ChE 304
Uniform Acceleration Example
Rotating Tank Problem
A 20 cm-diameter, 60-cm-high vertical cylindrical container is partially filled (to 50 cm) with a liquid whose density is
850 kg ë m3 . What is the rotational speed at which the liquid will start spilling over the top?
Assume: The increase in rotational speed is slow so the fluid acts as a rigid body. The bottom of the cylinder remains
covered with liquid.
Working in cylinderical coordinates Hr, q, zL
Centripetal acceleration is given as
ar = - rw2
where w is the angular velocity in radians ê s Hrad ê sL
¶∂ p
= r r w2
¶∂ r
¶∂ p
=0
¶∂ q
¶∂ p
= - rg
¶∂ z
In terms of the total derivative
¶∂ p
¶∂ p
dp =
dr +
dz
¶∂ r
¶∂ z
dp = r r w2 dr - r g dz
If we look at lines of constant pressure, at the surface p = 0 at all values of z
and we get Hr cancelsL
r w2 dr = g dz
rearranging and integrating yields z as
a function of r
z=
C1 +
1
g
2
‡ r w „ r + C1
r2 w2
2g
The constant of integration is evaluated at r = 0, C1 = hc (see figure on the right above). Substituting hc for C1 gives
2
RotatingTankProblem.nb
z = z ê. C1 Ø hc
r2 w2
+ hc
2g
Given that h0 is the height of the liquid in the stationary fluid, mass conservation coupled with incompressibility
means that the volume of the stationary and rotating fluid must be the same even if the shape of the free surface is
different. The volume of a cylindrical shell element is given by
dV = 2 p z r dr
Integrating between r = 0 and r = R and using the expression for z above
R
vol = ‡ 2 p z r „ r êê FullSimplify
0
p R4 w2
4g
+ p R2 hc
The original volume is given as
V0 = p R 2 h0
p R2 h0
Setting the two volumes equal and solving for hc .
ans = Solve@vol ã V0 , hc D êê Flatten
- R2 w2 + 4 g h0
:hc Ø
4g
>
Simplify the answer. Note: ans[[1,2]] is the right part (after the arrow) of equation ans
a1 = ans@@1, 2DD êê Simplify
-
R2 w2
4g
+ h0
Substituting the equation for hc into the equation for the free surface (above)
eq = z ê. hc -> a1 êê FullSimplify
I2 r2 - R2 M w2
4g
+ h0
Substituting the numbers from the problem
RotatingTankProblem.nb
ans2 = eq ê. :r Ø 0, h0 Ø 0.5 m, g Ø 9.81
m
s2
, R Ø 0.2 m>
0.5 m - 0.00101937 m s2 w2
Set the equation stored as ans2 equal to the difference between the surface of the fluid at rest and the height of the
container (z at R) and solve for w. The answer comes out in radians per second
ans3 = Solve@ans2 ã 0.1 m, wD êê Flatten
:w Ø -
19.8091
s
,wØ
19.8091
s
>
Changing to rpm. There are 2p radians in a full circle.
ans3@@2, 2DD
189.163
min
1
60 s
2 p min
3