Answered

1. Basically, the thermodynamic entropy limit must be greater than one because it has been equated that if it is less than one than entropy has decreased in the direction of the flow. The pipe diameter is a factor in these equations, which is most likely why it is less than one when you select 4'' piping instead of letting CAESAR select piping.

As far as the subsonic vent exit limit I believe is a ratio of velocities. This comparison has also been calculated.

I attached an article that helps clarify the fundamentals and background to the relief load synthesis, it is a tough read, but if understood, will explain the background of this utility..

This is taken straight from the users guide and may be helpful to you:

"Should CAESAR II Size the Vent Stack (Y/N)

Specifies whether or not the software sizes the vent stack. Select Y (for yes) for CAESAR II to calculate the length and diameter of the vent stack. The software sizing algorithm searches through a table of available inside pipe diameters starting at the smallest diameter until a vent stack ID is found that satisfies the thermodynamic criteria. The calculated inside diameter is automatically inserted into the input."

"Thermodynamic Entropy Limit/Subsonic Vent Exit Limit

The thermodynamic entropy limit or subsonic vent exit limit. These values should always be greater than one. If either value falls below 1.0, then the thermodynamic assumptions made regarding the gas properties are incorrect and the calculated thrust values should be disregarded."

2. For impulse loads the highest DLF generated will be 2, which means the dynamic load is twice force the static load.You can multiply the DLF by the dynamic load and apply it to the static load. I would recommend looking at ASME B31.3 NONMANDATORY APPENDIX II to see more information about Dynamic Amplification.

Thanks for the response. Just curious, is there a reason the highest DLF will be 2? Also, you stated that, "You can multiply the DLF by the __dynamic load__ and apply it to the static load"... just to clarify, "dynamic load" is synonymous with the __thrust__ calculated from the Relief Load Synthesis output? Also, did you mean B31.1 for the appendix reference? Couldn't find any non-mandatory appendices for B31.3. Thanks again!

Answer

I did mean B31.1. I found it to be a good reference.

The highest DLF for impulse loads will always be 2 because of the way dynamic systems react. A simple example of this is to think of a spring. If a weight is applied to a spring the spring will oscillate until it is static. Although the force on the spring is the weight while it is oscillating, the force on the spring when fully compressed(static) is 2W.

WX = KX^2 /2 (Work Equation)

This can also be presented by 2W = KX

When the spring is compressed, the force on the spring is twice the weight applied during the dynamic loading, hence a DLF of 2.

The thrust would be the dynamic load.

## Lonnie Davis

A few questions:

1. When I run a the Relief Load Synthesis to obtain thrust forces for relief valve piping, what does it mean when my Subsonic Vent Exit is less than 1.0 (and Thermodynamic Entropy limit)? I've looked around and can find the basis for these limits...is there a change is the equation of state? Please explain. I've attached two runs, one without and one with the auto C2 vent stack sizing selected. On the run where I've selected the 4" Sch. 40 vent stack, I get the Subsonic Vent Exit Limit less than 1.0.

2. What is the procedure for using the Dynamic Load Factor (DLF) generated? Am I to multiply this DLF by my Thrust force vector and rerun the static analysis as a "factored-up" static load?

Xavier Huerta-San Juansaid over 1 year agoI did mean B31.1. I found it to be a good reference.

The highest DLF for impulse loads will always be 2 because of the way dynamic systems react. A simple example of this is to think of a spring. If a weight is applied to a spring the spring will oscillate until it is static. Although the force on the spring is the weight while it is oscillating, the force on the spring when fully compressed(static) is 2W.

WX = KX^2 /2 (Work Equation)

This can also be presented by 2W = KX

When the spring is compressed, the force on the spring is twice the weight applied during the dynamic loading, hence a DLF of 2.

The thrust would be the dynamic load.