WEBVTT
Kind: captions
Language: en
00:00:02.939 --> 00:00:18.279
Hello friends welcome back to NPTEL online
certification course on Soil and Water Conservation
00:00:18.279 --> 00:00:22.260
Engineering, I am Rajendra Singh professor
in Agriculture and Food Engineering Department
00:00:22.260 --> 00:00:30.040
of IIT Kharagpur. We are in week 4 lecture
20 and today we will solve Problems on Broad
00:00:30.040 --> 00:00:33.870
Base Terrace.
Now, just to remind you of the course content
00:00:33.870 --> 00:00:41.230
that we covered in this week up till now in
lecture 16 we started the week with introducing
00:00:41.230 --> 00:00:43.129
terrace, in lecture 17 we covered the concepts
corresponding to bench terraces.
00:00:43.129 --> 00:00:46.860
In lecture 18 we used the concepts learnt
in lecture 17 to design bench terraces
00:00:46.860 --> 00:00:50.300
that is we solved problems dealing with the
design of bench terraces. In lecture 19 we
00:00:50.300 --> 00:00:55.539
dealt with broad based terraces and todays
lecture we will solve certain problems dealing
00:00:55.539 --> 00:01:02.030
with the design of broad based terraces.
Let us start with problem number one and this
00:01:02.030 --> 00:01:07.620
problem states that for a broad based terrace
constructed on 7 percent land slope, the channel
00:01:07.620 --> 00:01:12.490
depth is 0.3 meters and the width of terrace
is 6 meters. Compute the cut and fill heights
00:01:12.490 --> 00:01:15.850
and the slope ratios for the front slope and
back slope assuming a balanced cross section.
00:01:15.850 --> 00:01:22.470
So; that means, it is a broad based terrace,
a constructed on 7 percent land slope, channel
00:01:22.470 --> 00:01:28.650
depth and channel width of the terrace of
are already given. And we have to compute
00:01:28.650 --> 00:01:32.740
the cut and fill heights; at the slope ratios
for the front slope and the back slope and
00:01:32.740 --> 00:01:37.140
we have to assume a balanced cross section.
And balanced cross section means the cut is
00:01:37.140 --> 00:01:41.520
equal to fill that is this is the information
which is already given in this particular
00:01:41.520 --> 00:01:46.200
problem .
So, once again the data given are land slope
00:01:46.200 --> 00:01:55.370
S equals to 7 percent; channel depth, h equal
to 0.3 meters; terrace width, W is 6 meters
00:01:55.370 --> 00:02:01.140
and balanced cross section. So, we have been
given h is known to us, W is known to us and
00:02:01.140 --> 00:02:05.210
of course, the land slope which is here S
that is known to us.
00:02:05.210 --> 00:02:10.489
So, these are three things which are known
to us and of course, balance cross section
00:02:10.489 --> 00:02:14.440
so, that simply means; cut is equal to fill
that is also known to us so; that means,
00:02:14.440 --> 00:02:18.530
this cut is equal to this fill that is known
to us. And for this section when it is a or
00:02:18.530 --> 00:02:23.843
width of the when all the three widths are
same that when means when when the section
00:02:23.843 --> 00:02:27.890
is balanced or the cross section is balanced.
Then we know the relationship between cut
00:02:27.890 --> 00:02:32.610
fill, the channel depth and terrace width,
which is given by equation 3 which we saw
00:02:32.610 --> 00:02:38.380
in the previous lecture. And that reads as
C plus f equal to h plus S times W. Where
00:02:38.380 --> 00:02:48.550
C is cut f is fill C is cut f is fill h is
the depth of the channel and W is the terrace
00:02:48.550 --> 00:02:59.260
width and S is the land slope. Ah So, this
is what it is and as per this problem statement
00:02:59.260 --> 00:03:06.590
it is a balance cross section so, C is equal
to f that is cut is equal to f.
00:03:06.590 --> 00:03:11.900
So, from this relationship C equal to f is
h plus SW divided by 2 because, C plus
00:03:11.900 --> 00:03:19.849
you can also write 2 C equals to or 2 f equal
to h plus SW or in this case writing C equal
00:03:19.849 --> 00:03:27.700
to f equal to h plus SW by 2. And in this
we know the value of h, we know the value
00:03:27.700 --> 00:03:35.220
of S, we know the value of W, so; that means,
by just by putting the values of h S and W,
00:03:35.220 --> 00:03:38.960
we can get the values of C and f that is
cut and fill.
00:03:38.960 --> 00:03:47.990
And so, C equals to f which is equal to h
is 0.3, S is 7 percent; that means, 0.07 W
00:03:47.990 --> 00:03:52.739
is width of the terraces which is given a
6 meters. So, by putting the value you get
00:03:52.739 --> 00:04:05.250
C equal to f is equal to 0.36 meters.
So, that is cut fill is 0.36 meters so;
00:04:05.250 --> 00:04:19.190
that means, from the this is the bottom of
the channel, that is measured at the channel
00:04:19.190 --> 00:04:26.090
centerline the bottom of the channel channel
and this is the original land slope, the vertical
00:04:26.090 --> 00:04:30.380
difference between two is cut.
Similarly here the the terrace centerline
00:04:30.380 --> 00:04:36.620
the top of the ridge and the original land
slope at this particular vertical line, the
00:04:36.620 --> 00:04:42.360
vertical difference is the fill. So, cut and
fill value we have obtained in this problem
00:04:42.360 --> 00:04:46.350
now we have to obtain the slopes .
So, from the geometry for the front slope
00:04:46.350 --> 00:04:52.880
the change in elevation over W is equal to
h. So, this is front slope so that this is
00:04:52.880 --> 00:04:57.199
W and elevation change from here to here this
is h basically. So, front slope will be S
00:04:57.199 --> 00:05:03.760
f is W by h and W value is known h value is
known W is 6 h is 0.3 so; that means, it is
00:05:03.760 --> 00:05:10.889
20.
So, front slope ratio comes out to be 20 is
00:05:10.889 --> 00:05:19.660
to 1, that is 20 horizontal to 1 vertical.
So, we are always in this subject typically
00:05:19.660 --> 00:05:23.830
we represent slopes as horizontal is to vertical
ratio. For the back slope if you see; the
00:05:23.830 --> 00:05:30.680
change in elevation over W has two components,
one is of course, h h this h is already there,
00:05:30.680 --> 00:05:37.729
but the same time there is a elevation difference
of the original land slope basically.
00:05:37.729 --> 00:05:43.680
So, that we have to take into account over
the width which is same. So, it is h plus
00:05:43.680 --> 00:05:49.640
SW. So, that is why back slope is W divided
by h plus WS. So, 6 divided by 0.3 which is
00:05:49.640 --> 00:05:56.449
h W is 6 S is 0.07 and by calculating this
we get 8.33 S b equals to 8.33.
00:05:56.449 --> 00:06:03.980
So, back slope ratio is 8.33 is to 1 so, front
slope ratio is comes out to be 20 is to 1,
00:06:03.980 --> 00:06:09.850
back slope ratio comes out to be 8.33 is to
1. So, that is means cut and fill is equal
00:06:09.850 --> 00:06:17.360
to 0.36 meters, front slope ratio is 20 is
to 1 and back slope ratio is 8.33 is to 1.
00:06:17.360 --> 00:06:22.260
So, these are the answers which we have obtained
for this particular problem .
00:06:22.260 --> 00:06:30.100
Now, we go to second problem; which says that
design a 300 meter long variable graded terrace
00:06:30.100 --> 00:06:38.970
for a land having average slope of 3 percent.
The maximum permeable velocity of water in
00:06:38.970 --> 00:06:45.139
the terrace channel is 60 centimeter per second.
Intensity of 1 hour rainfall exopected during
00:06:45.139 --> 00:06:51.150
10 year recurrent interval is 10 centimeter
per hour. The region is less humid and soils
00:06:51.150 --> 00:06:55.430
are normal.
So; that means, we have to design a 300 meter
00:06:55.430 --> 00:06:59.630
long variable graded terrace for a landslope
of 3 percent. The maximum permeabile velocity
00:06:59.630 --> 00:07:07.070
is given here 60 centimeter per second.
Ah Intensity of 1 hour rainfall for 10 year
00:07:07.070 --> 00:07:13.180
reccurrence interval is given as 10 centimeter
per hour and the region identified as less
00:07:13.180 --> 00:07:18.190
humid and soils ae normal.
So, thats means remember we had a condition
00:07:18.190 --> 00:07:22.000
for vertical interval. So, that simply shows
that we can straightaway use the one of the
00:07:22.000 --> 00:07:23.360
equations recommended for hm graded terraces
.
00:07:23.360 --> 00:07:28.300
So, the data once again with the the symbol
length of variable graded terrace l small
00:07:28.300 --> 00:07:33.240
l is 300 meters, land slope S is 3 percent,
maximum permissible velocity of water which
00:07:33.240 --> 00:07:38.910
is nothing, but the non erosive velocity we
call is 60 centimeter per second and intensity
00:07:38.910 --> 00:07:42.690
of 1 hour 10 year rainfall is 10 centimeter
per second, the region is less humid and with
00:07:42.690 --> 00:07:45.970
normal soils .
So, first thing we need to estimate is vertical
00:07:45.970 --> 00:07:54.669
intervals and vertical intervals and for
this purpose we may use equation 2 previous
00:07:54.669 --> 00:08:00.580
lecture, which there was equation given for
vertical interval for less humid regions and
00:08:00.580 --> 00:08:06.200
normal soil . So, as per the relationship
vertical interval equals to S plus 6 by 10,
00:08:06.200 --> 00:08:14.289
where S is the land slope, the land slope
which is 3 percent in this case .
00:08:14.289 --> 00:08:23.860
So, putting the value of S we can calculate
the vertical interval as 3 plus 6 9 by 10
00:08:23.860 --> 00:08:31.740
that is 0.9 meters. And the relationship between
HI and VI we already know because, we know
00:08:31.740 --> 00:08:39.479
that HI and VI they are related with
reference to S percent land slope . So, this
00:08:39.479 --> 00:08:45.160
is VI this is HI so VI by HI equal to S by
100.
00:08:45.160 --> 00:08:54.250
So, this we have seen many times so, from
this relationship HI is equals to 100 VI by
00:08:54.250 --> 00:09:00.200
S. So, same here 100 VI by S and VI value
we already know S value we already know so,
00:09:00.200 --> 00:09:05.010
putting these values we get horizontal interval
equals to 30 meter. So, for this particular
00:09:05.010 --> 00:09:08.510
given conditions vertical interval is 0.9
meters and horizontal interval is 30 meters
00:09:08.510 --> 00:09:11.180
.
Now, step two is let the grades of variable
00:09:11.180 --> 00:09:16.520
graded terrace be we can use the standard
table we already saw table 2, in lecture 19
00:09:16.520 --> 00:09:18.680
0.5 0.4 0.2 and 0.2 for each quarter length
from outlet respectively.
00:09:18.680 --> 00:09:26.680
And this we can see we can also see
the table basically by looking into the table
00:09:26.680 --> 00:09:33.899
because here the channel grades are recommended
for for variable graded terraces, based on
00:09:33.899 --> 00:09:39.210
us NRCS recommendations. So, our terrace length
is 300 meters; that means, we fall here in
00:09:39.210 --> 00:09:44.410
this 260-360 range and then the recommended
grades are in percent 0.50 for lower quarter
00:09:44.410 --> 00:09:50.260
0.40 for second quarter 0.30 for third quarter
and 0.20 for the upper quarter.
00:09:50.260 --> 00:09:57.000
So, right from upstream to downstream
the slope is increasing point 0.20 0.30 0.40
00:09:57.000 --> 00:10:03.310
0.50 slowly . So, this is these are the recommended
grades which we can accept basically.
00:10:03.310 --> 00:10:13.750
And so; that means, these are the grades
which we have chosen. Next step three we need
00:10:13.750 --> 00:10:18.769
to determine the peak of runoff and obviously,
for this we use rational method we already
00:10:18.769 --> 00:10:24.330
know that whenever we need peak. Peak rate
of runoff we use rational method which is
00:10:24.330 --> 00:10:29.010
Q equals to CIA by 36.
Where C is the runoff coefficient, Q is is
00:10:29.010 --> 00:10:37.589
in cubic meter per second that is peak runoff
rate in cubic meter per second, C is the runoff
00:10:37.589 --> 00:10:47.720
coefficient, I is the rainfall intensity equal
to time concentration unit is centimeter per
00:10:47.720 --> 00:11:00.380
hour. And A is the drainage area and the unit
used is hectares and using this this is the
00:11:00.380 --> 00:11:06.170
rational formula using this we can use the
cultivate the peak rate of runoff .
00:11:06.170 --> 00:11:19.090
So, for this case, the drainage area that
is the terrace area which is drainage area
00:11:19.090 --> 00:11:25.700
is nothing, but multiplication of length of
the terrace and the horizontal interval, because
00:11:25.700 --> 00:11:32.630
the area between two terraces that is used
as a drainage area for the designing the
00:11:32.630 --> 00:11:39.220
graded channel. And that means a length
is given as 300 meters HI we have calculated
00:11:39.220 --> 00:11:43.020
as 30 meters.
So, the area comes out to be to be 9000 square
00:11:43.020 --> 00:11:49.220
meters and or because the unit required in
q I equals to CIA hectares. So, we are again
00:11:49.220 --> 00:11:53.960
converting this 9000 square meter into 0.9
hectare divided by 10000, where 10000 square
00:11:53.960 --> 00:11:58.579
meters equal to 1 hectare. So, it comes out
to be 0.9 hectare. Now we need to determine
00:11:58.579 --> 00:12:03.870
the time of concentration because I definition
says that it is peak rainfall intensity for
00:12:03.870 --> 00:12:07.300
duration equal to time of concentration.
And for time of concentration we use Kirpich
00:12:07.300 --> 00:12:13.550
formula that is the standard formula we use
and as per this formula T c is equal to 0.0195
00:12:13.550 --> 00:12:21.771
L to the power 0.77 S to the power minus minus
0.385 where L is the total flow length and
00:12:21.771 --> 00:12:31.910
S is the total slope over the flow length.
So, the total flow length typically is length
00:12:31.910 --> 00:12:37.660
time plus horizontal interval that every
time you have seen that in between two two
00:12:37.660 --> 00:12:42.250
graded terraces will be built here. And then
the second one is let us say here so; obviously,
00:12:42.250 --> 00:12:47.990
this is the horizontal interval this is the
horizontal interval between these two. So,
00:12:47.990 --> 00:12:55.990
this is the area that x is the terrace area.
So, the flow has to traverse this distance
00:12:55.990 --> 00:13:02.860
and along the along the channel perpendicular
longitudinal distance has traverse this length
00:13:02.860 --> 00:13:10.940
of the terrace.
So, that is why that is the total flow length
00:13:10.940 --> 00:13:16.010
is HI plus the length along the channel. So,
this is 300 plus 30 or 330 meters. And the
00:13:16.010 --> 00:13:25.670
total slope is elevation difference again
the elevation difference here, plus the grade
00:13:25.670 --> 00:13:38.250
of the channel we have to consider and the
total length for calculating the slope. So,
00:13:38.250 --> 00:13:45.550
H is the difference in elevation between most
remote point and outlet. So, most remote point
00:13:45.550 --> 00:13:50.270
could be considered the upstream of the previous
terrace.
00:13:50.270 --> 00:14:14.170
Now, coming to elevation difference between
adjacent terraces over HI is land slope is
00:14:14.170 --> 00:14:26.040
3 percent. So, 0.03 times 30 so, 0.9 meters
it comes out to be straight away. And over
00:14:26.040 --> 00:14:38.570
the channel length we have already decided
the slopes that is 0.5 percent this should
00:14:38.570 --> 00:14:56.300
this should be read as point 4 percent. So,
0.4 percent, then 0.3 percent and 0.2 percent
00:14:56.300 --> 00:15:08.000
so, from table we read point 0.2 0.3 0.4 and
0.5 percent.
00:15:08.000 --> 00:15:22.610
So, that is why the and and total 300 meters
so, the quarter length is 75 meters. So that
00:15:22.610 --> 00:15:37.730
means, 0.005 times 75 0.004 times 75 0.003
times 75 0.002 times 75 and then calculate
00:15:37.730 --> 00:15:53.690
slopes of these . So, the total elevation
difference is this plus sum of these four.
00:15:53.690 --> 00:16:00.990
So, all sums comes out to be 1.95 meters.
So, total elevation difference is 1.95 meters,
00:16:00.990 --> 00:16:04.459
total length is 330 meters slope over the
total flow length is 0.006 that is the slope
00:16:04.459 --> 00:16:10.720
we calculate .
So, that simply means that now putting the
00:16:10.720 --> 00:16:18.660
because, we know L now we know S. So, putting
the value in Kirpichs formula that is 0.0195
00:16:18.660 --> 00:16:26.550
330 to power 0.77 0.006 to the power minus
0.385 we get time of concentration equals
00:16:26.550 --> 00:16:33.050
to 12.15 meters for the given area.
Now, we have been given 1 hour intensity
00:16:33.050 --> 00:16:45.450
is 10 centimeter per hour and we have to obtain
the intensity corresponding to 12.15 meters.
00:16:45.450 --> 00:16:49.570
So, like in the previous class where we
solved problem on bench terraces and we use
00:16:49.570 --> 00:16:55.420
the standard nomograph. So, the same standard
nomograph we have to use here, which relates
00:16:55.420 --> 00:17:02.579
1 hour rainfall intensity, 1 hour rainfall
intensity to rainfall intensity for different
00:17:02.579 --> 00:17:07.669
durations. And the durations are given here
5 minutes 15 minutes 30 minutes 60 minutes
00:17:07.669 --> 00:17:13.489
120 minutes and so on.
So, for for our case it is 1 hour intensity
00:17:13.489 --> 00:17:17.229
is 10 centimeter per hour so; that means,
1 hour intensity. So, our value will lie on
00:17:17.229 --> 00:17:21.359
this horizontal line, then corresponding to
10 centimeter per hour and we have to look
00:17:21.359 --> 00:17:27.380
for 12.15 minutes. So, this is 5 minute line
so, this is one point of reference and this
00:17:27.380 --> 00:17:35.840
is 15 minutes, this is one point of reference.
And we have to find the this is 10 will be
00:17:35.840 --> 00:17:44.429
somewhere here so, 12.5 will be somewhere
here so more than 20 somewhere here. So, roughly
00:17:44.429 --> 00:17:50.830
we are approximately we are saying this value
is 22 centimeter per hour.
00:17:50.830 --> 00:17:56.850
So, we are assuming rather from this nomograph
we are reading the value that corresponding
00:17:56.850 --> 00:18:03.549
to time of concentration, the intensity, the
peak intensity is 22 centimeter per hour and
00:18:03.549 --> 00:18:11.690
that is the value we will use in our CIA by
36 .
00:18:11.690 --> 00:18:15.149
So; that means, now we can use the runoff
rational formula and for the we we assume
00:18:15.149 --> 00:18:22.370
C equals to 0.3 so, now, CIA by 36. So, C
we assumed at 0.3 I we have just now approximated
00:18:22.370 --> 00:18:27.700
at 22 the area we have calculated as 0.9 hectares.
So, putting the value we get Q value equals
00:18:27.700 --> 00:18:34.809
to 0.165 cubic meter per second, this is this
should be read a cubic meter per second not
00:18:34.809 --> 00:18:38.590
the m 3.
So, it is a cubic meter per second or cumec
00:18:38.590 --> 00:18:42.129
which is the unit of discharge, because here
we are calculating the peak runoff rate. Now,
00:18:42.129 --> 00:18:45.660
once we have calculated that discharge
which our terrace would be able to handle
00:18:45.660 --> 00:18:50.960
or grade channel should be able to handle.
Then we have to come to defining the terrace
00:18:50.960 --> 00:18:54.289
cross section . So, for terrace cross section
we need to select the cut, front and back
00:18:54.289 --> 00:18:57.479
slopes . So, for this again we can use the
standard table, that is table 3 provided in
00:18:57.479 --> 00:19:03.000
lecture number 19 that is the US SCS recommendation
and from this table we we will just see the
00:19:03.000 --> 00:19:06.369
table, the table is here.
So, from this table the land slope is ours
00:19:06.369 --> 00:19:11.110
is 3 percent so, closer to this 3 percent.
And for this the channel slopes are recommended
00:19:11.110 --> 00:19:17.970
as 6 is to 1 that is cut slope this is ridge
fore slope or what we are calling as front
00:19:17.970 --> 00:19:21.940
slope 8 is to 1 and back slope is 8 is to
1.
00:19:21.940 --> 00:19:31.249
So, 6 is to 1 8 is to 1 8 is to 1 are our
recommended slopes from the table. So that
00:19:31.249 --> 00:19:39.039
means, we can come back here and we we see
that using a standard table we get S c is
00:19:39.039 --> 00:19:46.610
equals to 6 is to 6 S f equals to 8 is to
1 and S b equals to 8 is to 1. So, slopes
00:19:46.610 --> 00:19:51.960
we have take taken a decision regarding .
So, now we need to choose the channel shape
00:19:51.960 --> 00:19:59.029
and determine it is dimensions. So, let us
say that we consider a triangular channel.
00:19:59.029 --> 00:20:04.460
So, triangular channel means; we are saying
that the depth of channel is h that is the
00:20:04.460 --> 00:20:12.789
depth of M bounding or depth of cut channel
is h. Here in this case so; obviously, the
00:20:12.789 --> 00:20:18.989
total cross sectional area that is the area
that will be utilized for water flow .
00:20:18.989 --> 00:20:28.809
So, here; that means, the half is to 6 is
to 1 times h because, this is the channel
00:20:28.809 --> 00:20:40.109
here this is h, this slope is 6 is to 1, this
slope is 8 is to 1. So, that simply means
00:20:40.109 --> 00:20:48.230
that this width is 8 h and this width is 6
h and so, two triangles. So, this triangle
00:20:48.230 --> 00:20:57.870
is half into 6 h into h and the other triangle
is half into 8 h into h and this is how we
00:20:57.870 --> 00:21:01.049
can calculate the total cross sectional area
which comes out to be 7 h square.
00:21:01.049 --> 00:21:08.409
Now, we can assume we assume a simple value
of h equals to 0.3 that is h value we are
00:21:08.409 --> 00:21:18.149
assuming. So, the cross sectional area A is
one 1.27 square meters, we also require the
00:21:18.149 --> 00:21:23.090
wetted perimeter because we have to calculate
the hydraulic radius.
00:21:23.090 --> 00:21:28.250
So that means, length of this and length of
this sum of these two because, we know these
00:21:28.250 --> 00:21:33.299
two values we can always calculate using Pythagoras
theorem. So, here S 1 plus S 2 that is
00:21:33.299 --> 00:21:39.220
h square plus 6 h square h square plus 8 h
square, here it will be 8 h square and if
00:21:39.220 --> 00:21:45.299
we solve this it will come out to be 14.1
h for 14.15 h or the value of h we are assuming
00:21:45.299 --> 00:21:51.259
0.3 so, wetted perimeter is 4.24 meters. So,
hydraulic radius will be ratio of A and P
00:21:51.259 --> 00:21:57.210
A value we calculated as 1.27 P value we calculated
as 4.24. So, value of hydraulic radius is
00:21:57.210 --> 00:21:58.259
0.3 meters .
So, from there Manning's equation can be
00:21:58.259 --> 00:22:02.409
used for calculating the velocity which is
n by n R R to the power 2 by 3 S to the power
00:22:02.409 --> 00:22:07.249
half. So, n value we are assuming as 0.04,
R we have already calculated and the grade
00:22:07.249 --> 00:22:12.129
is already given so, 0.0025. Ah So, using
this the flow velocity we get is 0.56 meter
00:22:12.129 --> 00:22:19.769
per second which is non erosive velocity
by as per recommendations. And Q of the channel
00:22:19.769 --> 00:22:25.840
is A times V A we are know V we know so, it
comes out to be 0.71 cubic meter per second.
00:22:25.840 --> 00:22:27.090
So, design channel has sufficient capacity
because, the capacity required were 0.165
00:22:27.090 --> 00:22:28.090
cubic meters per second wherein we are calculating
0.71 cubic meter per second. So, now, we can
00:22:28.090 --> 00:22:31.249
say that the channel has sufficient capacity
so, assuming 20 percent freeboard that is
00:22:31.249 --> 00:22:37.330
0.2 h ridge height will be 1.2 h or 0.36 meters.
So, this way we have designed the various
00:22:37.330 --> 00:22:39.720
cross sections various parameters of the
channel.
00:22:39.720 --> 00:22:44.649
So, the answers are vertical interval is 0.9
meters, horizontal interval is 30, terrace
00:22:44.649 --> 00:22:50.330
grades of 0.4 0.3 0.4 0.3 and 0.2 for each
quarter length, terrace flow depth is 0.3,
00:22:50.330 --> 00:23:02.149
slope ratios are 6 is to 1 8 is to 1 8 is
to 1, terrace ridge height is 0.36 and channel
00:23:02.149 --> 00:23:14.340
flow velocity is 0.71, terrace channel
flow capacity is 0.71 cubic meter per second.
00:23:14.340 --> 00:23:16.789
So, these are the various dimensions we have
already designed .
00:23:16.789 --> 00:23:27.679
Now, we go to the third channel which is also
a design channel, design a 800 meter long
00:23:27.679 --> 00:23:36.179
level broad terrace, for a land having average
slope of 4 percent, maximum expected rainfall
00:23:36.179 --> 00:23:43.729
having 10 year recurrence interval is 18 centimeter.
Infilteration capacity of the soil of the
00:23:43.729 --> 00:23:51.210
area is such that 40 percent of the rainfall
is absorbed in the field and the region is
00:23:51.210 --> 00:23:53.149
sub humid and has clay loam.
So, the data given are length, length of level
00:23:53.149 --> 00:23:58.889
broad terrace is l is 800 meters, land slope
is 4 percent, maximum expected rainfall having
00:23:58.889 --> 00:24:04.659
10 years recurrence interval is 18 centimeters,
infilteration capacity is 40 percent of rainfall
00:24:04.659 --> 00:24:08.289
and it is a sub humid region with clay loam
soil.
00:24:08.289 --> 00:24:13.919
So, we need to estimate vertical interval
for the purpose we have to use the same equation
00:24:13.919 --> 00:24:27.219
because, it is a less humid region and normal
soil. So, VI equals S plus 6 by 10 so, VI
00:24:27.219 --> 00:24:35.440
comes out to be 1 meter because S is 4percent
and HI once VI is known HI comes out to be
00:24:35.440 --> 00:24:40.299
25 meters .
So, now we need to determine the maximum runoff
00:24:40.299 --> 00:24:44.429
volume and so maximum expected rainfall having
10 years recurrence interval is 18 centimeters.
00:24:44.429 --> 00:24:48.570
And infilteration capacity 40 percent of rainfall
that simply means the runoff potential is
00:24:48.570 --> 00:24:51.820
60 percent of rainfall because, 40 percent
is absorbed in infilteration .
00:24:51.820 --> 00:24:58.090
So, the level terrace system needs to be designed
to handle, the maximum runoff potential which
00:24:58.090 --> 00:25:03.210
is 60 percent of the rainfall. So, it is 60
percent of the rainfall; that means, 0.6 or
00:25:03.210 --> 00:25:05.519
0.18 or 0.108 meters that is the maximum runoff
to be stored .
00:25:05.519 --> 00:25:10.870
Now, the storage area which is the maximum
runoff depth times horizontal interval so,
00:25:10.870 --> 00:25:19.989
putting these values we get 2.7 square meters.
And here also we need to choose a channel
00:25:19.989 --> 00:25:25.169
shape and determine it is dimension . So,
let us consider a trapezoidal section with
00:25:25.169 --> 00:25:31.989
cut slope, front slope, back slope with 6
is to 1 8 is to 1 and 8 is to 1.
00:25:31.989 --> 00:25:37.739
So, these ratio we are assuming here and a
trapezoidal section we are considering so;
00:25:37.739 --> 00:25:42.619
that means, the when we calculate the flow
area that we have to consider three sections
00:25:42.619 --> 00:25:49.229
a 1 a 2 and a 3 because slope here is 6 is
to 1 here it 8 is to 1 .
00:25:49.229 --> 00:25:57.590
So, the cross sectional area so, three a 1
a 2 a 3 the first one is triangle, half 6
00:25:57.590 --> 00:26:06.499
6 h into h, the second one is a rectangle
b times h and third one is rectangle half
00:26:06.499 --> 00:26:10.840
8 h times h so, it is bh plus 7 h square.
Since storage area required is 2.7 square
00:26:10.840 --> 00:26:18.239
meters, the cross section area of the channel
should be equal to the required storage area.
00:26:18.239 --> 00:26:29.039
So, assuming b equals to 0.4 meters, let us
say that bottom width is 40 centimeters or
00:26:29.039 --> 00:26:40.379
0.4 meters. And if we put in the in this bh
plus 7 h square is 2.7, if you put in this
00:26:40.379 --> 00:26:48.059
equation b equals to 0.4 we get 7 h square
plus 0.4 h equal to 2.7 or we can write 7
00:26:48.059 --> 00:26:59.900
h square plus 0.4 h minus 2.7 equals to 0.
So, quadratic equation we can use solve
00:26:59.900 --> 00:27:09.309
using a straight formula which we know minus
b plus minus square root of b square minus
00:27:09.309 --> 00:27:21.879
4 ac by 2 a. So, by putting the values of
b b this is b this is a and this is c. So,
00:27:21.879 --> 00:27:29.950
putting these values here we get the value
of h as 0.65 or 0.59. So, we select eiher
00:27:29.950 --> 00:27:34.059
of the two so, we select h equals to 0.65
meters .
00:27:34.059 --> 00:27:40.710
So, assuming 20 percent freeboard that is
0.2 h, ridge height comes out to be 0.78 and
00:27:40.710 --> 00:27:47.249
terrace width 8.78 6.24 meters. So, terrace
the answers are vertical interval is 1 meter,
00:27:47.249 --> 00:27:54.200
horizontal interval is 25 meters, flow depth
is 0.56 meters, channel bottom width 0.4 meters,
00:27:54.200 --> 00:28:02.039
slopes ratios are 6 is to 1 8 is to 1 and
8 is to o1, channel ridge height is 0.78 meters
00:28:02.039 --> 00:28:09.450
and channel width is 6.24 meters. So, these
are this is all various parameters of this
00:28:09.450 --> 00:28:19.230
level broad base terrace .
So, now, we take the last problem and that
00:28:19.230 --> 00:28:28.320
says the compute the depth of cut and fill
and side slope of a broad base terrace for
00:28:28.320 --> 00:28:33.929
30 centimeter depth of flow on 8 percent land
slope. Consider that the tillage operations
00:28:33.929 --> 00:28:50.879
are to be performed using machines and the
width of machine is 4.27 meters. So, we have
00:28:50.879 --> 00:28:55.389
been given the depth of flow, land slope and
the width of the machine .
00:28:55.389 --> 00:29:01.820
So, that simply means the data given are land
slope S equals to 8 percent, channel depth
00:29:01.820 --> 00:29:20.769
h equal to 0.3 meters that is here h is given
here, this is h, this is h, this is the this
00:29:20.769 --> 00:29:29.960
is h it is given here and machinerty width
is given as 4.27 meters. So, because machinery
00:29:29.960 --> 00:29:39.799
width width of terrace is typically kept W
equal to machinery width so, W comes out to
00:29:39.799 --> 00:29:46.529
be 4.27 meters.
So, from first problem we saw this equation
00:29:46.529 --> 00:29:57.129
C plus f equals h plus S times W, now now
putting the known value of h S and W because,
00:29:57.129 --> 00:30:13.539
h S and W is known, C plus f we come comes
out to be 0.68 meters. So, this is what we
00:30:13.539 --> 00:30:24.950
get C plus f equal to 0.18 0.64 meters .
And since we normally use a balanced cross
00:30:24.950 --> 00:30:56.289
section, where cut is equal to fill so; that
means, C equal to f will be 0.32 meters. And
00:30:56.289 --> 00:31:14.429
then front slope because front slope we know
the width is width and
00:31:14.429 --> 00:32:04.490
the fall is h which is 0.3 so, it is 4.27
by 0.3that is 14.2. So, front slope ratio
00:32:04.490 --> 00:32:15.929
comes out to be 14.2 is to 1, wherein for
the back slope when we calculate back slope
00:32:15.929 --> 00:32:33.149
the elavation difference is h plus WS. So,
here
00:32:33.149 --> 00:33:18.049
the slope ratio will be W divided by h plus
WS, W is 4.27, h is 0.3, W is 4.27, S is 0.08.
00:33:18.049 --> 00:33:24.749
So that means, it comes out to be 6.67 so
back slope ratio comes out to be 6.67 is to
00:33:24.749 --> 00:33:32.679
1. So that means, the cut and fill comes out
to be 0.32 meters, the front slope ratio comes
00:33:32.679 --> 00:33:40.019
out to be 14.2 is to 1 and backslope ratio
comes out to be 6.67 is to 1 .
00:33:40.019 --> 00:33:51.669
So, with this
we have seen that we can design broad base
00:33:51.669 --> 00:34:10.820
terraces, be it graded terraces or be it be
it a level terrace. And also we can find
00:34:10.820 --> 00:34:23.300
what should be the cut and fill for a given
condition and what should be the ratios
00:34:23.300 --> 00:34:43.710
the slope ratios. So, these are the problems,
but as I have said earlier also that there
00:34:43.710 --> 00:34:57.569
can be other forms of the problem. So, if
you solve some more problems for practice
00:34:57.569 --> 00:35:26.369
probably things will be even clearer, but
I think that with this much practice itself
00:35:26.369 --> 00:35:38.640
you
00:35:38.640 --> 00:35:58.359
will be able to design broad base terraces.
Thank you very much.