Chapter 11.1 : Flow Over Bodies- Drag and Lift - Notes, Engineering, Semester Notes | EduRev

: Chapter 11.1 : Flow Over Bodies- Drag and Lift - Notes, Engineering, Semester Notes | EduRev

 Page 1


Chapter 11 Flow Over Bodies: Drag and Lift 
 
Flow over Flat Plates 
 
11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity 
forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), 
the lower the thickness of the boundary layer.   
 
11-44C The friction coefficient represents the resistance to fluid flow over a flat plate. It is proportional to 
the drag force acting on the plate. The drag coefficient for a flat surface is equivalent to the mean friction 
coefficient. 
 
11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate.  
 
11-46C The average friction coefficient in flow over a flat plate is determined by integrating the local 
friction coefficient over the entire plate, and then dividing it by the length of the plate. Or, it can be 
determined experimentally by measuring the drag force, and dividing it by the dynamic pressure. 
 
11-47E Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 × 10
5
. 3 
The surface of the plate is smooth.  
Properties The density and kinematic viscosity of light oil at 75°F are ? = 55.3 lbm/ft
3
 and ? = 7.751×10
–3
 
ft
2
/s.  
Analysis Noting that L = 15 ft, the Reynolds number at the end of the plate is  
4
2 3
10 161 . 1
/s ft 10 751 . 7
ft) ft/s)(15 6 (
Re × =
×
= =
-
?
VL
L
 
Oil 
6 ft/s 
which is less than the critical Reynolds number. Thus we have laminar flow 
over the entire plate, and the average friction coefficient is determined from 
  01232 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1
5 . 0 4 5 . 0
= × × = =
- -
L f
C
L = 15 ft 
Noting that the pressure drag is zero and thus C for a flat plate, the 
drag force acting on the top surface of the plate per unit width becomes  
f D
C =
lbf 5.87 = ?
?
?
?
?
?
·
× × = =
2
2 3
2
2
ft/s lbm 32.2
lbf 1
2
ft/s) 6 )( lbm/ft 8 . 56 (
) ft 1 15 ( 01232 . 0
2
V
A C F
f D
?
 
The total drag force acting on the entire plate can be determined by multiplying the value obtained above 
by the width of the plate.  
Discussion The force per unit width corresponds to the weight of a mass of 5.87 lbm. Therefore, a person 
who applies an equal and opposite force to the plate to keep it from moving will feel like he or she is using 
as much force as is necessary to hold a 5.87 lbm mass from dropping.  
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-22
Page 2


Chapter 11 Flow Over Bodies: Drag and Lift 
 
Flow over Flat Plates 
 
11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity 
forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), 
the lower the thickness of the boundary layer.   
 
11-44C The friction coefficient represents the resistance to fluid flow over a flat plate. It is proportional to 
the drag force acting on the plate. The drag coefficient for a flat surface is equivalent to the mean friction 
coefficient. 
 
11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate.  
 
11-46C The average friction coefficient in flow over a flat plate is determined by integrating the local 
friction coefficient over the entire plate, and then dividing it by the length of the plate. Or, it can be 
determined experimentally by measuring the drag force, and dividing it by the dynamic pressure. 
 
11-47E Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 × 10
5
. 3 
The surface of the plate is smooth.  
Properties The density and kinematic viscosity of light oil at 75°F are ? = 55.3 lbm/ft
3
 and ? = 7.751×10
–3
 
ft
2
/s.  
Analysis Noting that L = 15 ft, the Reynolds number at the end of the plate is  
4
2 3
10 161 . 1
/s ft 10 751 . 7
ft) ft/s)(15 6 (
Re × =
×
= =
-
?
VL
L
 
Oil 
6 ft/s 
which is less than the critical Reynolds number. Thus we have laminar flow 
over the entire plate, and the average friction coefficient is determined from 
  01232 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1
5 . 0 4 5 . 0
= × × = =
- -
L f
C
L = 15 ft 
Noting that the pressure drag is zero and thus C for a flat plate, the 
drag force acting on the top surface of the plate per unit width becomes  
f D
C =
lbf 5.87 = ?
?
?
?
?
?
·
× × = =
2
2 3
2
2
ft/s lbm 32.2
lbf 1
2
ft/s) 6 )( lbm/ft 8 . 56 (
) ft 1 15 ( 01232 . 0
2
V
A C F
f D
?
 
The total drag force acting on the entire plate can be determined by multiplying the value obtained above 
by the width of the plate.  
Discussion The force per unit width corresponds to the weight of a mass of 5.87 lbm. Therefore, a person 
who applies an equal and opposite force to the plate to keep it from moving will feel like he or she is using 
as much force as is necessary to hold a 5.87 lbm mass from dropping.  
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-22
Chapter 11 Flow Over Bodies: Drag and Lift 
11-48 Air flows over a plane surface at high elevation. The drag force acting on the top surface of the plate 
is to be determined for flow along the two sides of the plate. v 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The surface of the plate is smooth. 
Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10
-5
 
kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 
 
3
3
 kg/m 9751 . 0
K) K)(298 /kg m kPa (0.287
 kPa 83.4
=
· ·
= =
RT
P
? 
Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 
 
6
5
3
10 531 . 2
s kg/m 10 849 . 1
m) m/s)(8 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
 11-23
which is greater than the critical Reynolds number. Thus we 
have combined laminar and turbulent flow, and the friction 
coefficient is determined to be   
003189 . 0
10 531 . 2
1742
-
) 10 531 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
6 5 / 1 6 5 / 1
=
× ×
= =
L L
f
C 
2.5 m 
8 m 
6 m/s 
Air
Noting that the pressure drag is zero and thus  for a flat plate, the 
drag force acting on the top surface of the plate becomes 
f D
C C =
N 1.12 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 003189 . 0
2
V
A C F
f D
?
 
(b) If the air flows parallel to the 2.5 m side, the Reynolds number is 
 
5
5
3
10 910 . 7
s kg/m 10 849 . 1
m) m/s)(2.5 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient is determined to be   
002691 . 0
10 910 . 7
1742
-
) 10 910 . 7 (
074 . 0
Re
1742
-
Re
074 . 0
5 5 / 1 5 5 / 1
=
× ×
= =
L L
f
C 
Then the drag force acting on the top surface of the plate becomes 
N 0.94 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 002691 . 0
2
V
A C F
f D
?
 
Discussion Note that the drag force is proportional to density, which is proportional to the pressure. 
Therefore, the altitude has a major influence on the drag force acting on a surface.  Commercial airplanes 
take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save 
fuel. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
Page 3


Chapter 11 Flow Over Bodies: Drag and Lift 
 
Flow over Flat Plates 
 
11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity 
forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), 
the lower the thickness of the boundary layer.   
 
11-44C The friction coefficient represents the resistance to fluid flow over a flat plate. It is proportional to 
the drag force acting on the plate. The drag coefficient for a flat surface is equivalent to the mean friction 
coefficient. 
 
11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate.  
 
11-46C The average friction coefficient in flow over a flat plate is determined by integrating the local 
friction coefficient over the entire plate, and then dividing it by the length of the plate. Or, it can be 
determined experimentally by measuring the drag force, and dividing it by the dynamic pressure. 
 
11-47E Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 × 10
5
. 3 
The surface of the plate is smooth.  
Properties The density and kinematic viscosity of light oil at 75°F are ? = 55.3 lbm/ft
3
 and ? = 7.751×10
–3
 
ft
2
/s.  
Analysis Noting that L = 15 ft, the Reynolds number at the end of the plate is  
4
2 3
10 161 . 1
/s ft 10 751 . 7
ft) ft/s)(15 6 (
Re × =
×
= =
-
?
VL
L
 
Oil 
6 ft/s 
which is less than the critical Reynolds number. Thus we have laminar flow 
over the entire plate, and the average friction coefficient is determined from 
  01232 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1
5 . 0 4 5 . 0
= × × = =
- -
L f
C
L = 15 ft 
Noting that the pressure drag is zero and thus C for a flat plate, the 
drag force acting on the top surface of the plate per unit width becomes  
f D
C =
lbf 5.87 = ?
?
?
?
?
?
·
× × = =
2
2 3
2
2
ft/s lbm 32.2
lbf 1
2
ft/s) 6 )( lbm/ft 8 . 56 (
) ft 1 15 ( 01232 . 0
2
V
A C F
f D
?
 
The total drag force acting on the entire plate can be determined by multiplying the value obtained above 
by the width of the plate.  
Discussion The force per unit width corresponds to the weight of a mass of 5.87 lbm. Therefore, a person 
who applies an equal and opposite force to the plate to keep it from moving will feel like he or she is using 
as much force as is necessary to hold a 5.87 lbm mass from dropping.  
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-22
Chapter 11 Flow Over Bodies: Drag and Lift 
11-48 Air flows over a plane surface at high elevation. The drag force acting on the top surface of the plate 
is to be determined for flow along the two sides of the plate. v 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The surface of the plate is smooth. 
Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10
-5
 
kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 
 
3
3
 kg/m 9751 . 0
K) K)(298 /kg m kPa (0.287
 kPa 83.4
=
· ·
= =
RT
P
? 
Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 
 
6
5
3
10 531 . 2
s kg/m 10 849 . 1
m) m/s)(8 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
 11-23
which is greater than the critical Reynolds number. Thus we 
have combined laminar and turbulent flow, and the friction 
coefficient is determined to be   
003189 . 0
10 531 . 2
1742
-
) 10 531 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
6 5 / 1 6 5 / 1
=
× ×
= =
L L
f
C 
2.5 m 
8 m 
6 m/s 
Air
Noting that the pressure drag is zero and thus  for a flat plate, the 
drag force acting on the top surface of the plate becomes 
f D
C C =
N 1.12 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 003189 . 0
2
V
A C F
f D
?
 
(b) If the air flows parallel to the 2.5 m side, the Reynolds number is 
 
5
5
3
10 910 . 7
s kg/m 10 849 . 1
m) m/s)(2.5 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient is determined to be   
002691 . 0
10 910 . 7
1742
-
) 10 910 . 7 (
074 . 0
Re
1742
-
Re
074 . 0
5 5 / 1 5 5 / 1
=
× ×
= =
L L
f
C 
Then the drag force acting on the top surface of the plate becomes 
N 0.94 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 002691 . 0
2
V
A C F
f D
?
 
Discussion Note that the drag force is proportional to density, which is proportional to the pressure. 
Therefore, the altitude has a major influence on the drag force acting on a surface.  Commercial airplanes 
take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save 
fuel. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
Chapter 11 Flow Over Bodies: Drag and Lift 
11-49 Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be 
determined for two different wind velocities. vEES 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The wall surface is smooth (the actual wall surface is usually very rough). 5 The wind 
blows parallel to the wall. 
Properties The density and kinematic viscosity of air at 1 atm and 5 °C are ? = 1.269 kg/m
3
 and ? = 
1.382×10
–5
 m
2
/s . 
Air 
55 km/h 
4 m 
10 m
Analysis The Reynolds number is 
 
7
2 5
10 105 . 1
/s m 10 382 . 1
m) m/s)(10 6 . 3 / 55 (
Re × =
×
= =
-
?
VL
L
 
which is greater than the critical Reynolds number. Thus we have 
combined laminar and turbulent flow, and the friction coefficient is 
002730 . 0
10 105 . 1
1742
-
) 10 105 . 1 (
074 . 0
Re
1742
-
Re
074 . 0
7 5 / 1 7 5 / 1
=
× ×
= =
L L
f
C 
Noting that the pressure drag is zero and thus C
f D
C = for a flat plate, the drag force acting on the wall 
surface is 
N 16.2 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 . 3 / 55 )( kg/m 269 . 1 (
) m 4 10 ( 00273 . 0
2
V
A C F
f D
?
 
(b) When the wind velocity is doubled to 110 km/h, the Reynolds number becomes 
 
7
2 5
10 211 . 2
/s m 10 382 . 1
m) m/s)(10 6 . 3 / 110 (
Re × =
×
= =
-
?
VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient and the drag force become   
002435 . 0
10 211 . 2
1742
-
) 10 211 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
7 5 / 1 7 5 / 1
=
× ×
= =
L L
f
C 
N 57.7 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 . 3 / 110 )( kg/m 269 . 1 (
) m 4 10 ( 002435 . 0
2
V
A C F
f D
?
 
Treating flow over the side wall of a house as flow over a flat plate is not quite realistic. When flow hits a 
bluff body like a house, it separates at the sharp corner and a separation bubble exists over most of the side 
panels of the house. Therefore, flat plat boundary layer equations are not appropriate for this problem, and 
the entire house should considered in the solution instead.  
Discussion Note that the actual drag will probably be much higher since the wall surfaces are typically very 
rough. Also, we can solve this problem using the turbulent flow relation (instead of the combined laminar-
turbulent flow relation) without much loss in accuracy. Finally, the drag force nearly quadruples when the 
velocity is doubled. This is expected since the drag force is proportional to the square of the velocity, and 
the effect of velocity on the friction coefficient is small. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-24
Page 4


Chapter 11 Flow Over Bodies: Drag and Lift 
 
Flow over Flat Plates 
 
11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity 
forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), 
the lower the thickness of the boundary layer.   
 
11-44C The friction coefficient represents the resistance to fluid flow over a flat plate. It is proportional to 
the drag force acting on the plate. The drag coefficient for a flat surface is equivalent to the mean friction 
coefficient. 
 
11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate.  
 
11-46C The average friction coefficient in flow over a flat plate is determined by integrating the local 
friction coefficient over the entire plate, and then dividing it by the length of the plate. Or, it can be 
determined experimentally by measuring the drag force, and dividing it by the dynamic pressure. 
 
11-47E Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 × 10
5
. 3 
The surface of the plate is smooth.  
Properties The density and kinematic viscosity of light oil at 75°F are ? = 55.3 lbm/ft
3
 and ? = 7.751×10
–3
 
ft
2
/s.  
Analysis Noting that L = 15 ft, the Reynolds number at the end of the plate is  
4
2 3
10 161 . 1
/s ft 10 751 . 7
ft) ft/s)(15 6 (
Re × =
×
= =
-
?
VL
L
 
Oil 
6 ft/s 
which is less than the critical Reynolds number. Thus we have laminar flow 
over the entire plate, and the average friction coefficient is determined from 
  01232 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1
5 . 0 4 5 . 0
= × × = =
- -
L f
C
L = 15 ft 
Noting that the pressure drag is zero and thus C for a flat plate, the 
drag force acting on the top surface of the plate per unit width becomes  
f D
C =
lbf 5.87 = ?
?
?
?
?
?
·
× × = =
2
2 3
2
2
ft/s lbm 32.2
lbf 1
2
ft/s) 6 )( lbm/ft 8 . 56 (
) ft 1 15 ( 01232 . 0
2
V
A C F
f D
?
 
The total drag force acting on the entire plate can be determined by multiplying the value obtained above 
by the width of the plate.  
Discussion The force per unit width corresponds to the weight of a mass of 5.87 lbm. Therefore, a person 
who applies an equal and opposite force to the plate to keep it from moving will feel like he or she is using 
as much force as is necessary to hold a 5.87 lbm mass from dropping.  
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-22
Chapter 11 Flow Over Bodies: Drag and Lift 
11-48 Air flows over a plane surface at high elevation. The drag force acting on the top surface of the plate 
is to be determined for flow along the two sides of the plate. v 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The surface of the plate is smooth. 
Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10
-5
 
kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 
 
3
3
 kg/m 9751 . 0
K) K)(298 /kg m kPa (0.287
 kPa 83.4
=
· ·
= =
RT
P
? 
Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 
 
6
5
3
10 531 . 2
s kg/m 10 849 . 1
m) m/s)(8 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
 11-23
which is greater than the critical Reynolds number. Thus we 
have combined laminar and turbulent flow, and the friction 
coefficient is determined to be   
003189 . 0
10 531 . 2
1742
-
) 10 531 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
6 5 / 1 6 5 / 1
=
× ×
= =
L L
f
C 
2.5 m 
8 m 
6 m/s 
Air
Noting that the pressure drag is zero and thus  for a flat plate, the 
drag force acting on the top surface of the plate becomes 
f D
C C =
N 1.12 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 003189 . 0
2
V
A C F
f D
?
 
(b) If the air flows parallel to the 2.5 m side, the Reynolds number is 
 
5
5
3
10 910 . 7
s kg/m 10 849 . 1
m) m/s)(2.5 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient is determined to be   
002691 . 0
10 910 . 7
1742
-
) 10 910 . 7 (
074 . 0
Re
1742
-
Re
074 . 0
5 5 / 1 5 5 / 1
=
× ×
= =
L L
f
C 
Then the drag force acting on the top surface of the plate becomes 
N 0.94 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 002691 . 0
2
V
A C F
f D
?
 
Discussion Note that the drag force is proportional to density, which is proportional to the pressure. 
Therefore, the altitude has a major influence on the drag force acting on a surface.  Commercial airplanes 
take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save 
fuel. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
Chapter 11 Flow Over Bodies: Drag and Lift 
11-49 Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be 
determined for two different wind velocities. vEES 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The wall surface is smooth (the actual wall surface is usually very rough). 5 The wind 
blows parallel to the wall. 
Properties The density and kinematic viscosity of air at 1 atm and 5 °C are ? = 1.269 kg/m
3
 and ? = 
1.382×10
–5
 m
2
/s . 
Air 
55 km/h 
4 m 
10 m
Analysis The Reynolds number is 
 
7
2 5
10 105 . 1
/s m 10 382 . 1
m) m/s)(10 6 . 3 / 55 (
Re × =
×
= =
-
?
VL
L
 
which is greater than the critical Reynolds number. Thus we have 
combined laminar and turbulent flow, and the friction coefficient is 
002730 . 0
10 105 . 1
1742
-
) 10 105 . 1 (
074 . 0
Re
1742
-
Re
074 . 0
7 5 / 1 7 5 / 1
=
× ×
= =
L L
f
C 
Noting that the pressure drag is zero and thus C
f D
C = for a flat plate, the drag force acting on the wall 
surface is 
N 16.2 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 . 3 / 55 )( kg/m 269 . 1 (
) m 4 10 ( 00273 . 0
2
V
A C F
f D
?
 
(b) When the wind velocity is doubled to 110 km/h, the Reynolds number becomes 
 
7
2 5
10 211 . 2
/s m 10 382 . 1
m) m/s)(10 6 . 3 / 110 (
Re × =
×
= =
-
?
VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient and the drag force become   
002435 . 0
10 211 . 2
1742
-
) 10 211 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
7 5 / 1 7 5 / 1
=
× ×
= =
L L
f
C 
N 57.7 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 . 3 / 110 )( kg/m 269 . 1 (
) m 4 10 ( 002435 . 0
2
V
A C F
f D
?
 
Treating flow over the side wall of a house as flow over a flat plate is not quite realistic. When flow hits a 
bluff body like a house, it separates at the sharp corner and a separation bubble exists over most of the side 
panels of the house. Therefore, flat plat boundary layer equations are not appropriate for this problem, and 
the entire house should considered in the solution instead.  
Discussion Note that the actual drag will probably be much higher since the wall surfaces are typically very 
rough. Also, we can solve this problem using the turbulent flow relation (instead of the combined laminar-
turbulent flow relation) without much loss in accuracy. Finally, the drag force nearly quadruples when the 
velocity is doubled. This is expected since the drag force is proportional to the square of the velocity, and 
the effect of velocity on the friction coefficient is small. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-24
Chapter 11 Flow Over Bodies: Drag and Lift 
11-50E Air flows over a flat plate. The local friction coefficients at intervals of 1 ft is to be determined and 
plotted against the distance from the leading edge. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The surface of the plate is smooth. 
Properties The density and kinematic viscosity of air at 1 atm and 70°F are ? = 0.07489 lbm/ft
3
 and ? = 
0.5913 ft
2
/h = 1.643×10
–4
 ft
2
/s . 
Analysis For the first 1 ft interval, the Reynolds number is  
 
5
2 4
10 522 . 1
/s ft 10 643 . 1
ft) ft/s)(1 25 (
Re × =
×
= =
-
?
VL
L
 
10 ft 
Air 
25 ft/s which is less than the critical value of 510
5
× . Therefore, the flow 
is laminar. The local friction coefficient is 
 001702 . 0
) 10 522 . 1 (
664 . 0
Re
664 . 0
5 . 0 5 5 . 0
,
=
×
= =
x f
C 
We repeat calculations for all 1-ft intervals. The results are  
x, ft Re  C
f
 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
1.522E+05 
3.044E+05 
4.566E+05 
6.088E+05 
7.610E+05 
9.132E+05 
1.065E+06 
1.218E+06 
1.370E+06 
1.522E+06 
0.001702 
0.001203 
0.000983 
0.004111 
0.003932 
0.003791 
0.003676 
0.003579 
0.003496 
0.003423 
rho=0.07489 "lbm/ft3" 
nu=0.5913/3600 "ft2/s" 
V=25  
“Local Re and C_f” 
Re=x*V/nu 
"f=0.664/Re^0.5" 
f=0.059/Re^0.2 
 
 
12 3 4 5 6 7 8 9 10
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
0.004
0.0045
x, ft  
f  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Discussion Note that the Reynolds number exceeds the critical value for x > 3 ft, and thus the flow is 
turbulent over most of the plate. For x > 3 ft, we used  for friction 
coefficient. Note that C
L L f
C 1742/Re - Re / 074 . 0
5 / 1
=
f
  decreases with Re in both laminar and turbulent flows.   
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-25
Page 5


Chapter 11 Flow Over Bodies: Drag and Lift 
 
Flow over Flat Plates 
 
11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity 
forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), 
the lower the thickness of the boundary layer.   
 
11-44C The friction coefficient represents the resistance to fluid flow over a flat plate. It is proportional to 
the drag force acting on the plate. The drag coefficient for a flat surface is equivalent to the mean friction 
coefficient. 
 
11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate.  
 
11-46C The average friction coefficient in flow over a flat plate is determined by integrating the local 
friction coefficient over the entire plate, and then dividing it by the length of the plate. Or, it can be 
determined experimentally by measuring the drag force, and dividing it by the dynamic pressure. 
 
11-47E Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 × 10
5
. 3 
The surface of the plate is smooth.  
Properties The density and kinematic viscosity of light oil at 75°F are ? = 55.3 lbm/ft
3
 and ? = 7.751×10
–3
 
ft
2
/s.  
Analysis Noting that L = 15 ft, the Reynolds number at the end of the plate is  
4
2 3
10 161 . 1
/s ft 10 751 . 7
ft) ft/s)(15 6 (
Re × =
×
= =
-
?
VL
L
 
Oil 
6 ft/s 
which is less than the critical Reynolds number. Thus we have laminar flow 
over the entire plate, and the average friction coefficient is determined from 
  01232 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1
5 . 0 4 5 . 0
= × × = =
- -
L f
C
L = 15 ft 
Noting that the pressure drag is zero and thus C for a flat plate, the 
drag force acting on the top surface of the plate per unit width becomes  
f D
C =
lbf 5.87 = ?
?
?
?
?
?
·
× × = =
2
2 3
2
2
ft/s lbm 32.2
lbf 1
2
ft/s) 6 )( lbm/ft 8 . 56 (
) ft 1 15 ( 01232 . 0
2
V
A C F
f D
?
 
The total drag force acting on the entire plate can be determined by multiplying the value obtained above 
by the width of the plate.  
Discussion The force per unit width corresponds to the weight of a mass of 5.87 lbm. Therefore, a person 
who applies an equal and opposite force to the plate to keep it from moving will feel like he or she is using 
as much force as is necessary to hold a 5.87 lbm mass from dropping.  
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-22
Chapter 11 Flow Over Bodies: Drag and Lift 
11-48 Air flows over a plane surface at high elevation. The drag force acting on the top surface of the plate 
is to be determined for flow along the two sides of the plate. v 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The surface of the plate is smooth. 
Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10
-5
 
kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 
 
3
3
 kg/m 9751 . 0
K) K)(298 /kg m kPa (0.287
 kPa 83.4
=
· ·
= =
RT
P
? 
Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 
 
6
5
3
10 531 . 2
s kg/m 10 849 . 1
m) m/s)(8 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
 11-23
which is greater than the critical Reynolds number. Thus we 
have combined laminar and turbulent flow, and the friction 
coefficient is determined to be   
003189 . 0
10 531 . 2
1742
-
) 10 531 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
6 5 / 1 6 5 / 1
=
× ×
= =
L L
f
C 
2.5 m 
8 m 
6 m/s 
Air
Noting that the pressure drag is zero and thus  for a flat plate, the 
drag force acting on the top surface of the plate becomes 
f D
C C =
N 1.12 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 003189 . 0
2
V
A C F
f D
?
 
(b) If the air flows parallel to the 2.5 m side, the Reynolds number is 
 
5
5
3
10 910 . 7
s kg/m 10 849 . 1
m) m/s)(2.5 6 ( ) kg/m 9751 . 0 (
Re × =
· ×
= =
-
µ
?VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient is determined to be   
002691 . 0
10 910 . 7
1742
-
) 10 910 . 7 (
074 . 0
Re
1742
-
Re
074 . 0
5 5 / 1 5 5 / 1
=
× ×
= =
L L
f
C 
Then the drag force acting on the top surface of the plate becomes 
N 0.94 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 )( kg/m 9751 . 0 (
) m 5 . 2 8 ( 002691 . 0
2
V
A C F
f D
?
 
Discussion Note that the drag force is proportional to density, which is proportional to the pressure. 
Therefore, the altitude has a major influence on the drag force acting on a surface.  Commercial airplanes 
take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save 
fuel. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
Chapter 11 Flow Over Bodies: Drag and Lift 
11-49 Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be 
determined for two different wind velocities. vEES 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The wall surface is smooth (the actual wall surface is usually very rough). 5 The wind 
blows parallel to the wall. 
Properties The density and kinematic viscosity of air at 1 atm and 5 °C are ? = 1.269 kg/m
3
 and ? = 
1.382×10
–5
 m
2
/s . 
Air 
55 km/h 
4 m 
10 m
Analysis The Reynolds number is 
 
7
2 5
10 105 . 1
/s m 10 382 . 1
m) m/s)(10 6 . 3 / 55 (
Re × =
×
= =
-
?
VL
L
 
which is greater than the critical Reynolds number. Thus we have 
combined laminar and turbulent flow, and the friction coefficient is 
002730 . 0
10 105 . 1
1742
-
) 10 105 . 1 (
074 . 0
Re
1742
-
Re
074 . 0
7 5 / 1 7 5 / 1
=
× ×
= =
L L
f
C 
Noting that the pressure drag is zero and thus C
f D
C = for a flat plate, the drag force acting on the wall 
surface is 
N 16.2 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 . 3 / 55 )( kg/m 269 . 1 (
) m 4 10 ( 00273 . 0
2
V
A C F
f D
?
 
(b) When the wind velocity is doubled to 110 km/h, the Reynolds number becomes 
 
7
2 5
10 211 . 2
/s m 10 382 . 1
m) m/s)(10 6 . 3 / 110 (
Re × =
×
= =
-
?
VL
L
 
which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, 
and the friction coefficient and the drag force become   
002435 . 0
10 211 . 2
1742
-
) 10 211 . 2 (
074 . 0
Re
1742
-
Re
074 . 0
7 5 / 1 7 5 / 1
=
× ×
= =
L L
f
C 
N 57.7 =
?
?
?
?
?
?
?
?
·
× × = =
2
2 3
2
2
m/s kg 1
N 1
2
m/s) 6 . 3 / 110 )( kg/m 269 . 1 (
) m 4 10 ( 002435 . 0
2
V
A C F
f D
?
 
Treating flow over the side wall of a house as flow over a flat plate is not quite realistic. When flow hits a 
bluff body like a house, it separates at the sharp corner and a separation bubble exists over most of the side 
panels of the house. Therefore, flat plat boundary layer equations are not appropriate for this problem, and 
the entire house should considered in the solution instead.  
Discussion Note that the actual drag will probably be much higher since the wall surfaces are typically very 
rough. Also, we can solve this problem using the turbulent flow relation (instead of the combined laminar-
turbulent flow relation) without much loss in accuracy. Finally, the drag force nearly quadruples when the 
velocity is doubled. This is expected since the drag force is proportional to the square of the velocity, and 
the effect of velocity on the friction coefficient is small. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-24
Chapter 11 Flow Over Bodies: Drag and Lift 
11-50E Air flows over a flat plate. The local friction coefficients at intervals of 1 ft is to be determined and 
plotted against the distance from the leading edge. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 The surface of the plate is smooth. 
Properties The density and kinematic viscosity of air at 1 atm and 70°F are ? = 0.07489 lbm/ft
3
 and ? = 
0.5913 ft
2
/h = 1.643×10
–4
 ft
2
/s . 
Analysis For the first 1 ft interval, the Reynolds number is  
 
5
2 4
10 522 . 1
/s ft 10 643 . 1
ft) ft/s)(1 25 (
Re × =
×
= =
-
?
VL
L
 
10 ft 
Air 
25 ft/s which is less than the critical value of 510
5
× . Therefore, the flow 
is laminar. The local friction coefficient is 
 001702 . 0
) 10 522 . 1 (
664 . 0
Re
664 . 0
5 . 0 5 5 . 0
,
=
×
= =
x f
C 
We repeat calculations for all 1-ft intervals. The results are  
x, ft Re  C
f
 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
1.522E+05 
3.044E+05 
4.566E+05 
6.088E+05 
7.610E+05 
9.132E+05 
1.065E+06 
1.218E+06 
1.370E+06 
1.522E+06 
0.001702 
0.001203 
0.000983 
0.004111 
0.003932 
0.003791 
0.003676 
0.003579 
0.003496 
0.003423 
rho=0.07489 "lbm/ft3" 
nu=0.5913/3600 "ft2/s" 
V=25  
“Local Re and C_f” 
Re=x*V/nu 
"f=0.664/Re^0.5" 
f=0.059/Re^0.2 
 
 
12 3 4 5 6 7 8 9 10
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
0.004
0.0045
x, ft  
f  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Discussion Note that the Reynolds number exceeds the critical value for x > 3 ft, and thus the flow is 
turbulent over most of the plate. For x > 3 ft, we used  for friction 
coefficient. Note that C
L L f
C 1742/Re - Re / 074 . 0
5 / 1
=
f
  decreases with Re in both laminar and turbulent flows.   
 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-25
Chapter 11 Flow Over Bodies: Drag and Lift 
11-51 Air flows on both sides of a continuous sheet of plastic. The drag force air exerts on the plastic sheet 
in the direction of flow is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re
cr
 = 5 ×10
5
.  3 
Air is an ideal gas. 4 Both surfaces of the plastic sheet are smooth.  5 The plastic sheet does not vibrate and 
thus it does not induce turbulence in air flow. 
Properties The density and kinematic viscosity of air at 1 atm and 60 °C are ? = 1.059 kg/m
3
 and ? = 
1.896×10
–5
 m
2
/s . 
Analysis The length of the cooling section is   
Air 
V
8
 = 3 m/s 
  m 0.5 = s) 2 ( m/s] ) 60 / 15 [(
sheet
= ? = t V L
15 m/min 
Plastic sheet 
 
The Reynolds number  is 
 
5
2 5
10 899 . 1
/s m 10 896 . 1
m) m/s)(1.2 (3
Re × =
×
= =
-
?
VL
L
 
which is less than the critical Reynolds number. Thus the flow is 
laminar. The area on both sides of the sheet exposed to air flow is  
  
2
m 1.2 = m) m)(0.5 2 . 1 ( 2 2 = = wL A
Then the friction coefficient and the drag force become 
 003048 . 0
) 10 899 . 1 (
328 . 1
Re
328 . 1
5 . 0 5 5 . 0
=
×
= =
L
f
C 
 N 0.0174 = = =
2
m/s) )(3 kg/m (1.059
) m 2 . 1 )( 003048 . 0 (
2
2 3
2
2
V
A C F
f D
?
 
Discussion Note that the Reynolds number remains under the critical value, and thus the flow remains 
laminar over the entire plate. In reality, the flow may be turbulent because of the motion of the plastic 
sheet. 
 
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution 
permitted only to teachers and educators for course preparation.  If you are a student using this Manual, you 
are using it without permission.   
11-26
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