From the information, let us list the parameters given to solve for the diameter of the pipe.
Given that:
= 2600 kg/m³
Using the standard conversion rates:
Since 1 lbf/in² = 6894.76 N/m²
∴
From the given information, let's start by determining the volumetric flow rate of the liquid in the pipe:
[tex]\mathbf{The \ volumetric \ flow \ rate \ ( Q) = \dfrac{mass \ flow \ rate}{density \ of \ the \ liquid}}[/tex]
[tex]\mathbf{Q = \dfrac{0.882 \ kg/s}{2600 \ kg/m^3}}[/tex]
Q = 0.00034 m³/s
In a cylindrical flow pipe, using the formula for the pressure drop to estimated the pipe diameter, we have:
[tex]\mathsf{\dfrac{\Delta P}{\rho g}= \dfrac{8fLQ^2}{\pi^2gd^5}} --- (1)[/tex]
where (f) can be computed as;
[tex]f = \dfrac{64}{\dfrac{\rho vd}{\mu}}[/tex]
[tex]f = \dfrac{64}{\dfrac{\rho Qd}{A\mu}}[/tex]
replacing the values from the above-listed parameters, we have:
[tex]f = \dfrac{64}{\dfrac{2600 \times 0.00034 \times d}{\dfrac{\pi}{4}(d)^2 \times 0.002}}[/tex]
[tex]f = \dfrac{64}{2600 \times 0.00034 \times d} \times \dfrac{\dfrac{\pi}{4}(d)^2 \times 0.002}{1}[/tex]
f = 0.1137d
From equation (1), Recall that:
[tex]\mathsf{\dfrac{\Delta P}{\rho g}= \dfrac{8fLQ^2}{\pi^2gd^5}}[/tex]
[tex]\mathsf{\dfrac{\Delta P}{\rho }= \dfrac{8fLQ^2}{\pi^2d^5}}[/tex]
Replacing the values, we have;
[tex]\mathsf{\dfrac{1261.74}{2600}= \dfrac{8\times 0.1173(d) \times (2784) \times (0.00034)^2}{\pi^2(d)^5}}[/tex]
[tex]\mathsf{0.48528= \dfrac{2.966\times 10^{-5}}{(d)^4}}[/tex]
[tex]\mathsf{d^4= \dfrac{2.966\times 10^{-5}}{0.48528}}[/tex]
[tex]\mathsf{d^4= 6.11193538 \times 10^{-5}}[/tex]
[tex]\mathbf{d = \sqrt[4]{ 6.11193538 \times 10^{-5}}}[/tex]
d = 0.0884 m
d = 88.4 mm
since 1 mm = 0.0393701 inch
∴
88.4 mm will be = 3.48 inches
Therefore, we can conclude that the diameter of the pipe = 3.48 inches
Learn more about volumetric flow rate here:
https://brainly.com/question/15119966?referrer=searchResults
