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.

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

Masoom said:   6 years ago
Thank you @Sunil.

Nagdive said:   7 years ago
Thank you @Sunil And @Hemanth.

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)

Pavani said:   7 years ago
According to me, it is 0.5.

Randos said:   7 years ago
Check the unit. @Sunil.

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)

Dheerak singh said:   8 years ago
V=√k(gd (s-1)).
=√4.5 (10*.005 (1.2-1)).

Ovizit Saha said:   8 years ago
Thanks @ Sunil Gupta.

RAVI YADAV said:   9 years ago
@ALL.

Use Neethu's formulae for solve this.

Thanks @Neethu.


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