XYLENE POWER LTD.

DESIGN FORMULAE

By Charles Rhodes, P.Eng., Ph.D.

FLUID PRESSURE CONTAINMENT:
Irrespective of thermal stress, the fluid pressure must always be safely contained. In this respect there are two formulae that are frequently used on this web site.

The first is Barlow's formula for calculating hoop stress in a cylinder containing a pressurized fluid. In Barlow's formula the applied force F is:
F = P D L
where:
P = pressure difference acrosss the cylinder wall;
D = inside diameter;
L = a unit length of the cylinder.

At force equilibrium:
P D L = S (2 W) L
or
P = S (2 W) / D

In any matter potentially involving personnel life safety the maximum acceptable value of S is Sy / 3, where:
Sy = cylinder material yield stress

Thus the maximum safe fluid pressure Pm in a cylinder is:
Pm = [2 Sy W] / [3 D]
 

The second important formula is to calculate the relationship between the pressure across a uniform disc that is supported at its perimeter and the maximum internal stress in the disk material.

The applied torque transferred to the edge of the disc by the pressure is given by:
Integral from R = 0 to R = (D / 2) of:
P 2 Pi R dR [(D / 2) - R)
= P 2 Pi (D / 2)(1 / 2) (D / 2)^2 - P 2 Pi (1 / 3) (D / 2)^3
= P 2 Pi (1 / 6) (D / 2)^3

Assume that the resisting torque is zero half way through the thickness of the disk and increases to reach a maximum at the disc surfaces. Let Sm = maximum stress at surface of disc.
Then S = Z Sm / (T / 2)
= 2 Z Sm / T

Then the resisting torque is:
Integral from Z = 0 to Z = (T / 2) of:
2 {S 2 Pi D dZ Z}
= Integral from Z = 0 to Z = (T / 2) of:
2 {(2 Z Sm / T) 2 Pi D dZ Z}
= [8 Sm Pi D / T] [Z^3 / 3]|Z = (T / 2) = [8 Sm Pi D / 3 T] [T / 2]^3
= [Sm Pi D / 3] [T]^2

At equilibrium the applied torque equals the resisting torque, or:
P 2 Pi (1 / 6) (D / 2)^3 = [Sm Pi D / 3] [T]^2
or P (D^2 / 8) = [Sm] [T]^2

In general, for life safety the maximum value of Sm is Sy / 3. Thus the maximum value of pressure P is:
P = 8 (Sy / 3) (T / D)^2
or rearranging:
T^2 = 3 D^2 P / 8 Sy
or T = D [3 P / 8 Sy]^0.5
 

CONSTRAINTS ON HEAT EXCHANGER TUBE LENGTH, OD and ID:
Each FNR Heat Transport Circuit involves five shell and tube type counter flow heat exchangers: the intermediate heat exchange bundle, the NaK-nitrate salt heat exchanger, the NaK-oil heat exchanger, the steam generator hot stage and the steam generator cool stage. These heat exchangers involve 5 m to 6 m long heat exchange tubes which are only supported at their ends. From a heat transport perspective, the smaller the tube diameter the better, because a smaller diameter tube allows a larger heat transfer surface area. However, there are practical limitations on the unsupported tube length and the tube OD and ID due to the torque on the tube ends exerted by the weight of the tube when the long axis of the heat exchange bundle is horizontal, such as during truck transport. It is shown herein that, due to lack of support along the tube length, it is impractical to make these tubes too small in diameter. This issue constrains the ultimate heat exchange surface area.

Define:
Rho = density of tube material
Ri = Tube inside radius
Ro = Tube outside radius
Pi = 3.14159265
X = distance along tube measured from the supporting end
L = tube length
g = gravitational acceleration
Sy = tube material yield stress
Z = height from tube center line
S= tube material stress at height Z fromthe tube center line.

The applied torque Ta exerted by the tube weight on its end is:
Ta = Integral from X = 0 to X = (L / 2) of:
Pi (Ro^2 - Ri^2) dX Rho g X
= Pi (Ro^2 - Ri^2) Rho g (1 / 2) (L / 2)^2
= Pi (Ro^2 - Ri^2) Rho g L^2 / 8

A circle is described by the formula:
Y^2 + Z^2 = R^2
or
Y^2 = R^2 - Z^2
or
Y = [R^2 - Z^2]^0.5

At any particular value of Z the value of Y on the inside surface of the tube is:
Yi = [Ri^2 - Z^2]^0.5
and the value of Y on the outside surface of the tube is:
Yo = [Ro^2 - Z^2]^0.5

Hence for Z < Ri:
Yo - Yi = {[Ro^2 - Z^2]^0.5 - [Ri^2 - Z^2]^0.5}

For Z > Ri, Yi = 0 which gives:
Yo = {[Ro^2 - Z^2]^0.5}

Assume that the tube material stress S(Z) varies linearly from S = 0 at Z = 0 to S = So at Z = Ro.

Hence the opposing torque To is given by:
To = Integral from Z = 0 to Z = Ri of:
4 (Yo - Yi) dZ So (Z / Ro) Z
+ Integral from Z = Ri to Z = Ro of:
4 Yo dZ So (Z / Ro) Z

Hence the opposing torque To is given by:
To = Integral from Z = 0 to Z = Ri of:
4 {[Ro^2 - Z^2]^0.5 - [Ri^2 - Z^2]^0.5} dZ So (Z / Ro) Z
+ Integral from Z = Ri to Z = Ro of:
4 {[Ro^2 - Z^2]^0.5} dZ So (Z / Ro) Z
 
= Integral from Z = 0 to Z = Ri of:
4 Ro^3 {[1 - (Z / Ro)^2]^0.5 - [(Ri / Ro)^2 - (Z / Ro)^2]^0.5} d(Z / Ro) So (Z / Ro) (Z / Ro)
+ Integral from Z = Ri to Z = Ro of:
4 Ro^3 {[1 - (Z / Ro)^2]^0.5} d(Z / Ro) So (Z / Ro) (Z / Ro)

Define:
V = Z / Ro
Then:
To = Integral from V = 0 to V = (Ri / Ro) of:
4 Ro^3 {[1 - V^2]^0.5 - [(Ri / Ro)^2 - V^2]^0.5} V^2 dV So
+ Integral from V = Ri / Ro to V = 1 of:
4 Ro^3 {[1 - V^2]^0.5} V^2 dV So

The maximum acceptable value of So is:
So = Sy / 3

The applied torque Ta must be less than the maximum opposing torque To. Ta < To
gives:
Pi Ro^2 (1 - (Ri / Ro)^2) Rho g L^2 / 8
< Integral from V = 0 to V = (Ri / Ro) of:
4 Ro^3 {[1 - V^2]^0.5 - [(Ri / Ro)^2 - V^2]^0.5} V^2 dV (Sy / 3)
+ Integral from V = (Ri / Ro) to V = 1 of:
4 Ro^3 {[1 - V^2]^0.5} V^2 dV (Sy / 3)

Hence the heat exchange tube outside radius Ro must be sufficiently large to ensure satisfaction of this inequality.

Try substitution:
V = sin U
{(1 - V^2)^0.5} = cos U
dV = cos U dU
giving:
Integral from V = 0 to V = 1 of:
{[1 - V^2]^0.5} V^2 dV = Integral from U = arc sin(0) to U = arc sin 1 of:
cos^2 U sin^2 dU = (1 / 8)[U - (sin 4 U) /4]|U = arc sin 1
- (1 / 8) [U - (sin 4 U) / 4]|U = arc sin 0
= (1 / 8)[U - (sin 4 U) /4]|U = (Pi / 2)
- (1 / 8) [U - (sin 4U) / 4]|U = 0
= (1 / 8)[(Pi / 2)]

Now consider:
Integral from V = 0 to V = (Ri / Ro) of:
[(Ri / Ro)^2 - V^2]^0.5} V^2 dV
= Integral from V = 0 to V = (Ri / Ro) of:
(Ri / Ro)[1 - (Ro / Ri)^2 V^2]^0.5 V^2 dV

Try substitution:
(Ro / Ri) V = sin U
V^2 = (Ri / Ro)^2 sin^2 U
[1 - (Ro / Ri)^2 V^2]^0.5 = cos U
dV = (Ri / Ro) cos U dU

Then:
Integral from V = 0 to V = (Ri / Ro) of:
(Ri / Ro)[1 - (Ro / Ri)^2 V^2]^0.5 V^2 dV

Thus the inequality:
Pi (1 - (Ri / Ro)^2) Rho g (L^2 / 8 Ro)(3 / 4 Sy)
< Integral from V = 0 to V = 1 of:
{[1 - V^2]^0.5 V^2 dV
- Integral from V = 0 to V = (Ri / Ro) of:
[(Ri / Ro)^2 - V^2]^0.5} V^2 dV
 
becomes:
Pi (1 - (Ri / Ro)^2) Rho g (L^2 / 8 Ro)(3 / 4 Sy)
< (1 / 8)[(Pi / 2)] - (Ri / Ro)^4 (1 / 8)(Pi / 2)
or
(1 - (Ri / Ro)^2) Rho g (L^2 / Ro)(3 / 4 Sy)
< [(1 / 2)]{1 - (Ri / Ro)^4}

This equation sets the minimum value of Ro for all of the heat exchange tubes in the FNR heat transport system.

Units check: (kg / m^3) (m / s^2) m^2 / (kg m / s^2 m^2) = m

Numerical evaluation:
Rho = 8030 kg / m^3 g = 9.8 m / s
L = 5 m
Sy = 117 X 10^6 Pa

3 Rho g L^2 / 2 Sy
= (3 X 8.03 X 10^3 X 9.8 X 25)m / (2 X 117 X 10^6) = 25.22 X 10^-3 m

Consider tubing which is (1 / 2) inch ID, (5 / 8) inch OD: Then: Ri / Ro = 0.8 (Ri / Ro)^2 = 0.64 (Ri / Ro)^4 = .4096 [1 - (Ri / Ro)^2] / [1 - (Ri / Ro)^4]
= [1 - 0.64] / [1 - 0.4096] = 0.36 / 0.5904 = 0.6097

Hence the corresponding minimum value of Ro is:
0.6097 X 25.22 mm = 15.37 mm.
Hence the contemplated:
(5 / 8) inch OD = 15.875 mm OD
is sufficient with 5 m long heat exchange tubes but is not adequate with 6 m long heat exchange tubes because as L^2 changes from 25 m^2 to 36 m^2 the effect of the L^2 term becomes over whelming. Hence with (1 / 2) inch ID, (5 / 8) inch OD the unsupported heat exchange tube length is limited to 5 m. It is necessary to substantially increase the tube diameter to overcome this issue. This constraint on tube geometry affects many aspects of the FNR heat transport system design.

In summary, for the steam generator the proposed exposed tubes are 0.500 inch ID, 0.625 inch OD, length = 5.00 m. The resulting tube wall thickness is:
(0.625 inch - 0.500 inch) / 2 = 0.0625 inch

An advantage of the smaller tube diameter is that it better withstands the external steam pressure.

In the intermediate heat exchanger, in the NaK-salt heat exchanger and in the NaK-oil heat exchanger it is necessary to increase the tube ID in order to provide the required cross sectional area for NaK flow. Increasing the tube ID to (3 / 4) inch greatly improves the cross section for NaK flow. The minimum required number of tubes to match the cross sectional area of the NaK piping becomes:
[12 inch / (3 / 4) inch]^2 = 256

Assume that the tube OD is 7 / 8 inch. Then:
Ri / Ro = 6 / 7
(Ri / Ro)^2 = 0.7346
(Ri / Ro)^4 = 0.53977
[1 - (Ri / Ro)^2] / [1 - (Ri / Ro)^4]
= [1 - 0.7346] / [1 - 0.53977]
= 0.2654 / 0.46023
= 0.576668

For L = 6 m:
3 Rho g L^2 / 2 Sy
= (3 X 8.03 X 10^3 X 9.8 X 36)m / (2 X 117 X 10^6) = 36.32 X 10^-3 m

Hence Ro must be greater than:
0.576668 X 36.32 mm = 20.944 mm

(7 / 8) inch X 25.4 mm / inch = 22.225 mm,<
which satisfies the inequality. Hence the intermediate heat exchange tubes can be 6 m long if they are (3 /4) inch ID. (7 / 8) inch OD.

In summary, without consideration of thermal stress, the tubes in the intermediate heat exchange bundle and in the sodium-salt heat exchanger are (7 / 8) inch OD, (3 / 4) inch ID and 6 m long. The tubes in the steam generator are (5 / 8) inch OD, (1 / 2) inch ID and are 5 m long.
 

This web page last updated May 1, 2023