Specifications

Chapter 5: General Installation Guidelines
Piping Design
(12/99)
5-17
Example Calculation
Suppose a customer requires a tank scale with a system accuracy of 0.1% of the
applied load. One pipe will be connected to the tank. To meet the system accuracy
requirement, the vertical force exerted by the pipe (F
P
) must be equal to or less
than 1% times the live load of the system. For this application, assume that the
live (net) load equals 25,000 pounds.
Use the resultant force formula to determine the maximum pipe force that you can
compensate for during calibration:
F
p
≤ 0.1 × 0.1 × 25,000 pounds
F
P
cannot be greater than 250 pounds maximum pipe force.
Use the pipe force equation to calculate the actual force exerted by a pipe with the
following characteristics:
D = 4 inches (Outside diameter of pipe)
d = 3.75 inches (Inside diameter of pipe)
∆h = 0.09 inch (Total deflection of pipe at the vessel)
E = 29 × 10
6
(Young’s modulus)
L = 60 inches (Length of pipe from the vessel to the first support point)
0.59 × (256 - 197.75) × 0.09 × 29,000,000
216,000
Since a pipe force of 415.27 pounds is greater than 250 pounds, it would not
satisfy the requirement for a 0.1% accuracy system. One solution is to increase
the length of the pipe from 60 inches to 80 inches. When you recalculate the pipe
force for a length of 80 inches, you get F
P
= 175.2 pounds, which is well below
the maximum of 250 pounds.
F
P
= = 415.27 pounds