major and minor losses in pipes
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Major and Minor Losses in Pipes project pptTRANSCRIPT
Major and Minor Losses in pipes
Roll No. : 110517Section : M6
PREVIEW
Head Losses Major Losses Minor Losses Example Problem Total Head Loss Friction Loss Extended Bernoulli’s equation
Head Losses
In the analysis of piping systems, pressure losses are commonly
expressed in terms of the equivalent fluid column height called
Head Losses (hL).
It also represents the additional height that the fluid needs to be
raised by a pump inorder to overcome the frictional losses in the
pipe
gd2
fLV
g
Ph
2avgL
L
3
4
5
Major losses
Physical problem is to relate pressure drop to fluid parameters and pipe geometry
Differential Pressure Gauge- measure ΔP
PipeD
V
L
ρ μ ε
),,,,,( DLVP
Using dimensional analysis we can show that
DD
LVD
V
P
,,
2
1 2
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Minor Losses
• Apart from major loss due to friction, there are also other forms of losses which are caused by
changes in internal pipe geometries and by fitted components.
• These types of losses are referred to as minor losses. There are four types of minor losses:
Sudden or gradual flow expansion and flow contraction, Entrance and exit flows to and from reservoirs or tanks, Bends, elbows, junctions and other fittings, Valves, including those completely opened or partially closed.
Piping systems include fittings, valves, bends, elbows, tees, inlets, exits, enlargements, and
contractions.
These components interrupt the smooth flow of fluid and cause additional losses because of flow
separation and mixing.
The minor losses associated with these components:
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KL is the loss coefficient.
— Is different for different components.
— Typically provided by manufacturers.g
VKhm 2
2
8
The loss coefficient of a component (such as the gate valve shown) is determined by measuring the pressure loss it causes and dividing it by the dynamic pressure in the pipe.
Minor losses are usually expressed in terms of the loss coefficient KL
Example Problem
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Ans: 169 kPa
Total Head Loss
Total head loss in a system is comprised of major losses (in the
pipe sections) and the minor losses (in the components)
If the entire piping system has a constant diameter, then
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i pipe sections
j components
• For laminar flow in rough pipes, the friction factor f is dominantly caused by viscous friction due to molecular interaction. Hence, we can use Eq. (8.12) for all occasions involving laminar flow.
• However, for turbulent flow, the profile at the core of the pipe is close to inviscid profile and the friction factor f is much due to the existence of viscous sublayer near the wall.
• Thus, if the wall surface is rougher, the resulting viscous sublayer is thicker. The roughness of a pipe is measured in length which is defined as equivalent roughness . The values of for typical pipes are listed in Table 8.1.
Table 8.1 Equivalent Roughness for Typical New Pipes
Friction Loss
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• For turbulent flow, the friction factor f can be obtained by using the graphical representation of the Colebrook formula which is the Moody chart as shown in Figure 8.11
Table 8.11: The moody chart
Friction Loss
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• In some texts, the same experimental data are refitted to a simpler form of correlation which can be solved directly with 2% error. This correlation is known as the Haaland formula which takes the following form:
• After knowing the friction factor f for the pipe, we can calculate the major head loss due to friction for a fluid flowing in the pipe. If fluid properties, and , pipe length and relative roughness of the pipe wall are all known, provided that other variables are also known, the problem can be one of the following types:
1. Determine pressure loss p or friction head loss hf, 2. Determine volumetric flow-rate Q or average velocity V, 3. Determine pipe diameter D. After knowing f, then hf can be calculated via Eq. (8.11),
Re
9.6
7.3
/log8.1
111.1
D
f
g
V
Dfh f 2
2
Friction Loss
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Extended Bernoulli’s Equation
B a s e d o n t h e p r e v i o u s d i s c u s s i o n , t h e p r e s s u r e d i s t r i b u t i o n a l o n g a p i p e s y s t e m c a n b e c h a r a c t e r i z e d b y u s i n g t h e e x t e n d e d B e r n o u l l i ' s e q u a t i o n :
2 21 1 2 2
1 2
1 2 1 2
A E
2 2
F o r a h o r i z o n t a l ( z = z ) , f u l l y - d e v e l o p e d ( V = V ) , s t r a i g h t p i p e s y s t e m w i t h n o e x t e r n a l p o w e r i n o r o u t
( h = h = 0 ) , t h e p r e s s u r e d r o p i s r e l a t e d t o t h e l o s s e s a l o n e ,
A E L
P V P Vg z g h g h g h g z
2 21 2
t h a t i s ,
L 1 1f r i c t i o n a l l o s s e s + m i n o r l o s s e s = f ( ) ( )2 2DL LP P P g h V K V
B o t h t h e l o s s c o e f f i c i e n t a n d t h e f r i c t i o n a l f a c t o r c a n b e d e t e r m i n e d u s i n g t h e e m p i r i c a l v a l u e s a v a i l a b l e i n s t a n d a r d i z e d t a b l e s o r c h a r t s i n c h a p t e r 8 . S p e c i a l n o t e : d u e t o t h e n o n - u n i f o r m i t y o f t h e v e l o c i t y p r o f i l e s i n s i d e a p i p e , t h e k i n e t i c e n e r g y t e r m s i n t h e e x t e n d e d B e r n o u l l i ' s e q u a t i o n s h o u l d b e m o d i f i e d t o i n c l u d e t h i s e f f e c t . T h e k i n e t i c e n e r g y c o e f f i c i e n t , , i s s o d e f i n e d i n c h a p t e r 8 . 6 . 1 . S e e e q u a t i o n s 8 . 2 8 a n d 8 . 2 9 t o g e t m o r e i n f o r m a t i o n .
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Courtesy: NIT Kurukshetra