Modeling Relief Valve (Pressure Safety Valve) Thrust Forces in START-PROF

START-PROF calculates support loads and stresses from relief valve discharge using static analysis methods.

The relief valve thrust load acts on elbow 28. Apply this force at the elbow end (node 29) in START-PROF.

Calculate the equivalent dynamic thrust force F using these methods:

ASME B31.1 Method

Calculate equivalent dynamic thrust force:

F = DLF ∙ F₁

Where:

DLF - dynamic load factor, based on piping natural period. Use DLF=2.0 if period is unknown.

F₁ - static reaction force (N):

Where:

W - mass flow rate (valve relieving capacity × 1.11), kg/sec

g_c = 1 - gravitational constant

P_a - atmospheric pressure, N/m²

A₁ - exit flow area, m²: A₁ = π(D-2t)²/4

V₁ - exit velocity (node 36), m/sec

P₁ - static pressure, N/m²

h₀ - stagnation enthalpy at safety valve inlet, MJ/kg

J = 999835.0529 m*N/MJ

a, b - constants from table below

Steam Condition	a, MJ/kg	b
Wet steam <90% quality	0.6769	11
Saturated steam ≥90% quality 1.05 kgf/cm² ≤ P₁ ≤ 70.31 kgf/cm²	1.9143	4.33
Superheated steam ≥90% quality 0.07 kgf/cm² ≤ P₁ ≤ 140.61 kgf/cm²	1.93291	4.33

Example project file: ReliefValve.ctp

Input data:

Calculation:

A₁ = π(D-2t)²/4 = 3.14159×(0.219-2×0.016)²/4 = 0.027465 m²

a = 1.9143 MJ/kg

b = 4.33

W = 48×1.11 = 53.28 kg/sec

Stagnation enthalpy for steam at 64 kgf/cm² and 538°C: h₀ = 3.506 MJ/kg

J = 999835.0529 m*N/MJ

P₁ = 53.28/0.027465×(4.33-1)/4.33×(2×(3.506-1.9143)×999835.0529/(2×4.33-1)/1)^0.5 = 961693 N/m²

V₁ = (2×999835.0529×(3.506-1.9143)/(2×4.33-1))^0.5 = 644.6 m/sec

F₁ = 53.28×644.77 + (961693-100000)×0.027465 = 58019 N

F = DLF × F₁ = 2×58019 = 116039 N

Create occasional force loading 1.1 in the operation mode editor:

For multiple relief valves, create additional modes (1.2, 1.3, etc.) if they operate independently.

Apply dynamic thrust force at node 29 for mode 1.1:

Support loads from relief valve thrust:

Stresses from relief valve thrust: