|Chapter 5||Dilute systems
Heywood Tables - on-line
On-line settling velocity by Stokes and Heywood Tables
Illustration of Camp Hazen model by an animated gif
(use the browser's back button to return)
Links - links
Answers to problems:
i) the correct answer is (a)
ii) the correct answer is (b)
iii) log(Px) for the 50, and above, particles are: 0.156, 0.457 and 1.457
log(U/Q) for the 50, and above, particles are: -1.072, -0.513 and 0.759
Settling velocities (by Heywood) are: 0.00243, 0.0088 and 0.165
and by Stokes, for all the sizes: 9.81e-7, 9.81e-5, 0.00245, 0.0098 and 0.98.
iv) Stokes' settling equation only accounts for viscous drag of the fluid around the particle, the Heywood Tables approach includes drag effects from eddies and turbulences as well as viscous drag. More drag means the particle will travel slower - hence Stokes' settling underestimates settling velocity if significant turbulence is present in the fluid around the particle.
v) the correct answer is (a), obtained by substituting Stokes' settling equation into the Particle Reynolds number and equating to 0.2; i.e. the conventional limit of streamline flow of the fluid around the particle.
i) A parallel trajectory to the critical, but will reach the tank base before the full tank length.
ii) the correct answer is (a)
iii) the correct answer is (c)
iv) the correct answer is (b)
v) the correct answer is (a)
vi) the correct answer is (b)
vii) square metres and it represents the plan area of the settling tank; i.e. the area when looking down on the top surface of the tank.
(viii) the correct answer is (b)
i) the correct answer is (b)
For the rest of this problem, see the linked file here.
chapter 2 | chapter 3 | chapter 4 | chapter 5 | chapter 6 | chapter 7 | chapter 8 | chapter 9 | chapter 10 | chapter 11 | chapter 12 | chapter 13 | chapter 14 | chapter 15
Links | FAQ