Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 1 (Q.No. 10)
10.
The width of a rectangular sewer is twice its depth while discharging 1.5 m/sec. The width of the sewer is
Discussion:
30 comments Page 1 of 3.
Gidgo said:
3 years ago
1.36 m is the right answer.
(1)
PREM ROSHAN said:
3 years ago
@Gopal Lal Kumawat.
2% is the right answer.
2% is the right answer.
Gurjant singh said:
4 years ago
@Saeed.
It's the value of d square. So, the answer given is correct.
It's the value of d square. So, the answer given is correct.
(2)
Saeed said:
4 years ago
Min v = 0.45.
b= 2d,
q = AV.
1.5 = b*d *0.45.
1.5= 2d*d*0.45 ,,d^2=1.5/0.9= 1.36.
d=1.36.
b= 2d,
q = AV.
1.5 = b*d *0.45.
1.5= 2d*d*0.45 ,,d^2=1.5/0.9= 1.36.
d=1.36.
(17)
Emon khan said:
5 years ago
Develop a relation between a diameter of the circular section of a sewer and a side of the rectangular sewer section having width as twice its depth, the three sides being wetted.
(5)
Jinz said:
6 years ago
Isn't Q supposed to be in m^3/sec?
In question, it's given 1.5 m/sec.
In question, it's given 1.5 m/sec.
(4)
Gayatri said:
6 years ago
Why and how to take V= 1.62 m/sec? @Jinish.
(1)
Jinish said:
6 years ago
Take v= 1.62 m/sec.
B=2 D.
Q=AxV.
A= B * D.
A= 2 *D * D = 2D^2.
So Q=AxV and Q = 1.5 m/sec.
1.5= 2D^2 * 1.62.
1.5/1.62= 2D^2.
1.5/ 1.62*2 = D^2,
D= .68m.
B= 2D = 2*0.68 = 1.36 m.
B=2 D.
Q=AxV.
A= B * D.
A= 2 *D * D = 2D^2.
So Q=AxV and Q = 1.5 m/sec.
1.5= 2D^2 * 1.62.
1.5/1.62= 2D^2.
1.5/ 1.62*2 = D^2,
D= .68m.
B= 2D = 2*0.68 = 1.36 m.
(6)
Amu bokato said:
6 years ago
Thank you @Sahu.
(2)
Monika Thakur said:
6 years ago
According to me, it is (Q/v) = 1.5 * 0.9 = 1.35.
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