# Modeling Relief Valve (Pressure Safety Valve) Thrust force

Start-Prof can estimate the support loads and stresses caused by relief valve discharge by static method.

The relief valve discharge thrust load acts on elbow 28. In Start-Prof it should be applied at the end of the elbow in 29 node. There’s a two methods to estimate the dynamic equivalent thrust force F:

• ASME B31.1 method
• Direct calculation of V1 и P1 by special software like Hydrosystem

## ASME B31.1 Method Equivalent dynamic thrust force can be estimated by equation:
F = DLF ∙ F1
where
DLF – dynamic load factor, depend on first natural period of piping. If period is unknown the DLF=2.0. F1 – static reaction force, kgf. May be computed by the following equation: where
W – mass flow rate (relieving capacity stamped on the valve by 1.11), kg/sec
gc=9.81 m/sec² – gravitational constant,
Pa – atmospheric pressure, kgf/sm²
A1 – exit flow area, sm²  A1=p(D-2t)²/4
V1 – exit velocity (node  36), m/sec P1 – static pressure, kgf/sm² h0 – stagnation enthalpy at the safety valve inletMJ/kg
J = 101970.408 m*kgf/MJ
a, b – constants according to table below

 Steam Condition a, MJ/kg b Wet steam <90% quality 0.6769 11 Saturated steam ≥90%  quality 1.05 kgf/sm² ≤ P1 ≤ 70.31 kgf/sm² 1.9143 4.33 Superheated steam ≥90%  quality 0.07 kgf/sm² ≤ P1 ≤ 140.61 kgf/sm² 1.93291 4.33

### Example

Example project file: ReliefValve.ctp. See how to open the piping model file Input data:

• Diameter of discharge pipe: 21.9 sm

• Wall thickness of discharge pipe: 1.6 sm

• product: Saturated steam

• relief valve relieving capacity: 48 kg/sec

• Steam pressure: 64 kgf/sm²

• Steam temperature: 538°C

Calculation:

A1=p(D-2t)²/4 = 3.14159(21.9-2∙1.6)²/4 = 274.65 sm²
a = 1,9143 MJ/kg
b = 4.33
W = 48∙1.11 = 53.28 kg/sec
stagnation enthalpy for steam at 70 kgf/sm² and 538°C h0 =  3,506 MJ/kg
P1=53.28/274.65*(4.33-1)/4.33*(2*(3.506-1.9143)*101970.408/(2*4.33-1)/9.81)^0.5 = 9.8 kgf/sm²
V(2*9.81*101970.408*(3.506-1.9143)/(2*4.33-1))^0.5 = 644.77 m/sec
F1 = 53.28*644.77/9.81 + (9.8-1)*274.65 = 5919 kgf
F = DLF ∙ F1 = 2*5919 = 11838 kgf If we have more relief valves, then we should add more additional modes (1.2, 1.3, etc.) if it’s open not simultaneously.

Apply dynamic thrust force at node 29 for mode 1.1:   