143 
 

American Academic Scientific Research Journal for Engineering, Technology,  and Sciences   

ISSN (Print) 2313-4410, ISSN (Online) 2313-4402 

 

http://asrjetsjournal.org/  

 

An Enhanced Model for the Hydrodynamic Drag Force on 

a Steadily Translating Circular Cylinder 

Monday Ekhator
a*

, Vincent Ele Asor
b
, Stephen Etebefia

c 

a
Department of Mathematics and Statistics, Alex Ekwueme Federal University, Ndufu- Alike, 

Nigeria 

b
Department of Mathematics, Michael Okpara University of Agriculture, Umudike, 

Nigeria 

c
Federal School of Statistics, Enugu, Nigeria 

a
Email: ekhatorogie@gmail.com , 

b
Email: Vincent.Asor@gmail.com , 

c
Email: steveoghene@gmail.com 

Abstract 

A model for the hydrodynamic drag force on a steadily translating circular cylinder was studied for Reynolds 

number, Re << 1. Though the literature appears vast especially by method of rigorous asymptotic analysis, this 

work attempts to remove the mathematical complexity of asymptotic analysis by solving the Navier – Stokes 

equations directly to obtain the fluid velocity and then proceed to obtain the drag force and finally the drag 

coefficient which is a function of the Reynolds number. The result of this work is in complete agreement with 

the results in literature. 

Key words: Drag force; Drag coefficient; Streamline; Stream function; Reynolds number. 

1. Introduction 

Fluid mechanics provides scholarship on how impermeable and rigid surfaces such as pipes and particles affect 

the flow of a fluid and the effect of the fluid on the particles. One of the effects of fluid on particles in which 

they flow is the drag exerted by the fluid on the particles. Drag, a frictional force also called fluid resistance 

which acts opposite to the relative motion of any object moving with respect to a surrounding fluid is an enemy 

of moving objects in a fluid, e.g. aircrafts, submarines, water droplets, etc., in the sense that it tends to reduce 

the speed of the objects as they move in the fluid. Its calculation becomes imperative in the design of aircrafts, 

ships, submarines, automobiles, towers and in the calculation of the rate of sedimentation of small particles as 

they fall through a fluid as in the sedimentation of blood cells.  Just as other forms of frictional force, drag can 

be greatly reduced by shaping a body to the streamlines of the fluid through which it is moving, hence modern 

cars, ships and aircrafts are made in such shapes as to lessen the drag between them and the fluid through which 

they move. 

------------------------------------------------------------------------ 

* Corresponding author.  

http://asrjetsjournal.org/


American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

144 
 

Stokes [1], in his pioneering work, found the drag force, FD acting on a sphere as 

𝐹𝐷 = 6𝜋𝜇𝑎𝑈𝑝                                                                (1) 

Where a is the radius of the sphere moving with speed, Up. The dimensionless quantity called drag coefficient 

CD was defined as 

𝐶𝐷 =
2𝐹𝐷

𝜌𝑈𝑝
2𝜋𝑎2 =

24

𝑅𝑒
                                                                                          (2) 

where the Reynolds number, 𝑅𝑒 is defined as 

𝑅𝑒 =
2𝑎𝑈𝑝𝜌

𝜇
.                                                                                                                 (3) 

The result of [1] was useful for Robert Millikan [2,3] who measured the electric charge on a droplet and showed 

that charge is quantised. It seemed [1] was probably not interested in the force that was acting on the cylinder 

and so did not determine it. A mathematical problem thus arose, because the flow tended to decrease so slowly 

that the far field boundary conditions could not be satisfied. The problem of viscous flow past a cylinder was 

thus created and remained unsolved until sixty years later when Lamb [4] suggested a solution, using the 

equations of Oseen [5] to derive the drag force on a cylinder for Re << 1. He showed that: 

𝐹𝐷 = 2𝜋𝜇𝐶 

=
4𝜋𝜇𝑈

1

2
−𝛾−log(

1

2
𝑘𝑎)

≈
4𝜋𝜇𝑈

ln(7.4 𝑅𝑒⁄ )
                                                                            (4) 

and the drag coefficient is 

𝐶𝐷 =
2𝐹𝐷

2𝜌𝑎𝑈𝑝
2 

=
8𝜋

𝑅𝑒 ln(7.4 𝑅𝑒⁄ )
                                                                                                        (5) 

The problem on the drag force acting on a circular cylinder has always been the logarithm term in the velocity 

distribution of the fluid velocity which diverges at large distances from the cylinder, thereby making the velocity 

of the fluid at large distances not finite. To solve this problem many authors have adopted the asymptotic 

method. In this method, solutions are obtained at both the vicinity of the cylinder and far from the cylinder. Both 

solutions are then matched to give a solution at any point in the flow field, the matched solution of the fluid 

velocity and stream-function is then used to compute for the drag force. 

While [4] analysis could not be described as an asymptotic analysis, it does involve the idea of smallness and 

expansion. Mutlu Sumer and Jorgen Fredsoe [6] calculated analytically, the drag force by a fluid, flowing past a 

stationary circular cylinder. Eames and Klettner [7] found the drag force on a circular cylinder by solving the 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

145 
 

Navier-Stokes equations directly as follows: 

Non-dimensionalization of the equation of motion gives 

∇̃. �̃� = 0.                                                                                                                (6) 

 

𝑅𝑒

2
(�̃�. ∇̃)�̃� = −∇̃𝑝 + ∇̃2�̃�                                                                                               (7) 

The x-component of the inertial term of (7) is 

(�̃�. ∇̃)�̃� ≈
𝜕𝑢

𝜕𝑥 ̃
                                                                                                             (8) 

Substituting (8) in (7) gives 

𝜕𝑢

𝜕𝑥 ̃
= −

2

𝑅𝑒
∇̃𝑝 +

2

𝑅𝑒
∇̃2�̃�                                                                                      (9) 

Taking the curl of each term of (9) gives 

𝑅𝑒

2

𝜕�̃�

𝜕𝑥 ̃
= ∇̃2�̃�,                                                                                                      (10) 

where �̃� is the vorticity of flow. 

Solving (10) using separation of variables gives 

�̃� = 𝑃𝑒(𝑅𝑒 4⁄ )𝑥 ̃ = 𝑃𝑒(𝑅𝑒 4⁄ )𝑟 ̃ cos 𝜃                                                                     (11) 

Substituting (11) in (10) and evaluating gives 

(𝑅𝑒 4⁄ )2𝑃 = ∇̃2𝑃                                                                                                           (12) 

At low Reynolds numbers, the flow is to leading order symmetric and a solution to (12) is of the  

form,  

𝑃 = 𝑃1(�̃�) sin 𝜃                                                                                                 (13) 

Substituting (13) in (12) gives an ODE 

𝑃1
′′ +

1

𝑟 ̃
𝑃1

′ − ((
𝑅𝑒

4
)

2

+
1

𝑟 ̃2
) 𝑃1 = 0                                                                                         (14) 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

146 
 

The solution to (14) that satisfies 𝑃1 → 0 as �̃� → ∞ is 

𝑃1 = 𝐶1𝐾1 (𝑅𝑒
𝑟 ̃

4
),                                                                       (15) 

where 𝐾1 is the modified Bessel function of the second kind. The stream function is the sum of the known free 

stream component (first term) and a component to be determined (second term). 

�̃� = �̃� sin 𝜃 + 𝑓1 (�̃�) sin 𝜃                                                              (16) 

 

�̃� = −
𝜕2�̃�

𝜕𝑟 ̃2
−

1

𝑟 ̃

𝜕�̃�

𝜕𝑟 ̃ 
−

1

𝑟 ̃2

𝜕2�̃�

𝜕𝜃2                                                            (17) 

 

By substituting (16) in (17) and considering the lowest order symmetric solution  

(�̃� ≅ 𝑃1(�̃�) sin 𝜃) another ODE emerges 

�̃�2𝑓1
′′ + �̃�𝑓1

′ − 𝑓1 = −𝐶1𝐾1�̃�2                                                                                  (18) 

Solving (18) subject to the boundary conditions 𝑓1(1) = −1 and 𝑓1
′(1) = −1 gives 

𝑓1(�̃�) = −�̃� −
𝐶1

2
�̃� ∫ �̃�

∞

1
𝐾1𝑑�̃� +

𝐶1

2�̃�
∫ �̃�2∞

1
𝐾1𝑑�̃�                                     (19) 

The far field condition, 𝑓1
′ → 0, �̃� → ∞ determines 𝐶1, 

𝐶1 =
2

− ∫ 𝐾1(𝑅𝑒
𝑟 ̃

4
)𝑑�̃�

∞
1

≅
𝑅𝑒

2𝐾0(
𝑅𝑒

4
)
                                                              (20) 

Substituting (20) in (19) and putting the result in (16) gives  

�̃� =
𝑅𝑒

4𝐾0�̃�
sin 𝜃 ∫ �̃�2∞

1
𝐾1𝑑�̃� −

𝑅𝑒

4𝐾0
�̃� sin 𝜃 ∫ �̃�

∞

1
𝐾1𝑑�̃�                               (21) 

The drag coefficient is given by 

 

𝐶𝐷 = ∫
2

𝑅𝑒

𝜕�̃�

𝜕�̃�

2𝜋

0
sin 𝜃 𝑑𝜃 − ∫

2

𝑅𝑒
�̃�𝑠

2𝜋

0
sin 𝜃𝑑𝜃                                           (22) 

 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

147 
 

The integral in (22) is the pressure stress evaluated at �̃� = 1 while the second integral is the viscous stress and 

both stresses contribute equally to the drag force. 

Substituting (21) in (17) and putting the result in (22) gives 

 

𝐶𝐷 =
2𝜋𝐶1

𝑅𝑒
(

𝑑𝐾1

𝑑𝑧
− 𝐾1)                                                                                (23) 

 

                  Using the expansion 𝐾1(𝑧) = {

1

𝑧
+

𝑧

2
log (

𝑧

2
) + ⋯ , 𝑧 ≪ 1

(
𝜋

2𝑧
)

1

2𝑒−𝑧 , 𝑧 ≫ 1
       

 

(23) becomes 

𝐶𝐷 ≅
8𝜋

𝑅𝑒𝑙𝑜𝑔(
7.4

𝑅𝑒
)
                                                                                               (24) 

2. Mathematical Formulation 

Consider the two dimensional steady incompressible viscous flow of a uniform stream 𝑢  in which a rigid 

circular cylinder of radius 𝑟𝑜, centre 𝑜, is translating with a constant velocity 𝑈. 

 

 

Figure 1: Flow around a translating circular cylinder. 

It is well established that the motion of fluids is governed by the famous Navier-Stokes equations. For a 

compressible Newtonian fluid, this yields 

   𝑟 
 

𝜃 
 

𝑜 𝑈 
 

𝑟𝑜 

 

𝑜 

 𝑈 𝜃 
 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

148 
 

 

𝜌 (
𝜕𝑢

𝜕𝑡
+ (𝑢 ∙ ∇)𝑢) = −∇𝑝 + 𝜇∇2𝑢 + 𝜌𝐹                                 (25) 

where 𝑢 is the fluid velocity, 𝑝 is the fluid pressure, 𝜌 is the fluid density, and 𝜇 is the fluid dynamic viscosity. 

The terms on the left correspond to the inertial forces, the first term on the right is the pressure forces, the 

second term is the viscous forces, and the last term is the applied external force on the fluid. When there are no 

applied forces on the fluid, and the flow being steady, 𝐹 = 0,    
𝜕𝑢

𝜕𝑡
= 0 , then, equation (25) becomes 

𝜌(𝑢 ∙ ∇)𝑢 = −∇𝑝 + 𝜇∇2𝑢                                               (26) 

    These equations are always solved together with the continuity equation 

𝜕𝜌

𝜕𝑡
+ ∇. (𝜌𝑢) = 0                                                                                         (27) 

                                          

Since the fluid is incompressible, 
𝜕𝜌

𝜕𝑡
= 0 and upon dividing by 𝜌, equation (27) reduces to 

∇. 𝑢 = 0                                                                                         (28) 

which when taken together leaves the Navier-Stokes equations to represent the conservation of momentum and 

the continuity equation to represent the conservation of mass. 

The boundary conditions for the solution of (26) and (28) are 

                       𝑢(𝑟, 𝜃) = −𝑈,   𝑟 = 𝑟0                                             

                                            𝑢(𝑟, 𝜃) = 0,   𝑟 → ∞                                            

  The stream function 𝜑(𝑟, 𝜃) for the flow is of the form 

𝜑(𝑟, 𝜃) = −𝑓(𝑟)𝑈 sin 𝜃                                                (29) 

 

where 𝑓(𝑟) is an unknown function. 

In polar coordinates 

 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

149 
 

                             ∇2≡
𝜕2

𝜕𝑟2 +
1

𝑟

𝜕

𝜕𝑟
+

1

𝑟2

𝜕2

𝜕𝜃2 

 

and in Cartesian coordinates (two dimensional) 

                                                  ∇2≡
𝜕2

𝜕𝑥2 +
𝜕2

𝜕𝑦2 

3. Method of Solution 

Since the motion is extremely slow so that 𝑅𝑒 ≪ 1, the inertia forces will be small compared with the viscous 

forces, so the inertia forces can be ignored at the vicinity of the cylinder [1]. 

  Equation (26) becomes 

0 = −∇𝑝 + 𝜇∇2𝑢                                                (30) 

  By taking the divergence of each term of (30), we have 

0 = −∇. (∇𝑝) + 𝜇∇2(∇. 𝑢)                                                            (31) 

 Since ∇. 𝑢 = 0, (31) becomes 

∇2𝑝 = 0                                                                                                                (32) 

      

Equation (32) shows that for very slow motion the pressure 𝑝 satisfies the Laplace's equation and it is therefore 

a harmonic function. The solutions of (32) are series termed circular harmonics of integral degree. Hence the 

solution of equation 32, Batchelor [8,9] is of the form 

𝑝 = −
1

𝑟
𝛼𝑈 cos 𝜃                                                                                              (33) 

               

Here 𝛼 is a constant. 

Consider the x-component of (30) 

 

𝜕𝑝

𝜕𝑥
= 𝜇(

𝜕2𝑢

𝜕𝑥2 +
𝜕2𝑢

𝜕𝑦2)                                                (34) 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

150 
 

 

Resolving  
𝜕𝑝

𝜕𝑥
  into its components in polar form (see figure 1), we have 

 

𝜕𝑝

𝜕𝑥
=

𝜕𝑝

𝜕𝑟
cos 𝜃 −

1

𝑟

𝜕𝑝

𝜕𝜃
sin 𝜃                                                            (35) 

                                                            

Differentiating (33) with respect to 𝑟 and 𝜃 respectively, we have 

  

𝜕𝑝

𝜕𝑟
=

1

𝑟2 𝛼𝑈 cos 𝜃  and   
𝜕𝑝

𝜕𝜃
=

1

𝑟
𝛼𝑈 sin 𝜃                                                                   (36) 

So, 

𝜕𝑝

𝜕𝑥
=

1

𝑟2 𝛼𝑈𝑐𝑜𝑠2𝜃 −
1

𝑟2 𝛼𝑈𝑠𝑖𝑛2𝜃                                                            (37) 

          

Now, resolving 𝑢 into its components, in polar form (see figure 1), we obtain   

 

𝑢 = 𝑢𝑟 cos 𝜃 − 𝑢𝜃 sin 𝜃                                                                     (38) 

 

𝑢𝑟 = −
1

𝑟

𝜕𝜑

𝜕𝜃
   and  𝑢𝜃 =

𝜕𝜑

𝜕𝑟
                                                                                       (39) 

                                                                

By equation (29), 

 

𝜕𝜑

𝜕𝜃
= −𝑓(𝑟)𝑈 cos 𝜃  and   

𝜕𝜑

𝜕𝑟
= −𝑓′(𝑟)𝑈 sin 𝜃                                                         (40) 

                   



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

151 
 

So, 

                   𝑢 = −
1

𝑟

𝜕𝜑

𝜕𝜃
cos 𝜃 −

𝜕𝜑

𝜕𝑟
sin 𝜃 

 

𝑢 =
1

𝑟
 𝑓(𝑟)𝑈𝑐𝑜𝑠2𝜃 + 𝑓′(𝑟)𝑈𝑠𝑖𝑛2𝜃                                                            (41) 

 

Expressing equation (34) in polar form, we have  

         

1

𝑟2 𝛼𝑈𝑐𝑜𝑠2𝜃 −
1

𝑟2 𝛼𝑈𝑠𝑖𝑛2𝜃 = 𝜇(
𝜕2

𝜕𝑟2 +
1

𝑟

𝜕

𝜕𝑟
+

1

𝑟2

𝜕2

𝜕𝜃2)
1

𝑟
 𝑓(𝑟)𝑈𝑐𝑜𝑠2𝜃 + 𝑓′(𝑟)𝑈𝑠𝑖𝑛2𝜃                                       (42) 

 

1

𝑟2 𝛼𝑈𝑐𝑜𝑠2𝜃 −
1

𝑟2 𝛼𝑈𝑠𝑖𝑛2𝜃 = 𝜇 (−
1

𝑟3 𝑓(𝑟) +
1

𝑟2 𝑓′(𝑟) +
1

𝑟
𝑓′′(𝑟)) 𝑈𝑐𝑜𝑠2𝜃 + 𝜇(

2

𝑟3  𝑓(𝑟) −
2

𝑟2 𝑓′(𝑟) +
1

𝑟
𝑓′′(𝑟) +

𝑓′′′(𝑟)) 𝑈𝑠𝑖𝑛2𝜃                                                                                                                                                   (43) 

 

Equating coefficients of  𝑈𝑐𝑜𝑠2𝜃  and  𝑈𝑠𝑖𝑛2𝜃 , we have the following ODEs 

 

𝑟2𝑓′′(𝑟) + 𝑟𝑓′(𝑟) −  𝑓(𝑟) =
𝛼

𝜇
𝑟                                                                              (44a) 

 

𝑟3𝑓′′′(𝑟) + 𝑟2𝑓′′(𝑟) − 2𝑟𝑓′(𝑟) + 2𝑓(𝑟) = −
𝛼

𝜇
𝑟                                                                        (44b) 

                                        

Solving the non-homogeneous Euler-Cauchy equation (44a) subject to the boundary conditions  

   

                                        
𝑓

𝑟
(𝑟0) = −1   and 𝑓′(𝑟0) = −1, gives  



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

152 
 

 

𝑓(𝑟) = (−1 −
𝛼

2𝜇
ln 𝑟0)𝑟 +

𝛼

4𝜇

𝑟0
2

𝑟
+

𝛼

2𝜇
𝑟 ln 𝑟 −

𝛼

4𝜇
𝑟                                                   (45) 

                     

Now 𝑓(𝑟) also satisfy the third order ODE (44b).  

 

Substituting for 𝑓(𝑟) in equation (29), the expression for the stream function (𝜑(𝑟, 𝜃)) becomes 

          

𝜑(𝑟, 𝜃) = (𝑟 +
𝛼

2𝜇
𝑟 ln 𝑟0 +

𝛼

4𝜇
𝑟 −

𝛼

4𝜇

𝑟0
2

𝑟
−

𝛼

2𝜇
𝑟 ln 𝑟) 𝑈 sin 𝜃                                              (46) 

 

𝜕𝜑(𝑟, 𝜃)

𝜕𝜃
= (𝑟 +

𝛼

2𝜇
𝑟 ln 𝑟0 +

𝛼

4𝜇
𝑟 −

𝛼

4𝜇

𝑟0
2

𝑟
−

𝛼

2𝜇
𝑟 ln 𝑟) 𝑈 cos 𝜃 

 

𝜕𝜑(𝑟, 𝜃)

𝜕𝑟
= (1 +

𝛼

2𝜇
ln 𝑟0 +

𝛼

4𝜇
(
𝑟0

𝑟
)2 −

𝛼

2𝜇
ln 𝑟 −

𝛼

4𝜇
)𝑈 sin 𝜃 

 

𝑢𝑟 = −
1

𝑟

𝜕𝜑(𝑟, 𝜃)

𝜕𝜃
 

= −(1 +
𝛼

2𝜇
ln 𝑟0 +

𝛼

4𝜇
−

𝛼

4𝜇
(

𝑟0

𝑟
)2 −

𝛼

2𝜇
ln 𝑟) 𝑈 cos 𝜃                                                (47) 

 

𝑢𝜃 =
𝜕𝜑(𝑟, 𝜃)

𝜕𝜃
 

= (1 +
𝛼

2𝜇
ln 𝑟0 +

𝛼

4𝜇
(

𝑟0

𝑟
)2 −

𝛼

2𝜇
ln 𝑟 −

𝛼

4𝜇
) 𝑈 sin 𝜃                                                    (48) 

 

Recall that 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

153 
 

𝑢(𝑟, 𝜃) = 𝑢𝑟 cos 𝜃 − 𝑢𝜃 sin 𝜃                                                                                                (49) 

 

Substituting (47) and (48) into (49) and simplifying, we have 

  

𝑢(𝑟, 𝜃) = −𝑈 +
𝛼𝑈

2𝜇
(ln 𝑟 − ln 𝑟0) +

𝛼𝑈

4𝜇
((

𝑟0

𝑟
)2 − 1) cos 2𝜃                                                     (50) 

4. The Drag Force 

The force on the cylinder due to pressure is 

 

                                       𝐹𝑝 = − ∫ 𝑝(𝑟0 𝑑𝜃) cos 𝜃
2𝜋

0
 

 

Using equation (15) and evaluating at 𝑟 = 𝑟0 

 

𝐹𝑝 =
1

2
𝛼𝑈 ∫ (1 +

2𝜋

0
cos 2𝜃)𝑑𝜃 =

1

2
𝛼𝑈(𝜃 +

sin 2𝜃

2
)0

2𝜋 = 𝜋𝛼𝑈                                       (51) 

                               

The force on the cylinder due to friction or shear stress is 

 

𝐹𝜏𝑟𝜃
= − ∫ 𝜏𝑟𝜃(𝑟0

2𝜋

0
 𝑑𝜃) sin 𝜃                                                                                          (52) 

 

𝜏𝑟𝜃 = 𝜇
𝜕𝑢𝜃

𝜕𝑟
= −

𝛼

2
(

1

𝑟
+

𝑟0
2

𝑟3 ) 𝑈 sin 𝜃                                                                                (53) 

 

 Substituting (53) into (52) and evaluating at 𝑟 = 𝑟0, we have 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

154 
 

  

𝐹𝜏𝑟𝜃
=

1

2
𝛼𝑈 ∫ (1 −

2𝜋

0
cos 2𝜃)𝑑𝜃 =

1

2
𝛼𝑈(𝜃 −

sin 2𝜃

2
)0

2𝜋 = 𝜋𝛼𝑈                                                    (54) 

The drag force is 

  

𝐹𝐷 = 𝐹𝑝 + 𝐹𝜏𝑟𝜃
=  𝜋𝛼𝑈 + 𝜋𝛼𝑈 = 2𝜋𝛼𝑈                                                                                    (55) 

The arbitrary constant 𝛼 has to be obtained from the far-field boundary condition. However, in equation (50), no 

value of 𝛼 will make 𝑢(𝑟, 𝜃) go to the constant value corresponding to the undisturbed flow since ln 𝑟 → ∞ as 

𝑟 → ∞. The inertia forces are as significant as the viscous forces at large distances from the cylinder and 

equation (50) thus represent the solution of the flow field at the vicinity of the cylinder and not at large values of 

 𝑟 (Oseen’s paradox). Clearly, some approximation to the equation of motion is needed for large 𝑟 and equation 

(50) must match with this approximate solution at large distance from the cylinder.  

 

The author [5] made an approximation to the equation of motion, which is  

 

𝜌(−𝑈. ∇𝑢 + 𝑢. ∇𝑢) = −∇𝑝 + 𝜇∇2𝑢                                                                                   (56) 

                                                                           

The two terms in the inertial force of equation (56) are of the same order near the cylinder, but the first term is 

dominant far from the cylinder. So, far from the cylinder equation (56) becomes 

 

𝜌(−𝑈. ∇𝑢) = −∇𝑝 + 𝜇∇2𝑢                                                                          (57) 

                         

Equation (57) and the continuity equation are known as the Oseen equations. The author [4] showed that 

equation (57) has a solution which approximates to equation (50) near the cylinder provided the constant in 

equation (50) is chosen as 

𝛼 =
2𝜇

ln(7.4 𝑅𝑒)⁄
                                                                                               (58) 

                                                       



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

155 
 

Substituting (58) into (55), we have 

                                        

                                                       𝐹𝐷 = 2𝜋𝑈 (
2𝜇

ln(7.4 𝑅𝑒)⁄
) 

=
4𝜋𝑈𝜇

ln(7.4 𝑅𝑒)⁄
                                                                                              (59) 

                                                                                                                             

                                

(59) is the drag force acting on the circular cylinder. The drag coefficient (𝐶𝐷) which is a function of the 

Reynolds number (𝑅𝑒) is defined generally as 

 

𝐶𝐷 =
2𝐹𝐷

𝜌𝑈2𝐴
                                                                                                       (60) 

 

where 𝐴 is the reference area of the object in the fluid. For this work 

𝐶𝐷 =
2𝐹𝐷

2𝑟0𝜌𝑈2                                                                                                      (61) 

        

Where 𝐴 = 2𝑟0        

Substituting (59) into (61), we have 

𝐶𝐷 =
8𝜋

𝑅𝑒 ln(7.4 𝑅𝑒)⁄
 ,     𝑅𝑒 ≪ 1                                                                         (62) 

5. Discussion 

Equation (62) is the drag coefficient (𝐶𝐷) of a fluid on a translating circular cylinder and this result agrees with 

the result in literature. 

 

                                            𝐶𝐷 =
8𝜋

𝑅𝑒 ln(7.4 𝑅𝑒)⁄
 , 𝑅𝑒 ≪ 1 



American Academic Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2022) Volume 86, No 1, pp143-156 

156 
 

The description of the flow field of a rigid body translating steadily and with a speed 𝑈 through an infinite body 

which is otherwise undisturbed depends only on a dimensionless quantity called the Reynolds number (𝑅𝑒). 

For 𝑅𝑒 ≪ 1, the flow is laminar and the drag coefficient varies inversely as the Reynolds number. This means 

that as the drag coefficient decreases, the speed of the circular cylinder increases in the fluid. 

6. Conclusion 

This paper addressed the analytic method of obtaining the drag coefficient of a fluid on a translating circular 

cylinder by solving the Navier-Stokes equations directly. The closure method of [5] applied by [4] and the 

asymptotic method used by some authors in finding the drag coefficient on a cylinder are also important 

technique. Inspired by [7], this paper can inspire more researchers to develop analytic methods in obtaining drag 

coefficient on objects.  

References 

[1] G. G. Stokes. “On the effect of the internal friction of fluids on the motion of pendulums.” Cambridge 

Philosophical Society, vol. 9, pp. 8-106, 1851. 

[2] R. A. Millikan. “The isolation of an ion, a precision instrument of its charge and the correction of Stokes' 

law.” Phys. Rev,, vol. 32, pp. 349-970, 1911. 

[3] R. A. Millikan. “On the elementary electrical charge and the Avogadro constant.” Phys. Rev., vol. 2, pp. 

109-430, 1913. 

[4] H. Lamb. “On the uniform motion of a sphere through a viscous fluid.” Philosophical Magazine, vol. 21, 

pp. 112-121, 1911. 

[5] C. W. Oseen. “Ueber die Stokes'sche formel und ueber einer verwandte Aufgabe in der  Hydrodynamik.” 

Ark. f.    Mat. Astr. Och. Fys., vol. 6, pp. 1-20, 1910. 

[6] B. Mutlu Sumer, and Jorgen Fredsoe. Hydrodynamics Around Cylindrical Structures. World Scientific Co. 

Pte. Ltd.,Revised Edition, 2006. 

[7] I. Eames, and C. A. Klettner. “Stokes’ and Lamb’s viscous drag Laws.” European Journal of Physics, vol. 

38, pp. 11, 2017. 

[8] G. K. Batchelor. An Introduction to Fluid Dynamics. Cambridge: Cambridge University Press, 1st edition, 

1967. 

[9] G. K. Batchelor. “Slender-body theory for particles of arbitrary cross-section in Stokes flow.” Journal of 

Fluid  Mechanics, vol. 44, pp. 419-440, 1970.