A 30% (by volume) suspension of spherical sand particles in a viscous oil has a hindered settling velocity of 4.44 μm/s. If the Richardson Zaki hindered settling index is 4.5, then the terminal velocity of a sand grain is
A. 0.90 μm/s
B. 1 μm/s
C. 22.1 μm/s
D. 0.02 μm/s
Answer: Option B
Solution (By Examveda Team)
The Richardson-Zaki equation relates the hindered settling velocity (Vh) to the terminal settling velocity (Vt) through the expression:Vh = Vt × (1 - φ)n
where φ is the volume fraction of solids (here, φ = 0.30) and n is the Richardson-Zaki index (given as 4.5).
First calculate (1 - φ):
1 - 0.30 = 0.70
Then raise it to the power of n:
(0.70)4.5 ≈ 0.2008
Rearranging the equation for terminal velocity gives:
Vt = Vh / (1 - φ)n
Substituting the values:
Vt = 4.44 μm/s / 0.2008 ≈ 22.1 μm/s
Therefore, the terminal velocity of a sand grain is 22.1 μm/s.
This Question Belongs to Chemical Engineering >> Mechanical Operations
We are given:
Hindered settling velocity
𝑢
=
4.44
𝜇
m/s
u=4.44 μm/s
Volume fraction of solids
𝜙
=
0.30
ϕ=0.30
Richardson-Zaki index
𝑛
=
4.5
n=4.5
We are to find the terminal settling velocity
𝑢
𝑡
u
t
(i.e., settling velocity in clear fluid with no other particles).
✅ Richardson-Zaki equation:
𝑢
=
𝑢
𝑡
(
1
−
𝜙
)
𝑛
u=u
t
(1−ϕ)
n
Where:
𝑢
u = hindered settling velocity
𝑢
𝑡
u
t
= terminal settling velocity
𝜙
ϕ = solids volume fraction
𝑛
n = Richardson-Zaki exponent (typically 4–5 for fine particles)
🔍 Step-by-step:
Given:
𝑢
=
4.44
𝜇
m/s
u=4.44 μm/s
𝜙
=
0.30
ϕ=0.30
𝑛
=
4.5
n=4.5
(
1
−
𝜙
)
𝑛
=
(
1
−
0.30
)
4.5
=
(
0.70
)
4.5
(1−ϕ)
n
=(1−0.30)
4.5
=(0.70)
4.5
First compute
0.70
4.5
0.70
4.5
:
0.70
4.5
≈
𝑒
4.5
⋅
ln
(
0.70
)
≈
𝑒
4.5
⋅
(
−
0.3567
)
=
𝑒
−
1.605
≈
0.2009
0.70
4.5
≈e
4.5⋅ln(0.70)
≈e
4.5⋅(−0.3567)
=e
−1.605
≈0.2009
Now solve for
𝑢
𝑡
u
t
:
𝑢
𝑡
=
𝑢
(
1
−
𝜙
)
𝑛
=
4.44
0.2009
≈
22.1
𝜇
m/s
u
t
=
(1−ϕ)
n
u
=
0.2009
4.44
≈22.1 μm/s
✅ Final Answer:
C. 22.1 μm/s
C. 22.1 μm/s
Please give me detailed solution