Analysis of a Cylindrical (Parallel) Thermowell (or Sample Quill)

It is relatively easy to work out exactly, the fundamental natural resonance of a cantilevered cylinder with a concentric bore. (i.e. parallel thermowell).

The moment of inertia of the thermowell is everywhere the same and equal to:

Eq. 3.1 where D is the outside diameter, and d is the bore diameter.

The cross section is:

Eq. 3.2

The general motion of a beam of uniform cross section in free motion is:

Eq. 3.3

Where y is the deflection of the beam (as a function of x and t).

Because we are looking at a single harmonic function, y(x,t) can be written as:

y(x,t) = y(x) cos(wt) where w = 2 p fn (fn = natural frequency)

The equation to be solved is now:

Eq. 3.4

If we call

The generalized solution of the y(x) equation is:

Eq. 3.5

Where the A's are constants which depend only on the boundary conditions.

In the case of the thermowell, we have certain boundary conditions to satisfy:

1. At x = 0 y is always 0 as the thermowell at this point is fixed.

   This implies Eq. 3.6

2. At x = 0, the slope is always 0 (dy/dx = 0) as the thermowell is not free to rotate.

   This implies Eq. 3.7

3. At x = L, the bending moment is always 0, hence ,

   this implies, taking into account the relations obtained at Steps 1 and 2:

   Eq. 3.8

4. at x = L, the shear force is also always 0, hence , taking into account all previous relations we finally get:

   Eq. 3.9

   Eq. 3.10

For this to be true regardless of A1 is we must have

Eq. 3.10

An infinity of values satisfy this equation, but the lowest value is:

zL = 1.875104

This corresponds to or in frequency:

Eq. 3.11

If we define we can write I and A as functions of D and d:

And the frequency formula becomes:

Eq. 3.12

This formula is exact for parallel thermowells:

In metric units, E is in Pa, L and D in metres, r is in kg/m3, fn is in Hz.