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
Discussion:
16 comments Page 2 of 2.
Anchal said:
9 years ago
Thanks @Sunil.
Baloch said:
9 years ago
Thanks @Sunil Gupta.
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.
V = 1 * 2.65/ (5 * 1.2) = 0.441666 = 0.45 m/s.
(6)
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.
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)
Saroj said:
10 years ago
How come this answer?
What formula use in solve this question please tell me?
What formula use in solve this question please tell me?
(1)
Saroj said:
10 years ago
How come answer please tell me someone?
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