Chapter 6, Problem 038 The velocity potential for a certain inviscid flow field

Question

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.

Explanation / Answer

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)

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