Fluid Mechanics
Component Head Loss
The Minor Loss Coefficient, K:-
Combined Head Loss Equation
(Total head loss)=(Pipe head Loss)+(Component
head loss)
ℎ𝐿 =
=
𝑣2
2∗𝑔
𝑝𝑖𝑝𝑒𝑠 𝑓
∗
∗
𝐿
𝐷
𝑝𝑖𝑝𝑒𝑠 𝑓
𝑣2
∗ ( )+ 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝐾
2∗𝑔
𝐿
∗ ( ) + 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝐾
𝐷
∗
𝑣2
( )
2∗𝑔
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
𝐴𝑝 = ∗ 𝑑𝑝2 =
∗ 0.152 = 0.01766𝑚2
4
4
𝑄
0.028
𝑣=
=
= 1.58𝑚/𝑠
𝐴𝑝 0.01766
𝑉 ∗ 𝐷 1.58 ∗ 0.15
𝑅𝑒 =
=
= 5925
−5
𝑣
4 ∗ 10
The flow is turbulent
Assume e=0
0.25
𝑓=
=
𝑒
5.74 2
(log10 (
+ 0.9 ))
3.7 ∗ 𝐷 𝑅𝑒
0.25
= 0.036
5.74
(log10 ( 0 + (
)))^2
0.9
5925
Head Loss:ℎ𝐿 =
𝑣2
2∗𝑔
∗
𝑝𝑖𝑝𝑒𝑠 𝑓
∗
𝐿
( )+
𝐷
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝐾
ℎ𝐿
1.582
=
2 ∗ 9.80
197
∗ 0.036 ∗
+ 2 ∗ 0.19 + 0.5 + 1
0.15
ℎ𝐿 = 6.255𝑚
𝑃1
ɣ
+ 𝑧1 +
𝑣12
2∗𝑔
+ ℎ𝑝 =
𝑃2
+
ɣ
𝑧2 +
𝑣22
2∗𝑔
+ ℎ𝑡 + ℎ𝐿
Where
Hp=ht=0
V1=v2=0
P1=P2=Patm=0
𝑧1 = 𝑧2 + ℎ𝐿 = 130 + 6.25 = 136.25𝑚
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 ∗ 𝐶𝑟𝑜𝑠𝑠 𝐴𝑟𝑒𝑎
𝐷ℎ =
𝑊𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
For Example and rectangular tube with L(Length)
and W(wide) the hydraulic diameter will be:4∗𝐿∗𝑊 2∗𝐿∗𝑊
𝐷ℎ =
=
2(𝐿 + 𝑊)
(𝐿 + 𝑤)
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
𝐴 = 𝐿 ∗ 𝑊 = 0.6 ∗ 0.3 = 0.18𝑚2
𝑄
2.5
13.89𝑚
𝑣= =
=
𝐴 0.18
𝑠
4∗𝐿∗𝑊
4 ∗ 0.6 ∗ 0.3
𝐷ℎ =
=
= 0.4𝑚
2 ∗ (𝐿 + 𝑊) 2 ∗ (0.6 + 0.3)
𝑉 ∗ 𝐷ℎ 13.89 ∗ 0.4
𝑅𝑒 =
=
= 368000
−6
𝑣
15.1 ∗ 10
The flow is turbulent
𝑒 = 0.000046𝑚
0.25
𝑓=
=
𝑒
5.74 2
(log10 (
+ 0.9 ))
3.7 ∗ 𝐷ℎ 𝑅𝑒
0.25
0.000046
5.74
(log10 ((
)+(
)))^2
0.9
3.7 ∗ 0.4
368000
= 0.015
𝐿
𝑣2
ℎ𝐿 = 𝑓 ∗
∗
𝐷ℎ
2∗𝑔
50
13.892
= 0.015
∗
= 18.6𝑚
0.4
2 ∗ 9.81
𝑃𝑙𝑜𝑠𝑠 = ρ ∗ 𝑔 ∗ ℎ𝐿 = 1.2 ∗ 9.814 ∗ 18.6 = 220𝑃𝑎
1000 ∗ 9.81 ∗ ℎ𝑤 = 220
ℎ𝑤 = 0.0224𝑚 = 2.24𝑐𝑚 = 0.883𝑖𝑛𝑐ℎ
Pumps
a centrifugal pump is a
machine that uses a
rotating set of blades
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.
ℎ𝐿1 = ℎ𝐿2
𝐿1
𝑣12
𝐿2
𝑣22
𝑓1 ∗
∗
= 𝑓2 ∗
∗
𝐷1
2∗𝑔
𝐷2
2∗𝑔
𝑉1 2 𝑓2 ∗ 𝐿2 ∗ 𝐷1
( ) =
𝑉2
𝑓1 ∗ 𝐿1 ∗ 𝐷2
𝑉1
𝑓2 ∗ 𝐿2 ∗ 𝐷1 1/2
=(
)
𝑉2
𝑓1 ∗ 𝐿1 ∗ 𝐷2
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 𝑓1 = 0.017 𝑎𝑛𝑑 𝑓2
= 0.0145 𝑎𝑠 𝑖𝑛𝑖𝑡𝑎𝑙 𝑔𝑢𝑒𝑠𝑠