Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 2 (Q.No. 29)
29.
The drainage area of a town is 12 hectares. Its 40% area is hard pavement (K = 0.85), the 40% area is unpaved streets (K = 0.20) and the remaining is wooded areas (K = 0.15). Assuming the time of concentration for the areas as 30 minutes and using the formula 
the maximum run off is
the maximum run off isDiscussion:
15 comments Page 1 of 2.
MILI said:
3 years ago
@Gomathi.
Absolutely correct! thanks.
Absolutely correct! thanks.
Anirban said:
5 years ago
Absolutely correct, well explained, Thanks @Gomathi.
Naveenmanick said:
8 years ago
Yes, correct @Gomathi.
(1)
Zahir said:
8 years ago
Quantity of Run-off, Q = (K*Pc*A)/36.
Where,
K = Run-off Coefficient.
Pc = Intensity of rainfall (cm/hr).
A = Drainage area (hec).
Formula for finding K :-
K = (A1K1 + A2K2 + A3K3) / A.
A1 = 40% of area = 4.8 hec, K1 = 0.85.
A2 = 40% of area = 4.8 hec, K2 = 0.20.
A3 = 20% of area = 2.4 hec, K3 = 0.15.
Therefore, K = (4.8*0.85 + 4.8*0.2 + 2.4 *0.15) / 12 = 0.45.
Formula for finding Pc :-
Pc = 900 /(t+60).
Given, time of concentration (t) =30 min and also use the given formula for finding Intensity of Rainfall (Pc).
Pc = 900 / (30+60) = 10 mm/hr = 1 cm/hr.
Now, Use the rational formula for finding Quantity of storm run-off (Q).
Q =28 A*pc*k
= 28*12*1*.45
= 151 lit/sec.
Or 0.151 cumecs.
Where,
K = Run-off Coefficient.
Pc = Intensity of rainfall (cm/hr).
A = Drainage area (hec).
Formula for finding K :-
K = (A1K1 + A2K2 + A3K3) / A.
A1 = 40% of area = 4.8 hec, K1 = 0.85.
A2 = 40% of area = 4.8 hec, K2 = 0.20.
A3 = 20% of area = 2.4 hec, K3 = 0.15.
Therefore, K = (4.8*0.85 + 4.8*0.2 + 2.4 *0.15) / 12 = 0.45.
Formula for finding Pc :-
Pc = 900 /(t+60).
Given, time of concentration (t) =30 min and also use the given formula for finding Intensity of Rainfall (Pc).
Pc = 900 / (30+60) = 10 mm/hr = 1 cm/hr.
Now, Use the rational formula for finding Quantity of storm run-off (Q).
Q =28 A*pc*k
= 28*12*1*.45
= 151 lit/sec.
Or 0.151 cumecs.
(5)
Zahir said:
8 years ago
How the answer in formula accure in mm/hr?
Sumanta said:
8 years ago
Thanks @Gomathi.
(1)
Gomathi said:
8 years ago
Quantity of Run-off, Q = (K*Pc*A)/36.
Where,
K = Run-off Coefficient.
Pc = Intensity of rainfall (cm/hr).
A = Drainage area (hec).
Formula for finding K :-
K = (A1K1 + A2K2 + A3K3) / A.
A1 = 40% of area = 4.8 hec, K1 = 0.85.
A2 = 40% of area = 4.8 hec, K2 = 0.20.
A3 = 20% of area = 2.4 hec, K3 = 0.15.
Therefore, K = (4.8*0.85 + 4.8*0.2 + 2.4 *0.15) / 12 = 0.45.
Formula for finding Pc :-
Pc = 900 /(t+60).
Given, time of concentration (t) =30 min and also use the given formula for finding Intensity of Rainfall (Pc).
Pc = 900 / (30+60) = 10 mm/hr = 1 cm/hr.
Now, Use the rational formula for finding Quantity of storm run-off (Q).
Q = (K*Pc*A)/36,
= (0.45*1*12)/36,
= 0.15 cumecs.
Where,
K = Run-off Coefficient.
Pc = Intensity of rainfall (cm/hr).
A = Drainage area (hec).
Formula for finding K :-
K = (A1K1 + A2K2 + A3K3) / A.
A1 = 40% of area = 4.8 hec, K1 = 0.85.
A2 = 40% of area = 4.8 hec, K2 = 0.20.
A3 = 20% of area = 2.4 hec, K3 = 0.15.
Therefore, K = (4.8*0.85 + 4.8*0.2 + 2.4 *0.15) / 12 = 0.45.
Formula for finding Pc :-
Pc = 900 /(t+60).
Given, time of concentration (t) =30 min and also use the given formula for finding Intensity of Rainfall (Pc).
Pc = 900 / (30+60) = 10 mm/hr = 1 cm/hr.
Now, Use the rational formula for finding Quantity of storm run-off (Q).
Q = (K*Pc*A)/36,
= (0.45*1*12)/36,
= 0.15 cumecs.
(3)
Leo Basil George said:
8 years ago
We know that Q for wet weather flow = AKR/360.
A- area of catchment in hacters.
K - impermeability factor,
R - runoff,
K= (A1K1+A2K2+A3K3)/A,
= (4.8x.85+4.8x.2+2.4 *.15)/12,
= .45.
Here runoff R is given as Ps = 900/(t+60).
= 900/(30+60),
= 10.
The actual R is determined by the formula,
R = 25.4xa/(t+b).
Where a =30, b= 10 for t = 5 to 20 min,
a= 40 , b=20 for t>20 min,
Q= 12x.45x10/360,
= .15.
A- area of catchment in hacters.
K - impermeability factor,
R - runoff,
K= (A1K1+A2K2+A3K3)/A,
= (4.8x.85+4.8x.2+2.4 *.15)/12,
= .45.
Here runoff R is given as Ps = 900/(t+60).
= 900/(30+60),
= 10.
The actual R is determined by the formula,
R = 25.4xa/(t+b).
Where a =30, b= 10 for t = 5 to 20 min,
a= 40 , b=20 for t>20 min,
Q= 12x.45x10/360,
= .15.
(1)
Vasim said:
9 years ago
Here, (900÷(30 * 60+60)) * (0.2+0.15) = 0.16.
Rani said:
9 years ago
Please explain the Problem.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers