Method of Superposition

Q. The ordinates of a 3-hour unit hydrograph for a catchment are given in table below

$\begin{array}{cccccc}\text{Time}\left(h\right)& 0& 1& 2& 3& 4\\ \text{Ordinate}{m}^3/s& 0& 5& 12& 25& 41\end{array}$

The ordinate of a 6 hour unit hydrograph for this catchment at the time instant of 4 hours would be______$m^3/s$

1. 23

Q. Ordinates of a 1-hour unit hydrograph at 1 hour intervals, starting from time t = 0 , are 0, 2, 6, 4, 2, 1 and 0 $m^3/s$

Ordinate of a 3-hour unit hydrograph for the catchment at t = 3 hours is

1. $2.0m^3/s$
2. $3.0m^3/s$
3. $4.0m^3/s$
4. $5.0m^3/s$

Q. A 3 hour triangular unit hydrograph (UH) has a time base of 30 hours with peak discharge of $90m^3/s$. For the same catchment 9 hour unit hydrograph was derived using 3 hour UH. The peak of 9 hour UH will

1. $75m^3/s$
2. $50m^3/s$
3. $90m^3/s$
4. $30m^3/s$

Q. An average rainfall of 16 cm occurs over a catchment during a period of 12 h with a uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 h) of the catchment rises linearly from 0 to 30 cumec in 6 h and then falls linearly from 30 to 0 cumec in the next 12 h. $\varphi -index$ of the catchment is known to be 0.5 cm/h. Base flow in the river is known to be 5 cumec.

Peak discharge of the resulting direct runoff hydrograph shall be

1. 150 cumec
2. 225 cumec
3. 230 cumec
4. 360 cumec

Q. The ordinates of a 2-hour unit hydrograph for a catchment are given as
$\begin{array}{cccccc}\text{Time(h)}& 0& 1& 2& 3& 4\\ \text{Ordinate}& 0& 5& 12& 25& 41\\ \left(m^3/s\right)& & & & & \end{array}$
The ordinate (in $m^3/s$) of a 4-hour unit hydrograph for this catchment at the time of 3 h would be

1. 15

Q. A 2 hr - unit hydrograph has a base period of 6 hr. The base period of 8 hr unit hydrograph is

1. 1
2. 14
3. 12
4. 8

Q. If the base period of 6-hour hydrograph of a basin is 84 hours, then a 12-hour unit hydrograph derived from this 6-hour unit hydrograph will have a base period of

1. 72 hours
2. 78 hours
3. 84 hours
4. 90 hours

Q. The 4-h UH for a catchment is tabulated below :

$\begin{array}{cc}\text{Time (hr)}& \text{4-h UH ordinate}\\ & \left(m^3/sec\right)\\ 0& 0\\ 2& 8\\ 4& 20\\ 6& 35\\ 8& 38\\ 10& 32\\ 12& 16\\ 14& 14\\ 16& 10\\ 18& 8\\ 20& 6\\ 22& 3\\ 24& 0\end{array}$

The peak ordinate (in $m^3/sec$) of 8-h UH will be :

1. 67
2. 33.5
3. 76
4. 58

Q. The ordinates of 2-h UH for a basin is as follows

$\begin{array}{ccccccccc}\text{Time (hr)}& 0& 2& 4& 6& 8& 10& 12& 14\\ \text{2-h UH}& 0& 5& 18& 28& 36& 42& 32& 16\\ \left(m^3/sec\right)& & & & & & & & \end{array}$

The ordinate of 4-h UH derived from above UH at 10th hour_____ (in $\frac{m^3}{sec}$) is

1. 72
2. 78
3. 39
4. 32

Q. The ordinate of 2 hour unit hydrograph at one hour time interval starting from time t = 0 are $0,10,35,40,33,20,15,13\text{and}0m^3/s$. The ordinate of 4 hour unit hydrograph at t = 4 hours would be ______$m^3/s$.

1. 34
2. 66
3. 32
4. 68