Dr.

Ratnakar Swain
(M. Tech. & Ph.D., IIT KGP)
Asst. Professor
Department of Civil Engineering
NIT Rourkela
 Continues

Circulation (𝚪𝚪): The line integral of the tangential

velocity about a closed path.
𝚪𝚪 = ∮ 𝒗𝒗 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽𝜽𝜽𝜽𝜽

Circulation around an elementary box:

𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝚪𝚪 = 𝒖𝒖∆𝒙𝒙 + 𝒗𝒗 + ∆𝒙𝒙 ∆𝒚𝒚 − 𝒖𝒖 + ∆𝒚𝒚 ∆𝒙𝒙 − 𝒗𝒗∆𝒚𝒚
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
= − ∆𝒙𝒙 ∆𝒚𝒚
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

2
Dr. R. Swain, Asst. Prof., NIT RKL
𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕 𝜴𝜴 : Circulation per unit area

𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 (𝚪𝚪) 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

𝛀𝛀 = = −
𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

Rotation (𝝎𝝎): ½ of Vorticity

𝟏𝟏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝝎𝝎 = −
𝟐𝟐 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
For irrotational flow: 𝝎𝝎 =0
For rotational flow:
𝟏𝟏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝝎𝝎𝒙𝒙 = − 𝜴𝜴𝒙𝒙 = 𝟐𝟐𝝎𝝎𝒙𝒙 = −
𝟐𝟐 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝒚𝒚 𝝏𝝏𝒛𝒛

𝟏𝟏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

𝝎𝝎𝒚𝒚 = − 𝜴𝜴𝒚𝒚 = 𝟐𝟐𝒘𝒘𝒚𝒚 = −
𝟐𝟐 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

𝟏𝟏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

𝝎𝝎𝒛𝒛 = − 𝜴𝜴𝒛𝒛 = 𝟐𝟐𝒘𝒘𝒛𝒛 = −
𝟐𝟐 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

Dr. R. Swain, Asst. Prof., NIT RKL 3

In vector form
𝟏𝟏 �
𝛀𝛀 = 𝛀𝛀𝒙𝒙 𝒊𝒊̂ + 𝛀𝛀𝒚𝒚 𝒋𝒋̂ + 𝛀𝛀𝒛𝒛 𝒌𝒌
�
𝝎𝝎 = 𝝎𝝎𝒙𝒙 𝒊𝒊̂ + 𝝎𝝎𝒚𝒚 𝒋𝒋̂ + 𝝎𝝎𝒛𝒛 𝒌𝒌
𝟐𝟐
𝑖𝑖 𝑗𝑗 𝑘𝑘
𝑖𝑖 𝑗𝑗 𝑘𝑘
𝟏𝟏 𝜕𝜕 𝟏𝟏 𝜕𝜕 𝜕𝜕 𝜕𝜕
𝜕𝜕 𝜕𝜕 𝛀𝛀 = = ∇ × 𝑣𝑣
𝝎𝝎 = = ∇ × 𝑣𝑣 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝟐𝟐 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝟐𝟐
𝑢𝑢 𝑣𝑣 𝑤𝑤
𝑢𝑢 𝑣𝑣 𝑤𝑤

Problem-1:
The velocity components of a fluid flow are
u=a+by-cz, v=d-bx-cz, w=f+cx-ey a, b, c, d, e, f are constant
→ show it is a possible case of fluid flow
→ is the fluid flow irrotational and if not, determine vorticity and rotation
Dr. R. Swain, Asst. Prof., NIT RKL 4
 Solution:
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
+ + = 𝟎𝟎 + 𝟎𝟎 + 𝟎𝟎 = 𝟎𝟎 ⇒ 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 , so possible case of fluid flow
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

𝒊𝒊 𝒋𝒋 𝒌𝒌 𝒊𝒊 𝒋𝒋 𝒌𝒌
𝝏𝝏 𝝏𝝏 𝝏𝝏 𝝏𝝏 𝝏𝝏 𝝏𝝏
∆ × 𝒗𝒗 = =
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝒖𝒖 𝒗𝒗 𝒘𝒘 𝒂𝒂 + 𝒃𝒃𝒃𝒃 − 𝒄𝒄𝒄𝒄 𝒅𝒅 − 𝒃𝒃𝒃𝒃 − 𝒄𝒄𝒄𝒄 𝒇𝒇 + 𝒄𝒄𝒄𝒄 − 𝒆𝒆𝒆𝒆
𝝏𝝏 𝝏𝝏 𝝏𝝏 𝝏𝝏 𝝏𝝏 𝝏𝝏
= 𝒊𝒊[ (𝒇𝒇 + 𝒄𝒄𝒄𝒄 − 𝒆𝒆𝒆𝒆)- (𝒅𝒅 − 𝒃𝒃𝒃𝒃 − 𝒄𝒄𝒄𝒄)]+ j[ 𝒂𝒂 + 𝒃𝒃𝒃𝒃 − 𝒄𝒄𝒄𝒄 − (𝒇𝒇 + 𝒄𝒄𝒄𝒄 − 𝒆𝒆𝒆𝒆)]+ k[ 𝒅𝒅 − 𝒃𝒃𝒃𝒃 − 𝒄𝒄𝒄𝒄 − (𝒂𝒂 +
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝒃𝒃𝒃𝒃 − 𝒄𝒄𝒄𝒄)] = −𝟐𝟐 𝒄𝒄𝒄𝒄 + 𝒃𝒃𝒃𝒃 ≠ 𝟎𝟎 ⇒ 𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓

Vorticity: 𝜴𝜴 = ∆ × 𝒗𝒗 = −𝟐𝟐 𝒄𝒄𝒄𝒄 + 𝒃𝒃𝒃𝒃 = 𝟐𝟐 𝒄𝒄𝟐𝟐 + 𝒃𝒃𝟐𝟐

Rotation: = 𝛀𝛀𝟐𝟐 = 𝟏𝟏𝟐𝟐 𝟐𝟐 𝒄𝒄𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒄𝒄𝟐𝟐 + 𝒃𝒃𝟐𝟐

5
Dr. R. Swain, Asst. Prof., NIT RKL
Potential Flow: The flow which is frictionless (Inviscid), incompressible and irrotational.
To characterize the flow whether it is rotational or irrotational and to describe the velocity field, two scalar
functions are used.

 Velocity Potential Function (𝝋𝝋)

 Stream Function (𝜓𝜓)
Velocity potential (𝝋𝝋)
The scalar function in space and time such that its negative gradient gives the velocity in that
direction in an irrotational flow.
𝝋𝝋 = 𝒇𝒇 𝒙𝒙, 𝒚𝒚, 𝒛𝒛, 𝒕𝒕
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝐮𝐮 = − 𝐯𝐯 = −
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
• Negative sign indicates, 𝝋𝝋 decreses with an increase in the value of x,y,z,t

Dr. R. Swain, Asst. Prof., NIT RKL 6

 Using 𝝋𝝋 in Continuity equation in 3-d
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
+ + = 𝟎𝟎
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏 𝝏𝝏𝝏𝝏

− + − + − = 𝟎𝟎
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝝏𝝏𝟐𝟐 𝝋𝝋 𝝏𝝏𝟐𝟐 𝝋𝝋 𝝏𝝏𝟐𝟐 𝝋𝝋
⇒ + 𝟐𝟐 + = 𝟎𝟎 (Laplace Equation)
𝝏𝝏𝒙𝒙𝟐𝟐 𝝏𝝏𝒚𝒚 𝝏𝝏𝒛𝒛𝟐𝟐

Using 𝝋𝝋 to evaluate rotational components

𝟏𝟏 𝟏𝟏 𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏 𝝏𝝏𝝏𝝏 𝟏𝟏 𝟏𝟏 𝝏𝝏𝟐𝟐 𝝋𝝋 𝝏𝝏𝟐𝟐 𝝋𝝋
𝝎𝝎𝒙𝒙 = 𝒘𝒘𝒚𝒚 − 𝒗𝒗𝒛𝒛 = − − − = 𝟎𝟎 𝝎𝝎𝒛𝒛 = 𝒗𝒗 − 𝒖𝒖𝒚𝒚 = − + = 𝟎𝟎
𝟐𝟐 𝟐𝟐 𝝏𝝏𝝏𝝏 𝝏𝝏𝒛𝒛 𝝏𝝏𝝏𝝏 𝝏𝝏𝒚𝒚 𝟐𝟐 𝒙𝒙 𝟐𝟐 𝝏𝝏𝒙𝒙 𝝏𝝏𝒚𝒚 𝝏𝝏𝒚𝒚 𝝏𝝏𝒙𝒙

𝟏𝟏 𝟏𝟏 𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏 𝝏𝝏𝝏𝝏
𝝎𝝎𝒚𝒚 = 𝒖𝒖𝒛𝒛 − 𝒘𝒘𝒙𝒙 = − − − = 𝟎𝟎 𝝎𝝎𝒙𝒙 = 𝝎𝝎𝒚𝒚 = 𝝎𝝎𝒛𝒛 = 𝟎𝟎,
𝟐𝟐 𝟐𝟐 𝝏𝝏𝒛𝒛 𝝏𝝏𝒙𝒙 𝝏𝝏𝒙𝒙 𝝏𝝏𝒛𝒛
⇒ 𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇 𝒊𝒊𝒊𝒊 𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊

 Hence, If 𝝋𝝋 satisfy Laplace equation, it represents the possible case of Potential flow i.e. Frictionless,
incompressible and irrotational flow.
7
Dr. R. Swain, Asst. Prof., NIT RKL
Stream function (𝝍𝝍):
It defined a incompressible flow in 2-d and 3-d (with axisymmetric), such that its gradient w.r.t.
any direction gives the velocity component at perpendicular to that direction.
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝝍𝝍 = 𝒇𝒇(𝒙𝒙, 𝒚𝒚, 𝒕𝒕) 𝒖𝒖 = 𝒗𝒗 = −
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

2-d continuity equation 𝒖𝒖𝒙𝒙 + 𝒗𝒗𝒚𝒚 = 𝟎𝟎

𝝏𝝏𝟐𝟐 𝝍𝝍 𝝏𝝏𝟐𝟐 𝝍𝝍
⇒ − = 𝟎𝟎  Hence possible case of fluid flow
𝝏𝝏𝝏𝝏𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏𝝏𝝏𝝏𝝏

Rotational component:
𝟏𝟏 𝟏𝟏 𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏 𝝏𝝏𝝏𝝏 𝟏𝟏 𝝏𝝏𝟐𝟐 𝝍𝝍 𝝏𝝏𝟐𝟐 𝝍𝝍
𝒘𝒘𝒛𝒛 = 𝒗𝒗𝒙𝒙 − 𝒖𝒖𝒚𝒚 = − − =− 𝟐𝟐
+
𝟐𝟐 𝟐𝟐 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝟐𝟐 𝝏𝝏𝒙𝒙 𝝏𝝏𝒚𝒚𝟐𝟐

For irrotational flow 𝒘𝒘𝒛𝒛 = 𝟎𝟎 𝝏𝝏𝟐𝟐 𝝍𝝍 𝝏𝝏𝟐𝟐 𝝍𝝍

⇒ + = 𝟎𝟎
𝝏𝝏𝒙𝒙𝟐𝟐 𝝏𝝏𝒚𝒚𝟐𝟐
⇒ ∆𝟐𝟐 𝝍𝝍 = 𝟎𝟎 = 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 8
Dr. R. Swain, Asst. Prof., NIT RKL
Properties of Stream Function:
o 𝝍𝝍 is constant along any streamline.
o For a continuous flow, the volume flow rate is zero across the closed path.
o Rate of change of 𝝍𝝍 with distance ∝ component of velocity normal to that direction.
o If two incompressible flow patterns are superimposed, the sum of 𝝍𝝍 gives the 𝝍𝝍 of another
flow pattern, i.e. 𝝏𝝏𝝍𝝍𝟏𝟏 𝝏𝝏𝝍𝝍𝟐𝟐 𝝏𝝏(𝝍𝝍𝟏𝟏 + 𝝍𝝍𝟐𝟐 )
+ =
𝝏𝝏𝒔𝒔 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏

Equipotential Line: The line joining the points of e𝐪𝐪𝐪𝐪𝐪𝐪𝐪𝐪 𝝋𝝋.

𝝋𝝋 is constant ⇒ d𝝋𝝋 = 𝟎𝟎

𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
⇒ 𝒅𝒅𝒅𝒅 + 𝒅𝒅𝒅𝒅 = 𝟎𝟎
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝒅𝒅𝒅𝒅 𝒖𝒖
⇒ =−
⇒ −𝒖𝒖𝒖𝒖𝒖𝒖 − 𝒗𝒗𝒗𝒗𝒗𝒗 = 𝟎𝟎 𝒅𝒅𝒅𝒅 𝒗𝒗
𝒖𝒖
⇒ 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒐𝒐𝒐𝒐 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = −
Dr. R. Swain, Asst. Prof., NIT RKL 𝒗𝒗 9
Relationship Between 𝝓𝝓 and 𝝍𝝍:
𝝏𝝏𝝏𝝏
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝒖𝒖
For constant - 𝝓𝝓 curve, 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 = = =
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝒗𝒗
𝝏𝝏𝝏𝝏

For constant -𝝍𝝍 curve,

𝝏𝝏𝝏𝝏
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 −𝒗𝒗
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 = = =
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝒖𝒖
𝝏𝝏𝝏𝝏

⇒ 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 𝒐𝒐𝒐𝒐 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒐𝒐𝒐𝒐 𝝋𝝋 & 𝝍𝝍 = −𝟏𝟏

Hence the 𝝋𝝋 and 𝝍𝝍 lines are perpendicular to each other.

Dr. R. Swain, Asst. Prof., NIT RKL 10

Flow net:
• The grid obtained by drawing a series of streamlines and equipotential line is called flow net.

• It is a simple graphical technique for studying 2-D irrotational flow, especially when
mathematical relation for 𝝋𝝋 and 𝝍𝝍 are not available.

Use of flow net:

 To determine the streamlines and equipotential

line.

 To determine quantity of seepage and upward lift

pressure below the hydraulic structure.

 To determine streamlines of the outlets.

 To determine the velocity and pressure

distribution for a given flow boundaries.

Dr. R. Swain, Asst. Prof., NIT RKL 𝝍𝝍 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 (𝝓𝝓 − 𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍)

11
𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 𝒏𝒏𝒏𝒏𝒏𝒏
 Problem-1: 𝒖𝒖 = 𝟐𝟐𝟐𝟐𝟐𝟐, 𝒗𝒗 = 𝒖𝒖𝟐𝟐 + 𝒙𝒙𝟐𝟐 − 𝒚𝒚𝟐𝟐
• Show that flow is possible.
• Derive the relative stream function.

𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
Solution: + = 𝟐𝟐𝟐𝟐 + −𝟐𝟐𝟐𝟐 = 𝟎𝟎
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
⇒ 𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇 𝒊𝒊𝒊𝒊 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑

𝝏𝝏𝝏𝝏
𝒖𝒖 = = 𝟐𝟐𝟐𝟐𝟐𝟐
𝝏𝝏𝝏𝝏
⇒ 𝒗𝒗 = 𝒂𝒂𝟐𝟐 + 𝒙𝒙𝟐𝟐 − 𝒚𝒚𝟐𝟐 = −𝒚𝒚𝟐𝟐 − 𝒇𝒇′ 𝒙𝒙
⇒ 𝝍𝝍 = � 𝟐𝟐𝟐𝟐𝟐𝟐 𝒅𝒅𝒅𝒅 = 𝒙𝒙𝒚𝒚𝟐𝟐 + 𝒇𝒇(𝒙𝒙) ⇒ 𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉 𝒇𝒇′ 𝒙𝒙 = − 𝒂𝒂𝟐𝟐 + 𝒙𝒙𝟐𝟐
𝟑𝟑
𝒙𝒙
𝝏𝝏𝝏𝝏 ⇒ 𝒇𝒇 𝒙𝒙 = − 𝒂𝒂𝟐𝟐 𝒙𝒙 + + 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄
⇒ − = −𝒚𝒚𝟐𝟐 − 𝒇𝒇′ 𝒙𝒙 𝟑𝟑
𝝏𝝏𝝏𝝏

𝟐𝟐
𝒙𝒙𝟑𝟑
𝟐𝟐 𝟐𝟐
∴ 𝝍𝝍 = 𝒙𝒙𝒚𝒚 − 𝒂𝒂 𝒙𝒙 + + 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄
𝟑𝟑
12
Dr. R. Swain, Asst. Prof., NIT RKL
 Problem-(2) In a 2d incompressible flow, the fluid velocity components are given by
𝒖𝒖 = 𝒙𝒙 − 𝟒𝟒𝟒𝟒 𝒗𝒗 = −𝒚𝒚 − 𝟒𝟒𝟒𝟒
Show that 𝝋𝝋 exists and determine its from as well as 𝝍𝝍

Solution: The velocity potential (𝝋𝝋) exists if flow is irrotational

⇒ 𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗 𝛀𝛀𝒛𝒛 = 𝟎𝟎

𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝛀𝛀𝒛𝒛 = − = −𝟒𝟒 + 𝟒𝟒 = 𝟎𝟎
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
⇒ 𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓 𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇 𝒂𝒂𝒂𝒂𝒂𝒂 𝝋𝝋 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝝏𝝏𝝍𝝍 𝝏𝝏𝝍𝝍
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅 + 𝒅𝒅𝒅𝒅
𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅 + 𝒅𝒅𝒅𝒅 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏
𝛛𝛛𝛛𝛛 = −𝐯𝐯𝐯𝐯𝐯𝐯 + 𝐮𝐮𝐮𝐮𝐮𝐮
= −𝒖𝒖𝒖𝒖𝒖𝒖 − 𝒗𝒗𝒗𝒗𝒗𝒗
= [− −𝒚𝒚 − 𝟒𝟒𝟒𝟒 𝒅𝒅𝒅𝒅 + 𝒙𝒙 − 𝟒𝟒𝟒𝟒 𝒅𝒅𝒅𝒅
= − 𝒙𝒙 − 𝟒𝟒𝟒𝟒 𝒅𝒅𝒅𝒅 − −𝒚𝒚 − 𝟒𝟒𝟒𝟒 𝒅𝒅𝒅𝒅
⇒ 𝝍𝝍 = 𝒙𝒙𝒙𝒙 + 𝟐𝟐𝒙𝒙𝟐𝟐 + 𝒙𝒙𝒙𝒙 − 𝟐𝟐𝒚𝒚𝟐𝟐 + 𝒄𝒄
= −𝒙𝒙𝒙𝒙𝒙𝒙 + 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 + 𝒚𝒚𝒚𝒚𝒚𝒚 + 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
= 𝟐𝟐 𝒙𝒙𝟐𝟐 − 𝒚𝒚𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒄𝒄
𝟏𝟏 𝟐𝟐
⇒ 𝝋𝝋 = 𝒚𝒚 − 𝒙𝒙𝟐𝟐 + 𝟖𝟖𝟖𝟖𝟖𝟖 + 𝒄𝒄
𝟐𝟐 ⇒ 𝝍𝝍 = 𝟐𝟐 𝒙𝒙𝟐𝟐 − 𝒚𝒚𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒄𝒄 13
Dr. R. Swain, Asst. Prof., NIT RKL