AverageEngineer
- 5
- 2
- TL;DR
- Trouble using formulas or overthinking...
I am in the process of designing a pumping/piping system for fun, have no experience in this field, but I enjoy learning. I have been using ANSI/ASME codes in the project quite a bit.
For the system I am using 4" nom. Schedule 40 316 SS pipe. The reason I selected the pipe is because the objective for this project is to pump sea water from the ocean to a boiler, water will transfer to a tank to cool then be pumped up a hill to another tank. The tank is suppose to provide water to the made up town of 25,000 civilians. I am only factoring in drinking water, not other uses to keep it simple.
The first leg of pipe will see a 90 degree bend. I looked in ASME B31.1-2001, (only ANSI/ASME text I could find online for free), on page 16 "104.1 Straight Pipe" it gives me two formulas for minimal wall thickness. I chose to use formula (3)' to determine min. wall thickness.
Here is the link to the ANSI/ASME https://www.nrc.gov/docs/ML0314/ML031470592.pdf
The variables are the following in the formula:
P = Design Pressure
D_o = Outer Diameter in my case 4.5" (I am using the info for the pipe here: https://titanium-stainless-steel.continentalsteel.com/item/stainless-steel-pipes/stainless-steel-pipe--type-316-schedule-40s/316-4000-4500-40s#Typical Mechanical Properties
S = Allowable stress in pipe material (PSI)
F = Joint Factor, E = 1.0 for seamless, E = 0.85 for ERW pipe
Y or y = Wall thickness coefficient in ASME B31.3 Table 304.1.1 for ferritic steel, y = 0.4
W = Weld joint strength reduction factor
A = Corrosion allowance typically 0.5
It took me awhile to figure out how to determine design pressure. I used the formula
P_design = 2 * S * t / D_outer * SF
where,
S = Material Strength (PSI)
SF = Safety Factor
My pressure at yield was 4,740 PSI and desired was 3160 PSI using SF = 1.5
On Table 102.4.5 it gives me 1.14 * t_m which gave me a radius of r = 0.745 inches with a 0.653 in wall thickness. This radius would be minimal, correct? Too me it seems like such a small radius, of course the wall thickness is quite large. What do you think?
For the system I am using 4" nom. Schedule 40 316 SS pipe. The reason I selected the pipe is because the objective for this project is to pump sea water from the ocean to a boiler, water will transfer to a tank to cool then be pumped up a hill to another tank. The tank is suppose to provide water to the made up town of 25,000 civilians. I am only factoring in drinking water, not other uses to keep it simple.
The first leg of pipe will see a 90 degree bend. I looked in ASME B31.1-2001, (only ANSI/ASME text I could find online for free), on page 16 "104.1 Straight Pipe" it gives me two formulas for minimal wall thickness. I chose to use formula (3)' to determine min. wall thickness.
Here is the link to the ANSI/ASME https://www.nrc.gov/docs/ML0314/ML031470592.pdf
The variables are the following in the formula:
P = Design Pressure
D_o = Outer Diameter in my case 4.5" (I am using the info for the pipe here: https://titanium-stainless-steel.continentalsteel.com/item/stainless-steel-pipes/stainless-steel-pipe--type-316-schedule-40s/316-4000-4500-40s#Typical Mechanical Properties
S = Allowable stress in pipe material (PSI)
F = Joint Factor, E = 1.0 for seamless, E = 0.85 for ERW pipe
Y or y = Wall thickness coefficient in ASME B31.3 Table 304.1.1 for ferritic steel, y = 0.4
W = Weld joint strength reduction factor
A = Corrosion allowance typically 0.5
It took me awhile to figure out how to determine design pressure. I used the formula
P_design = 2 * S * t / D_outer * SF
where,
S = Material Strength (PSI)
SF = Safety Factor
My pressure at yield was 4,740 PSI and desired was 3160 PSI using SF = 1.5
On Table 102.4.5 it gives me 1.14 * t_m which gave me a radius of r = 0.745 inches with a 0.653 in wall thickness. This radius would be minimal, correct? Too me it seems like such a small radius, of course the wall thickness is quite large. What do you think?
Last edited by a moderator: