Civil Engineering - Water Supply Engineering - Discussion

Discussion Forum : Water Supply Engineering - Section 3 (Q.No. 29)
29.
The ratio of maximum hourly consumption and average hourly consumption of the maximum day, is
1.2
1.5
1.8
2.4
2.7
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
31 comments Page 2 of 4.

Subha Annadurai said:   7 years ago
1.5 is the crt option.

Satish Verma said:   7 years ago
B option is the correct answer.

Priya said:   7 years ago
Maximum daily demand = 1.8 x average daily demand,
Maximum hourly demand of maximum day i.e. Peak demand,

= 1.5 x average hourly demand.
= 1.5 x Maximum daily demand/24.
= 1.5 x (1.8 x average daily demand)/24.
= 2.7 x average daily demand/24.
= 2.7 x annual average hourly demand.

Kartik said:   7 years ago
Thank you @Priya.

ONKAR said:   6 years ago
Given answer is correct. I agree with it.
(1)

Hassan said:   6 years ago
According to Good rich formula :

P=180 x t^-0.1.

Where t=time in days
P is in per cent.

For maximum hourly consumption t= 1/24 days,
So, P= 180 x (1/24),
= 247%.

Hence the ratio should be 2.4 and not 2.7.
Thus the correct option should be (D).

Dipu said:   6 years ago
B will be the right answer.

Satyaki said:   5 years ago
The ratio of max. hourly consumption =( 2.7 LPCD /24).
The ratio of avg. hourly consumption = (1.8 LPCD /24).
So, Ans is B

Dheeraj kataria said:   5 years ago
The question is totally wrong... How the ratio come 2.7?
Max. Daily demand = 1.8 of (Average daily demand)
Peak Hour consumption = 1.5 of (Max. Daily demand)
Or u can say that (1.8 * 1.5 = 2.7)
So the way that the question is asked is totally wrong.

ATUL KOUSHAL said:   5 years ago
B is correct. I too agree.


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