Chapter 6 Problem 038 The velocity potential for a certain i

Chapter 6, Problem 038 The velocity potential for a certain inviscid flow field is 8x2y 3 where p has the units of ft2/s w hen x and y are in feet pressure difference (in psi) between the points (1, Determine the 2) and (3, 3), here the coordinates are in feet, if the fluid is w ater and elevation changes are negligible P1 P2. ps LINK TO TEXT Question Attempts: 0 of 5 used SAVE FOR LATER. Copyright C 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved. y Policy I C 2000-2016 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons Inc.

Solution

ince 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/  +  V₁²/2g  +  z₁  =  P₂/  +  V₂²/2g  +  z₂    or    P₁  +  ½ V₁²  +   z₁  =  P₂ +  ½ V₂²  + z₂   

                u  =   / x  =  - 16 xy     and      v  =    / y  =  - 8x²  +  3 y²     in   ft/sec

    u₁  =  - 16 (1)(2) =  -32  ft/sec,     v₁  =  - 8 (1²) + 3(2²)  = 4 ft/sec   So  V₁²  =  (-32)² + 4² = 1040

    u₂  =  - 16 (3)(4) =  - 144 ft/sec,     v₁  =  - 8(3²) + 3(3²)  = -45 ft/sec    So   V₂²  =  (-144)2 +45^2  = 22761

    Now    z₁   =   z₂  .                  So    P₁  -  P₂  =    ½ V₂²    -  ½ V₁²   

    P₁  -  P₂  =    ½ (1.94)(22761)   -  ½ (1.94)(1040)   = 21059.67 lb/ft²    = 146.33 psi    (result)