Velocity Potential

 

Example 1   In a certain two-dimensional flow field, the constant velocity components are

 

                                                             u = - 4 ft/sec   and     v  =  - 2 ft/sec

Find the velocity potential.

 

In general      u  =  ∂φ / ∂x   and  v  =   ∂ φ / ∂y

 

   ∂φ / ∂y   =  - 2      Integrate with respect to y  gives     φ(x,y)  =   - 2y    +  f(x)

 

Now      ∂ φ / ∂x =   u   So  df/dx  =  - 4   and   f(x)  =  - 4x + C

 

Then   φ(x,y)  =   - 2y    +  f(x)   and       φ(x,y)  =   - 2y - 4x  +  C       (result)

 

 

 

Example 2   The velocity potential for a certain inviscid flow field is

 

                                 φ(x,y)  =  - ( 3x2y  -  y3 )

 

where  φ(x,y) has the units of  ft2/sec when x and y are in feet.  Find the pressure difference

(in psi) between points (1,2) and (4,4), where the coordinates are in feet, if the fluid is water,

and elevation changes are negligible.

 

Since the fluid is water it is incompressible.  For steady, inviscid, and irrotational (potential flow)

you can use the following forms of Bernoulli’s equation to evaluate the pressure difference.

 

 P1/γ  +  V12/2g  +  z1  =  P2/γ  +  V22/2g  +  z2    or    P1  +  ½ ρV12  +  γ z1  =  P2 +  ½ ρV22  + γ z2   

 

                u  =  ∂φ / ∂x  =  - 6 xy     and      v  =   ∂ φ / ∂y  =  - 3x2  +  3 y2     in   ft/sec

 

    u1  =  - 6 (1)(2) =  -12  ft/sec,     v1  =  - 3 (12) + 3(22)  =  9 ft/sec   So  V12  =  (-12)2 + 92  =  225

 

    u2  =  - 6 (4)(4) =  - 96  ft/sec,     v1  =  - 3 (42) + 3(42)  =  0 ft/sec    So   V22  =  (-96)2  =  9216

 

    Now   γ z1   =  γ z2  .                  So    P1  -  P2  =    ½ ρV22    -  ½ ρV12   

 

    P1  -  P2  =    ½ (1.94)(9216)   -  ½ (1.94)(225)   =  8,721.3  lb/ft2    =  60.6 psi    (result)

 

 


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