A 60 m straight run of 4-inch steel heating main (114.3 mm OD, 102.3 mm ID) is filled at 10 °C and operates at 90 °C. Steel grows 11.7 × 10⁻⁶ per kelvin; the anchors at both ends will not move, so a symmetric expansion loop must absorb the whole growth without the bending stress exceeding the piping code's 155 MPa allowance, with steel's modulus at 200 GPa. The water in the run expands too — take its average volumetric coefficient over the range as 4.5 × 10⁻⁴ per kelvin — and the compression tank must accept that swell.
Given
L₀ = 60 m — Anchored run length
ΔT = 80 C° — Fill to operating rise (10→90 °C)
α = 0.0000117 1/K — Steel linear expansion coefficient
D = 114.3 mm — Pipe outside diameter
Dᵢ = 102.3 mm — Pipe inside diameter
E = 200 GPa — Steel modulus of elasticity
S_a = 155 MPa — Allowable bending stress
β = 0.00045 1/K — Water volumetric expansion coefficient (mean)
Determine
(a)how much the run grows from fill to operating temperature
(b)the loop leg length that absorbs the growth within the stress allowance
(c)the water content of the run
(d)the expansion volume the compression tank must accept
Step 1 of 4(a) · solve for Change in length
Steel does not ask permission: 60 m warming 80 K grows 56 mm, roughly a millimetre per metre per 100 K. The classic error is measuring ΔT from room temperature instead of the coldest fill the run will ever see — the anchors experience the whole swing, not the comfortable part of it.
Rearranged for ΔL
ΔL=αL0ΔT
Your values, in your units
ΔL=(0.00001171/K)(60m)(80C∘)
Answer
ΔL=56.16mm
Carried onward at full precision, not this rounded figure.
The guided-cantilever formula turns 56 mm of growth into steel: legs of about 5.0 m, so the loop stands roughly 5 m out from the run. Note Δ sits under a square root — absorbing DOUBLE the movement needs only 41% longer legs, which is why one generous loop beats two grudging ones.
Switch from the steel to the water it carries: the 102.3 mm bore over 60 m holds 493 L. Use the inside diameter here — the 114.3 mm OD that sized the loop legs would overstate the water by 25%, and the two diameters doing two different jobs in one chain is exactly where that slip happens.
Rearranged for V
V=4πD2L
Your values, in your units
V=4π⋅(102.3mm)2⋅(60m)
Answer
V=493.17L
Carried onward at full precision, not this rounded figure.
The same 80 K that grew the steel grows the water: 493 L × 4.5e-4 × 80 ≈ 17.8 L of swell that must go somewhere soft. In a closed loop 'somewhere' is the compression tank's air cushion — undersize it and the relief valve becomes the expansion tank, one 30 L discharge at a time.
Rearranged for ΔV
ΔV=V0βΔT
493.17 Lcarried from step 3
Your values, in your units
ΔV=(0.493165m3)×(0.000451/K)×(80C∘)
Converted to base units
ΔV=(493.165L)×(0.000451/K)×(80C∘)
Answer
ΔV=17.754L
Carried onward at full precision, not this rounded figure.
From a 10 °C fill to 90 °C operation the main grows about 56 mm, needing expansion-loop legs of roughly 5.0 m; the run holds 493 L of water, which swells about 17.8 L — the volume the compression tank must accept.
Why this order
One temperature rise, two different expansions, and the chain deliberately handles them in that order: the steel first, because its growth is a force problem — 56 mm of thwarted expansion in anchored pipe generates roughly EαΔT ≈ 187 MPa of compressive stress, enough to buckle the run or shear the anchors, so the loop is not a refinement but the thing that makes the layout survivable. The guided-cantilever formula treats each loop leg as a beam bent sideways by the growth, and solving it for length answers the only question the fitter has: how far out does the loop stand? Then the water: the same ΔT expands the 493 L in the bore by about 3.6%, and because water is incompressible that 17.8 L must be accepted by the tank's air cushion or the pressure climbs to the relief setting. Cross-checks: steel's rule of thumb is ~1.2 mm per metre per 100 K, and 60 m × 80 K gives 56 mm ✓; water from the steam tables expands about 3.5% between 10 and 90 °C, and β·ΔT = 4.5e-4 × 80 = 3.6% ✓.
The mistakes here are all about which number goes where. The loop formula wants the OUTSIDE diameter — bending stress lives at the outer fibre — while the volume wants the INSIDE; swap them and both answers are quietly wrong in opposite directions. The ΔT must run from the coldest credible fill to the hottest operating case, not from the design room temperature. And β is not a constant for water: it is about 1e-4 near 10 °C and 7e-4 near 90 °C, which is why a mean value over the actual range (or a density-table lookup) is used — the 4.6e-4 handbook figure at 60 °C applied to a chilled-water loop would overstate the swell fourfold. Last, the loop legs come out of a formula that assumes both legs guided and square corners; long-radius elbows and a bit of leg friction help in reality, so the 5.0 m is honest and slightly conservative — the direction a stress number should err.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
