Major Loss

Author
annacamillegonzales 
Category
Documents

view
109 
download
4
Embed Size (px)
Transcript of Major Loss
Major loss  head loss or pressure loss  due to friction in ducts, pipes and tube Pressure and Pressure Loss According the Energy Equation for a fluid the total energy can be summarized as elevation energy, velocity energy and pressure energy. The Energy Equation can then be expressed as: p1 + v1 / 2 + g h1 = p2 + v2 / 2 + g h2 + ploss where p = pressure in fluid (Pa (N/m ), psi (lb/ft )) ploss = pressure loss (Pa (N/m ), psi (lb/ft )) = density of the fluid (kg/m , slugs/ft ) v = flow velocity (m/s, ft/s) g = acceleration of gravity (m/s , ft/s ) h = elevation (m, ft) For horizontal steady state flow v1 = v2 and h1 = h2,  (1) can be transformed to: ploss = p1  p2 (2)2 2 3 3 2 2 2 2 2 2
(1)
The pressure loss is divided in
major loss due to friction and minor loss due to change of velocity in bends, valves and similar.
The pressure loss in pipes and tubes depends on the flow velocity, pipe or duct length, pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct, and whether the flow us turbulent or laminar  the Reynolds Number of the flow. The pressure loss in a tube or duct due to friction, major loss, can be expressed as: ploss = (l / dh) ( v / 2) where ploss = pressure loss (Pa, N/m ) = friction coefficient l = length of duct or pipe (m) dh = hydraulic diameter (m) (3) is also called the D'ArcyWeisbach Equation. (3) is valid for fully developed, steady, incompressible flow. Head and Head Loss2 2
(3)
The Energy equation can be expressed in terms of head and head loss by dividing each term by the specific weight of the fluid. The total head in a fluid flow in a tube or a duct can be expressed as the sum of elevation head, velocity head and pressure head. p1 / + v1 / 2 g + h1 = p2 / + v2 / 2 g + h2 + hloss where hloss = head loss (m, ft) = g = specific weight (N/m , lb/ft ) For horizontal steady state flow v1 = v2 and h1 = h2,  (4) can be transformed to: hloss = h1  h2 where h = p / = head (m, ft) The head loss in a tube or duct due to friction, major loss, can be expressed as: hloss = (l / dh) (v / 2 g) where hloss = head loss (m, ft) Friction Coefficient  The friction coefficient depends on the flow  if it is2 3 3 2 2
(4)
(5)
(6)
laminar, transient or turbulent
and the roughness of the tube or duct. To determine the friction coefficient we first have to determine if the flow is laminar, transient or turbulent  then use the proper formula or diagram. Friction Coefficient for Laminar Flow For fully developed laminar flow the roughness of the duct or pipe can be neglected. The friction coefficient depends only the Reynolds Number  Re  and can be expressed as: = 64 / Re where Re = the dimensionless Reynolds number The flow is (7)
laminar when Re < 2300 transient when 2300 < Re < 4000 turbulent when Re > 4000
Friction Coefficient for Transient Flow If the flow is transient  2300 < Re < 4000  the flow varies between laminar and turbulent flow and the friction coefficient is not possible to determine. Friction Coefficient for Turbulent Flow For turbulent flow the friction coefficient depends on the Reynolds Number and the roughness of the duct or pipe wall. On functional form this can be expressed as: = f( Re, k / dh ) where k = absolute roughness of tube or duct wall (mm, ft) k / dh = the relative roughness  or roughness ratio Roughness for materials are determined by experiments. Absolute roughness for some common materials are indicated in the table below (8)
Absolute Roughness  k Surface 103 (m)
Copper, Lead, Brass, Aluminum (new)
0.001  0.002
PVC and Plastic Pipes
0.0015  0.007
Epoxy, Vinyl Ester and Isophthalic pipe
0.005
Stainless steel
0.015
Steel commercial pipe
0.045  0.09
Stretched steel
0.015
Weld steel
0.045
Galvanized steel
0.15
Rusted steel (corrosion)
0.15  4
New cast iron
0.25  0.8
Worn cast iron
0.8  1.5
Rusty cast iron
1.5  2.5
Sheet or asphalted cast iron
0.01  0.015
Smoothed cement
0.3
Ordinary concrete
0.3  1
Coarse concrete
0.3  5
Well planed wood
0.18  0.9
Ordinary wood
5
The friction coefficient   can be calculated by the Colebrooke Equation: 1/1/2
= 2,0 log10 [ (2,51 / (Re )) + (k / dh) / 3,72 ]
1/2
(9)
Since the friction coefficient   is on both sides of the equation, it must be solved by iteration. If we know the Reynolds number and the roughness  the friction coefficient   in the particular flow can be calculated. A graphical representation of the Colebrooke Equation is the Moody Diagram:
The Moody Diagram  The Moody diagram in a printable format.
With the Moody diagram we can find the friction coefficient if we know the Reynolds Number  Re  and the Relative Roughness Ratio  k / dh In the diagram we can see how the friction coefficient depends on the Reynolds number for laminar flow  how the friction coefficient is undefined for transient flow  and how the friction coefficient depends on the roughness ratio for turbulent flow. For hydraulic smooth pipes  the roughness ratio limits zero  and the friction coefficient depends more or less on the Reynolds number only. For a fully developed turbulent flow the friction coefficient depends on the roughness ratio only. Example  Pressure Loss in Air Ducts Air at 0 C is flows in a 10 m galvanized duct  315 mm diameter  with velocity 15 m/s. Reynolds number are expressed as: Re = dh v / where Re = Reynolds number v = velocity (10)o
= density = dynamic or absolute viscosity Reynolds number calculated: Re = (15 m/s) (315 mm) (10 m/mm ) (1.23 kg/m ) / (1.79 10 Ns/m ) = 324679 (kgm/s )/N = 324679 ~ Turbulent flow Turbulent flow indicates that Colebrooks equation (9) must be used to determine the friction coefficient  . With roughness   for galvanized steel 0.15 mm, the roughness ratio can be calculated: Roughness Ratio = / dh = (0.15 mm) / (315 mm) = 4.76 104 2 3 3 5 2
Using the graphical representation of the Colebrooks equation  the Moody Diagram  the friction coefficient   can be determined to: = 0.017 The major loss for the 10 m duct can be calculated with the DarcyWeisbach Equation (3) or (6): ploss = ( l / dh ) ( v / 2 ) = 0.017 ((10 m) / (0.315 m)) ( (1.23 kg/m ) (15 m/s) / 2 ) = 74 Pa (N/m )2 3 2 2