iit-madras, momentum transfer: july 2005-dec 2005

IIT-Madras, Momentum Transfer: July 2005-Dec 2005


Background1. Energy conservation equation

2

.2

P Vgh Const

If there is no friction

21Kinetic energy

2mV

2

What is ?2

V

21 Kinetic energy

2 Unit massV

2 Total energy

2 Unit mass

P Vgh


2. If there is frictional loss , then

Frictional loss

Unit mass

P

2 2 Frictional loss

2 2 Unit massinlet outlet

P V P Vgh gh

In many cases

outlet inleth h

outlet inletV V

Background


Q. Where are all frictional loss can occur ?

• in pipe, in valves, joints etc

• First focus on pipe friction

In pipe, Can we relate the friction to other properties ?

Flow properties

Fuid propertiesproperties

Background


Example for general case:

At the normal operating condition given following data

Shear stress = 2 Pa

250

50

0.1

1 /

valveP Pa

L m

r m

V m s

250 valveP Pa

50 L m

0 gauge pressure

Example

What should be the pressure at inlet ?


Solution : taking pressure balance

0inlet valve pipeP P P

2 * . 2piper P rL

Example (continued)

For pipe, Force balance

Hence we can find total pressure drop


We have said nothing about fluid flow properties

valve pipeP and P However , Normally we do not know the

Usually they depend on flow properties and fluid properties

?pipeP

21

2valveP K V

2

32Laminar flow . pipe

VP L

D

2Turbulent flow , , , , ,pipe nP f L V e D

Flow properties

Empirical


2

( )12

Define f DimensionlessV

In general we want to find

f is a measure of frictional loss

higher f implies higher friction

This is Fanning-Friction factor ff

Friction Factor: Definition


So we write

,......pipe nP f

,......pipe nP f f

22

1 .2

2

f rLV

r

2

.2 rL

r

2 .f LV

r

Friction factor

This is for pipe with circular cross section

2 .2

f LV

D


Here f is function of other parameters

For laminar flow , don’t worry about f , just use

2

32 VLP

D

For turbulent flow , Is it possible to get expression for shear ?

Friction factor: Turbulent Flow

Using log profile

1 2 log( )V K K Y

1 2 2log( )V

1 2 3log( )avV 0where K, , are depends on the , , ,....


Equation relating shear stress and average velocity,

and implicit nis i

Because original equation

*where

VV

V

*.yVy

* 0V

5.5 2.5ln( )V Y

Equation for Friction Factor


10

14log Re 0.4f

f

2

In the implicit equation itself,

1substitute for with , and we get

2f V

r R

V

y

2

21m

rV V

R

This is equivalent of laminar flow equation relating f and Re (for turbulent flow in a smooth pipe)

Equation for Friction Factor


2

2 mV rV

r R

2 m

r R

VV

r R

21. 22 av mf V V R

Friction Factor: Laminar Flow

2 2 4 81.2

m av avav

V V Vf V

R R D

2

16 16 16

Reav

av av

Vf

V D V D

1

2av mV VFor laminar flow


21.2valve avP K V

?pipeP

ReDV

Use of f is for finding effective shear stress and corresponding “head loss” or “ pressure drop”

What is ?valveP

K 0.5valve

In the original problem, instead of saying “normal operating condition” we say

Pressure drop using Friction Factor

Laminar or turbulent?

1 av

mV

s


For turbulent flow

10

14log Re 0.4f

f

We can solve for f, once you know f, we can get shear

21.2

f V


Once you know shear , we can get pressure drop

2 * . 2piper P rL

If flow is laminar , ( i.e. Re < 2300 ), we use 16

Ref


2 22

1 1 2. .2 2

rLP K V f V

r

21.2 pipeP K V P

22

1 2.2

rLP K V

r

And original equation becomes,

In above equation the value of f can be substitute from laminar and turbulent equation

Laminar flow – straight forward

Turbulent flow – iterative or we can use graph


0 gauge pressure


Determination of Q or D

Given a pipe (system) with known D and a specified flow rate (Q ~ V), we can calculate the pressure needed

i.e. is the pumping requirement

We have a pump: Given that we have a pipe (of dia D), what is flow rate that we can get?

OR

We have a pump: Given that we need certain flow rate, of what size pipe should we use?



We have a pump: Given that we have a pipe (of dia D), what is flow rate that we can get?

To find Q

i.e. To find average velocity (since we know D)

Two methods: (i) Assume a friction factor value and iterate (ii) plot Re vs (Re2f)

Method (i)

Assume a value for friction factor

Calculate Vav from the formula relating P and f

Calculate Re

Using the graph of f vs Re (or solving equation), re-estimate f; repeat



Method (ii) 2

2

1 2.2

rLP f V

r

22

P Df

L V

ReDV

2 22

2 22Re

2

D P Df

L

V

V

3 2

2 2

D P

L

From the plot of f vs Re,

plot Re vs (Re2f)

From the known parameters, calculate Re2f

From the plot of Re vs (Re2f), determine ReCalculate Vav


We take original example , assume we know p, and need to find V and Q

Let us say 2250

0.5

0.1

What is ?

P Pa

K

r

V

2

2 pipe

KP V P

2 5 22250 250 5*10V V f

22

2

2

K rLP V

r

2 21 2.

2 2

K LP V f V

r

Iteration 1: assume f = 0.001 gives V = 1.73m/s , Re = 3.5x105, f = 0.0034

Iteration 2: take f = 0.0034 gives V = 1.15m/s , Re = 2.1x105, f = 0.0037

Iteration 3: take f = 0.0037 gives V = 1.04 m/s , Re = 2.07x105, f = 0.0038


If flow is laminar, you can actually solve the equation

22250 250 40V V

22

322250 250

4

VLV

r

2

32pipe

VLP

D

240 40 4*2250*250

2*250V

2.92 /V m s


If you are given pressure drop and Q , we need to find D

221 2

. .2 2 / 2

V LP K f V

D

2

.2 pipe

VP K P

2

2

2.

2

V rLP K

r

2 2

2 2

2

2 2 / 24 4

K Q f Q LP

DD D

2 2

2 4 2 5

8 32K Q fL QP

D D

4 5

0.4 159.842250

f

D D


4 5

0.4 1.59842250

D D

52250 0.4 1.5984 0D D

0.24

0.69 /

Re 160000

0.0045

D

V m s

f

Iteration 1: Assume f = 0.01

Iteration 2: take f = 0.0045 and follow the same procedure

Solving this approximately (how?), we get

iit-madras, momentum transfer: july 2005-dec 2005

Documents

momentum transfer

friction slide

laminar flow slide

friction factor laminar

pipe friction

gauge pressure slide

definition slide

higher friction