Solutions Manual for
Fluid Mechanics
Seventh Edition in SI Units
Frank M. White
Chapter 5
Dimensional Analysis
and Similarity
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of The McGraw-Hill Companies, Inc.
(“McGraw-Hill”) and protected by copyright and other state and federal laws. By
opening and using this Manual the user agrees to the following restrictions, and if the
recipient does not agree to these restrictions, the Manual should be promptly returned
unopened to McGraw-Hill: This Manual is being provided only to authorized
professors and instructors for use in preparing for the classes using the affiliated
textbook. No other use or distribution of this Manual is permitted. This Manual
may not be sold and may not be distributed to or used by any student or other
third party. No part of this Manual may be reproduced, displayed or distributed in
any form or by any means, electronic or otherwise, without the prior written
permission of the McGraw-Hill.
© 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators
for course preparation. If you are a student using this Manual, you are using it without permission.
,5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is
approximately 2300 [see Eq. (6.2)], based upon the diameter and average velocity. If d = 5 cm
and the fluid is kerosene at 20°C, find the volume flow rate in m3/h which causes transition.
Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ = 0.00192 kg/m⋅s. The only
unknown in the transition Reynolds number is the fluid velocity:
ρ Vd (804)V(0.05)
Re tr ≈ 2300 = = , solve for Vtr = 0.11 m/s
µ 0.00192
π m3 m3
Then Q = VA = (0.11) (0.05)2 = 2.16E−4 × 3600 ≈ 0.78 Ans.
4 s hr
P5.2 A prototype automobile is designed for cold weather in Denver, CO (-10°C, 83 kPa).
Its drag force is to be tested in on a one-seventh-scale model in a wind tunnel at 150 mi/h and
at 20°C and 1 atm. If model and prototype satisfy dynamic similarity, what prototype
velocity, in mi/h, is matched? Comment on your result.
Solution: First assemble the necessary air density and viscosity data:
p 83000 kg kg
Denver : T = 263K ; ρ p = = = 1.10 3 ; µ p = 1.75 E − 5
RT 287(263) m m−s
p 101350 kg kg
Wind tunnel : T = 293K ; ρ m = = = 1.205 3 ; µ m = 1.80 E − 5
RT 287(293) m m−s
Convert 150 mi/h = 67.1 m/s. For dynamic similarity, equate the Reynolds numbers:
ρVL (1.10)Vp (7Lm ) ρVL (1.205)(67.1)(Lm )
Re p = |p = = Re m = |m =
µ 1.75E − 5 µ 1.80E − 5
Solve for Vprototype = 10.2 m / s = 22.8 mi / h Ans.
This is too slow, hardly fast enough to turn into a driveway. Since the tunnel can go no faster,
the model drag must be corrected for Reynolds number effects. Note that we did not need to
know the actual length of the prototype auto, only that it is 7 times larger than the model
length.
,2
P5.3 The transfer of energy by viscous dissipation is dependent upon viscosity µ,
thermal conductivity k, stream velocity U, and stream temperature To. Group these quantities,
if possible, into the dimensionless Brinkman number, which is proportional to µ.
Solution: Here we have only a single dimensionless group. List the dimensions, from Table
5.1:
µ k U To
{ML-1T-1} {MLT-3Θ-1} {LT-1} {Θ}
Four dimensions, four variables (MLTΘ) – perfect for making a pi group. Put µ in the
numerator:
Brinkman number = k a U b Toc µ1 yields Br = µ U 2 / (kTo ) Ans.
5.4 When tested in water at 20°C flowing at 2 m/s, an 8-cm-diameter sphere has a measured
drag of 5 N. What will be the velocity and drag force on a 1.5-m-diameter weather balloon
moored in sea-level standard air under dynamically similar conditions?
Solution: For water at 20°C take ρ ≈ 998 kg/m3 and µ ≈ 0.001 kg/m⋅s. For sea-level standard
air take ρ ≈ 1.2255 kg/m3 and µ ≈ 1.78E-5 kg/m⋅s. The balloon velocity follows from dynamic
similarity, which requires identical Reynolds numbers:
ρVD
Re model = |model = 998(2.0)(0.08) = 1.6E5 = Reproto = 1.2255Vballoon (1.5)
µ 0.001 1.78E−5
or Vballoon ≈ 1.55 m/s. Ans. Then the two spheres will have identical drag coefficients:
F 5N Fballoon
CD,model = 2 2
= 2 2
= 0.196 = CD,proto =
ρV D 998(2.0) (0.08) 1.2255(1.55)2 (1.5)2
Solve for Fballoon ≈ 1.3 N Ans.
5.5 An automobile has a characteristic length and area of 2.45 m and 5.57 m2, respectively.
When tested in sea-level standard air, it has the following measured drag force versus speed:
V, km/h: 32 64 96
Drag, N: 137.9 511.5 1107.6
The same car travels in Colorado at 104 km/h at an altitude of 3500 m. Using dimensional
analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag.
Solution: For sea-level air in SI units, take ρ ≈ 1.23 kg/m 3 and µ ≈ 1.79 ×10 −5 N ⋅ s/m 2 .
Convert the raw drag and velocity data into dimensionless form:
V (km/h): 32 64 96
CD = F/(ρV2L2): 0.236 0.219 0.211
ReL = ρVL/µ: 1.50E6 3.00E6 4.50E6
, 3
Drag coefficient plots versus Reynolds number in a very smooth fashion and is well fit (to ±1%)
by the Power-law formula CD ≈ 1.07ReL− 0.106.
(a) The new velocity is V = 29.06 m/s, and for air at 3500-m Standard Altitude
(Table A-6) take ρ = 0.8633 kg/m 3 and calculate µ from power-law
0.7
µ T
= , µ ≈ 1.68 ×10 −5 N ⋅ s/m 2
µ 0 T0
Then compute the new Reynolds number and use our power-law above to estimate drag coefficient:
ρVL ( 0.8633) ( 29.06 ) ( 2.45 )
ReColorado = = −5
= 3.66 ×10 6 , hence
µ 1.68 ×10
1.07 2 2
CD ≈ 0.106
= 0.2156, ∴ F = 0.2156 ( 0.8633)( 29.06 ) 2( 2.45 ) ( 8.0 ) = 943.5 N Ans. (a)
(3.66E6)
(b) The horsepower required to overcome drag is
Power = FV = ( 943.5 ) ( 29.06 ) = 27417.6 W ÷ 550 = 36.77 hp Ans. (b)
P5.6 The full-scale parachute in the chapter-opener photo had a drag force of
approximatly 4225 N when tested at a velocity of 19 km/h in air at 20°C and 1 atm. Earlier, a
model parachute of diameter 1.7 m was tested in the same tunnel. (a) For dynamic similarity,
what should be the air velocity for the model? (b) What is the expected drag force of the
model? (c) Is there anything surprising about your result to part (b)?
Solution: (a) From Table A.2 for air at 20C, ρ = 1.20 kg/m3 and µ = 1.8E-5 kg/m-s. (a) For
similarity, equate the Reynolds numbers:
ρ pV p D p (1.20)(5.28)(16.8) ρ V D (1.20)(Vm )(1.7)
Re p = = = 5.91E6 = Re m = m m m =
µp 1.8E − 5 µm 1.8E − 5
m km
Solve for Vm = 52.2 = 187.8 Ans.(a)
s h
(b) For similarity, the force coefficients will be equal:
Fp 4225 N Fm Fm
C Fp = = = 0.447 = =
ρ pV p2 D 2p 2
(1.20)(5.28) (16.8) 2
ρ mVm2 Dm2 (1.20)(52.2)2 (1.7)2
Solve for Fm = 4225 N Ans.(b)
(c) It might be surprising that the drag forces are exactly the same for model and prototype!
This is because, if ρ and µ are the same, the product VD is the same for both and the force is
proportional to (VD)2.
Fluid Mechanics
Seventh Edition in SI Units
Frank M. White
Chapter 5
Dimensional Analysis
and Similarity
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of The McGraw-Hill Companies, Inc.
(“McGraw-Hill”) and protected by copyright and other state and federal laws. By
opening and using this Manual the user agrees to the following restrictions, and if the
recipient does not agree to these restrictions, the Manual should be promptly returned
unopened to McGraw-Hill: This Manual is being provided only to authorized
professors and instructors for use in preparing for the classes using the affiliated
textbook. No other use or distribution of this Manual is permitted. This Manual
may not be sold and may not be distributed to or used by any student or other
third party. No part of this Manual may be reproduced, displayed or distributed in
any form or by any means, electronic or otherwise, without the prior written
permission of the McGraw-Hill.
© 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators
for course preparation. If you are a student using this Manual, you are using it without permission.
,5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is
approximately 2300 [see Eq. (6.2)], based upon the diameter and average velocity. If d = 5 cm
and the fluid is kerosene at 20°C, find the volume flow rate in m3/h which causes transition.
Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ = 0.00192 kg/m⋅s. The only
unknown in the transition Reynolds number is the fluid velocity:
ρ Vd (804)V(0.05)
Re tr ≈ 2300 = = , solve for Vtr = 0.11 m/s
µ 0.00192
π m3 m3
Then Q = VA = (0.11) (0.05)2 = 2.16E−4 × 3600 ≈ 0.78 Ans.
4 s hr
P5.2 A prototype automobile is designed for cold weather in Denver, CO (-10°C, 83 kPa).
Its drag force is to be tested in on a one-seventh-scale model in a wind tunnel at 150 mi/h and
at 20°C and 1 atm. If model and prototype satisfy dynamic similarity, what prototype
velocity, in mi/h, is matched? Comment on your result.
Solution: First assemble the necessary air density and viscosity data:
p 83000 kg kg
Denver : T = 263K ; ρ p = = = 1.10 3 ; µ p = 1.75 E − 5
RT 287(263) m m−s
p 101350 kg kg
Wind tunnel : T = 293K ; ρ m = = = 1.205 3 ; µ m = 1.80 E − 5
RT 287(293) m m−s
Convert 150 mi/h = 67.1 m/s. For dynamic similarity, equate the Reynolds numbers:
ρVL (1.10)Vp (7Lm ) ρVL (1.205)(67.1)(Lm )
Re p = |p = = Re m = |m =
µ 1.75E − 5 µ 1.80E − 5
Solve for Vprototype = 10.2 m / s = 22.8 mi / h Ans.
This is too slow, hardly fast enough to turn into a driveway. Since the tunnel can go no faster,
the model drag must be corrected for Reynolds number effects. Note that we did not need to
know the actual length of the prototype auto, only that it is 7 times larger than the model
length.
,2
P5.3 The transfer of energy by viscous dissipation is dependent upon viscosity µ,
thermal conductivity k, stream velocity U, and stream temperature To. Group these quantities,
if possible, into the dimensionless Brinkman number, which is proportional to µ.
Solution: Here we have only a single dimensionless group. List the dimensions, from Table
5.1:
µ k U To
{ML-1T-1} {MLT-3Θ-1} {LT-1} {Θ}
Four dimensions, four variables (MLTΘ) – perfect for making a pi group. Put µ in the
numerator:
Brinkman number = k a U b Toc µ1 yields Br = µ U 2 / (kTo ) Ans.
5.4 When tested in water at 20°C flowing at 2 m/s, an 8-cm-diameter sphere has a measured
drag of 5 N. What will be the velocity and drag force on a 1.5-m-diameter weather balloon
moored in sea-level standard air under dynamically similar conditions?
Solution: For water at 20°C take ρ ≈ 998 kg/m3 and µ ≈ 0.001 kg/m⋅s. For sea-level standard
air take ρ ≈ 1.2255 kg/m3 and µ ≈ 1.78E-5 kg/m⋅s. The balloon velocity follows from dynamic
similarity, which requires identical Reynolds numbers:
ρVD
Re model = |model = 998(2.0)(0.08) = 1.6E5 = Reproto = 1.2255Vballoon (1.5)
µ 0.001 1.78E−5
or Vballoon ≈ 1.55 m/s. Ans. Then the two spheres will have identical drag coefficients:
F 5N Fballoon
CD,model = 2 2
= 2 2
= 0.196 = CD,proto =
ρV D 998(2.0) (0.08) 1.2255(1.55)2 (1.5)2
Solve for Fballoon ≈ 1.3 N Ans.
5.5 An automobile has a characteristic length and area of 2.45 m and 5.57 m2, respectively.
When tested in sea-level standard air, it has the following measured drag force versus speed:
V, km/h: 32 64 96
Drag, N: 137.9 511.5 1107.6
The same car travels in Colorado at 104 km/h at an altitude of 3500 m. Using dimensional
analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag.
Solution: For sea-level air in SI units, take ρ ≈ 1.23 kg/m 3 and µ ≈ 1.79 ×10 −5 N ⋅ s/m 2 .
Convert the raw drag and velocity data into dimensionless form:
V (km/h): 32 64 96
CD = F/(ρV2L2): 0.236 0.219 0.211
ReL = ρVL/µ: 1.50E6 3.00E6 4.50E6
, 3
Drag coefficient plots versus Reynolds number in a very smooth fashion and is well fit (to ±1%)
by the Power-law formula CD ≈ 1.07ReL− 0.106.
(a) The new velocity is V = 29.06 m/s, and for air at 3500-m Standard Altitude
(Table A-6) take ρ = 0.8633 kg/m 3 and calculate µ from power-law
0.7
µ T
= , µ ≈ 1.68 ×10 −5 N ⋅ s/m 2
µ 0 T0
Then compute the new Reynolds number and use our power-law above to estimate drag coefficient:
ρVL ( 0.8633) ( 29.06 ) ( 2.45 )
ReColorado = = −5
= 3.66 ×10 6 , hence
µ 1.68 ×10
1.07 2 2
CD ≈ 0.106
= 0.2156, ∴ F = 0.2156 ( 0.8633)( 29.06 ) 2( 2.45 ) ( 8.0 ) = 943.5 N Ans. (a)
(3.66E6)
(b) The horsepower required to overcome drag is
Power = FV = ( 943.5 ) ( 29.06 ) = 27417.6 W ÷ 550 = 36.77 hp Ans. (b)
P5.6 The full-scale parachute in the chapter-opener photo had a drag force of
approximatly 4225 N when tested at a velocity of 19 km/h in air at 20°C and 1 atm. Earlier, a
model parachute of diameter 1.7 m was tested in the same tunnel. (a) For dynamic similarity,
what should be the air velocity for the model? (b) What is the expected drag force of the
model? (c) Is there anything surprising about your result to part (b)?
Solution: (a) From Table A.2 for air at 20C, ρ = 1.20 kg/m3 and µ = 1.8E-5 kg/m-s. (a) For
similarity, equate the Reynolds numbers:
ρ pV p D p (1.20)(5.28)(16.8) ρ V D (1.20)(Vm )(1.7)
Re p = = = 5.91E6 = Re m = m m m =
µp 1.8E − 5 µm 1.8E − 5
m km
Solve for Vm = 52.2 = 187.8 Ans.(a)
s h
(b) For similarity, the force coefficients will be equal:
Fp 4225 N Fm Fm
C Fp = = = 0.447 = =
ρ pV p2 D 2p 2
(1.20)(5.28) (16.8) 2
ρ mVm2 Dm2 (1.20)(52.2)2 (1.7)2
Solve for Fm = 4225 N Ans.(b)
(c) It might be surprising that the drag forces are exactly the same for model and prototype!
This is because, if ρ and µ are the same, the product VD is the same for both and the force is
proportional to (VD)2.
