Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
Answer:
t = 56.6 min
Explanation:
Fick's second law is used to calculate time required for diffusion
where
= 1.15%
= 0.18%
= 0.35%
x = 0.40 mm = 0.0004 n
therefore we ahave
![\frac{1.15-0.35}{1.15- 0.18} = erf[\frac{4\times 10^{-4}}{2\sqrt{1.28\times 10^{-11} t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1.15-0.35%7D%7B1.15-%200.18%7D%20%3D%20%20erf%5B%5Cfrac%7B4%5Ctimes%2010%5E%7B-4%7D%7D%7B2%5Csqrt%7B1.28%5Ctimes%2010%5E%7B-11%7D%20t%7D%7D%5D)
![0.8247 = erf [\frac{55.90}{\sqrt{t}}] = erf z](https://tex.z-dn.net/?f=0.8247%20%3D%20erf%20%5B%5Cfrac%7B55.90%7D%7B%5Csqrt%7Bt%7D%7D%5D%20%3D%20%20erf%20z)
from error function table we hvae following result
for erf z z
0.8209 0.95
0.8247 x
0.8427 1
therefore
x = 0.959
thus
t = 56.6 min
Pipelines are a useful means of transporting oil because they offer low maintenance and dependable transportation for a narrow but important range of products.
<h3>What is a pipeline?</h3>
A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.
The options are attached
a. Pipelines provide jobs for consumers because of the resurgence of exploration and drilling in North America.
b. Pipelines are versatile, carrying more ton-miles than any other mode of transport over more than 2 million miles of pipeline.
c. Pipelines have more locations than water carriers.
d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.
Thus, the correct option is d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.
Learn more about Pipelines
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Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
Q = 5.06 x 10⁻⁸ m³/s
