Engineering
Hydrology by Rajesh Srivastava and Ashu Jain
McGraw
Hill Education, New Delhi (2017)
Errata: (Readers are encouraged to mail us
either at rajeshs@iitk.ac.in or at ashujain@iitk.ac.in about any errors they
spot)
Chapter3:
Abstractions
[1] Page
72: "(a) Penman method" should not have "(a)"
[2] Page
73: "(b) Actual Evaporation" should be "3.4.3 Actual
Evaporation"
[3] Page
79: "3.5.2.2 Actual Evapotranspiration" should be "3.5.3 Actual
Evapotranspiration"
Chapter4: Runoff
[1] Page 118: Example 4.2; Line 1: Area of
catchment should be taken as 24 ha instead of 240 ha. As a result in the table
on page 119, intensities should be 54.5, 80.1, and 93.3 and the peak discharges
should be 1.5, 2.1, 2.5 for return periods of 1year, 10years, and 25years,
respectively. Thus, the modified table should look like this:
[2] Page 126: Example 4.3; Line 7: In
expression V_{R} = P * A, "P" should be replaced by
"ER".
Chapter5:
Hydrograph Analysis
[1] On pages 166167, Example 5.8: Please note
that the time to peak calculated should be from the centroid of ERH and not the
start of the ERH as shown. As such, t_{R}/2 (= 6/2 = 3 hours) needs to
be added to the timings. The table and Figure 5.11 (on page 167) should
therefore need to be modified as follows. Note that Q^{s} remain the
same.

[2] Page 183, Q 5.2, last line: "Jun
30" should be read as "June 30".
Chapter6:
Hydrograph Routing
[1] On pages 193194, Example 6.1: The entries
in columns [5] to [8] are incorrect due to an inadvertent error in calculating
elevation and outflow in the first row for t
= 3 hours, which gets accumulated at later times. The modified table with
correct entries in columns [5][8] is provided below with modified numbers in
redfont.
Table of calculations
for outflow hydrograph for EXAMPLE 6.1 on page 193194
As a result, some of the numbers in the two
paragraphs following this table (on page 194) get modified as follows (the
modified numbers are represented in red font).
The computations for the
routing of inflow hydrograph are carried out in a tabular form above. The first
two columns of the table represent the given inflow hydrograph. Column 3
gives the average inflow value at each time interval, (I_{1}+I_{2})/2
or (I_{avg}), in m^{3}/s.
Column 4 is the inflow volume in each time interval, (I_{avg}Δt),
in Mm^{3}, which is obtained by multiplying the entries in column 3
by 3 hours or 10,800 s. The first row against time t = 0 represents the initial conditions i.e. I = 10 m^{3}/s, h
= 63.000 m, and Q = 20 m^{3}/s.
For the given initial h = 63.000 m,
we have S = 3.88 Mm^{3}. Using the values of initial S and Q, we calculate (SQΔt/2) = 3.772 Mm^{3}. This is entered in column 5 in the
row for t = 3 hours. Now as per equation
(6.6), the RHS is calculated as the sum of column 4 and column 5. Therefore,
(S+QΔt/2) = 3.934 Mm^{3},
this is entered in column 6 in the row for t = 3. Now using the indicative storage graph prepared (Figure
6.1), we find out the values of elevation in the reservoir (h) and outflow (Q) corresponding to the value of (S+QΔt/2) = 3.934 Mm^{3} just
calculated, by using the arrows. Note that this step can also be easily
carried out by using linear interpolation of the (S+QΔt/2) v/s h and Q v/s h data. The values of h and Q are thus found as h =
This process is then continued
for each time step and the inflow hydrograph is thus routed through the
reservoir to calculate the outflow hydrograph. The outflow hydrograph is
presented in column 8 in the table above and the inflow and outflow
hydrographs are shown in the figure below. Note that the peak flow of 80 m^{3}/s
occurs at t = 9 hours in the inflow
hydrograph, while the peak flow of 
The Figure 6.2 also gets modified slightly as
follows.
Figure 6.2: Inflow and
outflow hydrographs
[2] Example 6.4 should be replaced by the
following, due to a formula error in it:
Example 6.4: Using the inflow and outflow hydrographs from a channel reach given below, determine Muskingum coefficients K and x.
Time (hr) 
0 
4 
8 
12 
16 
20 
24 
28 
32 
36 
40 
44 
48 
52 
56 
Inflow (m^{3}/s) 
42 
68 
116 
164 
194 
200 
192 
170 
150 
128 
106 
88 
74 
62 
54 
Outflow (m^{3}/s) 
42 
39 
44 
65 
98 
133 
159 
174 
175 
168 
156 
140 
122 
106 
90 
Solution: The calculation of the Muskingum coefficients (K and x) is carried out using the continuity equation (6.18). The computations are arranged in the following table. The first three columns present the inflow and outflow hydrograph data given in the example. Column 4 is the difference (IQ) of the average inflow and outflow values at the beginning and end of the time interval and column 5 is obtained by multiplying the average of column 4 by the time interval Δt = 4 hours. The column 5 represents incremental storage ΔS in m^{3}/shour. The column 6 is storage S which is obtained by accumulating the entries in column 5. Then, the next 4 columns represent the quantity {xI + (1x)Q} for different trial values of x. The values tried are x=0.1, x=0.2, x=0.3, and x=0.4.
The plots of storage S versus the quantity {xI + (1x)Q} are then drawn to ascertain if it shows up as a straight line. The value of x which results in this plot will be the correct value of x. As seen from the plots, the value of x=0.2 is the most appropriate as it gives a straight line with a very narrow scatter (in fact, a value between 0.2 and 0.3 would give a better straight line, but it is not really necessary to do further finetuning). The value of K is then obtained as the slope of the straight line in the S v/s {xI + (1x)Q} plot corresponding to x=0.2. Selecting the storages S1=400 m^{3}/shour and S2 = 800 m^{3}/shour, the corresponding values of the quantity {xI + (1x)Q} are read as 110 m^{3}/s and 175 m^{3}/s. Then, K = (800400)/(175110) = 6.15 hours. Therefore, the values of the Muskingum coefficients for the given data are K = 6.15 hours and x = 0.2.
Time 
Inflow, I 
Outflow 
(IQ) 
ΔS=[4]*Δt 
S=ΣΔS 
[x I + (1  x) Q} (m^{3}/s) 

(Hours) 
(m^{3}/s) 
(m^{3}/s) 
(m^{3}/s) 
(m^{3}/sh) 
(m^{3}/sh) 
x = 0.1 
x = 0.2 
x = 0.3 
x = 0.4 
[1] 
[2] 
[3] 
[4] 
[5] 
[6] 
[7] 
[8] 
[9] 
[10] 
0 
42 
42 
0.0 
0 
0 
42.0 
42.0 
42.0 
42.0 
4 
68 
44 
12.0 
48 
48 
46.4 
48.8 
51.2 
53.6 
8 
116 
63 
38.5 
154 
202 
68.3 
73.6 
78.9 
84.2 
12 
164 
100 
58.5 
234 
436 
106.4 
112.8 
119.2 
125.6 
16 
194 
140 
59.0 
236 
672 
145.4 
150.8 
156.2 
161.6 
20 
200 
175 
39.5 
158 
830 
177.5 
180.0 
182.5 
185.0 
24 
192 
190 
13.5 
54 
884 
190.2 
190.4 
190.6 
190.8 
28 
170 
189 
8.5 
34 
850 
187.1 
185.2 
183.3 
181.4 
32 
148 
175 
23.0 
92 
758 
172.3 
169.6 
166.9 
164.2 
36 
125 
155 
28.5 
114 
644 
152.0 
149.0 
146.0 
143.0 
40 
106 
136 
30.0 
120 
524 
133.0 
130.0 
127.0 
124.0 
44 
88 
117 
29.5 
118 
406 
114.1 
111.2 
108.3 
105.4 
48 
74 
98 
26.5 
106 
300 
95.6 
93.2 
90.8 
88.4 
52 
62 
82 
22.0 
88 
212 
80.0 
78.0 
76.0 
74.0 
56 
54 
70 
18.0 
72 
140 
68.4 
66.8 
65.2 
63.6 




Chapter7:
Groundwater
The hydraulic gradient, i, has been
defined as the ‘drop’ in hydraulic head per unit length. However, at several
places the sign convention has not been followed. The following corrections
need to be made:
[1] Page 242, in ‘hydraulic gradient along the
streamline, ¶h/¶s’, there should be a negative sign
[2] Page 244, in the expressions for q_{r}
and q_{q}, there should be negative signs
[3] Page 247, after Eq. (7.13), h_{2}−h_{1} should be replaced by h_{1}−h_{2} in all points and ‘negative’ should be changed
to ‘positive’ in the third bullet
[4] Page 249, first bullet should have negative
sign in the expression for i, and second bullet should have ‘negative’
in place of ‘positive’
Chapter9:
Statistical Methods in Hydrology
[1] On page 342, Last term in Eq. (9.75) should
have a negative sign ( k^{5}/3). This expression is an expansion of
the more compact form obtained using the WilsonHilferty transformation as
given below
K_{T}={[1+ k (z  k)]^{3 }
1}/3k
[2] On Page 343, The Log Pearson Type III table
has incorrect values of K_{T} (and, therefore, y_{T}
and x_{T}). The values should be as shown below:
T (Yrs) 
z 
K_{T} 
y_{T} 
x_{T} 
50 
2.0537 
2.0406 
1.6492 
44.6 
100 
2.3263 
2.3083 
1.6882 
48.8 
500 
2.8779 
2.8482 
1.7669 
58.5 
[3] On page 346, Eq. (9.78) denominator should
be 2n NOT n. Consequently, in Example 9.11 (a), Se will change to 2.638 and the
values of lower and upper confidence limits will change to 37.43 mm and 47.77
mm, respectively.
[4] On page 348, for Gumbel’s distribution,
correct value of Se is 6.0260 and the values of lower and upper confidence
limits will change to 41.10 mm and 64.72 mm, respectively. This will change the
conclusions drawn from this example.
[5] On page 348 and 349, for Log Pearson Type
III distribution, correct value of K_{T} is 2.3083, b is 5.2183, the
values of lower and upper frequency factors are 1.7723 and 3.1206, the values
of lower and upper limits of y are 1.6101 and 1.8066, and the confidence limits
will change to 40.75 mm and 64.06 mm, respectively.
[6] On page 360 in Q9.10, the exceedance
probabilities should be taken as 0.3, 0.1, and 0.03 in place of 0.03, 0.01, and
0.003, respectively.
[7] On page 360 in Q9.12, ȳ
should be taken as 3.6
instead of 4.6. Also, the subheading (d) should be read as (c).
[8] On page 360 in Q9.13, flood magnitudes for which return periods are
to be found shall be taken as 10,000 m^{3}/s, 9,000 m^{3}/s,
8,000 m^{3}/s, and 5,000 m^{3}/s, respective, instead of the
values given.