Derivation of a Simplified D'Arcy's Law Equation

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Homework Statement
Done for a research essay on physics models for water filtration, and what I am focusing on is the change in speed after water passes through a porous material
Relevant Equations
q = -k*∆p/(µ*L) Darcy’s Law (flux rate)
∆p = f(L/D)(𝜌V^2/2) Darcy-Weisbach equation
Re = ρVD/µ Reynolds Number equation
f = 64/Re Friction factor equation
By substituting the darcy-weisbach equation into darcy’s law we get
q = -kf/µL * (L/D) * (𝜌V^2/2)
This can be further simplified by substituting the equation for friction factor for laminar flow, f = 64/Re , with the equation for reynolds number, Re = ρVD/µ substituted in such that:
q = (-k/µL)(64µ/ρVD)*(L/D)(𝜌V^2/2)
Which can be simplified from crossing out variables into:
q =-32kV/D^2

Based on physics and research i've done on filtration mechanics this makes kinda perfect sense, but I haven't found any evidence of this substitution online.
 
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SpaceDuck127 said:
Relevant Equations: q = -k*∆p/(µ*L) Darcy’s Law (flux rate)
∆p = f(L/D)(𝜌V^2/2) Darcy-Weisbach equation
Re = ρVD/µ Reynolds Number equation
f = 64/Re Friction factor equation

By substituting the darcy-weisbach equation into darcy’s law we get
q = -kf/µL * (L/D) * (𝜌V^2/2)
This can be further simplified by substituting the equation for friction factor for laminar flow, f = 64/Re , with the equation for reynolds number, Re = ρVD/µ substituted in such that:
q = (-k/µL)(64µ/ρVD)*(L/D)(𝜌V^2/2)
Which can be simplified from crossing out variables into:
q =-32kV/D^2
Not my area but…

d’Arcy’s law is about a fluid passing through a porous material.

Reynold’s number is essentially about the transition between laminar and turbulent flow in a ‘free’ fluid.

Does it make sense to combine these when they apply to such different situations?

Beware of combining equations merely because they have some common parameters.
 
SpaceDuck127 said:
Homework Statement: Done for a research essay on physics models for water filtration, and what I am focusing on is the change in speed after water passes through a porous material
Relevant Equations: q = -k*∆p/(µ*L) Darcy’s Law (flux rate)
∆p = f(L/D)(𝜌V^2/2) Darcy-Weisbach equation
Re = ρVD/µ Reynolds Number equation
f = 64/Re Friction factor equation

By substituting the darcy-weisbach equation into darcy’s law we get
q = -kf/µL * (L/D) * (𝜌V^2/2)
This can be further simplified by substituting the equation for friction factor for laminar flow, f = 64/Re , with the equation for reynolds number, Re = ρVD/µ substituted in such that:
q = (-k/µL)(64µ/ρVD)*(L/D)(𝜌V^2/2)
Which can be simplified from crossing out variables into:
q =-32kV/D^2

Based on physics and research i've done on filtration mechanics this makes kinda perfect sense, but I haven't found any evidence of this substitution online.
What do you mean "change in speed after the water passes through a porous material"? Do you instead mean "as it is going through the porous material"?
 
Your analysis would work if you had an array of parallel pores running through your medium. Then, for each pore, you would have the Poiseulle equation: $$-\frac{dP}{dL}=\frac{128Q\mu}{\pi D^4 }=\frac{32\mu v}{D^2}$$where v is the pore velocity. The pore velocity is related to the superficial velocity q by $$q=\epsilon v$$where ##\epsilon## is the porosity. So, we have Darcy's law for such a medium being: $$-\frac{dP}{dL}=\frac{32q\mu}{\epsilon D^2}$$or $$q=-\frac{dP}{dL}\frac{\epsilon D^2}{32 \mu}$$So, the permeability for such a medium is $$k=\frac{\epsilon D^2 }{32}$$
 
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