Mechanical Operations

Q1: A suspension of glass beads in ethylene glycol has a hindered settling velocity of 1.7 mm/s, while the terminal settling velocity of a single glass bead in ethylene glycol is 17 mm/s. If the Richardson-Zaki hindered settling index is 4.5, the volume fraction of solids in the suspension is

A 0.1

B 0.4

C 0.6

D none of these

ANS:C - 0.6

To find the volume fraction of solids (ϕ\phiϕ) in the suspension, we can use the Richardson-Zaki equation for hindered settling: VhVt=ϕn\frac{V_{\text{h}}}{V_{\text{t}}} = \phi^nVt​Vh​​=ϕn where:

  • VhV_{\text{h}}Vh​ is the hindered settling velocity of the suspension,
  • VtV_{\text{t}}Vt​ is the terminal settling velocity of a single particle,
  • ϕ\phiϕ is the volume fraction of solids,
  • nnn is the Richardson-Zaki hindered settling index.
Given:
  • Hindered settling velocity of the suspension, Vh=1.7V_{\text{h}} = 1.7Vh​=1.7 mm/s
  • Terminal settling velocity of a single particle, Vt=17V_{\text{t}} = 17Vt​=17 mm/s
  • Richardson-Zaki hindered settling index, n=4.5n = 4.5n=4.5
First, calculate the ratio of hindered settling velocity to terminal settling velocity: VhVt=1.717=0.1\frac{V_{\text{h}}}{V_{\text{t}}} = \frac{1.7}{17} = 0.1Vt​Vh​​=171.7​=0.1 Now, use the Richardson-Zaki equation to find ϕ\phiϕ: 0.1=ϕ4.50.1 = \phi^{4.5}0.1=ϕ4.5 Taking the 4.5-th root of both sides to solve for ϕ\phiϕ: ϕ=(0.1)1/4.5\phi = (0.1)^{1/4.5}ϕ=(0.1)1/4.5 ϕ≈0.522\phi \approx 0.522ϕ≈0.522 Therefore, the volume fraction of solids in the suspension is approximately 0.522. However, since none of the provided options match exactly, we should consider the closest option. Given the options provided:
  • 0.1
  • 0.4
  • 0.6
The closest option to 0.522 is none of these. Thus, the volume fraction of solids in the suspension is approximately 0.522, which is not exactly matched by any of the given options.



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