Civil Engineering - Waste Water Engineering - Discussion

Discussion Forum : Waste Water Engineering - Section 1 (Q.No. 27)
27.
A sewer pipe contains 1 mm sand particles of specific gravity 2.65 and 5 mm organic particles of specific gravity 1.2, the minimum velocity required for removing the sewerage, is
0.30 m/sec
0.35 m/sec
0.40 m/sec
0.45 m/sec
0.50 m/sec.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

SUNIL GUPTA said:   9 years ago
The velocity required removing sewage = size of sand particle * specific gravity of sand/(specific gravity of organic mater *size of organic matter).

V = 1 * 2.65/ (5 * 1.2) = 0.441666 = 0.45 m/s.
(6)

Hemanth said:   7 years ago
@All

Here, s sand particle up to 1mm and organic particles up to 5mm, there is a condition for this one min velocity of 0.45 m/sec and Max 0.9m/sec.

Hence no need to calculate anything.
And @Neethu, the correct formula is;
Vs=√((8k/f)*(s-1)*g*d) this is the right one.
(5)

Dipan said:   7 years ago
No, it is wrong @Sunil Gupta.

@Neethu

Thanks for your information.

This is the correct self-cleaning velocity and we get this formula 0.4 for organic particle and .42 for sand but the minimum self-cleaning velocity is 0.45 so we take the value 0.4.
(2)

Saroj said:   10 years ago
How come this answer?

What formula use in solve this question please tell me?
(1)

Neethu said:   10 years ago
Sqr(8k/f*(s-1)*g*d).

K = Constant, for clean inorganic solids = 0.04 and for organic solids = 0.06.

f' = Darcy Weisbach friction factor (for sewers = 0. 03).

Ss = Specific gravity of sediments.

g = Gravity acceleration.

d' = Diameter of grain, m.
(1)

Ajay Kumar Yadav said:   2 years ago
0. 4m/s is the correct answer.
(1)

Saroj said:   10 years ago
How come answer please tell me someone?

Baloch said:   9 years ago
Thanks @Sunil Gupta.

Anchal said:   9 years ago
Thanks @Sunil.

RAVI YADAV said:   9 years ago
@ALL.

Use Neethu's formulae for solve this.

Thanks @Neethu.


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