Question

Calculate the pressure drop over a 100m length due to friction when a slurry made from 1.0-mm silica particles is pumped through a horizontal 6-cm diameter pipeline (smooth pipe) at 2.5 m/s. The slurry contains 25 per cent silica by volume. The density of silica is 2700 kg/m3, rhow = 1000 kg/m3, μw = 0.001 kg/ms. Use a value of 82 for Ω. Drag coefficient for these particles,CD, may be taken as 0.44 .


For the slurry/Pipe system in question 1, estimate the deposition velocity.

Answer 1

Answer:

The answer is "2.78".

Explanation:

Given values:

CD= 0.44

Formula:

$\bold{f_r= \frac{v^{2}}{ g(s-1)D}}$

g=9.8

s= 2.7

D= 0.06

$\to f_r=\frac{2.5^2}{9.8(2.7-1)0.06}$

        $=\frac{2.5 \times 2.5}{9.8 \times 1.7 \times 0.06 }\\\\=\frac{6.25}{.9996 }\\\\=6.252501$

$\phi = \frac{82\times v}{ \sqrt{cD} \times f_r^{-1.5}}$

   $= \frac{82 \times 0.25 }{ \sqrt{0.44} \times 6.25^{1.5}}\\\\=2.4285$

$\frac{\bigtriangleup P f_1 s_1}{L} = \frac{\bigtriangleup Pf_w}{L}(1+\phi)$

           $=\frac{2fwSwv^2 (1+2.4285)}{D}$

$Re= \frac{D \bar v Sw}{M_w}$

     $=\frac{0.06 \times 2.5 \times 1000 }{0.001}\\\\=\frac{150 }{0.001}\\\\= 150 \times 10^{3}\\\\= 1.50 \times 10^{5}$

$fw= 0.00389$

$\to \frac{\bigtriangleup P f_1 s_1}{L}$

$\to \frac{2 \times 0.00389 \times 1000 \times2.5^2 \times 3.4265}{0.06}\\\\\to 2.78$