fluid formula sheet

8/18/2019 fluid formula sheet

1/9

A special form of the Euler’s equation derived along afluid flow streamline is often called the BernoulliEquation

For steady state incompressible flow the Eulerequation becomes (1). f we integrate (1) along thestreamline it becomes (!). (!) can further be modifiedto (") by dividing by gravity.

Head of Flow

Equation (") is often referred to the head because allelements has the unit of length.

Dynamic Pressure

(!) and (") are two forms of the #ernoulli Equation for

steady state incompressible flow. f we assume thatthe gravitational body force is negligible$ (") can bewritten as (%). #oth elements in the equation have theunit of pressure and it&s common to refer the flowvelocity component as the dynamic pressure of thefluid flow (').

ince energy is conserved along the streamline$ (%)can be epressed as (*). +sing the equation we seethat increasing the velocity of the flow will reduce thepressure$ decreasing the velocity will increase thepressure.

,his phenomena can be observed ina venturi meter where the pressure is reduced in the

constriction area and regained after. t can also beobserved in a pitot tube wherethe stagnation pressure is measured. ,he stagnationpressure is where the velocity component is -ero.

Example - Bernoulli Equation andFlow from a an! through a small"rifice

iquid flows from a tan/ through a orifice close to thebottom. ,he #ernoulli equation can be adapted to astreamline from the surface (1) to the orifice (!) as

(e1)0

ince (1) and (!)&s heights from a common referenceis related as (e!)$ and the equation of continuity canbe epressed as (e")$ it&s possible to transform (e1)to (e%).

#ented tan!

A special case of interest for equation (e%) is whenthe orifice area is much lesser than the surface areaand when the pressure inside and outside the tan/ isthe same when the tan/ has an open surface or2vented2 to the atmosphere. At this situation the (e%)can be transformed to (e').

2,he velocity out from the tan/ is equal to speed of afreely body falling the distance h.2 also /nownas orricelli$s heorem%

Example - outlet velocity from a ventedtan!

,he outlet velocity of a tan/ with height 10 m can becalculated as

V 2 = (2 (9.81 m/s2 ) (10 m))1/2

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2/9

= 14 m/s

Pressuri&ed an!

f the tan/s is pressuri-ed so that product of gravityand height (g h) is much lesser than the pressuredifference divided by the density$ (e%) can betransformed to (e*).

,he velocity out from the tan/ depends mostly on thepressure difference.

Example - outlet velocity from apressuri&ed tan!

,he outlet velocity of a pressuri-ed tan/ where

p1 = 0.2 MN/m2

p2 = 0.1 MN/m2

A2 /A1 = 0.01

h = 10 m

can be calculated as

V 2 = ( (2/(1-(0.01)2 ) ((0.2 10 6 N/m2 ) - (0.1 10 6 N/m2 ))/

(1000 kg/m3 ) + (9.81 m/s2 )(10 m)))1/2

= 19.9 m/s

'oefficient of Discharge - Friction'oefficient

3ue to friction the real velocity wil l be somewhatlower than this theoretic eample. f we introducea friction coefficient c (coefficient of discharge)$ (e')can be epressed as (e'b).

,he coefficient of discharge can be determinedeperimentally. For a sharp edged opening it maybee as low as 0.6 . For smooth orifices it may beebetween 0.95 and 1.

he Energy Equation is a statement of the first law ofthermodynamics. ,he energy equation involvesenergy$ heat transfer and wor/. 4ith certainlimitations the mechanical energy equation can becompared to the #ernoulli Equation.

he (echanical Energy Equationin erms of Energy per )nit (ass

,he mechanical energy equation for a pump or afan can be written in terms of energy per unit mass0

pin / ρ + in2 / 2 + g hin + ! sh"#$ = p%&$ / ρ + %&$

2 / 2 + g

h%&$ + ! '%ss (1)

!h

p = s$"$ic pss&

ρ = *nsi$

= #'%! '%ci$

g = "cc'"$i%n %# g"i$

h = '"$i%n high$

! sh"#$ = n$ sh"#$ ng inn p &ni$ m"ss #% " p&mp,#"n % simi'"

! '%ss = '%ss *& $% #ic$i%n

,he energy equation is often used for incompressibleflow problems and is called the (echanical EnergyEquation or the Extended Bernoulli Equation.

,he mechanical energy equation for a turbine can bewritten as0

pin / ρ + in2 / 2 + g hin = p%&$ / ρ + %&$

2 / 2 + g h%&$ +

! sh"#$ + ! '%ss (2)

!h

! sh"#$ = n$ sh"#$ ng %&$ p &ni$ m"ss #% "$&in % simi'"

Equation (1) and (!) dimensions are

• ng p &ni$ m"ss (#$ 2 /s2 = #$ '/s'&g %

m2 /s2 = N m/kg)

Efficiency According to (1) a larger amount of loss ! '%ss resultin more shaft wor/ required for the same rise ofoutput energy. ,he efficiency of a pump or fanprocess can be epressed as0

= (! sh"#$ - ! '%ss ) / ! sh"#$ (3)

,he efficiency of a turbine process can beepressed as0

= ! sh"#$ / (! sh"#$ + ! '%ss ) (4)

he (echanical Energy Equationin erms of Energy per )nit#olume

,he mechanical energy equation for a pump or afan (1) can also be written in terms of energy perunit volume by multiplying (1) with fluid density ρ0

pin + ρ in2 / 2 + hin + ρ ! sh"#$ = p%&$ + ρ %&$

2 / 2 + h%&$ + ! '%ss (5)

!h

= ρ g = spci#ic !igh$

,he dimensions of equation (') are

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• ng p &ni$ %'&m (#$.'/#$ 3 = '/#$ 2 %

N.m/m3 = N/m2 )

he (echanical Energy Equationin erms of Energy per )nit*eight involves Heads

,he mechanical energy equation for a pump or afan (1) can also be written in terms of energy per

unit weight by dividing with gravity g 0

pin / + in2 / 2 g + hin + hsh"#$ = p%&$ / + %&$

2 / 2 g +h%&$ + h'%ss (6)

!h

= ρ g = spci#ic !igh$

hsh"#$ = ! sh"#$ / g = n$ sh"#$ ng h"* inn p &ni$m"ss #% " p&mp, #"n % simi'"

h'%ss = ! '%ss / g = '%ss h"* *& $% #ic$i%n

,he dimensions of equation (*) are

• ng p &ni$ !igh$ (#$.'/' = #$ % N.m/N =

m)

5ead is the energy per unit weight.

hsh"#$ can also be epressed as0

hsh"#$ = ! sh"#$ / g = sh"#$ / m g = sh"#$ / ()

!h

sh"#$ = sh"#$ p%!

m = m"ss #'%! "$

= %'&m #'%! "$

Example - Pumping *ater 4ater is pumped from an open tan/ at level -ero toan open tan/ at level 10 #$ . ,he pump

adds #%& h%sp%!s to the water when pumping 2#$ 3 /s.

ince in = %&$ = 0, pin = p%&$ = 0 and hin 6 7 equation(*) can be modified to0

hsh"#$ = h%&$ + h'%ss

or

h'%ss = hsh"#$ - h%&$ (8)

Equation (9) gives0

hsh"#$ = sh"#$ /

= (4 hp)(550 #$.'/s/hp) / (62.4 '/#$ 3 )(2 #$ 3 /s)

= 1.6 #$

• specific weight of water 62.4 '/#$ 3

• 1 hp ( ng'ish h%s p%! ) = 550 #$. '/s

:ombined with (8)0

h'%ss = (1.6 #$ ) - (10 #$)

= .6 #$

,he pump efficiency can be calculated from (")modified for head0

6 ((1.6 #$) - (.6 #$)) ; (1.6 #$)

6 0.58

,he aw of


4/9

And obvious the mass in a system increase if theinflow is higher than the outflow.

,he aw of


5/9

= #'%! '%ci$ (m/s, #$/s)

g = "cc'"$i%n %# g"i$ (m/s2 , #$/s2 )

h = '"$i%n (m, #$)

For hori-ontal steady state flow 1 = 2 and h1 = h2 $ (1) can be transformed to0

p'%ss = p1 - p2 (2)

,he pressure loss is divided in

• maor loss due to friction and

• minor loss due to change of velocity in bends$

valves and similar.

,he pressure loss in pipes and tubes depends on theflow velocity$ pipe or duct length$ pipe or ductdiameter$ and a friction factor based on the

roughness of the pipe or duct$ and whether the flowus turbulent or laminar the =eynolds >umber of theflow. ,he pressure loss in a tube or duct due tofriction$ ma?or loss$ can be epressed as0

p'%ss = (' / * h ) (ρ 2 / 2) (3)

!h

p'%ss = pss& '%ss (", N/m2 )

= #ic$i%n c%##icin$

' = 'ng$h %# *&c$ % pip (m)

* h = h*"&'ic *i"m$ (m)

(") is also called the D$.rcy-*eisbach Equation. (")is valid for fully developed$ steady$ incompressibleflow.

Head and Head ,oss,he Energy equation can be epressed in terms ofhead and head loss by dividing each term by

the specific weight of the fluid. ,he total head in afluid flow in a tube or a duct can be epressed as thesum of elevation head$ velocity head and pressurehead.

p1 / + 12 / 2 g + h1 = p2 / + 2

2 / 2 g + h2 + h'%ss (4)

!h

h'%ss = h"* '%ss (m, #$)

= ρ g = spci#ic !igh$ (N/m3, '/#$ 3 )

For hori-ontal steady state flow 1 = 2 and h1 = h2 $ (%) can be transformed to0

h'%ss = h1 - h2 (5)

!h

h = p / = h"* (m, #$)

,he head loss in a tube or duct due to friction$ ma?orloss$ can be epressed as0

h'%ss = (' / * h ) ( 2 / 2 g) (6)

!h

h'%ss = h"* '%ss (m, #$)

Friction 'oefficient - λ,he friction coefficient depends on the flow if it is

• laminar$

• transient or

• turbulent

and the roughness of the tube or duct.

,o determine the friction coefficient we first have todetermine if the flow is laminar$ transient or turbulent then use the proper formula or diagram.

Friction :oefficient for aminar Flow

For fully developed laminar flow the roughness of theduct or pipe can be neglected. ,he friction coefficient

depends only the =eynolds >umber 7 and canbe epressed as0

= 64 / 7 ()

!h

7 = $h *imnsi%n'ss 7n%'*s n&m

,he flow is

• laminar when 7 2300

• transient when 2300 7 4000

• turbulent when 7 4000

Friction :oefficient for ,ransient Flow

f the flow is transient !"77 @ =e @ %777 the flowvaries between laminar and turbulent flow and thefriction coefficient is not possible to determine.

Friction :oefficient for ,urbulent Flow

For turbulent flow the friction coefficient depends onthe =eynolds >umber and the roughness of the ductor pipe wall. n functional form this can beepressed as0

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6/9

= #( 7, k / * h ) (8)

!h

k = absolute roughness %# $& % *&c$ !"'' (mm, #$)

k / * h = the relative roughness - or roughnessratio

,he friction coefficient can be calculated bythe :olebroo/e Equation0

1 / 1/2 = -2,0 '%g 10 : (2,51 / (7 1/2 )) + (k / * h ) /

3,2 ; (9)

ince the friction coefficient is on both sides ofthe equation$ it must be solved by iteration. f we/now the =eynolds number and the roughness thefriction coefficient in the particular flow can becalculated.

A graphical representation of the :olebroo/eEquation is the (oody Diagram0

• ,he


7/9

!h $h *nsi$

ρ = m / V (3)

,he ndividual Cas :onstant 7 depends on theparticular gas and is related to the molecular weightof the gas.

Equation (1) can also be modified to

p1V 1 /? 1 = p2 V 2 /? 2 (4)

epressing the relationship between different statesfor a given mass of gas.

he 1deal 2as ,aw and the)niversal 2as 'onstant - R u

,he )niversal 2as 'onstant is independent of theparticular gas and is the same for all 2perfect2 gases.,he deal Cas aw can be epressed withthe )niversal 2as 'onstant0

p V = n 7 & ? (5)

!h

p = "s%'&$ pss& (N/m2 , '/#$ 2 )

V = %'&m (m3, #$ 3 )

n = is $h n&m %# m%'s %# g"s psn$

7 & = &nis"' g"s c%ns$"n$ (/m%'.%B, '#.#$/

('m%'.%7))

? = "s%'&$ $mp"$& ( %B, %7)

Example - he 1deal 2as ,aw A tan/ with volume of 1 ft" is filled with aircompressed to a gauge pressure of '7 psi. ,hetemperature in tan/ is 97 oF.

,he air density can be calculated with atransformation of the ideal gas law (!) to0

ρ = p / 7 ? (6)

ρ= :(50 '/in2 + 14, '/in2 )(144 in2 /#$ 2 );/:(116#$.'/s'&g.%7)((0 + 460) %7);

= 0,0102 s'&gs/#$ 3

,he weight of the air is the product of specificweight and the air volume. t can be calculated as0

! = ρ g V ()

! = (0,0102 s'&gs/#$ 3 )(32,2 #$/s2 )(1 #$ 3 )

= 0,32844 s'&gs.#$/s2

= 0,32844 '

0ote3,he deal Cas aw is accurate only at relatively lowpressures and high temperatures. ,o account for thedeviation from the ideal situation$ another factor isincluded. t is called the Cas :ompressibility Factor$or Dfactor. ,his correction factor is dependent onpressure and temperature for each gas considered.

,he ,rue Cas aw$ or the >ondeal Cas aw$

becomes0

V = C n 7 ? ()

!h

C = D"s


8/9

he 1ndividual 2as 'onstant of the2as (ixture

,he individual gas constant for the gas miture canbe epressed as0

Dm = (D1m1+ D2 m2 +..+ Dnmn ) / (m1 + m2 +..+mn ) (3)

he Density of the 2as (ixture,he density of a gas miture can be calculated as0

ρm = (ρ1 1 + ρ2 2 + .. + ρn n )/( 1 + 2 + .. + n ) (4)

!h

ρm = *nsi$ %# $h g"s miF$& (kg/m3, '/#$ 3 )

ρ1 .. ρn = *nsi$ %# "ch %# $h c%mp%nn$s (kg/m3,

'/#$ 3 )

1 + 2 + .. + n = %'&m sh" %# "ch %# $hc%mp%nn$s (m3, #$ 3 )

,he pressure indicates the normal force per unit areaat a given point acting on a given plane. ince thereis no shearing stresses present in a fluid at rest thepressure in a fluid is independent of direction.

For fluids liquids or gases at rest the pressuregradient in the vertical direction depends only on thespecific weight of the fluid.

5ow pressure changes with elevation can be

epressed as

*p = - *G (1)

!h

*p = ch"ng in pss&

*G = ch"ng in high$

= spci#ic !igh$

,he pressure gradient in vertical direction is negative the pressure decrease upwards.

4pecific *eightpecific 4eight can be epressed as0

= ρ g (2)

!h

= spci#ic !igh$

g = "cc'"$i%n %# g"i$

n general the specific weight is constant forfluids. For gases the specific weight varies withthe elevation.

4tatic Pressure in a FluidFor a incompressible fluid as a liquid the pressuredifference between two elevations can be epressedas0

p2 - p1 = - (G 2 - G 1 ) (3)

!h

p2 = pss& "$ '' 2

p1 = pss& "$ '' 1

G 2 = '' 2

G 1 = '' 1

(") can be transformed to0

p1 - p2 = (G 2 - G 1 ) (4)

%

p1 - p2 = h (5)

!h

h = G 2 - G 1 *i##nc in '"$i%n - $h *p$ *%!n#%m '%c"$i%n G 2 .

%

p1 = h + p2 (6)

Eample ressure in a Fluid

,he absoute pressure at water depth of 17 m can becalulated as0

p1 = h + p2

= (1000 kg/m3 ) (9.81 m/s2 ) (10 m) + (101.3 k")

= (98100 kg/ms2 % ") + (101300 ")

= 199.4 k"

!h

ρ = 1000 kg/m3

g = 9.81 m/s2

p2 = pss& "$ s&#"c '' = "$m%sphic pss&= 101.3 k"

,he gauge pressure can be calulated setting p2 = 0

p1 = h + p2

= (1000 kg/m3 ) (9.81 m/s2 ) (10 m)

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= 98.1 k"

he Pressure Head(*) can be transformed to0

h = (p2 - p1 ) / (6)

h epress the pressure head the height of a

column of fluid of specific weight required to givea pressure difference of ( p2 - p1 ).

Eample ressure 5ead

A pressure difference of 5 psi ('#/in2 ) is equivalent to

(5 '# /in2 ) (12 in/#$) (12 in/#$) / (62.4 '/#$ 3 )

= 11.6 #$ %# !"$

(5 '# /in2 ) (12 in/#$) (12 in/#$) / (84 '/#$ 3 )

= 0.85 #$ %# mc&

when specific weight of water is 62.4('/#$ 3 ) and specific weight of mercury is 84 ('/#$ 3 ).
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