Dr. MohsinSiddique Assistant Professor · 2018. 7. 11. · to mean velocity of flow (V) as long as the flow in laminar. i.e., But with increasing velocity, as the flow become turbulent

Dynamics of Viscous Fluid Flow in

Closed Pipe: Darcy-Weisbach equation for flow in pipes. Major and minor losses in pipe

lines.

Dr. Mohsin Siddique

Assistant Professor

1

FLUID MECHANICS

Steady Flow Through Pipes

2

� Laminar Flow:

flow in layers

Re4000 (pipe flow)


3

� Reynold’s Number(R or Re): It is ratio of inertial forces (Fi) to viscous forces (Fv) of flowing fluid

� For laminar flow: Re


4

� Hydraulic Radius (Rh) or Hydraulic Diameter: It is the ratio of area of flow to wetted perimeter of a channel or pipe

P

A

perimeterwetted

AreaRh ==

For Circular Pipe

( )( )

h

h

RD

D

D

D

P

AR

4

4

4/ 2

=

===π

π

For Rectangular pipe

DB

BD

P

ARh

2+==

B

D

Note: hydraulic Radius gives us indication for most economical section. More the Rh more economical will be the section.

ννh

h

VRVDR

4==

By replacing D with Rh, Reynolds’ number formulae can be used for non-circular sections as well.

Head Loss in Pipes

5

� Total Head Loss=Major Losses+ Minor Losses

� Major Loss: Due to pipe friction

� Minor Loss: Due to pipe fittings, bents and valves etc

Head Loss in Pipes due to Friction

6

The head loss due to friction in a given length of pipe is proportional to mean velocity of flow (V) as long as the flow in laminar. i.e.,

But with increasing velocity, as the flow become turbulent the head loss also varies and become proportion to Vn

Where n ranges from 1.75 to 2

Log-log plot for flow in uniform pipe (n=2.0 for rough wall pipe; n=1.75 for smooth wall pipe

VH f ∝

n

f VH ∝

Frictional Head Loss in Conduits of Constant

Cross-Section

7

� Consider stead flow in a conduit of uniform cross-section A. The pressure at section 1 & 2 are P1 & P2 respectively. The distance between the section is L. For equilibrium in stead flow,

∑ == 0maF

Figure: Schematic diagram of conduit

0sin 21 =−−− APPLWAP oτα

P= perimeter of conduit= Avg. shear stress

between pipe boundary and liquid

oτ

01221 =−

−−− PL

L

zzALAPAP oτγ

αsin12 =

−

L

zz

Frictional Head Loss in Conduits of Constant

cross-section

8

01221 =−

−−− PL

L

zzALAPAP oτγ

Dividing the equation by Aγ

( ) 01221 =−−−−A

PLzz

PP o

γ

τ

γγ

fLo hhA

PLPz

Pz ===

+−

+

γ

τ

γγ2

21

1

Therefore, head loss due to friction hf can be written as

h

oof

R

L

A

PLh

γ

τ

γ

τ==

Lhg

vz

P

g

vz

P+++=++

22

2

22

2

2

11

1

γγ

Remember !! For pipe flow

For stead flow in pipe of uniform diameter v1=v2

LhzP

zP

=

+−

+ 2

21

1

γγ

This is general equation and can be applied to any shape conduit having either Laminar or turbulent flow.

P

ARh =Q

Determining Shear Stress

9

� For smooth-walled pipes/conduits, the average shear stress at the wall is

� Using Rayleigh's Theorem of dimensional analysis, the above relation can be written as;

� Rewriting above equation in terms of dimension (FLT), we get

( ),,,, VRf ho ρµτ =

( )

=

ncb

a

T

L

L

FT

L

FTLK

L

F4

2

22

( )ncbaho VRk ... ρµτ =

LlengthR

L

Fareaforce

h

o

==

==2

/τ

( )22

4233

/.

////

/

−==

===

=

FTLmsN

LFTLaFLM

TLV

µ

ρ( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− =


10

� According to dimensional homogeneity, the dimension must be equal on either side of the equation, i.e.,

( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− =

)(20:

)(422:

)(1:

iiincbT

iincbaL

icbF

→−+=

→+−−=−

→+=Solving three equations, we get

1;2;2 −=−=−= ncnbna

� Substituting values back in above equation

( ) ( ) 22

122 ...... VVR

kVRkVRk

n

hnnnn

h

ncba

ho ρµ

ρρµρµτ

−

−−−

===

( ) 22 VRk neo ρτ−

=

� Setting = we get

2

2V

C fo ρτ = Where, Cf is coefficient of friction

( ) 2−neRk 2/fC


11

� Now substituting the equation of avg. shear stress in equation of head loss,

� For circular pipe flows, Rh=D/4

� Where, f is a friction factor. i.e.,

� The above equation is known as pipe friction equation and as Darcy-Weisbach equation.

� It is used for calculation of pipe-friction head loss for circular pipes flowing full (laminar or turbulent)

2/2VC fo ρτ =

h

of

R

Lh

γ

τ=

h

f

h

f

fgR

LVC

R

LVCh

22

22

==γ

ρ

g

V

D

Lf

g

V

D

LC

Dg

LVCh f

f

f22

442

4 222

===

( )Re4 fCf f ==

Friction Factor for Laminar and

Turbulent Flows in Circular Pipes

12

� Smooth and Rough Pipe

� Mathematically;

� Smooth Pipe

� Rough Pipe

� Transitional mode

Turbulent flow near boundary

=

=

v

e

δ

Roughness height

Thickness of viscous sub-layer

Smooth pipe

Rough pipe

f

D

fVv

Re

14.1414.14==

νδ

ve δ<

ve δ>

ve δ<

ve δ14>

vv e δδ 14≤≤

Friction Factor for Laminar and Turbulent Flows in

Circular Pipes

13

� For laminar flow

� For turbulent flow

Re

64=f2000Re <

51.2

Relog2

1 f

f=

9.6

Relog8.1

1=

f

25.0Re

316.0=f

7/1

max

=

or

y

u

u

Def /

7.3log2

1=

+−=

f

De

f Re

51.2

7.3

/log2

1

+

−=

Re

9.6

7.3

/log8.1

111.1

De

f

Colebrook Eq. for turbulent flow in all pipes

Halaand Eq. For turbulent flow in all pipes

Von-karman Eq. for fully rough flow

Blacius Eq. for smooth pipe flow

Seventh-root law

510Re3000 ≤≤

From Nikuradse experiments

Colebrook Eq. for smooth pipe flow

for smooth pipe flow

4000Re >

Friction Factor for Laminar and Turbulent Flows

in Circular Pipes

14

� The Moody chart or Moody diagram is a graph in non-dimensional form that relates the Darcy-Weisbach friction factor, Reynoldsnumber and relative roughness for fully developed flow in a circular pipe.

� The Moody chart is universally valid for all steady, fully developed,incompressible pipe flows.

Friction Factor for Laminar and Turbulent Flows

in Circular Pipes

15

� For laminar flow For non-laminar flow

eRf

64=

+−=

f

De

f Re

51.2

7.3

/log2

1 Colebrook eq.

Friction Factor for Laminar and

Turbulent Flows in Circular Pipes

16

� The friction factor can be determined by its Reynolds number (Re) and the Relative roughness (e/D) of the Pipe.( where: e = absolute roughness and D = diameter of pipe)

Problem Types

18

� Type 1: Determine f and hf,

� Type 2: Determine Q

� Type 3: Determine D

Problem

19

� Find friction factor for the following pipe

� e=0.002 ft

� D=1ft

� Kinematic Viscosity, ν=14.1x10-6ft2/s

� Velocity of flow, V=0.141ft/s

� Solution:

� e/D=0.002/1=0.002

� R=VD/ ν =1x0.141/(14.1x10-6)=10000

� From Moody’s Diagram; f=0.034___________

Re

51.2

7.3

/log2

1

=

+−=

f

f

De

f

Problem-Type 1

20

� Pipe dia= 3 inch & L=100m

� Re=50,000 ʋ=1.059x10-5ft2/s

� (a): Laminar flow:

� f=64/Re=64/50,000=0.00128

ftgD

fLVH Lf 0357.0

)12/3)(2.32(2

)12.2)(100(00128.0

2

22

===

sftVVVD

/12.210059.1

)12/3(50000Re

5=⇒

×=⇒=

−ν

Problem-Type 1

21


� Re=50,000 ʋ=1.059x10-5ft2/s

� (b): Turbulent flow in smooth pipe: i.e.: e=0

0209.0

50000

51.2

7.3

0log2

Re

51.2

7.3

/log2

1

=

+−=

+−=

f

ff

De

f

ftgD

fLVH Lf 582.0

)12/3)(2.32(2

)12.2)(100(0209.0

2

22

===

Problem-Type 1

22


� Re=50,000 ʋ=1.059x10-5ft2/s

� (c): Turbulent flow in rough pipe: i.e.: e/D=0.05

0720.0

50000

51.2

7.3

05.0log2

Re

51.2

7.3

/log2

1

=

+−=

+−=

f

ff

De

f

ftgD

fLVH Lf 01.2

)12/3)(2.32(2

)12.2)(100(0720.0

2

22

===

Problem-Type 1

23

hL=?

memDmL 0005.0;25.0;1000 ===

smsmQ /10306.1;/051.0 263 −×== ν

( ) ( )002.025.0/0005.0/

10210306.1/25.0039.1/ 56

==

×=××== −

De

VDR ν

0.0245f

Diagram sMoody' From

=

mgD

fLVhL 39.5

2

2

==

smAQV /039.1/ ==Q

Problem-Type 2

24

gDfLVhL 2/2=

Problem-Type 2

25

� For laminar flow For non-laminar flow

eRf

64=

+−=

f

De

f Re

51.2

7.3

/log2

1 Colebrook eq.

Problem-Type 3

26

Problem

27

gD

flVh

f

De

f

Lf2

Re

51.2

7.3

/log2

1

2

=

+−=

Problem

28

Problem

29

MINOR LOSSES

30

� Each type of loss can be quantified using a loss coefficient (K). Losses are proportional to velocity of flow and geometry of device.

� Where, Hm is minor loss and K is minor loss coefficient. The value of K is typically provided for various types/devices

� NOTE: If L > 1000D minor losses become significantly less than that of major losses and hence can be neglected.

g

VKHm

2

2

=

Minor Losses

31

� These can be categorized as

� 1. Head loss due to contraction in pipe

� 1.1 Sudden Contraction

� 1.2 Gradual Contraction

� 2. Entrance loss

� 3. Head loss due to enlargement of pipe

� 3.1 Sudden Enlargement

� 3.2 Gradual Enlargement

� 4. Exit loss

� 5. Head loss due to pipe fittings

� 6. Head loss due to bends and elbows

Minor Losses

32

� Head loss due to contraction of pipe (Sudden contraction)

� A sudden contraction (Figure) in pipe usually causes a marked drop in pressure in the pipe because of both the increase in velocity and the loss of energy of turbulence.

g

VKH cm

2

2

2=

Head loss due to sudden contraction is

Where, kc is sudden contraction coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe

Minor Losses

33

� Head loss due to enlargement of pipe (Gradual Contraction)

� Head loss from pipe contraction may be greatly reduced by introducing a gradual pipe transition known as a confusor as shown Figure.

g

VKH cm

2'

2

2=

Head loss due to gradual contraction is

Where, kc’ is gradual contraction

coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe

Minor Losses

34

� Entrance loss

� The general equation for an entrance head loss is also expressed in terms of velocity head of the pipe:

� The approximate values for the entrance loss coefficient (Ke) for different entrance conditions are given below

g

VKH em

2

2

=

Minor Losses

35

� head loss due to enlargement of pipe (Sudden Enlargement)

� The behavior of the energy grade line and the hydraulic grade line in the vicinity of a sudden pipe expansion is shown in Figure

The magnitude of the head loss may be expressed as

( )g

VVHm

2

2

21 −=

Minor Losses

36

� head loss due to enlargement of pipe (Gradual Enlargement)

� The head loss resulting from pipe expansions may be greatly reduced by introducing a gradual pipe transition known as a diffusor

The magnitude of the head loss may be expressed as

( )g

VVKH em

2

2

21 −=

The values of Ke’ vary with the diffuser angle (α).

Minor Losses

37

� Exit Loss

� A submerged pipe discharging into a large reservoir (Figure ) is a special case of head loss from expansion.

( )g

VKH dm

2

2

=

Exit (discharge) head loss is expressed as

where the exit (discharge) loss coefficient Kd=1.0.

Minor Losses

38

� Head loss due to fittings valves

� Fittings are installed in pipelines to control flow. As with other losses in pipes, the head loss through fittings may also be expressed in terms of velocity head in the pipe:

g

VKH fm

2

2

=

Minor Losses

39

� Head loss due to bends

� The head loss produced at a bend was found to be dependent of the ratio the radius of curvature of the bend (R) to the diameter of the pipe (D). The loss of head due to a bend may be expressed in terms of the velocity head as

� For smooth pipe bend of 900, the values of Kb for various values of R/D are listed in following table.

g

VKH bm

2

2

=

Minor Losses

40

Numerical Problems

41

Numerical Problems

42

Thank you

� Questions….

� Feel free to contact:

43

Fluid Dynamics:

(ii) Hydrodynamics: Different forms of energy in a flowing

liquid, head, Bernoulli's equation and its application, Energy

line and Hydraulic Gradient Line, and Energy Equation

Dr. Mohsin Siddique

Assistant Professor

1

Fluid Mechanics

Forms of Energy

2

� (1). Kinetic Energy: Energy due to motion of body. A body of mass, m, when moving with velocity, V, posses kinetic energy,

� (2). Potential Energy: Energy due to elevation of body above an arbitrary datum

� (3). Pressure Energy: Energy due to pressure above datum, most usually its pressure above atmospheric

2

2

1mVKE =

mgZPE =

m andV are mass and velocity of body

Z is elevation of body from arbitrary datumm is the mass of body

hγ=PrE !!!

Forms of Energy

3

� (4). Internal Energy: It is the energy that is associated with the molecular, or internal state of matter; it may be stored in many forms, including thermal, nuclear, chemical and electrostatic.

HEAD

4

� Head: Energy per unit weight is called head

� Kinetic head: Kinetic energy per unit weight

� Potential head: Potential energy per unit weigh

� Pressure head: Pressure energy per unit weight

g

VmgmV

Weight

KE

2/

2

1head Kinetic

22 =

== mgWeight =Q

( ) ZmgmgZWeight

PE=== /head Potential

γ

P

Weight==

PrEhead Pressure

TOTAL HEAD

5

� TOTAL HEAD

= Kinetic Head + Potential Head + Pressure Head

g

VPZ

2HHead Total

2

++==γ

g

V

2

2

γ

PZ

Bernoulli’s Equation

6

� It states that the sum of kinetic, potential and pressure heads of a fluid particle is constant along a streamline during steady flow when compressibility and frictional effects are negligible.

� i.e. , For an ideal fluid, Total head of fluid particle remains constant during a steady-incompressible flow.

� Or total head along a streamline is constant during steady flow when compressibility and frictional effects are negligible.

21

22

22

12

11

2

22

2Head Total

HH

g

VPZ

g

VPZ

consttg

VPZ

=

++=++

=++=

γγ

γ

1

2

Pipe

Derivation of Bernoulli’s Equation

7

� Consider motion of flow fluid particle in steady flow field as shown in fig.

� Applying Newton’s 2nd Law in s-direction on a particle moving along a streamline give

� Where F is resultant force in s-direction, m is the mass and as is the acceleration along s-direction.

ss maF =

Assumption:Fluid is ideal and incompressibleFlow is steadyFlow is along streamlineVelocity is uniform across the section and is equal to mean velocityOnly gravity and pressure forces are acting

ds

dVV

dtds

dsdV

dsdt

dsdV

dt

dVas ====

Eq(1)

Eq(2)

Fig. Forces acting on particle along streamline


8

W=weight of fluidWsin( )= component acting along s-directiondA= Area of flowds=length between sections along pipe

θ

( ) θsinWdAdpPPdAFs −+−=

Substituting values from Eq(2) and Eq(3) to Eq(1)

Eq(3)

( )ds

dVmVWdAdpPPdA =−+− θsin

ds

dVdAdsV

ds

dzgdAdsdpdA ρρ =−−

( )gdAdsmgW ρ==

ds

dz=θsin

Cancelling dA and simplifying

VdVgdzdp ρρ =−−

Note that 2

2

1dVVdV =

2

2

1dVgdzdp ρρ =−−

Eq(4)

Eq(5)

Fig. Forces acting on particle along streamline


9

� Dividing eq (5) by

� Integrating

� Assuming incompressible and steady flow

� Dividing each equation by g

ρ

02

1 2 =++ dVgdzdp

ρ

conttdVgdzdp

=

++∫

2

2

1

ρ

conttVgzP

=++ 2

2

1

ρ

conttg

Vz

g

P=++

2

2

ρ

� Hence Eq (9) for stead-incompressible fluid assuming no frictional losses can be written as

Eq (6)

Eq (7)

Eq (8)

Eq (9)

( ) ( )21

22

22

12

11

Head TotalHead Total

22

=

++=++g

VPZ

g

VPZ

γγ

Above Eq(10) is general form of Bernoulli’s Equation

Eq (10)

Energy Line and Hydraulic Grade line

10

� Static Pressure :

� Dynamic pressure :

� Hydrostatic Pressure:

� Stagnation Pressure: Static pressure + dynamic Pressure

Hg

Vz

P=++

2

2

γ

Head TotalheadVelocity headElevation head Pressure =++

P

gZρ

2/2Vρ

conttV

gzP =++2

2

ρρ

Multiplying with unit weight,γ,

stagPV

P =+2

2

ρ


11

� Measurement of Heads

� Piezometer: It measures pressure head ( ).

� Pitot tube: It measures sum of pressure and velocity heads i.e.,

g

VP

2

2

+γ

γ/P

What about measurement of elevation head !!


12

� Energy line: It is line joining the total heads along a pipe line.

� HGL: It is line joining pressure head along a pipe line.


13

Energy Equation for steady flow of any fluid

14

� Let’s consider the energy of system (Es) and energy of control volume(Ecv) defined within a stream tube as shown in figure. Therefore,

� Because the flow is steady, conditions within the control volume does not change so

� Hence

in

CV

out

CVCVs EEEE ∆−∆+∆=∆

in

CV

out

CVs EEE ∆−∆=∆

0=∆ CVE

Eq(1)

Eq(2)

Figure: Forces/energies in fluid flowing in streamt ube

Energy Equation for steady flow of any

fluid

15

� Now, let’s apply the first law of thermodynamics to the fluid system which states ” For steady flow, the external work done on any system plus the thermal energy transferred into or out of the system is equal to the change of energy of system”

( )

( ) inCVout

CV

s

EEshaftworklowwork

Eshaftworklowwork

∆−∆=++

∆=++

=+

ferredheat transf

ferredheat transf

energy of changeferredheat transdone work External

� Flow work: When the pressure forces acting on the boundaries move, in present case when p1A1 and p2A2 at the end sections move

through ∆s1 and ∆s2, external work is done. It is referred to as flow work.

msAsApp

mg

sAp

sAp

sApsAp

∆=∆=∆

−∆=

∆−∆=∆−∆=

222111

2

2

1

1

222

2

2111

1

1111111

workFlow

workFlow

ρργγ

γγ

γγ

Q Steady flow

Eq(3)

Eq(4)

Eq(5)


fluid

16

� Shaft work: Work done by machine, if any, between section 1 and 2

( ) ( ) mm

m

hmghsA

thdt

dsAtime

weight

energy

time

weight

∆=∆=

∆

==

111

111

Shaft work

Shaft work

γ

γ

� Where, hm is the energy added to the flow by the machine per unit weight of flowing fluid. Note: if the machine is pump, which adds energy to the fluid, hm is positive and if the machine is turbine, which remove energy from fluid, hm is -ve

� HeatTransferred: The heat transferred from an external source

into the fluid system over time interval ∆t is

( ) ( ) HH

H

QmgQsA

tQdt

dsA

∆=∆=

∆

=

111

111

ferredHeat trans

ferredHeat trans

γ

γ

� Where, QH is the amount of energy put into the flow by the external heat source per unit weight of flowing fluid. If the heat flow is out of the fluid, the value QH is –ve and vice versa

Eq(6)

Eq(7)


fluid

17

� Change in Energy: For steady flow during time interval ∆t, the weight of fluid entering the control volume at section 1 and leaving at section 2 are both equal to g∆m . Thus the energy (Potential+Kinetic+Internal) carried by g∆m is;

( )

( )

++∆=∆

++=∆

++

=∆

++∆=∆

++=∆

++

=∆

2

2

22

2

2

222222

2

22

222

1

2

11

1

2

111111

2

11

111

2

22

2

22

Ig

VzmgE

Ig

VzdsAtI

g

Vz

dt

dsAE

Ig

VzmgE

Ig

VzdsAtI

g

Vz

dt

dsAE

out

CV

out

CV

in

CV

in

CV

α

αγαγ

α

αγαγ

α is kinetic energy correction factor and ~ 1

Eq(8)

Eq(9)


fluid

18

� Substituting all values from Eqs. (5),(6), (7), (8), & (9) in Eq(4)

( ) inCVout

CV EEshaftworklowwork ∆−∆=++ ferredheat transf

( ) ( )

++∆−

++∆=∆+∆+

−∆ 1

2

112

2

22

2

2

1

1

22I

g

VzmgI

g

VzmgQmghmg

ppmg Hm αα

γγ

++−

++=++

− 1

2

112

2

22

2

2

1

1

22I

g

VzI

g

VzQh

ppHm αα

γγ

+++=++

++− 2

2

22

2

21

2

11

1

1

22I

g

Vz

pQhI

g

Vz

pHm α

γα

γ

This is general form of energy equation, which applies to liquids, gases, vapors and to ideal fluids as well as real fluids with friction, both incompressible and compressible. The only restriction is that its for steady flow.

Eq(10)

Energy Equation for steady flow of

incompressible fluid

19

� For incompressible fluids

� Substituting in Eq(10), we get

( )122

22

2

2

11

1

22II

g

Vz

pQh

g

Vz

pHm −+

++=++

+−

γγ

γγγ == 21

( ) Hm QIIg

Vz

ph

g

Vz

p−−+

++=+

+− 12

2

22

2

2

11

1

22 γγ

Lm hg

Vz

ph

g

Vz

p+

++=+

+−

22

2

22

2

2

11

1

γγ Eq(11)

( ) HL QIIh −−= 12Q

� Where hL=(I2-I1)-QH= head loss. It equal to is gain in internal energy minus any heat added by external source.

� Hm is head removed/added by machines. It can also be referred to head loss due to pipe fitting, contraction, expansion and bends etc in pipes.

Energy Equation for steady flow of

incompressible fluid

20

� In the absence of machine, pipe fitting etc, Eq(11) can be written as

� When the head loss is caused only by wall or pipe friction, hLbecomes hf, where hf is head loss due to friction

Lhg

Vz

p

g

Vz

p+

++=

++

22

22

22

21

11

γγEq(12)

Power

21

� Rate of work done is termed as power

Power=Energy/time

Power=(Energy/weight)(weight/time)

� If H is total head=total energy/weight and γQ is the weight flow ratethen above equation can be written as

Power=(H)(γQ)= γQH

In BG:

Power in (horsepower)=(H)(γQ)/550

In SI:

Power in (Kilowatts)=(H)(γQ)/1000

1 horsepower=550ft.lb/s

Reading Assignment

22

� Kinetic energy correction factor

� Limitation of Bernoulli’s Equation

� Application of hydraulic grade line and energy line

NUMERICALS

23

� 5.2.1

24

� 5.2.3

25

� 5.3.2

26

� 5.3.4

27

� 5.3.6

28

� 5.9.6

Momentum and Forces in Fluid Flow

29

� We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted by the air moving over the wing. A jet of water from a hose exerts a force on whatever it hits.

� In fluid mechanics the analysis of motion is performed in the same way as in solid mechanics - by use of Newton’s laws of motion.

� i.e., F = ma which is used in the analysis of solid mechanics to relate applied force to acceleration.

� In fluid mechanics it is not clear what mass of moving fluid we should use so we use a different form of the equation.

( )dt

mdma s

VF ==∑

Momentum and Forces in Fluid Flow

30

� Newton’s 2nd Law can be written:

� The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force.

� The symbols F and V represent vectors and so the change in momentum must be in the same direction as force.

It is also termed as impulse momentum principle

( )dt

md sVF =∑

=

=∑

mV

F Sum of all external forces on a body of fluid or system s

Momentum of fluid body in direction s

( )smddt VF =∑

Impact of a Jet on a Plane

31

Impact of a Jet on a Plane

32

Thank you

� Questions….

� Feel free to contact:

33

Unit 4 Fluid Dyanamics.pdf (p.1-43)unit 4 part 2fluiddynamics.pdf (p.44-76)

Documents

Dr. MohsinSiddique Assistant Professor · 2018. 7. 11. · to mean velocity of flow (V) as long as the flow in laminar. i.e., But with increasing velocity, as the flow become turbulent