Solution
Given that:
v =
Q
A
=
Q
π D 2
4
=
0.9
m
3
s
π
0.5 2
4
m 2
= 4.584
m
s
Reynolds’ number is:
Re =
ρDv
μ
=
1.2
Kg
m 3 · 0.5 · 4.584
m
s
1.86 × 10 −5 N
s
m 2
= 147,870.96
With Re calculated, it can be verified that the conditions to apply the Haaland
expression are fulfilled:
Re = 4000−10
8
ε
D
=
0.0015 × 10
−3
500 × 10 −3 = 3 × 10
−6
ε
D
belongs to the interval : (0 − 0.05)
Now, you can estimate f from the Haaland equation:
1
√
f
= −1.8 log
6.9
Re
+
ε
D
3.71
1.11
= −1.8 log
6.9
147,870.96
+
3 × 10 −6
3.71
1.11
f = 0.016466
If the duct has the same cross section, which implies the same speed, but there is
a difference in height, then:
P 1 − P 2
ρg
= z 2 − z 1 + f
L
D h
v
2
2g
Taking into account the definition of flow, then v =
Q
S
, so substituting:
P 1 − P 2
ρg
= 0 − z 1 + f L
Q
π D 2
4
2
D2g
23
Solution
Given that:
v =
Q
A
=
Q
π D 2
4
=
0.9
m
3
s
π
0.5 2
4
m 2
= 4.584
m
s
Reynolds’ number is:
Re =
ρDv
μ
=
1.2
Kg
m 3 · 0.5 · 4.584
m
s
1.86 × 10 −5 N
s
m 2
= 147,870.96
With Re calculated, it can be verified that the conditions to apply the Haaland
expression are fulfilled:
Re = 4000−10
8
ε
D
=
0.0015 × 10
−3
500 × 10 −3 = 3 × 10
−6
ε
D
belongs to the interval : (0 − 0.05)
Now, you can estimate f from the Haaland equation:
1
√
f
= −1.8 log
6.9
Re
+
ε
D
3.71
1.11
= −1.8 log
6.9
147,870.96
+
3 × 10 −6
3.71
1.11
f = 0.016466
If the duct has the same cross section, which implies the same speed, but there is
a difference in height, then:
P 1 − P 2
ρg
= z 2 − z 1 + f
L
D h
v
2
2g
Taking into account the definition of flow, then v =
Q
S
, so substituting:
P 1 − P 2
ρg
= 0 − z 1 + f L
Q
π D 2
4
2
D2g