Upon what variables would the rate of filtration of a suspension of fine solid particles be expected to depend? Consider the flow through unit area of filter medium and express the variables in the form of dimensionless groups.
It is found that the filtration rate is doubled if the pressure difference is doubled. What would be the effect of raising the temperature of filtration from 293 to 313 K?

Question

The viscosity of the liquid is given by:

[latex]\mu=\mu_0(1-0.015(T-273))[/latex]

where μ is the viscosity at a temperature T K and [latex]\mu_0[/latex] is the viscosity at 273 K.

Accepted Answer

The volume flow of filtrate per unit area, u m³/m²s, will depend on the fluid density, ρ; fluid viscosity, μ; particle size, d; pressure difference across the bed, ΔP, and the voidage of the cake, e or: [latex]v=f( ho, \mu, d, \Delta P, e)[/latex]. The dimensions of each of these variables are [latex]u= L / T , ho= M / L ^3, \mu= M / L T , d= L , \Delta P= M / L T ^2[/latex] and e = dimensionless. There are 6 variables and 3 fundamental dimensions and hence [latex](6-3)=3[/latex] dimensionless groups. Taking, d, ρ and μ as the recurring set, then:

[latex]d \equiv L , \quad L =d[/latex]

[latex] ho \equiv M / L ^3, \quad M = ho L ^3= ho d^3[/latex]

[latex]\mu \equiv M / L T , \quad T = M / L \mu= ho d^3 /(d \mu)= ho d^2 / \mu[/latex]

Thus: dimensionless group 1: [latex]u T / L =u ho d^2 /(\mu d)=d u ho / \mu[/latex]

dimensionless group 2: [latex]\Delta P L T ^2 / M =\Delta P d\left( ho d^2 / \mu ight)^2 / ho d^3=\Delta P ho d^2 / \mu^2[/latex]

and the function is: [latex]\underline{\underline{\Delta P ho d^2 / \mu^2=f(d u ho / \mu)}}[/latex]

This may be written as: [latex]\Delta P ho d^2 / \mu^2=K(d u ho / \mu)^n[/latex]

Since the filtration rate is doubled when the pressure difference is doubled, then:

[latex]u \propto \Delta P[/latex] and [latex]n=1, \Delta P ho d^2 / \mu^2=K d u ho / \mu[/latex]

and: [latex]u=(1 / K) \Delta P d / \mu,[/latex] or [latex]u \propto 1 / \mu[/latex]

[latex]\left(\mu_{293} / \mu_{313} ight)=\mu_0(1-0.015(293-273)) / \mu_0(1-0.015(313-273))=(0.7 / 0.4)=1.75[/latex]

∴ [latex]\left(v_{313} / v_{293} ight)=\left(\mu_{293} / \mu_{313} ight)=1.75[/latex]

and the [latex]\underline{\underline{ ext{filtration rate will increase by 75 \%.}}}[/latex]

Question 1.P.19: Upon what variables would the rate of filtration of a suspen...

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