Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 1 (Q.No. 35)
35.
A water treatment plant is required to process 28800 m3/d of raw water (density = 1000 kg/m3, kinematic viscosity = 10-6 m2/s). The rapid mixing tank imparts tank a velocity gradient of 900s-1 to blend 35 mg/l of alum with the flow for a detention time of 2 minutes. The power input (W) required for rapid mixing is
Discussion:
4 comments Page 1 of 1.
Huidrom said:
6 years ago
Dynamic viscosity = kinematic viscosity * Density.
RuchitaMalunjkar said:
7 years ago
Please explain the answer briefly.
Kirubakara Raj said:
9 years ago
Thanks for your explanation @Samala.
SAMALA LAXMAN said:
1 decade ago
G = sqrt(power input/volume*dynamic viscosity).
Here,
G = 900/sec.
Q = 28800m^3/day = 0.33m^3/sec.
t = 2 min = 120 sec.
Volume = Q*t= 40m^3.
Dynamic viscosity = kinematic viscosity/density.
If we substitute all these values in above formula we will get power = 32400W.
Here,
G = 900/sec.
Q = 28800m^3/day = 0.33m^3/sec.
t = 2 min = 120 sec.
Volume = Q*t= 40m^3.
Dynamic viscosity = kinematic viscosity/density.
If we substitute all these values in above formula we will get power = 32400W.
(1)
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