### Document

```Fluid Mechanics
The Minor Loss Coefficient, K:-
ℎ =
=
2
2∗

∗
∗

2
∗ ( )+
2∗

∗ ( ) +

∗
2
( )
2∗
Example
If oil (ν = 4 × 10-5 m2/s; S = 0.9) flows from the
upper to the lower reservoir at a rate of 0.028
m3/s in the 15 cm smooth pipe, what is the
elevation of the oil surface in the upper
reservoir?
Solution
Minor head loss coefficients:entrance = Ke = 0.5
bend = Kb = 0.19
outlet = KE = 1.0
Π
3.14
= ∗ 2 =
∗ 0.152 = 0.017662
4
4

0.028
=
=
= 1.58/
0.01766
∗  1.58 ∗ 0.15
=
=
= 5925
−5

4 ∗ 10
The flow is turbulent
Assume e=0
0.25
=
=

5.74 2
(log10 (
+ 0.9 ))
3.7 ∗
0.25
= 0.036
5.74
(log10 ( 0 + (
)))^2
0.9
5925
2
2∗
∗

∗

( )+

ℎ
1.582
=
2 ∗ 9.80
197
∗ 0.036 ∗
+ 2 ∗ 0.19 + 0.5 + 1
0.15
ℎ = 6.255
1
ɣ
+ 1 +
12
2∗
+ ℎ =
2
+
ɣ
2 +
22
2∗
+ ℎ + ℎ
Where
Hp=ht=0
V1=v2=0
P1=P2=Patm=0
1 = 2 + ℎ = 130 + 6.25 = 136.25
Non-Round Conduits
If the conduit is not a pipe it is a square, triangle
or any other shapes we replace the diameter by
the hydraulic diameter.
4 ∗
ℎ =

For Example and rectangular tube with L(Length)
and W(wide) the hydraulic diameter will be:4∗∗ 2∗∗
ℎ =
=
2( + )
( + )
Example
Air (T = 20°C and p = 101 kPa absolute) flows at a
rate of 2.5 m3/s in a horizontal, commercial
steel, HVAC duct. (Note that HVAC is an acronym
for heating, ventilating, and air conditioning.)
What is the pressure drop in inches of water per
50 m of duct?
Solution
=  ∗  = 0.6 ∗ 0.3 = 0.182

2.5
13.89
= =
=
0.18

4∗∗
4 ∗ 0.6 ∗ 0.3
ℎ =
=
= 0.4
2 ∗ ( + ) 2 ∗ (0.6 + 0.3)
∗ ℎ 13.89 ∗ 0.4
=
=
= 368000
−6

15.1 ∗ 10
The flow is turbulent
= 0.000046
0.25
=
=

5.74 2
(log10 (
+ 0.9 ))
3.7 ∗ ℎ
0.25
0.000046
5.74
(log10 ((
)+(
)))^2
0.9
3.7 ∗ 0.4
368000
= 0.015

2
ℎ =  ∗
∗
ℎ
2∗
50
13.892
= 0.015
∗
= 18.6
0.4
2 ∗ 9.81
= ρ ∗  ∗ ℎ = 1.2 ∗ 9.814 ∗ 18.6 = 220
1000 ∗ 9.81 ∗ ℎ = 220
ℎ = 0.0224 = 2.24 = 0.883ℎ
Pumps
a centrifugal pump is a
machine that uses a
situated within a housing
to add energy to a flowing
fluid.
Pump Curve
Example
A pump is to be used to transfer crude oil (! = 2 ×
10!4 lbf-s/ft2, 1 = 0*86) from the lower tank to the
upper tank at a flow rate of 100 gpm. The loss
coefficient for the check valve is 5.0. The loss
coefficients for the elbow and the inlet are
0.9 and 0.5, respectively. The 2-in. pipe is made
from commercial steel (e = 0*002in.) and is 40 ft
long. The elevation distance between the liquid
surfaces in the tanks is 10 ft. The pump efficiency is
80%. Find the power required to operate the
pump.
Solution
Pipes in Parallel
The Losses is equal in both branches as the
energy different in point (1) and point (2) is
equal according to Energy equation.
ℎ1 = ℎ2
1
12
2
22
1 ∗
∗
= 2 ∗
∗
1
2∗
2
2∗
1 2 2 ∗ 2 ∗ 1
( ) =
2
1 ∗ 1 ∗ 2
1
2 ∗ 2 ∗ 1 1/2
=(
)
2
1 ∗ 1 ∗ 2
Example
A piping system consists of parallel pipes as shown
in the following diagram. One pipe has an internal
diameter of 0.5 m and is 1000 m long. The other
pipe has an internal diameter of 1 m and is 1500 m
long. Both pipes are made of cast iron
(e = 0.26 mm). The pipes are transporting water at
20C (ρ= 1000 kg/m3, v= 10^-6 m2/s). The total flow
rate is 4 m3/s. Find the flow rate in each pipe
and the pressure drop in the system. There is no
elevation change. Neglect minor losses.
Solution
Take 1 = 0.017  2
= 0.0145
```