Question

Calculate the weight of steel saved if a 250 mile crude pipeline (class 2 location) operating at 1440 psia MAOP upgrades its steel from grade b (35,000 psi) to x-70. 16” Pipeline

Answer 1

Given data:

Length of the crude pipeline, L = 250 miles = 250*5280 ft.

Class location = class 2

Maximum allowable operating pressure (MAOP), = 1440 psia.= 1425.3 psig

Existing steel pipe grade = grade b (35000 psi)

Upgraded steel pipe grade = x-70

The diameter of the pipeline, d_o = 16"

Solution:

For class 2 location, design factor = 0.6

So, Maximum working pressure, p_w is,

p_w = 2375.5 psig

As per Barlow's formula,

Where, p_w = maximum working pressure,

= yield strength

t = thickness of the pipe

d_o = Outside diameter of the pipe

First, calculate the thickness of the pipe for grade b.

here, p_w = 2375.5 psig, = 35000 psi, d_o = 16"

Substitute the values of p_w, , d_o in Eq. (1)

t = 0.543"

The available thickness of the pipe just above the calculated thickness, t_a = 0.656" (schedule 60)

Weight of the pipe per ft for sch 60 = 107.5 lb/ft.

Weight of the pipeline for 250-mile length = 107.5*250*5280 lb. = 141.9*10⁶ lb

We know 1 tonne = 2204.62 lb

Weight of the pipeline grade b for 250-mile length =(141.9*10⁶)/2204.62 = 64364.8 tonne

Now calculate the thickness of the pipe of grade x-70.

here, p_w = 2375.5 psig, = 70,000 psi, d_o = 16"

Substitute the values of p_w, , d_o in Eq. (1)

t = 0.271"

The available thickness of the pipe just above the calculated thickness, t_a = 0.321" (schedule 20)

Weight of the pipe per ft for sch 20 = 52.27 lb/ft.

Weight of the pipeline for 250-mile length = 52.27*250*5280 lb. = 68.10*10⁶ lb

We know 1 tonne = 2204.62 lb

Weight of the pipeline grade x-70 for 250-mile length =(141.9*10⁶)/2204.62 = 30889.68 tonne

The weight of steel saved = (Weight of the pipe grade b) - (Weight of the pipe grade x-70)

= 64364.8 - 30889.68

= 33475.12 tonne

Hence, The weight of the steel saved = 33475.12 tonne