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Centrifugal Pumps
An introduction to Centrifugal Pumps
A centrifugal pump converts the input power to kinetic energy in the liquid by accelerating the liquid by
a revolving device - an impeller. The most common type is the volute pump. Fluid enters the pumpthrough the eye of the impeller which rotates at high speed. The fluid is accelerated radially outwardfrom the pump chasing. A vacuum is created at the impellers eye that continuously draws more fluidinto the pump.
The energy created by the pump is kinetic energy accordingthe Bernoulli Equation.The energytransferred to the liquid corresponds to the velocity at the edge or vane tip of the impeller. The fasterthe impeller revolves or the bigger the impeller is the higher will the velocity of the liquid energytransferred to the liquid be. This is described by the Affinity !aws.
Pressure and Head
"f the discharge of a centrifugal pump is pointed straight up into the air the fluid will pumped to a
certain height - or head - called the shut off head. This ma#imum head is mainly determined by theoutside diameter of the pump$s impeller and the speed of the rotating shaft. The head will change asthe capacity of the pump is altered.
The kinetic energy of a liquid coming out of an impeller is obstructed by creating a resistancein theflow. The first resistance is created by the pump casing which catches the liquid and slows it down.%hen the liquid slows down the kinetic energy is converted to pressure energy.
it is the resistance to the pump$s flow that is read on a pressure gauge attached to the
discharge line
A pump does not create pressure it only creates flow. &ressure is a measurement of the resistance toflow.
"n'ewtonian fluids (non-viscous liquids like water or gasoline) the term headis used to measure thekinetic energy which a pump creates. *ead is a measurement of the height of the liquid column thepump creates from the kinetic energy the pump gives to the liquid.
the main reason for using head instead of pressure to measure a centrifugal pump$s energy is
that the pressure from a pump will change if the specific gravity (weight)of the liquid changesbut the head will not
The pump$s performance on any 'ewtonian fluidcan always be described by using the term head.
Different Types of Pump Head
Total +tatic *ead - Total head when the pump is not running
Total ,ynamic *ead (Total +ystem *ead) - Total head when the pump is running
+tatic +uction *ead - *ead on the suction side with pump off if the head is higher than the
pump impeller
+tatic +uction !ift - *ead on the suction side with pump off if the head is lower than the pump
impeller
+tatic ,ischarge *ead - *ead on discharge side of pump with the pump off
,ynamic +uction *ead!ift - *ead on suction side of pump with pump on
,ynamic ,ischarge *ead - *ead on discharge side of pump with pump on
The head is measured in either feet or meters and can beconverted to common unitsfor pressure aspsi or bar.
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it is important to understand that the pump will pump all fluids to the same height if the shaft is
turning at the same rpm
The only difference between the fluids is the amount of power it takes to get the shaft to the properrpm. The higher thespecific gravityof the fluid the more power is required.
entrifugal &umpsare /constant head machines/
'ote that the latter is not a constant pressure machine since pressure is a function of head anddensity. The head is constant even if the density (and therefore pressure) changes.
The head of a pumpin metric units can be e#pressed in metric units as0
h = (p2- p1)/( g) + v22/(2 g) (1)
where
h = total head developed (m)
p2= pressure at outlet (N/m2)
p1= pressure at inlet (N/m2)
= density (kg/m3)
g = aeleration o! gravity ("#$1) m/s2
v2= veloity at the outlet (m/s)
*ead described in simple terms
a pump$s vertical discharge /pressure-head/ is the vertical lift in height - usually measured in
feet or m of water - at which a pump can no longer e#ert enough pressure to move water. At thispoint the pump may be said to have reached its /shut-off/ head pressure. "n the flow curve chartfor a pump the /shut-off head/ is the point on the graph where the flow rate is 1ero
Bernoulli Equation
A statement of the conservation of energy in a form useful for solvingproblems involving fluids. For a non-viscous incompressible fluid in
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steady flo! the sum of pressure potential and "inetic energies perunit volume is constant at any point
A special form of the Euler2s equationderived along a fluid flow streamline is often called the BernoulliEquation
!iquid flows from a tank through a orifice close to the bottom. The Bernoulli equation can be adaptedto a streamline from the surface (3) to the orifice (4) as (e3)0
For steady state incompressible flow the Euler equation becomes (3). "f we integrate (3) along thestreamline it becomes (4). (4) can further be modified to (5) by dividing by gravity.
Head of Flo!
Equation (5) is often referred to the headbecause all elements has the unit of length.
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#ynamic Pressure
(4) and (5) are two forms of the Bernoulli Equation for steady state incompressible flow. "f we assumethat the gravitational body force is negligible (5) can be written as (6). Both elements in the equationhave the unit of pressure and it$s common to refer the flow velocity component as the dynamicpressureof the fluid flow (7).
+ince energy is conserved along the streamline (6) can be e#pressed as (8). 9sing the equation wesee that increasing the velocity of the flow will reduce the pressure decreasing the velocity willincrease the pressure.
This phenomena can be observed in a venturimeterwhere the pressure is reduced in the constrictionarea and regained after. "t can also be observed in a pitottubewhere the stagnationpressure ismeasured. The stagnation pressure is where the velocity component is 1ero.
E$ample - Bernoulli Equation and Flo! from a %an" through a small &rifice
!iquid flows from a tank through a orifice close to the bottom. The Bernoulli equation can be adaptedto a streamline from the surface (3) to the orifice (4) as (e3)0
+ince (3) and (4)$s heights from a common reference is related as (e4) and the equation of continuitycan be e#pressed as (e5) it$s possible to transform (e3) to (e6).
'ented tan"
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A special case of interest for equation (e6) is when the orifice area is much lesser than the surfacearea and when the pressure inside and outside the tank is the same - when the tank has an opensurface or /vented/ to the atmosphere. At this situation the (e6) can be transformed to (e7).
/The velocity out from the tank is equal to speed of a freely body falling the distance h./ - also knownas %orricelli(s %heorem.
Example - outlet velocity from a vented tank
The outlet velocity on a tank were
h = 1% m
can be calculated as
&2= '2 "$1 1%*1/2= 1 m/s
Pressuri)ed %an"
"f the tanks is pressuri1ed so that product of gravity and height (g h) is much lesser than the pressuredifference divided by the density (e6) can be transformed to (e8).
The velocity out from the tank depends mostly on the pressure difference.
Example - outlet velocity from a pressurized tank
The outlet velocity of a pressuri1ed tank where
p1= %2 ,N/m2# p2= %1 ,N/m
22/1= %%1# h = 1% m
can be calculated as
&2= '(2/(1-(%%1)
2
) ( (%2 - %1)1%
.
/11%
3
+ "$1 1%)*
1/2
= 1"" m/s
Coefficient of #ischarge - Friction Coefficient
,ue to friction the real velocity will be somewhat lower than this theoretic e#amples. "f we introduce afriction coefficient(coefficient of discharge) (e7) can be e#pressed as (e7b).
The coefficient of discharge can be determined e#perimentally. For a sharp edged opening it may beas low as :.8. For smooth orifices it may bee between :.;7 and 3.
Hydraulic Pump Po!er
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The ideal hydraulic power to drive a pump depends on the mass flow rate the liquid density and thedifferential height
- either it is the static lift from one height to an other or the friction head loss component of the system- can be calculated as
h= 0 g h / 3. 1%
.
(1)
where
h= power (k)
0 = !low apaity (m3/h)
= density o! !luid (kg/m3)
g = gravity ("$1 m/s2)
h = di!!erential head (m)
*haft Pump Po!er
The shaft power - the power required transferred from the motor to the shaft of the pump - depends onthe efficiency of the pump and can be calculated as
s= h/ (2)
where
s= sha!t power (k)
= pump e!!iieny
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&nline Pump Calculator - *+-units
The calculator below can used to calculate the hydraulic and shaft power of a pump0
3570 - !low apaity (m3/h)
3::: - density o! !luid (kg/m3)
;.
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w = (p2- p1) / (1)
where
w= spei!i work (Nm/kg = 4/kg = m2/s2)
p=pressure(N/m2)
=density (kg/m3)
*pecific ,or" of a %urbine
+pecific work of a turbine with an incompressible fluid can be e#pressed as0
w = (p1- p2) / (2)
*pecific ,or" of a Compressor
A compressor works with compressible fluids and the specific work for an isentropic compressorprocess can be e#pressed with the help of0
p1v15= p2v2
5(3)
where
v= volume (m3)
5=p/ v-ratio o! spei!i heats(4/kg6)
+pecific work0
w=5 / (5 -1) 7 81'( p2/ p1)((5-1)/5)- 1*()
where
7= individual gas onstant(4/kg6)
8= asolute temperature (6)
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*pecific ,or" of a as %urbine
A gas turbine e#pands a compressible fluid and the specific work can be e#pressed as0
w=5 / (5 -1) 7 81'1 - ( p2/ p1)((5-1)/5)*(9)
Head in %urbomachines
The specific work can on basis ofthe energy equationbe e#pressed withthe head as0
w = g h(.)
where
h= head (m)
g= aeleration o! gravity(m/s2)
Transformed to e#press head0
h = w / g(:)
E$ample - *pecific ,or" of a ,ater Pump
A water pump works between 3 bar (3 3:7'm4) and 3: bar (3: 3:7'm4). The specific work can becalculated with (3)0
w = (p2- p1) / =( (1% 1%9N/m2) - (1 1%9N/m2) ) / (1%%% kg/m3)
= "%% Nm/kg
,ividing by acceleration of gravity the headcan be calculated using (=)0
hwater= ("%% Nm/kg) / ("#$1 kg/s2)
= "1#: (m) water olumn
E$ample - *pecific ,or" of an Air Compressor
An air compressor works with air at 4: o compressing the air from 3 bar absolute (3 3:7'm4) to 3:bar (3: 3:7'm4). The specific work can be e#pressed with (6)0
w=5 / (5 -1) 7 81'( p2/ p1)((5-1)/5)- 1*
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= ( (1# 4/kg6)/ (1# - 1 4/kg6) ) (2$.#" 4/kg 6) (2:3 + 2% 6) '( (1% 1%9N/m2) / (1 1%9N/m2) )((1# - 14/kg6)/(1# 4/kg6))- 1 *
= 2:2%% Nm/kg
where
5air= 1# (4/kg6) - ratio o! spei!i heat air
7air= 2$." (4/kg 6) - individual gas onstant air
,ividing by acceleration of gravity the headcan be calculated using (=)0
hair= (2:2%% Nm/kg) / ("#$1 kg/s2)
= 2:"91 (m) air olumn
+pecific speed is a number characteri1ing the type of impeller in a unique and coherent manner.+pecific speed are determined independent of pump si1e and can be useful comparing different pumpdesigns. The specific speed identifies the geometrically similarity of pumps.
+pecific speed is dimensionless and are e#pressed as
Ns= ; 01/2/ h3/(1)
where
Ns= spei!i speed
; = pump sha!t rotational speed (rpm)
0 = !low rate (m3/h# l/s# m3/min# ritish gpm) at >est ?!!iieny oint (>?)
h = head rise (m# !t)
&nline *pecific *peed Pump Calculator
The calculator below can used to calculate pump specific speed0
3=8:; - pump sha!t rotational speed (rpm)
37::0 - !low rate (m3/h# l/s# m3/min# ritish gpm)
3::h =head rise (m# !t)
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ote/%hen comparing pumps and their documentation be aware of the units used.
Typical values for specific speed - Ns- for different designs in 9+ units (gpm)
radial flow - 7:: @ Ns@ 6::: - typical for centrifugal impeller
pumps with radial vanes - double and single suction. Francisvane impellers in the upper range
mi#ed flow - 4::: @ Ns@
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The hydraulic efficiency can be e#pressed as0
h= w / (w + wl) (1)
where
h=hydrauli e!!iieny
w = spei!i work !rom the pump or !an
wl= spei!i work lost due to hydrauli e!!ets
3echanical 6oss and 3echanical Efficiency
echanical components - as transmission gear and bearings - generates a mechanical loss thatreduces the power transferred from the motor shaft to the pump or fan impeller.
The mechanical efficiency can be e#pressed as0
m= ( - l) / (2)
where
m= mehanial e!!iieny
= power trans!erred !rom the motor to the sha!t
l= power lost in the transmission
'olumetric 6oss and 'olumetric Efficiency
,ue to leakage of fluid between the back surface of the impeller hub plate and the casing or throughother pump components - there is a volumetric lossreducing the pump efficiency.
The volumetric efficiency can be e#pressed as0
v= 0 / (0 + 0l) (3)
where
v= volumetri e!!iieny
0 = volume !low out o! the pump or !an
0l= leakage volume !low
%otal 6oss and &verall Efficiency
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The overall efficiency is the ratio of power actually gained by the fluid to the shaft power supplied. Theoverall efficiency can be e#pressed as0
= hmv()
where
= overall e!!iieny
The losses in the pump or fan converts to heat transferred to the fluid and the surroundings. As a ruleof thumb the temperature increase in a fan transporting air is appro#imately 3 o.
E$ample - Hydraulic Efficiency for a Pump
An inline water pump works between pressure 3 bar (3 3:7'm4) and 3: bar (3: 3:7'm4).,ensity ofwater is 3::: kgm5. The hydraulic efficiency h> :;3.
The actual water head(water column) can be calculated as0
h = (p2- p1) / @
= (p2- p1) / g
= ((1% 1%9N/m2) - (1 1%9N/m2)) / (1#%%% kg/m3) ("$1 m/s2)
= "1: m - water olumn
The pump must be constructed for the specific work0
w= g h / h
= ("$1 m/s2) ("1: m) / %"1
= "$$. m2/s2
The construction or designheadis0
h = w/ g
= ("$$. m2/s2) / ("$1 m/s2)
= 1%%$ m - water olumn
Po!er ained by Fluid
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The power gained by the fluid from a pump or fan can be e#pressed as0
= m w (1)
where
= power
m = mass !low rate
w = spei!i work
*pecific ,or"
+pecific work - w- can be e#pressed0
w = g h (2)
where
h = head
g = aeleration o! gravity
3ass Flo! 7ate
ass flow rate - m -can be e#pressed0
m = A (3)
where
=density
A = volume !low rate
ombining (3) (4) and (5) the power gained by the fluid from a pump or fan can be e#pressed as0
= A g h ()
%ith specific weight e#pressed as0
@ = g (9)
where
@ =spei!i weight
equation (6) can be modified so the power gained by the fluid from a pump or fan can be e#pressedas0
= @ A h (.)
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+inceheadcan be e#pressed as
h = (p2- p1) / @ (:)
equation (6) can be modified so the power gained by the fluid from a pump or fan can be e#pressedas0
= A (p2- p1) ($)
Example - Head Rise of a Inline Pump
An inline water pump works between measured pressure 3 bar (3 3:7'm4) and 3: bar (3: 3:7'm4).,ensity of wateris 3::: kgm5. The volume flow is measured to 3 3:-5m5s.
The actual water head (water column) can be calculated using (=)0
h = (p2- p1) / @
= (p2- p1) / g
= ((1% 1%9N/m2) - (1 1%9N/m2)) / (1#%%% kg/m3) ("$1 m/s2)
= "1: m - water olumn
The power gained by the fluid can be calculated using equation (6)0
= A g h
= (1#%%% kg/m3) (1 1%-3m3/s) ("$1 m/s2) ("1: m)
= $"". kgm2/s3()
= %" k
%hen aviscous fluidis handled by a centrifugal pump
brake horsepower requirement increases
the head generated is reduced
capacity is reduced
efficiency of pump is reduced and theBest Efficiency &oint-
BE&- is moved
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The head flow and capacity at other viscosities than used in the original documentation can bemodifying with coefficients.
Flo!
0v= 00 (1)
where
0v= !low ompensated !or visosity (m3/h# gpm)
0= visosity !low oe!!iient
0 = original !low aording pump urve (m3
/h# gpm)
Head
hv= hh (1)
where
hv= head ompensated !or visosity (m# !t)
h= visosity head oe!!iient
h = original head aording pump urve (m# !t)
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Efficiency
Bv= BB (1)
where
Bv= e!!ieny ompensated !or visosity
B= visosity e!!iieny oe!!iient
B = original e!!iieny aording pump urve
Po!er - *+ units
v= 0vhvvg / (3. 1%. Bv) (1)
where
v= power ompensated !or visosity (k)
v= density o! visous !luid (kg/m3)
g = aeleration o! gravity ("$1 m/s2)
Po!er - +mperial units
v= 0vhvC / (3".% Bv) (1)
where
v= power ompensated !or visosity (hp)
C =spei!i gravityo! visous !luid
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%he *ystem Curve
A fluid flow system can in general be characteri1ed with the *ystem Curve- a graphical presentationofthe Energy Equation.
The system head visuali1ed in the +ystem urve is a function of the elevation - the static head in thesystem and the maor and minor lossesand can be e#pressed as0
h = dh + hl(1)
where
h = system head
dh = h2- h1= elevation (stati) head - di!!erene etween inlet and outlet o! the system
hl=head loss
A generic e#pression of head loss is0
hl= k 02(2)
where
0 = !low rate
k =onstant desriing the total system harateristis- inluding all maDor and minor losses
"ncreasing the constant - k- by closing some valves reducing the pipe si1e or similar - will increase
the head loss and move the system curve upwards. The starting point for the curve - at no flow will bethe same.
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Pump Performance Curve
The pump characteristic is normally described graphically by the manufacturer as a pump performancecurve. The pump curve describes the relation between flowrate and head for the actual pump. Ctherimportant information for proper pump selection is also included - efficiency curves '&+*rcurvepump curves for several impeller diameters and different speedsand power consumption.
"ncreasing the impeller diameter or speed increases the head and flow rate capacity - and the pumpcurve moves upwards.
The head capacity can be increased by connecting two or more pumps in series or the flow ratecapacity can be increased by connecting two or morepumps in parallel.
*election of Pump
A pump can be selected by combining the+ystem urve and the &ump urve0
The operating point is where the system curve and the actual pump curve intersect.
Best Efficiency Point - BEP
The best operating conditions will in general be close to the best efficiencypoint - BE&.
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+pecial consideration should be taken for applications where the system conditions change frequentlyduring operation. This is often the situation for heating and air conditioning system or water supplysystems with variable consumption and modulating valves.
Carry &ut
%hen a pumps operates in the far right of its curve with poor efficiency - the pumps carry out.
*hutoff Head
+hutoff head is the head produced when the pump operates with fluid but with no flow rate.
Churn
A pump is in churn when it operates at shutoff head or no flow.
For a fluid flow process involving a pump or fan the overall efficiencyis related to the
hydraulic
mechanical
volumetric
loss in the pump or fan.
Hydraulic 6oss and Hydraulic Efficiency
*ydraulic loss relates to the construction of the pump or fan and is caused by the friction between thefluid and the walls acceleration and retardation of the fluid and the change of the fluid flow direction.
The hydraulic efficiency can be e#pressed as0
h= w / (w + wl) (1)
where
h=hydrauli e!!iieny
w = spei!i work !rom the pump or !an
wl= spei!i work lost due to hydrauli e!!ets
3echanical 6oss and 3echanical Efficiency
echanical components - as transmission gear and bearings - generates a mechanical loss thatreduces the power transferred from the motor shaft to the pump or fan impeller.
The mechanical efficiency can be e#pressed as0
m= ( - l) / (2)
where
m= mehanial e!!iieny
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= power trans!erred !rom the motor to the sha!t
l= power lost in the transmission
'olumetric 6oss and 'olumetric Efficiency
,ue to leakage of fluid between the back surface of the impeller hub plate and the casing or throughother pump components - there is a volumetric lossreducing the pump efficiency.
The volumetric efficiency can be e#pressed as0
v= 0 / (0 + 0l) (3)
where
v= volumetri e!!iieny
0 = volume !low out o! the pump or !an
0l= leakage volume !low
%otal 6oss and &verall Efficiency
The overall efficiency is the ratio of power actually gained by the fluid to the shaft power supplied. Theoverall efficiency can be e#pressed as0
= hmv()
where
= overall e!!iieny
The losses in the pump or fan converts to heat transferred to the fluid and the surroundings. As a ruleof thumb the temperature increase in a fan transporting air is appro#imately 3 o.
E$ample - Hydraulic Efficiency for a Pump
An inline water pump works between pressure 3 bar (3 3:7'm4) and 3: bar (3: 3:7'm4).,ensity ofwater is 3::: kgm5. The hydraulic efficiency h> :;3.
The actual water head(water column) can be calculated as0
h = (p2- p1) / @
= (p2- p1) / g
= ((1% 1%9N/m2) - (1 1%9N/m2)) / (1#%%% kg/m3) ("$1 m/s2)
= "1: m - water olumn
The pump must be constructed for the specific work0
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w= g h / h
= ("$1 m/s2) ("1: m) / %"1
= "$$. m2/s2
The construction or designheadis0
h = w/ g
= ("$$. m2/s2) / ("$1 m/s2)
= 1%%$ m - water olumn
Pressure and Pressure 6oss
According the Energy Equation for a fluidthe total energy can be summari1ed as elevation energyvelocity energy and pressure energy. The Energy Equation can then be e#pressed as0
p1+ v12/ 2 + g h1= p2+ v2
2/ 2 + g h2+ ploss(1)
where
p =pressure in !luid (a (N/m2)# psi (l/!t2))
ploss= pressure loss (a (N/m2)# psi (l/!t2))
=density o! the !luid(kg/m3# slugs/!t3)
v = !low veloity (m/s# !t/s)
g = aeleration o! gravity(m/s2# !t/s2)
h = elevation (m# !t)
For hori1ontal steady state flow v1= v2and h1= h2 - (3) can be transformed to0
ploss= p1- p2(2)
The pressure loss is divided in
ma8or lossdue to friction and
minor lossdue 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 ductdiameter and a friction factor based on the roughness of the pipe or duct and whether the flow usturbulent or laminar -the Deynolds 'umber of the flow. The pressure loss in a tube or duct due tofriction maor loss can be e#pressed as0
ploss= E (l / dh) ( v2/ 2) (3)
where
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ploss= pressure loss (a# N/m2)
E = !rition oe!!iient
l = length o! dut or pipe (m)
dh=hydrauli diameter(m)
(5) is also called the #arcy-,eisbach Equation. (5) is valid for fully developed steadyincompressible flow.
Head and Head 6oss
The Energy equation can be e#pressed in terms of head and head loss by dividing each term by thespecific weightof the fluid. The total head in a fluid flow in a tube or a duct can be e#pressed as thesum of elevation headvelocity headand pressure head.
p1/ @ + v12/ 2 g + h1= p2/ @ + v2
2/ 2 g + h2+ hloss()
where
hloss= head loss (m# !t)
@ = g = spei!i weight(N/m3# l/!t3)
For hori1ontal steady state flow v1= v2and h1= h2 - (6) can be transformed to0
hloss= h1- h2(9)
where
h = p / @ = head (m# !t)
The head loss in a tube or duct due to friction maor loss can be e#pressed as0
hloss= E (l / dh) (v2/ 2 g) (.)
where
hloss= head loss (m# !t)
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%he Friction Coefficient -
The friction coefficient depends on the flow - if it is
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 orturbulent - then use the proper formula or diagram.
The riction !oefficient for "aminar lo#
For fully developed laminar flow the roughness of the duct or pipe can be neglected. The frictioncoefficient depends only the Deynolds 'umber - 7e- and can be e#pressed as0
E= . / 7e (:)
where
7e = the dimensionless 7eynolds numer
The flow is
laminar when De @ 45::
transient when 45:: @ De @ 6:::
turbulent when De 6:::
The riction !oefficient for Transient lo#
"f the flow is transient - 45:: @ De @ 6::: - the flow varies between laminar and turbulent flow and thefriction coefficient is not possible to determine.
The riction !oefficient for Tur$ulent lo#
For turbulent flow the friction coefficient depends on the Deynolds 'umber and the roughness of theduct or pipe wall. Cn functional form this can be e#pressed as0
E = !( 7e# k / dh) ($)
where
k= relative rou%hnesso! tue or dut wall (mm# !t)
k / dh= the rou%hness ratio
Delative roughness for materials are determined by e#periments. Delative roughness for somecommon materials can be found in the table below
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+urface
Doughness -k
# 3:-5 m feet
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opper !ead Brass Aluminum (new) :.::3 - :.::4 5.55 - 8.= 3:-8
& and &lastic &ipes :.::37 - :.::= :.7 - 4.55 3:-7
Epo#y inyl Ester and "sophthalic pipe :.::7 3.= 3:-7
+tainless steel :.:37 7 3:-7
+teel commercial pipe :.:67 - :.:; 3.7 - 5 3:-6
+tretched steel :.:37 7 3:-7
%eld steel :.:67 3.7 3:-6
Galvani1ed steel :.37 7 3:-6
Dusted steel (corrosion) :.37 - 6 7 - 355 3:-6
'ew cast iron :.47 - :.< < - 4= 3:-6
%orn cast iron :.< - 3.7 4.= - 7 3:-5
Dusty cast iron 3.7 - 4.7 7 -
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%ell planed wood :.3< - :.; 8 - 5: 3:-6
Crdinary wood 7 38.= 3:-5
The friction coefficient -E- can be calculated by the olebrooke Equation0
1 / E1/2= -2#% log1%' (2#91 / (7e E1/2)) + (k / dh) / 3#:2 * (")
+ince the friction coefficient -E- is on both sides of the equation it must be solved by iteration. "f weknow the Deynolds number and the roughness - the friction coefficient -E- in the particular flow can becalculated.
A graphical representation of the olebrooke Equation is the 3oody #iagram0
The oody ,iagram- The oody diagram in a printable format.
%ith the oody diagram we can find the friction coefficient if we know the 7eynolds umber- 7e-and the
7oughness 7atio- k / dh.
"n the diagram we can see how the friction coefficient depends on the Deynolds number for laminarflow - how the friction coefficient is undefined for transient flow - and how the friction coefficientdepends on the roughness ratio for turbulent flow.
For hydraulic smooth pipes - the roughness ratio limits 1ero - and the friction coefficient depends moreor less on the Deynolds number only.
For a fully developed turbulent flow the friction coefficient depends on the roughness ratio only.
E$ample - Pressure 6oss in Air #ucts
Air at : o is flows in a 3: m galvani1ed duct - 537 mm diameter - with velocity 37 ms.
Deynolds number are e#pressed as0
7e = dhv / B (1%)
where
7e = 7eynolds numer
v = veloity
=density
B = dynami or asolute visosity
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Deynolds number calculated0
7e = ( 123 kg/m319 m/s 319 mm 1%-3m/mm ) / 1:" 1%-9Ns/m2
= 32.:" (kgm/s2)/N
= 32.:" F 8urulent !low
Turbulent flow indicates that olebrooks equation (;) must be used to determine the friction coefficient-E-.
%ith roughness - G-for galvani1ed steel :37 mm the roughness ratio can be calculated0
7oughness 7atio = G / dh
= %19 mm / 319 mm
= :. 1%-
9sing the graphical representation of the olebrooks equation - the oody ,iagram - the frictioncoefficient -E- can be determined to0
E> :.:37
The maor loss for the 3: m duct can be calculated with the ,arcy-%eisbach Equation (5) or (8)0
hloss= E ( l / dh) ( v2/ 2 )
= %#%19 (1% m / %#319 m) ( 1#23 kg/m3(19 m/s)2/ 2 )
= .9 a (N/m2)
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