Part II
Fundamentals of Piping Flexibility
A. Thermal expansion and restraint
A pipe that carries a hot fluid grows. If it is free to grow, it does so without stress. If it is not, the growth is turned into force, and the force into stress in the pipe and load on everything that holds it. Piping flexibility is the study of that conversion, and the design of lines that keep it within bounds.
This part teaches the engineering discipline and the assumptions behind a flexibility calculation. It does not claim that every method is implemented in SWBPIPE. Statements headed Today identify current routes and limitations, with links to the status of this edition. Part III explains how the program builds and solves a supported model.
1.Free growth.
A straight pipe of length L, warmed uniformly by ΔT and free at one end, lengthens by
where α is the coefficient of thermal expansion. For carbon steel, α is about 12 × 10−6 /K near ambient and rises with temperature. For austenitic stainless steel, it is about 16 × 10−6 /K. A carbon steel run 30 m long, warmed by 150 K (270 °F), grows 54 mm (2.1 in). A stainless line of the same length grows about a third more.
A line of any shape keeps its shape and grows in size when uniformly warmed and free to expand. Its free end moves by αΔT times the straight-line vector from the fixed point to that end. This movement is independent of the pipe route. The movement to absorb between two anchors depends only on their positions. The loops between them determine the required force.
2.Full restraint.
Hold both ends of the same straight run fixed and warm it. It cannot lengthen, so the anchors must push it back by the whole of its free growth. The mechanical strain is equal and opposite to the thermal strain, and the axial stress is
compressive, about 52 ksi, with E the elastic modulus of carbon steel near ambient. That is well above the yield strength of ordinary carbon steel pipe. Notice what is absent from equation (2): the length, the diameter and the wall. A straight run between anchors reaches this stress whatever its size. For a DN200 line (219.1 mm outside diameter, 8.18 mm wall, metal area 5420 mm2) the anchor force is 360 MPa × 5420 mm2 ≈ 1.95 MN (about 440 kip). No anchor, nozzle or pipe rack is built for that. A heavier wall does not help: it raises the force in proportion and leaves the stress where it was.
3.Mean and instantaneous coefficients.
The coefficient of expansion is not constant. The instantaneous coefficient is the slope of the length–temperature curve at one temperature. The mean coefficient is the total strain from a reference temperature divided by the whole temperature change. Growth equals the mean coefficient times that change.
The instantaneous value rises with temperature. Multiplying its hot value by the whole change overstates growth. The B31 codes tabulate total expansion from 21 °C (70 °F), in millimeters per meter (inches per 100 feet). With e(T) as that total strain, growth between two temperatures is
A flexibility program should ask for installed and operating temperatures and obtain the total expansion between them. Entering a coefficient can cause the error described above. A mean coefficient from 21 °C also gives an error if applied to a temperature change that starts elsewhere.
4.Relief by bending.
A straight run cannot absorb its growth through direct stress, so the line must bend. Turn the run through a right angle. The growth of each leg then becomes a sideways deflection of the other. A pipe is much less stiff in bending than in direct stress. Thus, the same movement requires a small fraction of the force.
In an L-bend, the short leg bends to absorb growth of the long leg. In a Z-bend, the offset leg absorbs growth of the two parallel legs. A U-loop absorbs growth on either side through bending of its two legs.
5.The guided cantilever.
A leg that absorbs a sideways movement Δ at one end, with both ends prevented from rotating, bends as a guided cantilever. The force and moment at each end are
with I the second moment of area and L the length of the leg. Divide the moment by the section modulus Z = I/(Do/2), and
The force falls with the cube of leg length, and stress falls with its square. Doubling a leg divides stress by four and end force by eight. Wall thickness does not appear in the stress expression. Diameter is in the numerator. For the same movement, the required leg length increases in proportion to the square root of pipe diameter.
Take the DN200 line of section 2, 30 m long, whose 54 mm of growth is taken by a leg at right angles to it. The section has I = 30.19 × 106 mm4 and Z = 275.6 × 103 mm3. With E = 200 000 MPa:
| Leg length L | End force F | End moment M | Bending stress M/Z |
|---|---|---|---|
| 6 m | 18.1 kN | 54.3 kN·m | 197 MPa |
| 8.4 m | 6.6 kN | 27.7 kN·m | 100 MPa |
| 12 m | 2.3 kN | 13.6 kN·m | 49 MPa |
Set the 18 kN of the 6 m leg beside the 1.95 MN of the straight run between anchors. The movement is the same; the shape has cut the force a hundredfold.
The guided cantilever is a sizing tool. It ignores corner rotation and extra bend flexibility. Both effects relieve the leg, so the estimate usually overstates force. It gives nominal straight-pipe stress. It does not give the higher bend stress discussed in section E. Use it to choose a layout; analyze the layout you choose.
6.Flexibility, not strength.
Most structural design concerns strength: find the load, then make the member strong enough to carry it. Thermal loading differs. The load is the force needed to impose a movement. That load increases with stiffness. A stronger pipe pushes harder.
The design problem concerns flexibility. The layout must absorb movement with forces and stresses that the pipe, supports and equipment can carry.
Equation (5) is worth working backwards. To keep the nominal bending stress in the DN200 leg to 100 MPa, L = √(3EDoΔ/σ) = √(3 × 200 000 × 219.1 × 54/100) mm ≈ 8.4 m. A layout engineer who knows this can place a loop on a plot plan before any analysis is run.