0.1 Derivation of Added Mass around a Sphere

2.016 Hydrodynamics

Professor A.H. Techet by TA B. P. Epps

0.1 Derivation of Added Mass around a Sphere

When a body moves in a fluid, some amount of fluid must move around it. When the body accelerates, so too must the fluid. Thus, more force is required to accelerate the body in the fluid than in a vacuum. Since force equals mass times acceleration, we can think of the additional force in terms of an imaginary added mass of the object in the fluid.

One can derive the added mass of an object by considering the hydrodynamic force acting on it as it accelerates. Consider a sphere of radius, R, accelerating at rate U/t = U . We find the hydrodynamic force in the x-direction by integrating the pressure over the area projected in the x-direction:

Fx = p dAx

where

? dAx = cos dA, dA = 2rds, r = R sin , ds = Rd

? p = -

t

+

1 2

||2

,

by

unsteady

Bernoulli's

equation

=

U

cos

R3 2r2

,

for

axisymmetric

flow

around

a

sphere

t

|r=R

=

U

cos

R3 2r2

=

U

cos

R 2

1 2

||2|r=R

=

1 2

|(-U

cos

R3 r3

,

-U

sin

R3 2r3

)|2

=

1 2

[U

2

cos2

+

1 4

U

2

sin2 ]

so

Fx =

0

-

+ 1 ||2 t 2

cos 2R2 sin d

=

-

U

cos

R

+

1 (U 2

cos2

+

1U2

sin2

)

cos 2R2 sin d

0

22

4

= - ? 2R2 ? U R

sin cos2 d - ? 2R2 ? 1 U 2

[sin

cos3

+

1

sin3

cos

]d

20

20

4

=2/3

=0

= - 2 R3U 3

where U is the acceleration of the body, and the negative sign indicates that the force is in the negative

x-direction, opposing the acceleration. Thus, the body must exert this extra force, and the apparent added

mass is

ma

=

2 R3 3

1

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 in the x-direction by integrating the pressure over the area projected in the xdirection:

Fx = P dAx

where

? dAx = cos dA dA = Lds

ds = Rd

? P = -

t

+

1 2

||2

,

by

unsteady

Bernoulli's

equation

=

U

R2 r

cos

,

for

flow

around

a

cylinder

t

|r=R

=

U

R2 r

cos

=

U R

cos

1 2

||2|r=R

=

1 2

|(-U

R2 r2

cos , -U

R2 r2

sin )|2

=

1 2

U

2

so

Fx =

2 0

-

+ 1 ||2 t 2

cos RLd

= 2 - U R cos + 1 U 2) cos RLd

0

2

= - ? RL ? U R

2 cos2 d - ? RL ? 1 U 2

2

cos d

0

20

=

=0

= -R2LU

where a = U is the acceleration of the body, and the negative sign indicates that the force is in the negative

x-direction, opposing the acceleration. Thus, the body must exert this extra force, and the apparent added

mass is

ma = R2L

2

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download