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

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