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2.016 HydrodynamicsProfessor A.H. Techet

by TA B. P. Epps

0.1 Derivation of Added Mass around a SphereWhen a body moves in a fluid, some amount of fluidmust move around it. When the body accelerates, sotoo must the fluid. Thus, more force is required to ac-celerate the body in the fluid than in a vacuum. Sinceforce equals mass times acceleration, we can think ofthe additional force in terms of an imaginary addedmass of the object in the fluid.

One can derive the added mass of an object byconsidering the hydrodynamic force acting on it as itaccelerates. Consider a sphere of radius, R, acceler-

ating at rate U/t = U. We find the hydrodynamicforce in the x-direction by integrating the pressureover the area projected in the x-direction:

Fx =

p d Ax

where

d Ax = cos dA, dA = 2rds, r = R sin , ds = Rd

p = t +

1

2||2

= Ucos R3

2r2 , for axisymmetric flow around a sphere

t |r=R = Ucos

R3

2r2= Ucos R

2

1

2||2|r=R =

1

2|(Ucos R

3

r3 ,Usin R3

2r3)|2 = 1

2[U2 cos2 + 1

4U2 sin2 ]

so

Fx =

0

t+

1

2||2

cos 2R2 sin d

=

0

Ucos

R

2+

1

2(U2 cos2 +

1

4U2 sin2 )

cos 2R2 sin d

= 2R2 UR

2

0

sin cos2 d

=2/3

2R2 1

2U2

0

[sin cos3 +1

4sin3 cos ]d

=0

= 2

3R3U

where U is the acceleration of the body, and the negative sign indicates that the force is in the negativex-direction, opposing the acceleration. Thus, the body must exert this extra force, and the apparent addedmass is

ma =2

3R3

1

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0.2 Derivation of Added Mass around a Cylinder

Similarly, for a cylinder of radius, R,and length, L, accelerating at rate U.We find the hydrodynamic force inthe x-direction by integrating the pres-sure over the area projected in the x-direction:

Fx =

Pd Ax

where

d Ax = cos dA

dA = Lds

ds = Rd

P = t + 12 ||2, by unsteady Bernoullis equation = UR

2

r cos , for flow around a cylinder

t |r=R = U

R2

r cos = UR cos

1

2||2|r=R =

1

2|(UR

2

r2 cos ,UR2

r2 sin )|2 = 1

2U2

so

Fx =

2

0

t+

1

2||2

cos RLd

= 2

0 UR cos +

1

2U2

) cos RLd= RL UR

2

0

cos2 d =

RL 1

2U2

2

0

cos d =0

= R2LU

where a = U is the acceleration of the body, and the negative sign indicates that the force is in the negativex-direction, opposing the acceleration. Thus, the body must exert this extra force, and the apparent addedmass is

ma = R2L

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