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Method notes / Plenara: API 661 air-cooler headers

Engine

End plates, stays and partitions, MAWP and the test condition

UG-34 for the welded end plates, the stayed sketches for partitions that act as stays, how MAWP and the required thickness are found, and what the hydrotest check does.

End plates

The welded flat end plates are not part of the App 13 frame. 13-4(f) sends them to UG-34 with C = 0.20 for every attachment. The spans are the inside spans of the condition, d the smaller and D the larger, and E is the efficiency of a butt joint within the plate.

Z = min(3.4 − 2.4·d/D, 2.5) t_req = d·√(Z·C·P/(S·E)) + CA MAWP = S·E·(t5 − CA)² / (Z·C·d²)UG-16: net thickness t5 − CA of at least 1.5 mm (1/16 in, 1.5875 mm, in a US-customary design). Only welded, unstayed end plates are covered. On a stayed box the end plate is checked across the whole section without credit for the stays.

End plate of COMPRESS 50183 (vector V-C3)

Inputs (in, psi), corrodedd = 4.125, D = 9.625, P = 150, S = 18100, t5 = 0.5, CA = 0.0625
Z3.4 − 2.4 · 4.125 / 9.625 = 2.3714
t_req4.125 · √(2.3714 · 0.2 · 150 / 18100) = 0.2586 in
t_req + CA0.3211 in, against t5 = 0.5 in: passes
MAWP, corroded18100 · 0.4375² / (2.3714 · 0.2 · 4.125²) = 429.3 psi

0.3211 in is the value in the fixture. The 429.3 psi is calculated here from the formula, not taken from a fixture.

Stays and partitions

A pass partition welded to the tubesheet and plugsheet acts as a stay on pair B. With one partition (sketch 7, 13-9(b)) or two at the third points (sketch 8, 13-9(c)), and with K = (I2/I1)·α and h the clear compartment:

Sketch 7: Λ7 = (2 + K·(5 − α²))/(1 + 2K) G7 = (1 + 2·α²·K)/(1 + 2K) Sm(A) = P·h·(4 − Λ7)/(4·t1) Sm(stay) = P·h·Λ7/(2·t3)
M_Q = P·h²·G7/12 M_N = P·(−3·H² + 2·h²·G7)/24 M_junction = P·h²·(1 + K·(3 − α²))/(12·(1 + 2K))Sketch 8 has the same form with Λ8 = (6 + K·(11 − α²))/(3 + 5K) and G8 = (3 + 5·α²·K)/(3 + 5K). Stay bars use the same bending with Sm(stay) = 2·P·h·p·Λ/(π·d²) and a UG-47(a) pitch limit.

Sketch (7), one stay plate: FV5 at the FE dimensions

Inputs (mm, MPa)H = 210, h = 85, t1 = t2 = t3 = 10, P = 1
α = K210 / 85 = 2.4706
G7, Λ75.2447, −0.1224
Sm(A), Sm(B)8.76, 10.50 MPa
Sm(stay)1 · 85 · (−0.1224) / 20 = −0.52 MPa (compression)
Sb at N, Q, junction M (6·M/t²)−141.28, 189.47, −40.55 MPa

These match the App 13 reference column of FV5 in the validation note. The 10 mm thicknesses and 1 MPa are test inputs, not a header design.

Pair B is one member continuous over the stay junctions, with each compartment following the statics formula between its end moments. Sketch (7) is exact whatever the stay stiffness, because the central stay does not rotate. Sketch (8) is exact for pin-jointed stays only; welded stays of real stiffness shift the junction moment (the frame oracle, V-OR-6). With unequal compartments (13-9(f)) the default is the sketch (7) or (8) model with every compartment the largest one; the engine also offers the 13-9(f)(1) rigid-frame analysis, and by default compares the two. Where the envelope sits more than 2 % below the frame it raises a warning. The code value remains the code result.

MAWP and required thickness

For sketches (1) to (3) every stress is proportional to P, so the pressure that brings the worst row to its allowable is exact: MAWP = P · allowable/|stress| over all rows, then the lower of that and the end plate. Sketch (4) is not linear, because the allowable spacing and the effective width fall as 1/√P; its MAWP is found by bisection on the whole rebuilt frame. The required thickness t_req is the smallest corroded thickness of one plate group for which the rows pass, with the ligament efficiencies held at their as-built values (the COMPRESS method).

MAWP of the E4.12.1 box (vector V-PTB-1), calculated here

Governing row(ST)M outer, pair B: 30196.6 psi against 1.5 · 20000 = 30000 psi
MAWP from the rows400 · 30000 / 30196.6 = 397.4 psi
End plate MAWP20000 · 0.625² / (1.5368 · 0.2 · 7.375²) = 467.3 psi
MAWP397.4 psi, governed by pair B at midspan

The fixture records that this one row fails by 0.7 %. The MAWP figures are derived here from that stress and the rule above, not read from a fixture.

Test condition

App 13 has no test rule; 13-16 sends the test to UG-99(b) and UG-100. The engine takes P_T = 1.3·MAWP·LSR for hydrostatic and 1.1·MAWP·LSR for pneumatic, with LSR the lowest S_test/S_design over the plates, end plates and any added component such as the tubes. The stresses are recomputed on the new (uncorroded) geometry with the static head added. The allowables for the test are not in App 13: the default policy compress90 uses 0.9·Sy on membrane and 1.5·0.9·Sy on membrane plus bending, multiplied by E(L).

COMPRESS 50183 test (vector V-C2)

LSR from the SA-214 tubes11400 / 10500 = 1.0857
P_T given (COMPRESS)211.712 psi
Allowables, Sy = 38000 psi0.9 · 38000 = 34200 psi; 1.5 · 34200 = 51300 psi
End plate at test, d = 4, D = 9.5, t5 = 0.5Z = 2.3895, σ = Z·C·P_T·d²/t5² = 6475 psi, against 51300

The policy div2-hydro (0.95·Sy membrane; 1.43·Sy, else 2.43·Sy − 1.5·Pm, on membrane plus bending) is marked [UNVERIFIED] in allowables.ts. The default is compress90.

Limits. Stay-plate-to-box welds are not checked. A pass partition with a pressure difference between passes is not covered (13-9(a), UG-19). Partitions with tubesheet and plugsheet of different thickness are blocked on the page, not assessed. External pressure (13-14) is refused. Nozzle loads (13-4(d)) and openings (13-4(j)) are reminders only.

Related notes

Next: The FE cross-check and what is verified against what