You are on page 1of 9

# NPTEL Course Developer for Fluid Mechanics Dr.

Niranjan Sahoo
IIT-Guwahati

Lecture-4

## Typical Model Studies

1. Flow through Closed Conduits
When there is a fluid flow through closed conduits (such as pipe flows), the dominant
forces are inertial and viscous because there is no fluid interface. Compressibility effects
can be neglected for low Mach numbers less than 0.3. In these classes of problems,
geometric similarity between the model and prototype must be maintained. Geometric

## similarity characteristics are described by series of length terms l1 , l2 , l3 ......., li and l ,

where l is some particular length dimension of the system. It leads to series of pi terms
of the form as,
li
Pi = where i = 1, 2,.....
l
In addition, other parameter of importance is the surface roughness (e) . The

## corresponding pi term representing the surface roughness is e l .

For low Mach number flows, Reynolds number must also match. Complete similarity
requirements are as follows.
lim li e m e i r m .Vm .lm r .V .l
= ; = ; = (1)
lm l lm l mm m
In order to find the pressure differential per unit length, the other dependent pi term is
expressed as,
Dp
P1 =
r .V 2
The prototype pressure drop can be obtained from the relation as,
2
r V
Dp = . .Dpm (2)
r m Vm

Example: Water flows through a large valve having 0.8m diameter at a rate 85m 3/s. It is
to be tested in a geometrically similar model of 10cm diameter with water as working
fluid. Determine the required flow rate in the model.

1
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

Solution: In order to ensure the dynamic similarity, the Reynolds number must match for
the model and the prototype i.e.
( Re ) model = ( Re ) prototype
Vm .Dm V p .Dp
=
nm np

Since, water is used as the working fluid for the model and the prototype, so n m = n p .
Hence,
Vm D p
=
V p Dm

The flow rate for the model and prototype can be related as,
[ p 4] .Dm2
Qm D p Dm
= =
[ p 4] .Dp2
Q p Dm Dp

Hence,
Dm 0.1
Qm = Q p . = 85 = 10.6 m3 s
Dp 0.8
2. Flow around Immersed Bodies
Typical examples that fall under this category are the flow around aircraft, automobiles
etc. The general formulations for these classes of problems are expressed in terms of
dependent pi terms i.e.
l e r .V .l
Dependent pi terms = f i , , (3)
l l m
For complete similarity requirements Eq. (1) must be satisfied. The parameter of interest

## in this type of problems is the drag coefficient ( CD ) and is defined by,

FD
CD =
1
r .V 2 .l 2
2
If the similarity conditions are met, then

2
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

FD FD
1 = 1 (4)
r .V 2 .l 2 r .V 2 .l 2
2
model 2
prototype

Example-1: In order to estimate the drag force on an airplane that cruises at 100m/s,
wind-tunnel test is carried out on a 1: 15 scale model. The airplane cruises at an altitude
where there is 10% drop in standard atmospheric pressure. The wind tunnel test is also
carried out at 100m/s and the measured drag force in one of the test is 5N. Determine the
required air pressure in the tunnel and drag force on the prototype (assuming the same air
temperature for the model and prototype).
Solution: For a geometrically model and prototype, the Reynolds number must match i.e.
( Re ) model = ( Re ) prototype
r m .Vm .Dm r p .V p .D p
=
mm mp
Vm l 1
In this case, =1 ; m = , so that
Vp l p 15

r m mm V p l p m m m
= . . = . ( 1) . ( 10 ) = 15. m
r p m p Vm lm m p mp

It shows that same fluid with r m = r p ; m m = m p cannot be used to maintain the Reynolds
number similarity. However, one possible solution is to use same fluid (air) for which

mm = m p , but increase the pressure to increase the density to a limit such that variation in

rm
= 15
rp

## Using the ideal gas equation of state, p = r .R.T , we get,

r m pm
= (Assuming constant temperature)
r p pp
The prototype cruises at an altitude where there is 10% drop in standard atmosphere i.e.
at a pressure of 0.9atm. So, required pressure in the wind tunnel will be,

3
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

pm = 15 0.9 = 13.5atm.

Using Eq. (4), the drag force on the prototype is estimated as,
2 2
r V p l p
( FD ) prototype =m
r p .
. . ( FD ) model = 15. ( FD ) model = 75N
Vm lm

## Example-2: A sonar transducer is a prototype sphere 0.3m-diameter towing at 9km/hr in

seawater at 50C. The drag on this transducer is to be predicted from wind tunnel data on a
model of 15cm diameter. Determine the required test speed in the wind tunnel. If the drag
of the model at test condition is 25N, estimate the drag on the prototype.
Solution:
D p = 0.3m Dm = 15cm

V p = 9km/hr Fp Vm

Fm = 25N

## Water at 50C Air

In order to estimate test speed and drag on the prototype, the following condition must be
satisfied, i.e.
F r .V .D
= g (Example II, Lecture 2)
r .V .D
2 2
m
For ensuring the dynamic similarity, the test should run such that
( Re ) model = ( Re ) prototype
For seawater at 50C, r p = 1000 kg m , n p = 1.57 10 m s ; At prototype conditions,
3 -6 2

V p = 9 km hr = 2.5 m s . So,

V p .D p
( Re ) prototype = = 4.77 105
np
The model must be tested at same Reynolds number, i.e.
Vm .Dm
( Re ) model = = 4.77 105
nm

4
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

## r m = 1.2 kg m3 , n m = 1.5 10-5 m 2 s . So,

nm
Vm = ( Re ) model . = 47.7 m s
Dm
At these conditions, the model and prototype are dynamic similar. Hence,
Fm Fp
=
r m .Vm .Dm r p .V p2 .Dp2
2 2

so that
r p Vp2 Dp2
Fp = Fm . . 2 . 2 = 229N
r m Vm Dm

## 3. Flow with a Free Surface

The flows in canals, rivers, spillways, around ships are some of the examples involving a
free surface. In these flow patterns, following forces and numbers are of important. They
are,
Gravitational force (Froude number)
Surface tension (Weber number)
Inertia force (Reynolds number)
Thus general formulation for problems involving the flow with a free surface is expressed
as,
l e r .V .l V r .V 2 .l
Dependent pi terms = j i , , , , (5)
l l m g .l s

Since gravity is the driving force, so Froude number similarity must be maintained i.e.
Vm Vp
=
g m .lm g p .l p

The model and prototype is expected to operate in same gravitational field. So,
Vm l
= m (6)
Vp lp

In many of the practical problems, involving free surface flows, both surface tension
and viscous effects are small and strict adherence to Weber and Reynolds number is not
required.

5
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

Example: A spillway model of 1:10 scale is constructed to study the flow characteristics
for a prototype dam of width 10m and to carry water at a flow rate of 60m 3/s. Determine
the required model width and flow rate. What operating time for the model corresponds
to a 24hr period in prototype? The effects of surface tension and viscosity may be
neglected.
Solution: If wm , wp are the width of the model and prototype respectively, then
wm 1 10
= , i.e. wm = = 0.67m
wp 15 15

Since surface tension and viscosity are insignificant, so there is no need for considering
Weber number and Reynolds number. However, the Froude number must match i.e.
Vm l
= m
Vp lp

## The flow rate is given by,

Qm Vm . Am lm 2
= =
Q p V p . Ap
l p
so that,
5
1 2
Qm = 60 = 0.19 m3 s
10
The time scale can be obtained as,
Vp l p tm
= .
Vm t p lm

i.e.
t m V p lm l
= . = m
lm Vm l p lp

so that

6
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

1
tm = 24hr = 7.6hr
10
4. Turbo-machinery models
There are certain situations of practical interest in which there may be more than one
dependent parameter. So, scaling is done for the dimensional groups with multiple
dependent parameters. A typical example is centrifugal pump. The performance
parameters of interest for a centrifugal pump include the pressure rise (or head)
developed, the power input, and machine efficiency measured under specific operating
conditions. These performance curves are usually generated by varying independent

parameters. The independent variables are volume flow rate ( Q ) , angular speed ( w ) ,

## impeller diameter ( D ) , and fluid properties like density ( r ) , viscosity ( m ) . Dependent

variables include the performance quantities.
The dependence of head ( h ) developed and power ( P ) can be written in terms of
independent parameters as,
h = f ( Q, r , w , D , m )
(18)
P = j ( Q, r , w , D , m )

## Using Pi theorem, the dimensionless parameters can be written as,

h Q rw D 2
= g1 3 , (19)
w 2 D2 w D m

P Q rw D 2
= g 2 , (20)
rw 3 D5 w D
3
m
The dimensionless parameters are,
Q
Flow coefficient =
w.D 3
P
Power coefficient =
r .w 3 .D 5

r .w.D 2
Reynolds number =
m
Complete similarity in pump performance tests requires identical flow coefficients
and Reynolds number. Thus form Eqs. (19) and (20), when

7
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

Q1 Q2
= (21)
w1.D1 w2 .D23
3

It follows that,
h1 h
= 22 2 (22)
w1 .D1 w 2 .D2
2 2

and
P1 P2
= (23)
r1.w1 .D1 r 2 .w 23 .D25
3 5

These useful scaling relationships are known as pump laws. If operating conditions of
one pump are known, the operating conditions for any geometrically similar machine can
be found by changing D , w in Eqs. (21) to (23).
Another dimensionless parameter is the specific speed which is defined as the
speed required for a pump to produce unit head at unit volume flow rate i.e.
w.Q 0.5
Ns = (24)
h 0.75
A constant specific speed describes all operating conditions of geometrically similar
pumps with similar flow conditions.
Example: A centrifugal pump has a specific speed of 5000 (units of gallons/min, rpm and
m) when operated at 1170rpm and volume flow rate of 300gallons/min. In order to
increase flow rate, it is fitted with same size motor operating at 1750rpm. Determine the
flow rate, head developed, power required by the pump at this conditions. Show that
specific speed remains constant at this speed.
Q1 Q2
Solutions: From pump laws, =
w1.D1 w2 .D23 , so that
3

3
w2 D 2 1750 3

Q2 = Q1 . = 300 ( 1) = 449 gallons min

w1 .D1
3
1170

## The pump head can be calculated from specific speed, i.e.

w.Q 0.5
Ns = or,
h 0.75

8
NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo
IIT-Guwahati

1.333 1.333
w1.Q10.5
1170 300
h1 = =
= 6.462m

N 5000
s
Again from pump laws for head developed,
2 2 2
w D 1750

( 1) = 14.45m
3
h2 = h1. 2 . 2 = 6.462
w1 D1 1170

Pump power at 1170rpm is given by,
P1 = r .g.Q1.H1 = 1000 9.81300 6.309 10 -5 6.462 = 1.2kW (1gallon/min =

6.30910-5m3/s)
From pump laws, pump power at 1750rpm is,
3 5 2
r w2 D2 1750

. . = 1.2 ( 1) ( 1) = 4.23kW
5
P2 = P1. 2
r1 w1 D1 1170

Specific speed at 1750rpm is given by,
w.Q 0.5 1750 449
Ns = = = 5003
( 14.45)
0.75
h0.75
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.scribd.com/document/353915190/lecture-4", "fetch_time": 1566167662000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027314130.7/warc/CC-MAIN-20190818205919-20190818231919-00010.warc.gz", "warc_record_offset": 955715547, "warc_record_length": 62039, "token_count": 3377, "char_count": 10772, "score": 3.953125, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "latest", "language": "en", "language_score": 0.8726406693458557, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00022-of-00064.parquet"}