© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–1. Investigate if each ratio is dimensionless. a) rV 2 >p,
b) Lr>s, c) p>V 2L, d) rL3 >Vm.
Solution
a)
rV
=
p
b)
Lr
=
s
c)
d)
a
M L 2
ba b
L3 T
= 1
M
a
b
LT 2
M
b
T2
L3
= 2
M
L
a 2b
T
rL3
=
Vm
Ans.
no
Ans.
(L)a
M
b
M
LT 2
=
= 4
2
2
V L
L
L
a b L
T
p
yes
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
2
a
M
b ( L3 )
L3
= T2
L
M
a ba
b
T LT
a
no
Ans.
no
Ans.
Ans:
a) yes
b) no
c) no
d) no
846
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–2. Use inspection to arrange each of the following three
variables as a dimensionless ratio: a) L, t, V, b) s, EV, L,
c) V, g, L.
Solution
L
a b(T)
T
= 1
L
b)
EVL
=
s
c)
a
Ans.
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a)
Vt
=
L
M
b(L)
LT 2
= 1
M
T2
Ans.
L 2
a
b
T
V2
=
= 1
gL
L
a 2 b(L)
T
Ans.
Ans:
Vt
a)
L
EVL
b)
s
V2
c)
gL
847
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–3. The pressure change that occurs in the aortic artery
during a short period of time can be modeled by the equation
∆p = ca ( mV>2R)1>2, where m is the viscosity of blood, V is
its velocity, and R is the radius of the artery. Determine the
M, L, T, dimensions for the arterial coefficient ca.
Solution
1
Table 8–1 are
mV 2
b given in
2R
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
The dimensions for the physical variables in the equation ∆p = ca a
M
LT 2
M
=
LT
Pressure change, ∆p ML-1T -2 =
Viscosity, m ML-1T -1
Velocity, V LT -1 =
L
T
Radius, R L
Thus, dimensional homogeneity requires
( M>LT )( L>T )
M
= ca £
§
2
L
LT
1
2
1
M
M 2
= ca a
b
2
LT
LT 2
1
ca = a
M 2
b
LT 2
Ans.
Ans: 1
M 2
a
b
LT 2
848
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–4. Determine the Mach number for a jet flying at
800 mi>h at an altitude of 10 000 ft. The velocity of sound in
air is determined from c = 2kRT, where the specific heat
ratio for air is k = 1.40. Note, 1 mi = 5280 ft.
Solution
c = 2kRTabs
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Referring to the table in Appendix A, at an altitude of 10 000 ft, T = 23.34° F. Also,
R = 1716 ft # lb>slug # R. Here, Tabs = (23.34° F + 460) R = 483.34 R. The speed
of sound through the air is then
= 21.40 ( 1716 ft # lb>slug # R ) (483.34 R)
= 1077.58 ft>s
The velocity of the jet is
V =
1h
800 mi 5280 ft
ba
b = 1173.3 ft>s
a
1h
1 mi
3600 s
The Mach number is
M =
1173.3 ft>s
V
=
= 1.089 = 1.09
c
1077.58 ft>s
Ans.
849
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–5. Determine the F, L, T dimensions of the following
terms. a) Q>rV, b) rg>p, c) V 2 >2g, d) rgh.
Solution
b)
c)
d)
Q
=
rV
rg
=
p
L3
b
T
2
FT
L
a 4 ba b
T
L
a
=
L6
FT 2
Ans.
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a)
a
FT 2
L
ba 2 b
4
1
T
L
=
L
F
a 2b
L
Ans.
L 2
a b
T
V
=
= L
2g
L
a 2b
T
2
rgh = a
Ans.
FT 2
L
F
ba 2 b(L) = 2
T
L
L4
Ans.
Ans:
L6
a)
FT 2
1
b)
L
c) L
F
d) 2
L
850
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–6. Determine the M, L, T dimensions of the following
terms. a) Q>rV, b) rg>p, c) V 2 >2g, d) rgh.
Solution
b)
c)
c)
rg
p
=
a
Ans.
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a)
L3
b
Q
T
L5
=
=
rV
M
M L
a 3 ba b
T
L
a
M L
ba b
1
L3 T 2
=
L
M
a
b
LT 2
Ans.
L 2
a b
T
V
=
= L
2g
L
a 2b
T
2
rgh = a
Ans.
M
L
M
ba 2 b(L) =
3
T
LT 2
L
Ans.
Ans:
L5
M
1
b)
L
c) L
M
d)
LT 2
a)
851
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–7. Show that the Weber number is dimensionless using
M, L, T dimensions and F, L, T dimensions. Determine its
value for water at 70°F flowing at 8 ft>s for a characteristic
length of 2 ft. Take sw = 4.98 1 10 - 3 2 lb>ft.
Solution
Using the M - L - T base dimensions,
We =
rV L
=
s
a
M L 2
M
ba b (L)
a 2b
3
T
T
L
=
= 1
M
M
a 2b
a 2b
T
T
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
2
Using the F - L - T base dimensions,
2
We =
rV L
=
s
a
FT 2 L 2
F
ba b (L)
a b
T
L
L4
=
= 1
F
F
a b
a b
L
L
From the table in Appendix A, r = 1.937 slug>ft 3 at T = 70° F. Substituting
numerically,
We =
rV 2L
=
s
( 1.937 slug>ft 3 )( 8 ft>s ) 2(2 ft)
= 49.8 ( 103 ) 3 4.98 ( 10-3 ) 4 lb>ft
Ans.
Ans:
49.8 ( 103 )
852
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Solution
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
*8–8. The Womersley number is often used to study blood
circulation in biomechanics when there is pulsating flow
through a circular tube of diameter d. It is defined as
Wo = 12 d22pfr>m, where f is the frequency of the pressure
in cycles per second. Like the Reynolds number, Wo is a
ratio of inertia and viscous forces. Show that this number is
dimensionless.
The dimensions for the physical variables for the Womersley number are
Diameter,
Frequency,
Density,
Viscosity,
Then,
Wo =
d
L
f
T -1
r
ML-3
m
ML-1T -1
d 2pfr
2A m
= L
= L
= L
A
( T -1 )( ML-3 )
ML-1T -1
1 M
1
a ba 3 ba b(L)(T)
T
M
L
B
1
= 1
A L2
Thus Wo is a dimensionless number.
853
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Solution
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
8–9. The Womersley number is a dimensionless parameter
that is used to study transient blood flow through the arteries
during heartbeats. It is a ratio of transient to viscous forces
and is written as Wo = r 22pfr>m, where r is the vessel
radius, f is the frequency of the heartbeat, m the apparent
viscosity, and r is the density of blood. Research has shown
that the radius r of the aorta of a mammal can be related to
its mass m by r = 0.0024m0.34, where r is in meters and m is in
kilograms. Determine the Womersley number for a horse
having a mass of 350 kg and heartbeat rate of 30 beats per
minute (bpm), and compare it to that of a rabbit having a
mass of 2 kg and heartbeat rate of 180 bpm. The viscosities of
blood for the horse and rabbit are mh = 0.0052 N # s>m2 and
mr = 0.0040 N # s>m2, respectively. The density of blood for
both is rb = 1060 kg>m3. Plot this variation of Womersley
number (vertical axis) with mass for these two animals. The
results should show that transient forces increase as the size
of the animal increases. Explain why this happens.
The Womersley number expressed in terms of mass is
Wo = 0.0024m0.34
For the horse, fh = a30
2pfr
A m
beats 1 min
ba
b = 0.5 beats>s
min
60 s
( Wo ) h = ( 0.0024m0.34 )
( Wo ) h = 1.9206m0.34 For m = 350 kg,
B
32p(0.5) rad>s 4 ( 1060 kg>m3 )
0.0052 N # s>m2
(1)
( Wo ) h = 1.9206 ( 3500.34 ) = 14.1
For the rabbit, fr = a180
Ans.
beats 1 min
ba
b = 3 beats>s
min
60 s
( Wo ) r = ( 0.0024m0.34 )
( Wo ) r = 5.3640m0.34 B
32p(3) rad>s 4 ( 1060 kg>m3 )
0.0040 N # s>m2
(2)
For m = 2 kg,
( Wo ) r = 6.79
Ans.
The plots of the Womersley number vs mass for the horse and rabbit are shown
in Fig. a and b, respectively. From these plots, we notice that Womersley number
increases with mass.
854
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–9. Continued
For the horse Eq (1),
m(kg)
0
50
100
150
200
250
300
350
400
( Wo ) h
0
7.26
9.19
10.55
11.64
12.55
13.36
14.07
14.73
(Wo)h
15
15
10
10
5
5
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
(Wo)h
m(kg)
0
50
100
0
50
100
For the rabbit Eq (2)
0
0.5
4.24
(Wo)r
200
250
300
350
400
150
200
(a)
250
300
350
400
m(kg)
(a)
( Wo(Wo)
)r r 0
m(kg)
150
1.0
1.5
2.0
2.5
3.0
3.5
4.0
5.36
6.16
6.79
7.32
7.79
8.21
8.59
1.5
2.0
2.5
3.0
3.5
2.0
(b)
2.5
10
10
5
5
m(kg)
0
0
0.5
0.5
1.0
1.0
1.5
4.0
m(kg)
3.0
3.5
4.0
(b)
855
Ans:
6.79
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–10. Express the group of variables L, μ, r, V as a
dimensionless ratio.
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f (L, m, r, V) = 0. Using the M - L - T system,
L
m
ML-1T -1
r
ML-3
V
LT -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
L
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, m, r, and V are chosen as m = 3 repeating variables
and L will become the q variable. Thus, the Π term is
Π = marbV cL
Π = ( MaL-aT -a )( MbL-3b )( LcT -c ) (L) = Ma + bL-a - 3b + c + 1T -a - c
M:
0 = a + b
L:
0 = - a - 3b + c + 1
T:
0 = -a - c
Solving a = - 1, b = 1, c = 1
Thus,
rVL
m
Ans.
m
is dimensionless
rVL
Ans.
Π = m-1r1V 1L =
or
This is the Reynolds number. But also
Π
-1
=
Ans:
m
rVL
or
m
rVL
856
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–11. Express the group of variables p, g, D, r as a
dimensionless ratio.
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f (p, V, D, r) = 0. Using the M - L - T system,
ML-1T -2
g
LT -2
D
L
r
ML-3
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
p
Here, three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, p, g, and r are chosen as m = 3 repeating variables
and D will become the q variable. Thus, the Π term is
Π = p ag brcD
Π = ( MaL-aT -2a )( L bT -2b )( McL-3c ) (L) = M a + cL-a + b - 3c + 1T -2a - 2b
M:
0 = a + c
L:
0 = - a + b - 3c + 1
T:
0 = - 2a - 2b
Solving, a = - 1, b = 1, c = 1
Thus,
Π = p-1g1r1D =
rgD
p
Ans.
Also,
Π
-1
=
p
is dimensionless
rgD
Ans.
Ans:
rgD
p
or
p
rgD
857
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–12. The force of buoyancy F is a function of the volume
V of a body and the specific weight g of the fluid. Determine
how F is related to V and g.
Solution
F
F
g
FL-3
V
L3
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Physical Variables. There are n = 3 variables and the unknown function is
f(F, g, V) = 0. Using the F - L - T system,
Here, only two base dimensions are used, so that m = 2. Thus, there is
n - m = 3 - 2 = 1 Π term
Dimensional Analysis. Here, F and g are chosen as m = 2 repeating variables and V
will become the q variable. Thus, the Π term is
Π = F ag bV
Π = F a ( F bL-3b ) L3 = F a + bL-3b + 3
F:
0 = a + b
L:
0 = - 3b + 3
Solving, a = - 1 and b = 1. Thus,
Π = F -1g 1V =
gV
F
The function can be written as
fa
gV
b = 0
F
Solving for F using this function
Ans.
F = kgV
where k is a constant to be determined by experiment.
858
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–13. Show that the hydrostatic pressure p of an
incompressible fluid can be established using dimensional
analysis by realizing that it depends upon the depth h in the
fluid and the fluid’s specific weight g.
Solution
Physical Variables. There are n = 3 variables and the unknown function is
f (p, g, h) = 0. Using the F - L - T system.
p
FL-2
g
FL-3
h
L
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Here, two base dimensions are used, so that m = 2. Thus, there is
n - m = 3 - 2 = 1 Π term
Dimensional Analysis. Here, p and g are chosen as m = 2 repeating variables and h
will become the q variable. Thus, the Π term is
Π = p ag bh
Π = ( F aL-2a )( F bL-3b ) (L) = F a + bL-2a - 3b + 1
0 = a + b
0 = - 2a - 3b + 1
Solving, a = - 1 and b = 1. Thus,
Π = p-1g 1h =
gh
p
The function can be written as
fa
gh
b = 0
p
Solving for p using this function,
p = kgh
(Q.E.D.)
where k is a constant to be determined by experiment.
859
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–14. Establish Newton’s law of viscosity using dimensional
analysis, realizing that shear stress t is a function of the fluid
viscosity μ and the angular deformation du>dy. Hint:
Consider the unknown function as f(t, μ, du, dy).
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f(t, m, du, dy) = 0. Here, t has the same dimensions as pressure p. Using the
M - L - T system,
ML-1T -2
m
ML-1T -1
du
LT -1
dy
L
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
t
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Note: Obtaining n 7 m was the reason for treating du and dy as two separate
variables.
Dimensional Analysis. Here t, m and du are chosen as m = 3 repeating variables and
dy will become the q variable. Thus, the Π term is
Π = tambducdy
Π = ( MaL-aT -2a )( MbL-bT -b )( LcT -c )( L ) = Ma + bL-a + b - c + 1T -2a - b - c
M:
0 = a + b
L:
0 = -a - b + c + 1
T:
0 = - 2a - b - c
Solving a = 1, b = - 1, c = - 1. Thus,
Π = t1m-1du -1dy =
The function can be written as
fa
t dy
a b
m du
t dy
b = 0
m du
Solving for t using this function,
du
dy
where k is a constant to be determined by experiment.
Ans.
t = km
Ans:
t = km
860
du
dy
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–15. The period of oscillation t, measured in seconds, of a
buoy depends upon its cross-sectional area A, and its mass
m, and the specific weight g of the water. Determine the
relation between t and these parameters.
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f (A, g, m, t) = 0. The dimensions for t is T. Using the M - L - T system,
A = L2
g = ML-2T -2
m = M
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, A, g, and m are chosen as m = 3 repeating variables
and t will become the q variable. Thus, the Π term is
Π = Aa g b mc t
Π = ( L2a )( MbL-2bT -2b ) (Mc)(T)
L:
0 = 2a - 2b
M:
0 = b + c
T:
0 = - 2b + 1
Solving, a = 1>2, b = 1>2, c = - 1>2. Thus,
1
1
1
Π = A2 g 2 M - 2 t = t
gA
Am
The function can be written as
f at
gA
b = 0
Am
Solving for t from this function
t = k
m
A gA
Ans.
where k is a constant to be determined by experiment.
Ans:
t = k
861
m
A gA
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–16. Laminar flow through a pipe produces a discharge Q
that is a function of the pipe’s diameter D, the change in
pressure ∆p per unit length, ∆p> ∆x, and the fluid viscosity, μ.
Determine the relation between Q and these parameters.
D
Solution
Physical Variables. There are n = 4 variables and the unknown function is
∆p
f aQ, D,
, mb = 0. Using the M - L - T system,
∆x
L3T -1
D
L
∆p
∆x
ML-2T -2
m
ML-1T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Q
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, Q,
and D becomes the q variable.
Π = Qaa
∆p
, and m are chosen as m = 3 repeating variables
∆x
∆p b c
b m D = ( L3aT -a )( MbL-2bT -2b )( McL-cT -c )( L ) = Mb + cL3a - 2b - c + 1T -a - 2b - c
∆x
M:
0 = b + c
L:
0 = 3a - 2b - c + 1
T:
0 = - a - 2b - c
Solving, a = - 1>4, b = 1>4, c = -1>4. Thus,
∆p 14
1
∆p 4 -1
Π = Q a
b m 4D = ° ∆x ¢ D
∆x
Qm
- 14
Therefore, the function can be written as
∆p 14
f ≥ ° ∆x ¢ D ¥ = 0
Qm
Solving for Q,
Q = kc
D4 ∆p
a
b d
m ∆x
Ans.
where k is a constant to be determined by experiment.
862
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–17. The speed of sound V in air is thought to depend on
the viscosity m, the density r, and the pressure p. Determine
how V is related to these parameters.
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f(V, m, r, p) = 0. Using the M - L - T system,
LT -1
m
ML-1T -1
r
ML-3
p
ML-1T -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
V
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, V, m, and r are chosen as m = 3 repeating variables and
p will become the q variable. Thus, the Π term is
Π = V ambrcp
Π = ( LaT -a )( MbL-bT -b )( McL-3c )( ML-1T -2 ) = Mb + c + 1La - b - 3c - 1T -a - b - 2
M:
0 = b + c + 1
L:
0 = a - b - 3c - 1
T:
0 = -a - b - 2
Solving, a = - 2, b = 0, c = - 1. Thus,
Π = V -2m0r-1p =
p
V 2r
The function can be rewritten as
fa
p
V 2r
b = 0
Solving for V using this function,
V = k
p
Ar
Ans.
where k is a constant to be determined by experiment. Notice that V is independent
of m.
Ans:
V = k
863
p
Ar
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–18. The flow Q of gas through the pipe is a function of
the density r of the gas, gravity g, and the diameter D of the
pipe. Determine the relation between Q and these
parameters.
D
Solution
Q
L3T -1
r
ML-3
g
LT -2
D
L
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Physical Variables. There are n = 4 variables and the unknown function is
f(Q, r, g, D) = 0. Using the M - L - T system,
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, Q, r, and g are chosen as m = 3 repeating variables and
D will become the q variable. Thus the Π term is
Π = QarbgcD
Π = ( L3aT -a )( MbL-3b )( LcT -2c ) (L) = MbL3a - 3b + c + 1 + T -a - 2c
M:
0 = b
L:
0 = 3a - 3b + c + 1
T:
0 = - a - 2c
2
1
Solving, a = - , b = 0, c = . Thus,
5
5
1
2
1
Π = Q -5r0g5D =
g5D
2
Q5
The function can be rewritten as
1
f°
g5D
2
Q5
¢ = 0
Solving for Q using this function,
2
1
Q5 = k′g5D
Q = k2gD5
Ans.
where k is a constant to be determined by experiment. Notice that Q is independent
of r.
Ans:
Q = k2gD5
864
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–19. The velocity V of the stream flowing from the side of
the tank is thought to depend upon the liquid’s density
r, the depth h, and the acceleration of gravity g. Determine
the relation between V and these parameters.
h
d
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f (V, r, g, h) = 0. Using the M - L - T system,
V = LT -1
h = L
g = LT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
r = ML-3
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, r, h, and g are chosen as m = 3 repeating variables and
V will become the q variable. Thus the Π term is
Π = r ah bgcV
Π = ( MaL-3a )( Lb )( LcT -2c )( LT -1 ) = MaL-3a + b + c + 1T -2c - 1
M:
0 = a
L:
0 = - 3a + b + c + 1
T:
0 = - 2c - 1
1
1
Solving, a = 0, b = - , c = - . Thus,
2
2
1
1
Π = r0h - 2g - 2V =
V
2gh
The function can be written as
f = a
V
2gh
b = 0
Solving for V from this function
Ans.
V = k2gh
where k is a constant to be determined by experiment. Notice that V is independent
of r.
Ans:
V = k2gh
865
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–20. The pressure p within the soap bubble is a function
of the bubble’s radius r and the surface tension s of the
liquid film. Determine the relation between p and these
parameters.
r
Solution
p
FL-2
s
FL-1
r
L
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Physical Variables. There are n = 3 variables and the unknown function is
f (p, s, r) = 0. Using the F - L - T system,
Here, only two base dimensions are used, so that m = 2. Thus, there is
n - m = 3 - 2 = 1 Π term
Dimensional Analysis. Here, p and s are chosen as m = 2 repeating variables and r
will become the q variable. Thus the Π term is
Π = p asbr
Π = ( F aL-2a )( F bL-b ) (L) = F a + bL-2a - b + 1
F:
0 = a + b
L:
0 = - 2a - b + 1
Solving, a = 1 and b = - 1. Thus,
Π = p1s -1r =
pr
s
The function can be written as
fa
pr
b = 0
s
Solving for p from this function
s
p = k r
Ans.
where k is a constant to be determined by experiment.
866
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–21. The velocity c of a wave on the surface of a liquid
depends upon the wave length l, the density r, and the
surface tension s of the liquid. Determine the relation
between c and these parameters. By what percent will c
decrease if the density of the liquid is increased by a factor
of 1.5?
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f(l, r, s, c) = 0. Using the F - L - T system,
L
r
FT 2L-4
s
FL-1
c
LT -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
l
Here, all three base dimensions are used, so that m = 3. Thus, there is
n - m = 4 - 3 = 1 Π term
Dimensional Analysis. Here, l, r, and s are chosen as m = 3 repeating variables
and c will become the q variable. Thus the Π term is
Π = la rb sc c
Π = ( La )( F bT 2bL-4b )( F cL-c )( LT -1 ) = F b + cLa - 4b - c + 1T 2b - 1
F:
0 = b + c
L:
0 = a - 4b - c + 1
T:
0 = 2b - 1
1
1
1
, b = , and c = - . Thus,
2
2
2
rl
1 1
1
Π = l2 r 2 s -2 c = c
As
Solving, a =
The function can be written as
f ac
Solving for c,
rl
b = 0
As
c = k
s
A rl
Ans.
where k is a constant to be determined by experiment.
, of decrease = a1 -
1
b * 100 = 18.4,
A 1.5
Ans.
Ans:
s
A rl
18.4%
c = k
867
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–22. The discharge Q over the weir A depends upon the
width b of the weir, the water head H, and the acceleration
of gravity g. If Q is known to be proportional to b, determine
the relation between Q and these variables. If H is doubled,
how does this affect Q?
H
A
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f (Q, H, g, b) = 0. Using the M - L - T system,
L3T -1
H
L
g
LT -2
b
L
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Q
Here, only two base dimensions are used, so that m = 2. Thus, there are
n - m = 4 - 2 = 2 Π terms
Dimensional Analysis. Here, g and H are chosen as m = 2 repeating variables. Thus
the q variables are Q for Π1 and b for Π2.
Π1 = gaHbQ = ( LaT -2a )( Lb )( L3T -1 ) = La + b + 3T -2a-1
L:
0 = a + b + 3
T:
0 = - 2a - 1
1
5
Solving, a = - and b = - . Thus,
2
2
5
1
Π1 = g - 2H - 2Q =
Q
2gH5
Π2 = gcHdb = ( LcT -2c )( Ld )( L ) = Lc + d + 1T -2c
L:
0 = c + d + 1
T:
0 = - 2c
Solving, c = 0 and d = - 1. Thus,
b
H
The function can be written as
Π = g0H -1b =
2
Q
b
¢ = 0
2gH H
Solving for Q,
f°
5
,
Q = 2gH5f1a
b
b
H
Since Q is proportional to b, this function becomes
Q = kb2gH3
Ans.
where k is a constant to be determined by experiment.
If H is doubled, Q increases by 223 = 2.83 times
868
Ans.
Ans:
Q = kb2gH3
increases by 2.83 times
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–23. The capillary rise of a fluid along the walls of the
tube causes the fluid to rise a distance h. This effect depends
upon the diameter d of the tube, the surface tension s, the
density r of the fluid, and the gravitational acceleration g.
Determine the relation between h and these parameters.
d
h
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (h, d, s, r, g) = 0. Using the M - L - T system,
L
d
L
s
MT -2
r
ML-3
g
LT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
h
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, d, r, and g are chosen as m = 3 repeating variables.
Thus the q variables are s for Π1 and h for Π2.
Π = d arbges = ( La )( MbL-3b )( LcT -2c )( MT -2 ) = La - 3b + cMb + 1T -2c - 2
1
M:
0 = b + 1
L:
0 = a - 3b + c
T:
0 = - 2c - 2
Solving, a = - 2, b = - 1, and c = - 1. Thus,
s
Π1 = d -2r-1g -1s =
rd 2g
Π2 = d erfgih = ( Le )( MfL-3f )( LiT -2i ) (L) = MfLe - 3f + i + 1T -2i
M:
0 = f
L:
0 = e - 3f + i + 1
T:
0 = - 2i
Solving, e = -1, f = 0, and i = 0. Thus,
h
d
The function can be written as
Π = d -1r0g0h =
2
f1a
s h
, b = 0
rd 2g d
Solving for h,
h
s
= fa 2 b
d
rd g
h = df a
s
b
rd 2g
Ans.
Notice that since d appears in the argument of f , we cannot say that h is proportional
to d (which in any case we already know from Ch. 1 is not true). We have, however,
established that for a given d, h is the same for all scenarios where s>rg is the same.
869
Ans:
h = df a
s
b
rd 2g
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–24. The torsional resistance T of the thrust bearing
depends upon the diameter D of the shaft, the axial force F,
the shaft rotation v, and the viscosity m of the lubricating
fluid. Determine the relation between T and these
parameters.
F
D
v
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (T, D, F, v, m) = 0. Using the F - L - T system,
FL
L
F
v
m
T -1
FTL-2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
T
D
F
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, D, F, and m are chosen as m = 3 repeating variables.
Thus the q variables are T for Π1 and v for Π2.
Π = DaF bmcT = ( La )( F b )( F cT cL-2c ) (FL) = F b + c + 1La - 2c + 1T c
1
F:
L:
T:
0 = b + c + 1
0 = a - 2c + 1
0 = c
Solving, a = - 1, b = - 1, and c = 0. Thus,
T
FD
Π = D-1F -1m0T =
1
Π = DeF fmgv = ( Le )( F f )( F gT gL-2g )( T -1 ) = F f + gLe - 2gT g - 1
2
F:
L:
T:
0 = f + g
0 = e - 2g
0 = g - 1
Solving, e = 2, f = - 1, and g = 1. Thus,
Π = D2F -1m1v =
2
mD2v
F
The function can be written as
f1a
T mD2v
,
b = 0
FD
F
Using this function to solve for T,
mD2v
T
= fa
b
FD
F
T = FDf a
mD2v
b
F
Ans.
Notice that since F and D appear in the argument of f, we cannot say that T is
­proportional to F or to D. However, we have established that for given F and D, T is
the same for all scenarios where mv is the same.
870
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–25. The thickness d of the boundary layer for a fluid
passing over a flat plate depends upon the distance x from
the plate’s leading edge, the free-stream velocity U of the
flow, and the density r and viscosity m of the fluid.
Determine the relation between d and these parameters.
U
u
y
x
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (d, x, U, r, m) = 0. Using the M - L - T system,
L
L
LT -1
r
ML-3
m
ML-1T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
d
x
U
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, x, U, and r are chosen as m = 3 repeating variables.
Thus the q variables are d for Π1 and m for Π2.
Π = xaU brcd = ( La )( LbT -b )( McL-3c ) (L) = McLa + b - 3c + 1T -b
1
0 = c
0 = a + b - 3c + 1
0 = -b
M:
L:
T:
Solving, a = - 1, b = 0, and c = 0. Thus,
Π = x -1V 0r0d =
1
d
x
Π = xdU erfm = ( Ld )( LeT -e )( MfL-3f )( ML-1T -1 ) = Mf + 1Ld + e - 3f - 1T -e - 1
2
0 = f + 1
0 = d + e - 3f - 1
0 = -e - 1
M:
L:
T:
Solving, d = -1, e = - 1, and f = -1. Thus,
Π = x -1U -1r-1m =
2
m
rUx
or
Π =
2
rUx
= Re
m
The function can be written as
d
f1a , Reb = 0
x
Solving for d,
d
= f (Re)
x
Ans.
d = xf (Re)
871
Ans:
d = xf (Re)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–26. The discharge Q from a turbine is a function of the
generated torque T, the angular rotation v of the turbine, its
diameter D, and the liquid density, r. Determine the relations
between Q and these parameters. If Q varies linearly with T,
how does it vary with the turbine’s diameter D?
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f(Q, T, v, D, r) = 0. Using the F - L - T system,
L3T -1
FL
T -1
L
FT 2L-4
Q
T
v
D
r
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Since Q is proportional to T, v, D, and r are chosen as m = 3
repeating variables. Thus, the q variables are Q for Π1 and T for Π2.
Π1 = vaDbrcQ = ( T -a )( Lb )( F cT 2cL-4c )( L3T -1 ) = F cLb - 4c + 3T -a + 2c - 1
F:
0 = c
L:
0 = b - 4c + 3
T:
0 = - a + 2c - 1
Solving, a = - 1, b = - 3, and c = 0. Thus,
Π1 = v-1D-3r0Q =
Q
vD3
Π2 = vdDer fT = ( T -d )( Le )( F fT 2fL-4f ) (FL) = F f + 1Le - 4f + 1T -d + 2f
F:
0 = f + 1
L:
0 = e - 4f + 1
T:
0 = - d + 2f
Solving, d = -2, e = - 5, and f = -1. Thus,
Π2 = v-2D-5r-1T =
T
v2D5r
The function can be written as
f1a
Q
3
,
T
2
vD v D5r
Solving for Q,
Q
vD3
= fa
b
T
v2D5r
b
T
b
v2D5r
Since Q is proportional to T,
Q = vD3f a
Q = vD3k a
T
2
5
vDr
b = k
T
vrD2
Ans.
2
where k is a constant to be determined by experiment. Q is inversely proportional to D .
872
Ans:
Q = k
T
vrD2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–27. The speed c of a water wave is a function of the wave
length l, the acceleration of gravity g, and the average depth
of the water h. Determine the relation between c and these
parameters.
l
c
h
Solution
Physical Variables. There are n = 4 variables and the unknown function is
f(c, l, g, h) = 0. Using the M - L - T system,
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
LT -1
L
LT -2
L
c
l
g
h
Here, only two base dimensions are used, so that m = 2. Thus, there are
n - m = 4 - 2 = 2 Π terms
Dimensional Analysis. Here, g, and l are chosen as m = 2 repeating variables. Thus
the q variables are c for Π1 and h for Π2.
Π1 = galbc = ( LaT -2a )( Lb )( LT -1 ) = La + b + 1T -2a - 1
L:
0 = a + b + 1
T:
0 = -2a - 1
1
1
Solving, a = - and b = - . Thus,
2
2
1
1
c
Π1 = g -2l-2c =
2gl
Π1 = gcldh = ( LcT -2c )( Ld ) L = Lc + d + 1T -2c
L:
0 = c + d + 1
T:
0 = -2c
Solving, d = -1 and c = 0. Thus,
Π2 = g0l-1h =
h
l
Therefore, the function is
c h
f1a
, b = 0
2gl l
Solving for c,
c
2gl
h
= fa b
l
l
c = 2glf a b
h
Note: In fact the equation is C =
result just obtained.
Ans.
gl
2ph
tan a
b , which is consistent with the
l
B 2p
873
Ans:
l
c = 2gl f a b
h
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–28. The torque T developed by a turbine depends upon
the depth h of water at the entrance, the density of the water
r, the discharge Q, and the angular velocity of the turbine v.
Determine the relation between T and these parameters.
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f(T, h, r, Q, v) = 0. Using the F - L - T system,
FL
L
FT 2L-4
L3T -1
T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
T
h
r
Q
v
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, r, Q, and v are chosen as m = 3 repeating variables.
Thus the q variables are T for Π1 and h for Π2.
Π1 = raQbvcT = ( F aT 2aL-4a )( L3bT -b )( T -c ) (FL) = F a + 1L-4a + 3b + 1T 2a - b - c
F:
0 = a + 1
L:
0 = - 4a + 3b + 1
T:
0 = 2a - b - c
5
1
Solving, a = - 1, b = - , and c = - . Thus,
3
3
5
1
T
Π1 = r-1Q -3v - 3T =
5
1
rQ 3v 3
Π2 = rdQevfh = ( F dT 2dL-4d )( L3eT -e )( T -f ) (L) = F dL-4d + 3e + 1T 2d - e - f
F:
0 = d
L:
0 = - 4d + 3e + 1
T:
0 = 2d - e - f
1
1
Solving, d = 0, e = - , and f = . Thus,
3
3
1
3
1
1
v
Π2 = r0Q -3v 3h = a b h
Q
Therefore, the function is
f1 £
1
T
5
3
rQ v
Solving for T,
1
3
,a
v 3
b h§ = 0
Q
1
T = rQ3v3f 3 a
5
1
v 3
b h4
Q
Ans.
874
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–29. The drag force FD on the jet plane is a function of the
speed V, the characteristic length L of the plane, and the
density r and viscosity m of the air. Determine the relation
between FD and these parameters.
V
FD
L
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f(FD, V, L, r, m) = 0. Using the M - L - T system,
MLT -2
LT -1
L
ML-3
ML-1T -1
FD
V
L
r
m
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, r, V, and L are chosen as m = 3 repeating variables.
Thus, the q variables are FD for Π1 and m for Π2.
Π1 = raV bLcFD = ( MaL-3a )( LbT -b )( Lc )( MLT -2 ) = Ma + 1L-3a + b + c + 1T -b - 2
M:
0 = a + 1
L:
0 = -3a + b + c + 1
T:
0 = -b - 2
Solving, a = - 1, b = - 2, and c = - 2. Thus,
Π1 = r-1V -2L-2FD =
FD
rV 2L2
Π2 = rdV eLfm = ( MdL-3d )( LeT -e )( Lf )( ML-1T -1 ) = Md + 1L-3d + e + f - 1T -e - 1
M:
0 = d + 1
L:
0 = -3d + e + f - 1
T:
0 = -e - 1
Solving, d = -1, e = - 1, and f = -1. Thus,
Π2 = r-1V -1L-1m =
m
rVL
or
rVL
= Re
m
Π2 =
Therefore, the function can be written as
f1 °
FD
rV 2L2
, Re¢ = 0
Solving forFD,
FD
rV 2L2
= f (Re)
FD = rV 2L2f (Re)
Ans.
875
Ans:
FD = rV 2L2f (Re)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–30. The time t needed for ethyl ether to drain from the
pipette is thought to be a function of the fluid’s density r
and viscosity m, the nozzle’s diameter d, and gravity g.
Determine the relation between t and these parameters.
d
Solution
t = f(r, m, d, g) or g(t, r, m, d, g) = 0. Thus, n = 5. Using the M - L - T system
given in table 8–1
Time, t T
r ML-3
Viscosity,
m ML-1 T -1
Diameter,
d, L
Gravity,
g, LT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Density,
Here, m = 3 since three base dimensions M, L and T are involved. Thus, there are
n - m = 5 - 3 = 2 Π terms. r, d and g are chosen as m = 3 repeating variables
since collectively they contain all three base dimensions as required. The first Π
term using t as g the variable is
Π1 = rad bgct
= ( Ma L-3a )( Lb )( Lc T -2c )( T )
= Ma L-3a + b + c T -2c + 1
Thus, for
M:
a=0
L:
- 3a + b + c = 0
T:
- 2c + 1 = 0
1
1
Solving, a = 0, b = - and c = . Then
2
2
g
1 1
Π1 = r0d -2 g 2t =
t
Ad
The second Π term using m as g variable is
Π2 = rdd egt m
= ( Md L-3d )( Le )( Lf T -2f )( ML-1T -1 )
= Md + 1 L-3d + e + f - 1 T -2f - 1
Thus, for
M:
d + 1 = 0
L:
- 3d + e + f - 1 = 0
T:
- 2f - 1 = 0
876
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–30. Continued
3
1
Solving, d = -1, e = - and f = - . Then
2
2
m
3
1
Π2 = r-1d - 2g - 2 m =
3 1
rd 2g2
Then
g
m
t,
¢ = 0
A d rd 32 g12
Solving for
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
f1 °
g
t in this equation,
Ad
g
m
t = f ° 3 1¢
Ad
rd 2 g2
t =
m
d
f ° 3 1¢
Ag
rd 2 g2
Ans.
Ans:
t =
877
m
d
f ° 3 1¢
Ag
rd 2 g2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–31. The head loss hL in a pipe depends upon its diameter
D, the velocity of flow V, and the density r and viscosity m
of the fluid. Determine the relation between hL and these
parameters.
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (hL, D, V, r, m) = 0. Using the M – L – T system,
L
D
L
V
LT -1
r
ML-3
m
ML-1T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
hL
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, D, V, and r are chosen as m = 3 repeating variables.
Thus, the q variables are hL for Π1 and m for Π2.
Π1 = DaV brchL = ( La )( LbT -b )( McL-3c ) (L) = McLa + b - 3c + 1T -b
M:
0 = c
L:
0 = a + b - 3c + 1
T:
0 = -b
Solving, a = - 1, b = 0, and c = 0. Thus,
Π1 = D-1V 0r0hL =
hL
D
Π2 = DdV er fm = ( Ld )( LeT -e )( MfL-3f )( ML-1T -1 ) = Mf + 1Ld + e - 3f - 1T -e - 1
M:
0 = f + 1
L:
0 = d + e - 3f - 1
T:
0 = -e - 1
Solving, d = -1, e = -1, and f = - 1. Thus,
Π2 = D-1V -1r-1m =
m
rVD
or
Π2 =
rVD
= Re
m
Therefore the function can be written as
f1 a
hL
, Reb = 0
D
Solving for hL,
hL
= f (Re)
D
Ans.
hL = Df (Re)
878
Ans:
hL = Df (Re)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–32. The pressure difference ∆p of air that flows through
a fan is a function of the diameter D of the blade, its angular
rotation v, the density r of the air, and the flow Q. Determine
the relation between ∆p and these parameters.
D
Solution
Physical Variables. There are n = 6 variables and the unknown function is
f (∆p, D, v, r, Q) = 0. Using the F – L – T system,
FL-2
D
L
v
T -1
r
FT 2L-4
Q
L3T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
∆p
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, D, v, and r are chosen as m = 3 repeating variables.
Thus, the q variables are ∆p for Π1 and Q for Π2.
Π1 = Davbrc ∆p = ( La )( T -b )( F cT 2cL-4c )( FL-2 ) = F c + 1La - 4c - 2T -b + 2c
F:
0 = c + 1
L:
0 = a - 4c - 2
T:
0 = -b + 2c
Solving, a = - 2, b = - 2, and c = - 1. Thus,
Π1 = D-2v-2r-1 ∆p =
∆p
rv2D2
Π2 = Ddver fQ = ( Ld )( T -e )( F fT 2fL-4f )( L3T -1 ) = F fLd - 4f + 3T -e + 2f - 1
F:
0 = f
L:
0 = d - 4f + 3
T:
0 = -e + 2f - 1
Solving, d = -3, e = - 1, and f = 0. Thus,
Π2 = D-3v-1r0Q =
Q
vD3
Therefore, the function can be written as
f1 a
∆p
Q
,
b = 0
rv2D2 vD3
Solving for ∆p,
∆p
2
2
rv D
= fa
Q
vD3
b
∆p = rv2D2f a
Q
vD3
Ans.
b
879
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–33. The period of time t between small water waves is
thought to be a function of the wave length l, the water
depth h, gravitational acceleration g, and the surface tension
s of the water. Determine the relation between t and these
parameters.
Solution
t
T
l
L
h
L
g
LT -2
s
MT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Physical Variables. There are n = 5 variables and the unknown function is
f (t, l, h, g, s) = 0. Using the M – L – T system,
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, l, g, and s are chosen as m = 3 repeating variables.
Thus, the q variables are t for Π1 and h for Π2.
Π1 = lagbsct = ( La )( LbT -2b )( McL-2c ) (T) = McLa + bT -2b - 2c + 1
M:
0 = c
L:
0 = a + b
T:
0 = - 2b - 2c + 1
1
1
Solving, a = - , b = , and c = 0. Thus,
2
2
1
1
Π1 = l-2g2s0t = t
g
Al
Π2 = ldgesfh = ( Ld )( LeT -2e )( MfL-2f ) (L) = MfLd + e + 1T -2e - 2f
M:
0 = f
L:
0 = d + e + 1
T:
0 = - 2e - 2f
Solving, d = -1, e = 0, f = 0. Thus,
Π2 = l-1g0s0h =
h
l
Therefore, the function can be written as
f1at
Solving for t,
t
g h
, b = 0
Al l
g
h
= fa b
Al
l
t =
l h
fa b
Ag l
Ans.
Ans:
t =
880
l h
fa b
Ag l
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–34. The drag force FD on the square plate held normal
to the wind depends upon the area A of the plate and the air
velocity V, density r, and viscosity m. Determine the relation
between FD and these parameters.
FD
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (FD, V, r, m, A) = 0. Using the M – L – T system,
MLT -2
V
LT -1
r
ML-3
m
ML-1T -1
A
L2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
FD
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, V, r, and A are chosen as m = 3 repeating variables.
Thus, the q variables are FD for Π1 and m for Π2.
Π1 = V arbAcFD = ( LaT -a )( MbL-3b )( L2c )( MLT -2 ) = Mb + 1La - 3b + 2c + 1T -a - 2
M:
0 = b + 1
L:
0 = a - 3b + 2c + 1
T:
0 = -a - 2
Solving, a = - 2, b = - 1, and c = - 1. Thus,
FD
Π1 = V -2r-1A-1FD =
rV 2A
Π2 = V dreAfm = ( LdT -d )( MeL-3e )( L2f )( ML-1T -1 ) = Me + 1Ld - 3e + 2f - 1T -d - 1
M:
0 = e + 1
L:
0 = d - 3e + 2f - 1
T:
0 = -d - 1
1
. Thus,
2
m
m
1
Π2 = V -1r-1A-2m =
=
1
rVL
2
rVA
Solving, d = -1, e = - 1, and f =
or
rVL
= Re
m
Π2 =
Therefore, the function can be written as
f1 a
FD
rV 2A
, Reb = 0
Solving for FD,
FD
rV 2A
= f (Re)
FD = rV 2Af (Re)
Ans.
881
Ans:
FD = rV 2Af (Re)
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–35. The thrust T of the propeller on a boat depends
upon the diameter D of the propeller, its angular velocity v,
the speed of the boat V, and the density r and viscosity m of
the water. Determine the relation between T and these
parameters.
V
T
Solution
Physical Variables. There are n = 6 variables and the unknown function is
f(T, D, v, V, r, m) = 0. Using the F – L – T system,
T
F
D
L
T
v
-1
V
LT -1
r
FT 2L-4
m
FTL-2
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 6 - 3 = 3 Π terms
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Dimensional Analysis. Here, r, v, and D are chosen as m = 3 repeating variables.
Thus the q variables are T for Π1, m for Π2, and V for Π3.
Π1 = ravbDcT = ( F aT 2aL-4a )( T -b )( Lc )( F ) = F a + 1L-4a + cT 2a - b
F:
0 = a + 1
L:
0 = - 4a + c
T:
0 = 2a - b
Solving, a = - 1, b = - 2, and c = -4. Thus,
Π1 = r-1v-2D-4T =
T
rv2D4
Π2 = rdveDfm = ( F dT 2dL-4d )( T -e )( Lf )( FTL-2 ) = F d + 1L-4d + f - 2T 2d - e + 1
F:
0 = d + 1
L:
0 = - 4d + f - 2
T:
0 = 2d - e + 1
Solving, d = -1, e = - 1, and f = -2. Thus,
Π2 = r-1v-1D-2m =
m
rvD2
Π3 = rgvhDiV = ( F gT 2gL-4g )( T -k )( Li )( LT -1 ) = F gL-4g + i + 1T 2g - h - 1
F:
0 = g
L:
0 = - 4g + i + 1
T:
0 = 2g - h - 1
Solving, g = 0, h = - 1, and i = -1. Thus,
Π3 = r0v-1D-1V =
V
vD
Therefore, the function is
f1 a
m
T
V
,
,
b = 0
2 vD
2 4
rv D rvD
Solving for T,
m
T
V
= fa
,
b
2 4
rvD2 vD
rv D
T = rv2D4f a
Ans:
V
,
b
rvD2 vD
m
Ans.
882
T = rv2D4f a
V
b
rvD vD
m
2
,
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–36. The power P of a blower depends upon the
impeller diameter D, its angular velocity v, the discharge Q,
and the fluid density r and viscosity m. Determine the
relation between P and these parameters.
Solution
Physical Variables. There are n = 6 variables and the unknown function is
f( p, D, v, Q, r, m) = 0. Using the M – L – T system,
p
ML2T -3
D
L
v
T -1
Q
L3T -1
r
ML-3
m
ML-1T -1
Here, all three base dimensions are used, so that m = 3. Thus, there are
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
n - m = 6 - 3 = 3 Π terms
Dimensional Analysis. Here, D, v, and m are chosen as m = 3 repeating variables.
Thus, the q variables are P for Π1, Q for Π2, and r for Π3.
Π1 = DavbmcP = ( La )( T -b )( McL-cT -c )( ML2T -3 ) = Mc + 1La - c + 2T -b - c - 3
M:
0 = c + 1
L:
0 = a - c + 2
T:
0 = -b - c - 3
Solving, a = - 3, b = - 2, and c = - 1. Thus,
P
D3v2m
Π1 = D-3v-2m-1P =
Π2 = DdvemfQ = ( Ld )( T -e )( MfL-fT -f )( L3T -1 ) = M fLd - f + 3T -e - f - 1
M:
0 = f
L:
0 = d - f + 3
T:
0 = -e - f - 1
Solving, d = -3, e = - 1, and f = 0. Thus,
Π2 = D-3v-1m0Q =
Q
D3v
Π3 = D v m r = ( L )( T -h )( MiL-iT -i )( ML-3 ) = Mi + 1Lg - i - 3T -h - i
g h i
g
M:
0 = i + 1
L:
0 = g - i - 3
T:
0 = -h - i
Solving, g = 2, h = 1, and i = - 1. Thus,
Π3 = D2v1m-1r =
rD2v
m
Therefore, the function can be written as
Q rD2v
P
,
,
b = 0
D3v2m D3v m
Solving for P,
f1 a
Q rD2v
P
=
f
,
a
b
D3v2m
D3v m
P = D3v2mf a
rD2v
b
D3v m
Q
Ans.
,
883
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–37. The discharge Q of a pump is a function of the
impeller # diameter D, its angular velocity v, the power
output W, and the density r and viscosity m of the fluid.
Determine the relation between Q and these parameters.
Solution
Physical Variables.
There are n = 6 variables and the unknown function is
#
f(Q, D, v, W, r, m) = 0. Using the M – L – T system,
L3T -1
L
T -1
ML2T -3
ML-3
ML-1T -1
Q
D
v
#
W
r
m
Here, all three base dimensions are used, so that m = 3. Thus, there are
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
n - m = 6 - 3 = 3 Π terms
Dimensional Analysis. Here, D, v, and P are chosen as m = 3 repeating variables.
Thus, the q variables are Q for Π1, r for Π2, and m for Π3.
#
Π1 = DavbW cQ = ( La )( T -b )( McL2cT -3c )( L3T -1 ) = McLa + 2c + 3T -b - 3c - 1
0 = c
0 = a + 2c + 3
0 = - b - 3c - 1
M:
L:
T:
Solving, a = - 3, b = - 1, and c = 0. Thus,
#
Π1 = D-3v-1W 0Q =
Q
D3v
#
Π2 = Ddve W fr = ( Ld )( T -e )( MfL2fT -3f )( ML-3 ) = Mf + 1Ld + 2f - 3T -e - 3f
0 = f + 1
0 = d + 2f - 3
0 = - e - 3f
M:
L:
T:
Solving, d = 5, e = 3, and f = -1. Thus,
#
rD5v3
#
W
Π2 = D5v3W -1r =
#
Π3 = DgvhW im = ( Lg )( T -h )( MiL2iT -3i )( ML-1T -1 ) = Mi + 1Lg + 2i - 1T -h - 3i - 1
0 = i + 1
0 = g + 2i - 1
0 = - h - 3i - 1
M:
L:
T:
Solving, g = 3, h = 2, and i = -1. Thus,
D3v2m
P
Therefore, the function can be written as
#
Π3 = D3v2W -1m =
f1 a
rD5v3 D3v2m
# ,
# b = 0
W
Dv W
Q
3
,
Solving for Q,
Q
D3v
= fa
rD5v3 D3v2m
# ,
# b
W
W
Ans:
5 3
3 2
rD v D v m
# b
Q = D vf a # ,
W
W
3
Ans.
884
Q = D3vf a
rD5v3 D3v2m
# , # b
W
W
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–38. As the ball falls through a liquid, its velocity V is a
function of the diameter D of the ball, its density rb, and the
density r and viscosity m of the liquid, and the acceleration
due to gravity g. Determine the relation between V and
these parameters.
D
V
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f(V, D, rb, r, m) = 0. Using the M – L – T system,
LT -1
D
L
rb
ML-3
r
ML-3
m
ML-1T -1
g
LT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
V
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 6 - 3 = 3 Π terms
Dimensional Analysis. Here, D, r, and m are chosen as m = 3 repeating variables.
Thus, the q variables are V for Π1, rb for Π2, g for Π3.
Π1 = DarbmcV = ( La )( MbL-3b )( McL-cT -c )( LT -1 ) = Mb + cLa - 3b - c + 1T -c - 1
M:
0 = b + c
L:
0 = a - 3b - c + 1
T:
0 = -c - 1
Solving, a = 1, b = 1, and c = -1. Thus,
Π1 = D1r1m-1V =
rVD
m
Π2 = Ddremfrb = ( Ld )( MeL-3e )( MfL-fT -f )( ML-3 ) = Me + f + 1Ld - 3e - f - 3T -f
M:
0 = e + f + 1
L:
0 = d - 3e - f - 3
T:
0 = -f
Solving, d = 0, e = - 1, and f = 0. Thus,
Π2 = D0r-1m0rb =
rb
r
Π3 = Dhrimjg = ( Lh )( MiL-3i )( MjL-jT -j )( LT -2 )
= Mi + jLh - 3i - j + 1T -j - 2
M:
0 = i + j
L:
0 = h - 3i - j + 1
T:
0 = -j - 2
885
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–38. Continued
Solving, h = 3, i = 2, j = - 2. Thus
Π3 = D3r2m-2g =
D3e 2g
m2
Therefore, the function can be written as
rVD rb D3r2g
b = 0
, ,
m
r
m2
Solving for V,
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
fa
rb D3r2g
rVD
b
= f1 a ,
m
r
m2
V =
m
rb D3r2g
fa ,
b
r
rD
m2
Ans.
Ans:
V =
886
m
rb D3r2g
fa ,
b
r
rD
m2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–39. The change in pressure ∆p in the pipe is a function
of the density r and the viscosity m of the fluid, the pipe
diameter D, and the velocity V of the flow. Establish the
relation between ∆p and these parameters.
D
V
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (∆p, m, D, r, V) = 0. Using the M – L – T system,
ML-1T -2
m
ML-1T -1
D
L
r
ML-3
V
LT -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
∆p
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, D, r, and V are chosen as m = 3 repeating variables.
Thus the q variables are ∆p for Π1 and m for Π2.
Π1 = DarbV c ∆p = ( La )( MbT -3b )( LCT -C )( ML-1T -2 ) = Mb + 1La - 3b + c - 1T -c - 2
M:
0 = b + 1
L:
0 = a - 3b + c - 1
T:
0 = -c - 2
Solving, a = 0, b = - 1, and c = - 2. Thus,
Π1 = D0r-1V -2 ∆p =
∆p
rV 2
Π2 = DdreV fm = ( Ld )( MeL-3e )( LfT -f )( ML-1T -1 ) = Me + 1Ld - 3e + f - 1T -f - 1
M:
0 = e + 1
L:
0 = d - 3e + f - 1
T:
0 = -f - 1
Solving, d = -1, e = - 1, and f = - 1. Thus,
Π2 = D-1r-1V -1m =
m
rVD
or
Π2 =
rVD
= Re
m
Therefore, the function can be written as
f1a
∆p
rV 2
, Reb = 0
∆p = rV 2f (Re)
Ans.
Ans:
∆p = rV 2f (Re)
887
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–40. The drag force FD on the automobile is a function
of its velocity V, its projected area A into the wind, and the
density r and viscosity m of the air. Determine the relation
between FD and these parameters.
v
FD
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (FD, V, A, r, m) = 0. Using the M – L – T system,
MLT -2
V
LT -1
A
L2
r
ML-3
m
ML-1T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
FD
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, V, A, and r are chosen as m = 3 repeating variables.
Thus, the q variables are FD for Π1, and m for Π2.
Π1 = V aAbrcFD = ( LaT -a )( L2b )( McL-3c )( MLT -2 ) = Mc + 1La + 2b - 3c + 1T -a - 2
M:
0 = c + 1
L:
0 = a + 2b - 3c + 1
T:
0 = -a - 2
Solving, a = - 2, b = - 1, and c = -1. Thus,
Π1 = V -2A-1r-1FD =
FD
rAV 2
Π2 = V gAhrim = ( LgT -g )( L2h )( MiL-3i )( ML-1T -1 ) = Mi + 1Lg + 2h - 3i - 1T -g - 1
M:
0 = i + 1
L:
0 = g + 2h - 3i - 1
T:
0 = -g - 1
1
Solving, g = - 1, h = - , and i = - 1. Thus,
2
m
1
Π3 = V -1A-2r-1m =
1
2
rVA
=
m
rVL
or
Π2 =
rVL
= Re
m
Therefore, the function can be written as
f1 a
FD
L
,
, Reb = 0
rAV 2 2A
Solving for FD in this equation.
FD
rV 2L2
= f (Re)
FD = rV 2L2 3f (Re) 4
Ans.
888
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–41. When an underwater explosion occurs, the pressure
p of the shock wave at any instant is a function of the mass
of the explosive m, the intial pressure p0 formed by
the explosion, the spherical radius r of the shock wave, and
the density r and the bulk modulus EV of the water.
Determine the relation between p and these parameters.
r
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (p, m, p0, r, r, EV) = 0. Using the M - L - T system,
ML-1T -2
m
M
p0
ML-1T -2
r
L
r
ML-3
EV
ML-1T -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
p
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 6 - 3 = 3 Π terms
Dimensional Analysis. Here, p0, m, and r are chosen as m = 3 repeating variables.
Thus, the q variables are p for Π1, r for Π2, and EV for Π3.
Π1 = p0ambr cr = ( MaL-aT -2a )( Mb )( Lc )( ML-1T -2 ) = Ma + b + 1L-a + c - 1T -2a-2
M:
0 = a + b + 1
L:
0 = -a + c - 1
T:
0 = - 2a - 2
Solving, a = - 1, b = 0, and c = 0. Thus,
Π1 = p0-1m0r 0p =
p
p0
Π2 = p0dmer fr = ( MdL-dT -2d )( Me )( L f )( ML-3 ) = Md + e + 1L-d + f - 3T -2d
M:
0 = d + e + 1
L:
0 = -d + f - 3
T:
0 = - 2d
Solving, d = 0, e = - 1, and f = 3. Thus,
Π2 = p00m-1r 3r =
rr 3
m
Π3 = p0gmhr iEV = ( MgL-gT -2g )( Mh )( Li )( ML-1T -2 ) = Mg + h + 1L-g + i - 1T -2g - 2
M:
0 = g + h + 1
L:
0 = -g + i - 1
T:
0 = - 2g - 2
889
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–41. Continued
Solving, g = - 1, h = 0, and i = 0. Thus,
Π3 = p0-1 m0r 0EV =
EV
p0
Therefore, the function can be written as
p rr 3 EV
f1 a ,
,
b =0
p0 m p0
p
rr 3 EV
= fa
,
b
p0
m p0
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Solving for p using this function,
p = p0 f a
rr 3 EV
,
b
m p0
Ans.
Ans:
p = p0 f a
890
rr 3 EV
, b
m p0
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–42. The drag force FD acting on a submarine depends
upon the characteristic length L of the vessel, the velocity V
at which it is traveling, and the density r and viscosity m of
the water. Determine the relation between FD and these
parameters.
Solution
Physical Variables. There are n = 5 variables and the unknown function is
f (F, L,V, r, m) = 0. Using the M - L - T system,
MLT -2
L
L
V
LT -1
r
ML-3
m
ML-1T -1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
FD
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 5 - 3 = 2 Π terms
Dimensional Analysis. Here, V, r, and m are chosen as m = 3 repeating variables.
Thus, the q variables are FD for Π1 and L for Π2.
Π1 = V arbmcFD = ( LaT - a )( MbL-3b )( McL-cT -c )( MLT -2 ) = Mb + c + 1La - 3b - c + 1T -a - c - 2
M:
0 = b + c + 1
L:
0 = a - 3b - c + 1
T:
0 = -a - c - 2
Solving, a = 0, b = 1, and c = -2. Thus,
Π1 = V 0r1m-2FD =
Π2 = V dremfL =
FDr
m2
( LdT -d )( MeL-3e )( MfL-fT -f )( L ) = Me + fLd - 3e - f + 1T -d - f
M:
0 = e + f
L:
0 = d - 3e - f + 1
T:
0 = -d - f
Solving, d = 1, e = 1, and f = - 1. Thus,
Π2 = V 1r1m-1L =
rVL
= Re
m
The function can be written as
f1a
FDr
m2
, Reb = 0
Solving for FD,
FDr
m2
= f (Re)
FD =
m2
f (Re)
r
Ans.
891
Ans:
FD =
m2
f (Re)
r
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
#
8–43. The power W supplied by a pump is thought to be a
function of the discharge Q, the change in pressure ∆p
between the inlet and outlet, and the density r of the fluid.
Use the Buckingham Pi theorem to establish a general
relation between these parameters so that an experiment
may be performed to determine this relationship.
Solution
#
#
W = f (Q, ∆p, r) or g(W, Q, ∆p, r) = 0. Thus, n = 4. Using the M - L - T
system given in Table 8–1,
#
W
ML2T -3
Discharge,
Q
L3T -1
Change in pressure,
∆p
ML-1T -2
Density,
r
ML-3
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Power,
Here, m = 3 since three base dimensions M, L and T are involved. Thus, there is
n - m = 4 - 3 = 1 Π term. r, Q and ∆p are chosen as m = 3 repeating variables
since collectively
they contain all three base dimensions as required. The only Π
#
term using W as the q variable is
#
Π = raQb ∆pcW
= ( MaL-3a )( L3bT -b )( McL-cT -2c )( ML2T -3 )
= Ma + c + 1L-3a + 3b - c + 2T -b - 2c - 3
Thus, for
M:
a + c + 1 = 0
L:
- 3a + 3b - c + 2 = 0
T:
- b - 2c - 3 = 0
Solving a = 0, b = -1 and c = - 1. Then
#
Π = r0Q -1 ∆p-1W =
#
W
Q∆p
Thus, the general relation between the given physical variables is
#
Ans.
W = CQ∆p
where C is a dimensionless constant to be determined from experiment.
Ans:
#
W = CQ∆p
892
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–44. The diameter D of oil spots made on a sheet of
porous paper depends upon the diameter d of the squirting
nozzle, the height h of the nozzle from the surface, the
velocity V of the oil, and its density r, viscosity m, and
surface tension s. Determine the dimensionless ratios that
define this process.
d
V
h
Solution
D = f (d, h, V, r, m, p) or g(D, d, h, V, r, m, s) = 0. Thus, n = 7 using the
M - L - T system given in Table 8–1,
D
L
Diameter of the nozzle,
d
L
Height,
h
L
Velocity,
V
LT -1
r
ML-3
m
ML-1T -1
s
MT -2
Density,
Viscosity,
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Diameter of the spot,
Surface tension,
Here, m = 3 since three base dimensions M, L and T are involved. Thus, there are
n - m = 7 - 3 = 4 Π terms. r, V and h are chosen as m = 3 repeating variables
since collectively they contain all three base dimensions as required. The first Π
term, using m as the q variable, is
Π1 = raV bhcm
= ( MaL-3a )( LbT -b )( Lc )( ML-1T -1 )
= Ma + 1L-3a + b + c - 1T -b - 1
Thus, for
M:
a + 1 = 0
L:
- 3a + b + c - 1 = 0
T:
-b - 1 = 0
Solving a = - 1, b = - 1 and c = -1. Then
Π1 = r-1V -1h-1m =
m
rVh
rVh
1
=
is the Reynolds number. The second Π term, using s as
m
Π1
the q variable, is
Recognize that
Π2 = rdV ehfs
= ( MdL-3d )( LeT -e )( Lf )( MT -2 )
Thus for
M:
d + 1 = 0
L:
- 3d + e + f = 0
T:
-e - 2 = 0
893
D
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–44. Continued
Solving, d = -1, e = - 2 and f = - 1. Then
Π2 = r-1V -2h-1s =
Recognize that
q variable, is
s
rV 2h
rV 2h
1
=
is the Weber number. The third Π term, using D as the
s
Π2
Π3 = r gV hhiD
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
= ( MgL-3g )( LhT -h )( Li )( L )
= MgL-3g + h + i + 1T -h
Thus, for
M:
g = 0
L:
- 3g + h + i + 1 = 0
T:
-h = 0
Solving, g = 0, h = 0 and i = - 1. Then
Π3 = r0V 0h-1D =
D
h
Since q variable d for fourth Π term has the same dimension as D, Hence
Π4 =
d
h
Thus, the functional relation is
g aRe, We,
D d
, b = 0
h h
Ans.
894
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–45. Mist from an aerosol produces droplets having a
diameter d, which is thought to depend upon the diameter
of the nozzle D, the surface tension s of the droplets, the
velocity V at which the droplets are ejected, and the density
r and viscosity m of the air. Determine the relation between
d and these parameters.
Solution
Physical Variables. There are n = 6 variables and the unknown function is
f (d, D, V, r, m, s) = 0. Using the M - L - T system,
L
D
L
V
LT -1
r
ML-3
m
ML-1T -1
s
MT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
d
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 6 - 3 = 3 Π terms
Dimensional Analysis. Here, D, V, and r are chosen as m = 3 repeating variables.
Thus, the q variables are d for Π1, m for Π2, and s for Π3.
Π1 = DaV brcd = ( La )( LbT -b )( McL-3c ) (L) = McLa + b - 3c + 1T -b
M:
0 = c
L:
0 = a + b - 3c + 1
T:
0 = -b
Solving, a = - 1, b = 0, and c = 0. Thus,
Π1 = D-1V 0r0d =
d
D
Π2 = DdV er fm = ( Ld )( LeT -e )( MfL-3f )( ML-1T -1 ) = Mf + 1Ld + e - 3f - 1T - e - 1
M:
0 = f + 1
L:
0 = d + e - 3f - 1
T:
0 = -e - 1
Solving, d = -1, e = - 1, and f = -1. Thus,
Π2 = D-1V -1r-1m =
m
rVD
or
Π2 =
rVD
= Re
m
895
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–45. Continued
Π3 = DgV hris = ( Lg )( LhT
-h
M:
0 = i + 1
L:
0 = g + h - 3i
T:
0 = -h - 2
)( MiL-3i )( MT -2 ) = Mi + 1Lg + h - 3iT -h - 2
Solving, g = - 1, h = - 2, and i = -1. Thus,
or
Π3 =
s
rV 2D
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Π3 = D-1V -2r-1s =
rV 2D
= We
s
Therefore, the function can be written as
d
f1a , Re, Web = 0
D
Solving for d,
d
= f ( Re, We )
D
d = Df ( Re, We ) Ans.
Ans:
d = Df ( Re, We )
896
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–46. Fluid flow depends upon the viscosity m, bulk
modulus EV , gravity g, pressure p, velocity V, density r,
surface tension s, and a characteristic length L. Determine
the dimensionless groupings for these eight variables.
Solution
Physical Variables. There are n = 8 variables and the unknown function is
f (m, EV, g, p, V, r, s, L) = 0. Using the M - L - T system,
ML-1T -1
V
LT -1
EV
ML-1T -2
r
ML-3
g
LT -2
s
MT -2
P
ML-1T -2
L
L
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
m
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 8 - 3 = 5 Π terms
Dimensional Analysis. Here, V, L, and r are chosen as m = 3 repeating variables.
Thus, the q variables are m for Π1, EV for Π2, g for Π3, P for Π4, and s for Π5.
Π1 = V aLbrcm = ( LaT -a )( Lb )( McL-3c )( ML-1T -1 ) = Mc + 1La + b - 3c - 1T -a - 1
M:
0 = c + 1
L:
0 = a + b - 3c - 1
T:
0 = -a - 1
Solving, a = - 1, b = - 1, and c = -1. Thus,
Π1 = V -1L-1r-1m =
or
Π1 =
m
rVL
rVL
= Re
m
Π2 = V dLerfEV = ( LdT -d )( Le )( MfL-3f )( ML-1T -2 ) = Mf + 1Ld + e - 3f - 1T -d - 2
M:
0 = f + 1
L:
0 = d + e - 3f - 1
T:
0 = -d - 2
Solving, d = -2, e = 0, and f = - 1. Thus,
Π2 = V -2L0r-1EV =
EV
rV 2
or
Π2 =
rV 2
= M
EV
897
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–46. Continued
Π3 = V gLhrip = ( LgT
-g
)( Lh )( MiL-3i )( ML-1T -2 ) = Mi + 1Lg + h - 3i - 1T -g - 2
M:
0 = i + 1
L:
0 = g + h - 3i - 1
T:
0 = -g - 2
Solving, g = - 2, h = 0, and i = -1. Thus,
p
Π3 = V -2L0r-1p =
= Eu
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
rV 2
Π4 = V jLkrlg = ( L jT -j )( Lk )( MlL-3l )( LT -2 ) = MlL j + k - 3l + 1T -j - 2
M:
0 = l
L:
0 = j + k - 3l + 1
T:
0 = -j - 2
Solving, j = - 2, k = 1, and l = 0. Thus,
Π4 = V -2L1r0g =
or
Π4 =
V
2gL
gL
V2
=
2gL
V
= Fr
Π5 = V mLnrps = ( LmT -m )( Ln )( MpL-3p )( MT -2 ) = Mp + 1Lm + n - 3pT -m - 2
M:
0 = p + 1
L:
0 = m + n - 3p
T:
0 = -m - 2
Solving, m = -2, n = - 1, and p = - 1. Thus,
Π5 = V -2L-1r-1s =
s
rV 2L
or
Π5 =
rV 2L
= We
s
Therefore, the function can be written as
Ans.
f (Re, M, Eu, Fr, We) = 0
Ans:
f (Re, M, Eu, Fr, We) = 0
898
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–47. The discharge Q over a small weir depends upon the
water head H, the width b and height h of the weir, the
acceleration of gravity g, and the density r, viscosity m, and
surface tension s of the fluid. Determine the relation between
Q and these parameters.
H
h
Solution
Physical Variables. There are n = 8 variables and the unknown function is
f (Q, H, b, h, g, r, m, s) = 0. Using the M - L - T system,
L3T -1
g
LT -2
H
L
r
ML-3
b
L
m
ML-1T -1
h
L
s
MT -2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Q
Here, all three base dimensions are used, so that m = 3. Thus, there are
n - m = 8 - 3 = 5 Π terms
Dimensional Analysis. Here, r, g, and H are chosen as m = 3 repeating variables.
Thus, the q variables are Q for Π1, b for Π2, h for Π3, m for Π4, and s for Π5.
Π1 = ragbHcQ = ( MaL-3a )( LbT -2b )( Lc )( LbT -1 ) = MaL-3a + b + c + 3T -2b - 1
M:
0 = a
L:
0 = - 3a + b + c + 3
T:
0 = - 2b - 1
1
5
Solving, a = 0, b = - , and c = - . Thus,
2
2
Q
5
1
Π1 = r0g -2H -2Q =
2gH5
Π2 = rdgeHfb = ( MdL-3d )( LeT -2e )( L f ) (L) = MdL-3d + e + f + 1T -2e
M:
0 = d
L:
0 = - 3d + e + f + 1
T:
0 = - 2e
Solving, d = 0, e = 0, and f = -1. Thus,
Π2 = r0g0H -1b =
b
H
Π3 = rhgiH jh = ( MhL-3h )( LiT -2i )( Lj ) (L) = MhL-3h + i + j + 1T -2i
M:
0 = h
L:
0 = - 3h + i + j + 1
T:
0 = - 2i
Solving, h = 0, i = 0, and j = -1. Thus,
Π3 = r0g0H -1h =
h
H
899
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–47. Continued
Π4 = rkglHmm = ( MkL-3k )( LlT -2l )( Lm )( ML-1T -1 ) = Mk + 1L-3k + l + m - 1T -2l - 1
M:
0 = k + 1
L:
0 = -3k + l + m - 1
T:
0 = -2l - 1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
1
3
Solving, k = -1, l = - , and m = - . Thus,
2
2
m
3
1
Π4 = r-1g - 2H - 2m =
r2gH3
Π5 = rngpHqs = ( MnL-3n )( LpT -2p )( Lq )( MT -2 ) = Mn + 1L-3n + p + qT -2p - 2
M:
0 = n + 1
L:
0 = -3n + p + q
T:
0 = -2p - 2
Solving, n = -1, p = - 1, and q = - 2. Thus,
Π5 = r-1g -1H -2s =
s
rgH2
Therefore, the function can be written as
f1 °
Q
m
b h
s
, ,
,
¢ = 0
2
2gH H H r2gH3 rgH
5
,
Solving for Q,
Q
2gH5
= f°
m
b h
s
, ,
,
¢
H H r2gH3 rgH2
Q = 2gH5f °
m
b h
s
, ,
,
¢
H H r2gH3 rgH2
Ans.
Ans:
Q = 2gH5 f °
900
m
b,h,
, s ¢
3
H H r2gH rgH2
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–48. If water flows through a 50-mm-diameter pipe at
2 m>s, determine the velocity of carbon tetrachloride
flowing through a 60-mm-diameter pipe so that they both
have the same dynamic characteristics. The temperature of
both liquids is 20°C.
50 mm
Solution
Since inertia and viscous forces are predominant, the Reynolds number must be
the same for both cases. From Appendix A, rte = 1590 kg>m3, rw = 998.3 kg>m3,
mte = 0.958 ( 10-3 ) N # s>m2, and mw = 1.00 ( 10-3 ) N # s>m2 at 20° C . This requires
a
rVD
rVD
b = a
b
m w
m te
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
( 998.3 kg>m3 )( 2 m>s ) (0.05 m) ( 1590 kg>m3 ) Vte(0.06 m)
=
1.00 ( 10-3 ) N # s>m2
0.958 ( 10-3 ) N # s>m2
Ans.
V = 1.00 m>s
901
60 mm
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–49. In order to test the flow over the surface of an
airplane wing, a model is built to a scale of 1>15 and is
tested in water. If the airplane is designed to fly at 350
mi>h, what should the velocity of the model be in order
to maintain the same Reynolds number? Is this test
realistic? Take the temperature of both the air and water
to be 60°F.
350 mi/h
Solution
From Appendix A, for water nm = 12.2 ( 10-6 ) ft 2 >s and for air np = 0.158 ( 10-3 ) ft 2 >s
VL
VL
at 60º F. Since, n = m>r, the Reynolds number can be written as Re =
=
.
n
m>r
Thus,
VL
VL
b = a
b
n m
n p
Vm = a
= £
n m Lp
ba
bV
n p Lm p
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
12.2 ( 10-6 ) ft 2 >s
0.158 ( 10-3 ) ft 2 >s
§a
15
b ( 350 mi>h )
1
Ans.
= 405.38 mi>h = 405 mi>hr
No, since the velocity is too large for a water tunnel.
Ans:
405 mi>hr
902
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–50. The model of a river is constructed to a scale of 1>60. If
the water in the river is flowing at 38 ft>s, how fast must the
water flow in the model?
Solution
Since the inertia and gravitational forces are predominant in the river flow, the
Froude numbers for both the model and the prototype must be same.
( Fr ) p = ( Fr ) m
V
2gL
b
m
= a
Vm = Vp
V
2gL
b
p
Lm
1
= ( 38 ft>s )
= 4.906 ft>s = 4.91 ft>s
A Lp
A 60
Ans.
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Ans:
4.91 ft>s
903
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–51. Water flowing through a 100-mm-diameter pipe is
used to determine the loss in pressure when gasoline flows
through a 75-mm diameter pipe at 3 m>s. If the pressure
loss in the pipe transporting water is 8 Pa, determine
the pressure loss in the pipe transporting the gasoline.
Take ng = 0.465 1 10 - 6 2 m2 >s and nw = 0.890 1 10 - 6 2 m2 >s,
rg = 726 kg>m3 , rw = 997 kg>m3.
Solution
For the Reynolds number,
VD
VD
b = a
b
n w
n g
Vw = Vg a
0.890 ( 10-6 ) m2 >s
nw Dg
75 mm
ba
b = (3 m>s) a
ba
b
-6
2
ng Dw
0.465 ( 10 ) m >s 100 mm
= 4.306 m>s
For the Euler number,
a
∆p
rv
2
b = a
w
∆p
rv2
b
∆pg = ∆pw a
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
g
rg
rw
∆pg = 2.83 Pa
ba
Vg 2
Vw2
b = (8 Pa)a
726 kg>m3
3
997 kg>m
ba
3 m>s
4.306 m>s
b
2
Ans.
Ans:
2.83 Pa
904
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–52. The effect of drag on a model airplane is to be
tested in a wind tunnel with a wind speed of 200 mi>h. If a
similar test is performed underwater in a channel, what
should the speed of the water be to achieve the same result
when the temperature is 60°F?
200 mi/h
Solution
Since the viscous and inertia forces are predominant, the Reynolds numbers for
both cases must be the same. Since n = m>r, the Reynold’s numbers can be written
VL
VL
as Re =
=
. Thus,
n
m>r
VL
VL
b = a
b
n w
n a
Vw = a
n w La
ba b ( Va )
n a Lw
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
La
b = 1. From Appendix A,
Lw
ft 2 >s at 60° F. Thus,
Since the same model is used for both cases, a
nw = 12.2 ( 10-6 ) ft 2 >s and na = 0.158 ( 10-3 )
Vw = c
0.0122 ( 10-3 ) ft 2 >s
0.158 ( 10-3 ) ft 2 >s
d (1)(200 mi>h)
Ans.
= 15.44 mi>hr = 15.4 mi>h
905
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–53. When a 100-mm-diameter sphere travels at 2 m>s in
water having a temperature of 15°C, the drag force is 2.80 N.
Determine the velocity and drag force on a 150-mm-diameter
sphere traveling through water under similar conditions.
Solution
Since the Reynolds number involves V, it will be used to determine the velocity of
the 150-mm diameter sphere. Since v = m>r, the Reynold’s numbers can be written
VD
VD
as Re =
=
. Thus,
m>r
v
VD
VD
b = a
b
v 2
v 1
V2 = a
v2 D1
ba bV
v1 D2 1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Since both spheres move in the same medium (water at 15º C), a
V2 = (1)a
v2
b = 1. Thus,
v1
100 mm
b ( 2 m>s ) = 1.333 m>s = 1.33 m>s
150 mm
Ans.
F
F
∝ 2 , the Euler numbers, which involve p, can be used
A
D
to determine the drag force on the 150-mm diameter sphere. This gives
Subsequently, since p =
a
a
p
rV 2
2
p
rV 2
b
1
F
F
b = a 2 2b
rD2V 2 2
rD V 1
F2 = a
Here,
b = a
r2 D2 2 V2 2
ba b a b F1
r1 D1
V1
r2
= 1 since both spheres move in the same medium (water at 15º C). Thus,
r1
F2 = (1)a
150 mm 2 1.333 m>s 2
b a
b (2.80 N)
100 mm
2 m>s
Ans.
= 2.80 N
Ans:
V2 = 1.33 m>s
F2 = 2.80 N
906
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–54. In order to determine the formation of waves around
obstructions in a river, a model having a scale of 1>10 is used.
If the river flows at 6 ft>s, determine the speed of the water
for the model.
Solution
For the river, the inertia and gravitational forces are predominant. Thus, the equality
of the Froude number will be used.
V
2gL
b
m
Vm =
= a
V
2gL
b
p
Lm
Vp
B Lp
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
=
1
( 6 ft>s ) = 1.897 ft>s = 1.90 ft>s
A 10
Ans.
Ans:
1.90 ft>s
907
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–55. The optimum performance of mixing blades 0.5 m in
diameter is to be tested using a model one-fourth the size of
the prototype. If the test of the model in water reveals the
optimum speed to be 8 rad>s, determine the optimum
angular speed of the prototype when it is used to mix ethyl
alcohol. Take T = 20°C.
0.25 m
0.25 m
Solution
rVD
vD
and n = m>r, the Reynolds number can be written as Re =
=
m
2
(vD>2)(D)
vD2
=
. Thus,
m>r
2n
Since V =
vD2
vD2
b = a
b
2n p
2n m
vp = a
np
nm
ba
Dm 2
b vm
Dp
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
From Appendix A, np = 1.51 ( 10-6 ) m2 >s and nm = 1.00 ( 10-6 ) m2 >s . Thus,
vp = c
1.51 ( 10-6 ) m2 >s
1.00 ( 10
= 0.755 rad>s
-6
)
1 2
d a b (8 rad>s)
4
Ans.
Ans:
0.755 rad>s
908
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–56. The flow of water around the structural support is
1.2 m>s when the temperture is 5°C. If it is to be studied
using a model built to a scale of 1>20, and using water at a
temperature of 25°C, determine the velocity of the water
used with the model.
1.2 m/s
Solution
a
VL
VL
b = a
b
n m
n p
Using Appendix A
Vm = Vp a
n m Lp
ba
b
n p Lm
0.898 ( 10-6 ) m2 >s
ba
20
b
1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
= 1.2 m>s a
= 14.2 m>s
1.52 ( 10-6 ) m2 >s
Ans.
909
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–57. A model of a ship is built to a scale of 1>20. If the
ship is to be designed to travel at 4 m>s, determine the
speed of the model in order to maintain the same Froude
number.
Solution
V
2gL
Vm =
=
b
m
= a
Lm
Vp
B Lp
V
2gL
b
p
1
( 4 m>s ) = 0.8944 m>s = 0.894 m>s
A 20
Ans.
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Ans:
0.894 m>s
910
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–58. The flow around the airplane flying at an altitude of
10 km is to be studied using a wind tunnel and a model that
is built to a 1>15 scale. If the plane has an air speed of
800 km>h, what should the speed of the air be inside the
tunnel? Is this reasonable?
800 km/h
Solution
The air flow around the airplane causes the inertia and viscous forces to be
predominant. Thus, the Reynolds number will be used. Since n = m>r, the Reynolds
VL
VL
number can be written as Re =
=
.
n
m>r
a
VL
VL
b = a
b
n m
n p
Lp
ba
nm
bV
np p
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Vm = a
Lm
From Appendix A, np = 35.25 ( 10-6 ) m2 >s at an altitude of 10 km and
nm = 14.61 ( 10-6 ) m2 >s at ground level. Thus,
Vm = a
-6
2
15 14.61 ( 10 ) m >s
bc
d (800 km>h)
1
35.25 ( 10-6 ) m2 >s
Vm = 4973.61 km>h = 4.97 Mm>h
Ans.
No, since a wind speed of Vm = 4.97 Mm>h is extremely difficult to achieve. Also, it
is greater than the speed of sound, and so the results would not be valid.
Ans:
4.97 Mm>h
911
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–59. The model of an airplane has a scale of 1>30. If the
drag force on the prototype is to be determined when the
plane is flying at 600 km>h, find the speed of the air in a
wind tunnel for the model if the air has the same temperature
and pressure. Is it reasonable to do this test?
Solution
Since the air flow around the airplane causes the inertia and viscous forces to be
predominant, the Reynolds number will be used. Since n = m>r, the Reynolds
VL
VL
number can be written as Re =
=
.
n
m>r
VL
VL
b = a
b
n m
n p
Vm = a
n m Lp
ba
bV
n p Lm p
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Since the air has the same pressure and temperature in both cases, a
Vm = (1)a
30
b(600 km>h)
1
nm
b = 1. Thus,
np
Ans.
= 18 000 km>h = 18 Mm>h
No, Vm = 18 Mm>h is too fast to achieve, and since it is far greater than the speed of
sound the results would not be valid even if the test could be done.
Ans:
18 Mm>h
912
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–60. The resistance of waves on a 250-ft-long ship is tested
in a channel using a model that is 15 ft long. If the ship travels
at 35 mi>h, what should be the speed of the model to resist
the waves?
35 mi/h
Solution
The wave force on the ship causes the inertia and gravitational forces to be
predominant. Thus, the Froude number will be
V
2gL
Vm =
Here,
b
m
V
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
= a
Lm
V
A Lp p
2gL
b
p
Lm
15 ft
3
=
. Thus,
=
Lp
250 ft
50
Vm =
3
(35 mi>h)
A 50
Ans.
= 8.573 mi>h = 8.57 mi>h
913
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–61. A model of a submarine is built to a scale of 1>25
and tested in a wind tunnel at an airspeed of 150 mi>h. What
is the intended speed of the prototype if it is in water at the
same temperature of 60°F?
Solution
a
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
The flow of fluid around the submarine causes the inertia and viscous forces to be
predominant. Thus, the Reynolds number will be used. Since n = m>r, the Reynolds
VL
VL .
number can be written as Re =
=
n
m>r
VL
VL
b = a
b
n p
n m
Vp = Vma
Lm n p
ba b
Lp n m
From Appendix A, nm = 0.158 ( 10-3 ) ft 2 >s and np = 12.2 ( 10-6 ) ft 2 >s. Thus,
Vp = ( 150 mi>h ) a
12.2 ( 10-6 ) ft 2 >s
1
b£
§
25 0.158 ( 10-3 ) ft 2 >s
Ans.
Vp = 0.4633 mi>h = 0.463 mi>h
Ans:
0.463 mi>h
914
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–62. The flow of water around the bridge pier is to be
studied using a model built to a scale of 1>15. If the river
flows at 0.8 m>s, determine the corresponding velocity of
the water in the model at the same temperature.
0.8 m/s
Solution
Water flow of fluid around the pier causes the inertia and viscous forces to be
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
predominant. Thus, the Reynolds number will be used. Since n = m>r, the Reynolds
number can be written as Re =
a
VL
VL
=
.
n
m>r
VL
VL
b = a
b
n m
n p
Vm = a
n m Lp
ba
bV
n p Lm p
Since water is used for the model and the prototype, a
Vm = (1)a
15
b ( 0.8 m>s )
1
= 12 m>s
nm
b = 1. Thus,
np
Ans.
Ans:
12 m>s
915
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–63. The resistance created by waves on a 100-m-long
ship is tested in a channel using a model that is 4 m long. If
the ship travels at 60 km>h, what should be the speed of
the model?
Solution
°
n
2gL
Vm =
Here,
¢
m
= °
Lm
Vp
B Lp
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
The wave force on the ship causes the inertia and gravitational forces to be
predominant. Thus, the Froude number will be
n
2gL
¢
p
Lm
4m
1
=
=
. Thus,
Lp
100 m
25
Vm =
1
( 60 km>h )
A 25
= 12 km>h
Ans.
Ans:
12 km>h
916
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–64. The velocity of water waves in a channel are
studied in a laboratory using a model of the channel onetwelfth its actual size. Determine the velocity of waves in
the channel if they have a velocity of 6 m>s in the model.
6 m/s
Solution
°
n
2gL
Vp =
Here,
Lp
Lm
=
¢ = °
p
Lp
B Lm
Vm
12
. Thus,
1
Vp =
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
The motion of the wave causes the inertia and gravitational forces to be predominant.
Thus, the equality of the Froude numbers will require
n
2gL
¢
m
12
( 6 m>s )
A1
Ans.
= 20.78 m>s = 20.8 m>s
917
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–65. A model of a submarine is built to determine the
drag force acting on its prototype. The length scale is 1>100,
and the test is run in water at 20°C, with a speed of 8 m>s. If
the drag force on the model is 20 N, determine the drag
force on the prototype if it runs in water at the same speed
and temperature. This requires that the drag coefficient
CD = 2FD >rV 2L2 be the same for both the model and the
prototype.
8 m/s
Solution
The requirement is
2FD
rV 2L2
b = a
p
( FD ) p = a
rp
rm
2FD
rV 2L2
ba
b
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Vp
Vm
2
m
b a
Lp
Lm
2
b ( FD ) m
Since the model and prototype run with the same speed in water having the same
rp
Vp
Lp
100
=
= 1. Here,
= a
b . Thus,
temperature,
rm
Lm
1
Vm
( FD ) p = (1)(1)2 a
100 2
b ( 20 N )
1
= 200 ( 103 ) N = 200 kN
Ans.
Ans:
200 kN
918
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–66. A model of a plane is built to a scale of 1>15 and is
tested in a wind tunnel. If the plane is designed to travel at
800 km>h at an altitude of 5 km, determine the required
density of the air in the wind tunnel so that the Reynolds
and Mach numbers are the same. Assume the temperature
is the same in both cases and the speed of sound in air at
this temperature is 340 m>s.
Solution
Using the Mach number,
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
V
V
a b = a b
c m
c p
The speed of sound in air having the same temperature is the same, cm = cp. Thus,
Vm = VP
Using the Reynolds number,
a
rVL
rVL
b = a
b
m m
m p
rm = a
mm VP LP
ba ba
br
mP Vm Lm p
For air at the same temperature, mm = mp. From Appendix A, rp = 0.7364 kg>m3 at
an altitude of 5 km. Thus,
rm = (1)(1)a
15
b ( 0.7364 kg>m3 ) = 11.046 kg>m3 = 11.0 kg>m3
1
Ans.
Note: The result is not reasonable, since the value of rm is not possible with air in
realistic conditions.
Ans:
11.0 kg>m3
919
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–67. The motion of water waves in a channel are to be
studied in a laboratory using a model one-twelfth the size of
the channel. Determine the time for a wave in the channel
to travel 10 m if it takes 15 seconds for the wave to travel
this distance in the model.
Solution
The motion of the wave causes the inertia and gravitational forces to be predominant,
so the Froude number will be used.
V
2gL
Vm =
Here, Vm =
b
m
= a
Lm
B Lp
Vp
V
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
2gL
b
p
sp
sm
10 m
10 m
=
= 0.6667 m>s and Vp =
=
. Then,
tm
15 s
tp
tp
0.6667 m>s =
1 10 m
a
b
A 12 t p
Ans.
t p = 4.330 s = 4.33 s
Ans:
4.33 s
920
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–68. It is required that a pump be designed for use in a
chemical plant such that it delivers 0.8 m3 >s of benzene with
a pressure increase of 320 kPa. What is the expected flow and
pressure increase produced by a model one-sixth the size of
the prototype? If the model produces a power output of
900 kW, what would be the power output of the prototype?
Solution
Since the viscous force is the dominant force, then Reynolds number similitude must
be achieved.
rVL
rVL
b = a
b
m m
m p
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Since benzene will be used for both model and prototype, rm = rp and mm = mp.
Then
VmLm = VpLp
Lp
Vm
=
Vp
Lm
(1)
Since Q = VA and A has the dimension of L2, Eq. 1 becomes
Qm >Lm2
Qp >Lp2
=
Lp
Lm
2
Lp
Q m Lp
a 2b = a
b
Q p Lm
Lm
Qm = Qp a
Lp
Lm
ba
Here, Qp = 0.8 m3 >s. and
Lm 2
Lm
b = Qp a
b
Lp
Lp
Lm
1
= . Then
Lp
6
1
Qm = ( 0.8 m3 >s ) a b = 0.133 m3 >s
6
Ans.
For the pressure comparison, Euler number similitude should be used since the
pressure change is involved.
a
∆P
∆P
b = a 2b
2
rV m
rV p
Since rm = rp,
(∆P)m
Vm2
=
(∆P)p
Vp 2
(∆P)m = (∆P)pa
Vm 2
b
Vp
921
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–68. Continued
Substituting Eq 1 into this equation,
Lp
Lm
b
Lp
2
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
(∆P)m = (∆P)pa
Here (∆P)p = 320 kPa and
Lm
= 6,
(∆P)m = (320 kPa ) ( 62 ) = 11520 kPa = 11.5 MPa
As shown earlier,
#
W = CQ∆p
So that
#
#
Wp
Ans.
Wm
=
Qp(∆p)p Qm(∆p)m
and
#
Here,
#
Wp = Wm a
Qp
Qm
ba
(∆p)p
(∆p)m
#
b = Wm a
Lp
Lm
ba
# Lm
Lm 2
b = Wma
b
Lp
Lp
#
1
W = 900 kW a b = 150 kW
6
Ans.
922
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–69. If the jet plane can fly at Mach 2 in air at 35°F,
determine the required speed of wind generated in a wind
tunnel at 65°F and used on a model built to a scale of 1>25.
Hint: Use Eq. 13–24, c = 2kRT, where k = 1.40 for air.
Solution
For air, Appendix A gives R = 1716 ft # lb>slug # R. Thus,
2k RTm
Vm
2Tm
=
=
Vp
Vp
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Vm
2k RTp
2Tp
Vp = 2 ( 2kRT ) p
Vp = 221.40 ( 1716 ft # lb>slug # R ) (35° + 460) R
= 2181.0 ft>s = 2.18 ( 103 ) ft>s
Vm
265° + 460
=
Ans.
2181.0
235° + 460
Vm = 2246.1 ft>s = 2.25 ( 103 ) ft>s
Ans.
Ans:
2.25 ( 103 ) ft>s
923
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–70. The drag coefficient on an airplane is defined by
CD = 2FD >rV 2L2. If the drag force acting on the model of a
plane tested at sea level is 0.3 N, determine the drag force on
the prototype, which is 15 times larger and is flying at
20 times the speed of the model at an altitude of 3 km.
Solution
The requirement is
2FD
2 2
rV L
b = a
p
2FD
rV 2L2
b
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
( FD ) p = ( FD ) m = a
m
rp
rm
ba
Vp
Vm
Lp
Lm
b
2
From Appendix A, rp = 0.9092 kg>m at an altitude of 3 km and rm = 1.225 kg>m3
at sea level. Thus
( FD ) p = (0.3 N)a
3
2
b a
0.9092 kg>m3
1.225 kg>m3
ba
= 20.04 ( 103 ) N = 20.0 kN
20Vm 2 15Lm 2
b a
b
Vm
Lm
Ans.
Ans:
20.0 kN
924
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–71. The model of a hydrofoil boat is to be tested in a
channel. The model is built to a scale of 1>20. If the lift
produced by the model is 7 kN, determine the lift on the
prototype. Assume the water temperature is the same in
both cases. This requires Euler number and Reynolds
number similarity.
Solution
For the Reynolds number,
rVL
rVL
b = a
b
m p
m m
Vp
Vm
= a
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
rm mp Lm
ba ba
b
rp mm Lp
At the same temperature, rp = rm and mm = mp. Thus,
Vp
Vm
=
Lm
1
=
Lp
20
F>L2
F
F
,
the
Euler
number
can
be
written
as
Eu
=
=
.
2
2
L
rV
rV 2L2
The Euler number gives
Since p = F>A =
a
F
F
b = a 2 2b
2 2
rV L p
rV L m
Fp = a
rp
rm
= (1)a
ba
Vp
Vm
2
b a
Lp
Lm
2
b Fm
1 2 20 2
b a b (7 kN)
20
T
= 7 kN
Ans.
Ans:
7 kN
925
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–72. The model of a boat is built to a scale of 1>50.
Determine the required kinematic viscosity of the water in
order to test the model so that the Froude and Reynolds
numbers remain the same for the model and the prototype.
Is this test practical if the prototype operates in water at
T = 20°C?
Solution
For the Froude number
V
2gL
b
m
= a
Lm
Vm
=
Vp
B Lp
V
2gL
For the Reynolds number,
a
b
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
p
VL
VL
b = a
b
n p
n m
3
nm = a
Vm Lm
Lm Lm
Lm 2
ba
bnp =
a
bnp = a
b np
Vp
Lp
Lp
B Lp Lp
nm = a
1 2
b 3 1.00 ( 10-6 ) m2 >s 4
50
From Appendix A, np = 1.00 ( 10-6 ) m2 >s. at T = 20° C . Thus,
3
nm = 2.828 ( 10-9 ) m2 >s = 2.83 ( 10-9 ) m2 >s
Ans.
No, this value is too low to be practical.
926
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–73. If an airplane flies at 800 mi>h at an altitude of
5000 ft, what should its speed be so that it has the same
Mach number when it is at 15 000 ft? Assume the air has the
same bulk modulus. Use Eq. 13–25, c = 2EV >r.
800 mi/h
RES
CUE
RES
CUE
51204271
Solution
For the Mach number,
V2 = a
c2
bV
c1 1
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
V
V
a b = a b
c 2
c 1
EV
r1
EV
c2
B r2
Since c =
and EV is constant, then
=
=
. Thus,
c1
A r2
B r
EV
B r1
V2 =
r1
V
A r2 1
From Appendix A, r1 = 2.043 ( 10-3 ) slug>ft 3 at an altitude of 5000 ft and
r2 = 1.495 ( 10-3 ) slug>ft 3 at an altitude of 15 000 ft . Thus,
V2 = £
2.043 ( 10-3 ) slug>ft 3
B 1.495 ( 10-3 ) slug>ft 3
§ ( 800 mi>h )
Ans.
= 935.20 mi>h = 935 mi>h
Ans:
935 mi>h
927
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–74. A 60-ft-long “check dam” on a river provides a
means of collecting debris that flows downstream. If the
flow over the dam is 8000 ft 3 >s, and a model of this dam is
to be built to a scale of 1>20, determine the flow over the
model and the depth of water that flows over its crest.
Assume that the water temperature for the prototype and
the model is the same. The volumetric flow over the dam
can be determined using Q = CD 2gLH 3>2, where CD is
the coefficient of discharge, g the acceleration of gravity, L
is the length of the dam, and H is the height of the water
above the crest. Take CD = 0.71.
60 ft
Solution
Since the gravitational force is the dominant force, then Froude number similitude
must be achieved.
V
2gL
b
m
Here gm = gp. Thus
Vm
2Lm
=
= a
V
2gL
Vp
2Lp
1
b
p
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
a
Lm 2
Vm
= a
b
Vp
Lp
Since V = Q>A and A has a dimension of L2, Eq 1 becomes
Qm >Lm2
Qp >Lp 2
(1)
1
= a
Lm 2
b
Lp
1
Q m Lp 2
Lm 2
a
b = a
b
Q p Lm
Lp
1
Qm
Lm 2 Lm 2
= a
ba
b
Qp
Lp
Lp
5
Qm
Lm 2
= a
b
Qp
Lp
Qm = Qp a
Here
5
Lm 2
b
Lp
Lm
1
=
and Qp = 8000 ft 3 >s, then
Lp
20
5
Qm
1 2
= ( 8000 ft >s ) a b = 4.47 ft 3 >s
20
3
Ans.
The height over the dam is
3
3
Q = CD 2g LH2; 8000 ft 3 >s = 0.71232.2(60 ft) ( Hp2 )
Hp = 10.31 ft
Here
Hm
Lm
Hm
1
=
; =
Hp
Lp
10.31 ft
20
Ans.
Hm = 0.515 ft
Ans:
Qm = 4.47 ft 3 >s
Hm = 0.515 ft
928
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–75. A ship has a length of 180 m and travels in the sea
where rs = 1030 kg>m3. A model of the ship is built to a 1
>60 scale, and it displaces 0.06 m3 of water such that its hull
has a wetted surface area of 3.6 m2. When tested in a towing
tank at a speed of 0.5 m>s, the total drag on the model was
2.25 N. Determine the drag on the ship and its corresponding
speed. What power is needed to overcome this drag? The
drag due to viscous (frictional) forces can be determined
using (FD)f = 1 12 rV 2A 2 CD, where CD is the drag coefficient
determined from CD = 1.328> 2Re for Re 6 106 and
CD = 0.455> 1log 10Re2 2.58 for 106 6 Re 6 109. Take r =
1000 kg>m3 and n = 1.00 1 10 - 6 2 m2 >s.
Solution
Using the scale,
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Lm
Lm
1
1
=
; =
Lm = 3 m
Lp
60
180 m
60
For Froude number similitude,
a
V
2gL
b
m
Since g is a constant,
= a
1
Lp 2
VP
= a
b
Vm
Lm
Here,
Lp
Lm
Vp = a
Lp
Lm
1
2
V
2gL
b
p
b Vm
= 60 and Vm = 0.5 m>s . Then
1
Vp = ( 602 )( 0.5 m>s ) = 3.873 m>s = 3.87 m>s
Ans.
Next, we will compute the frictional drag force. Here, Vm = 0.5 m>s, Lm = 3 m and
nm = 1.00 ( 10-6 ) m2 >s . Then
(Re)m =
(0.5 m>s)(3 m)
VmLm
=
= 1.5 ( 10-6 )
nm
1.00 ( 10-6 ) m2 >s
Since 106 6 (Re)m 6 109,
( CD ) m =
3( FD ) f 4 m
0.455
3 log 10 ( Re ) m 4
2.58
=
0.455
3 log 101.5 ( 10 ) 6 4 2.58
= 4.1493 ( 10-3 )
1
1
= a rm Vm2 Am b ( CD ) m = c ( 1000 kg>m3 )( 0.5 m>s ) 2 ( 3.6 m2 ) d 3 4.1493 ( 10-3 ) 4
2
2
= 1.867 N
929
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–75. Continued
Thus, the drag force due to the wave action on the model is
3 ( FD ) g 4 m
= FD -
3 ( FD ) f 4 m
= 2.25 N - 1.867 N = 0.3828 N
T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)
Using the equation given in the text
3 ( FD ) g 4 p
=
=
=
3 ( FD ) g 4 m a
3 ( FD ) g 4 m a
3 ( FD ) g 4 m a
rp
rm
rp
rm
rp
rm
= ( 0.3828 N ) °
ba
ba
ba
Vp
Vm
Lp
Lm
Lp
Lm
2
b a
ba
b
Lm
Lp
Lm
3
1030 kg>m3
1000 kg>m3
= 85.17 ( 103 ) N
Lp
b
b
2
2
¢ ( 603 )
The frictional drag force on the prototype must be determined. Here, Vp = 3.873 m>s,
Lp = 180 m and np = 1.00 ( 10-6 ) m2 >s . Then
(Re)p =
VpLp
np
Since 106 6 (Re)p 6 109,
( CD ) p =
Here, Ap = Am a
3 ( FD ) f 4 p
=
( 3.873 m>s )( 180 m )
= 0.6971 ( 109 )
1.00 ( 10-6 ) m2 >s
0.455
3 log 10 ( Re ) p 4 2.58
Lp
Lm
2
=
0.455
3 log 100.6971 ( 109 ) 4 2.58
= 1.6434 ( 10-3 )
b = ( 3.6 m2 )( 602 ) = 12960 m2. Then
1
1
= a rp Vp2 Ap b ( CD ) p = c ( 1030 kg>m3 )( 3.873 m>s ) 2 ( 12960 m2 ) d 3 1.6434 ( 10-3 ) 4
2
2
= 164.53 ( 103 ) N
Thus, the total drag force is
( FD ) p = 3 ( FD ) g 4 p + 3 ( FD ) f 4 p = 85.17 ( 103 ) N + 164.53 ( 103 ) N
= 249.70 ( 103 ) N = 250 kN
The power is
#
W = ( FD ) pVp =
3 249.70 ( 103 ) N 4 ( 3.873 m>s )
Ans.
Ans:
Vp = 3.87 m>s
= 967.09 ( 103 ) W
= 967 kN
Ans.
( FD ) p = 250 kN
#
W = 967 kN
930