hydrology and IRR

Question 1

Gumbel’s extreme value distribution applied on design flood discharge

Answer 1

Xt = Xm + (Yt -Yn) / Sn
Yt = -ln(ln(T / (T-1)))
Yn = 0.577
Sn = 1.2825

Question 2

Slope area method

Answer 2

1. find average conveyance K= (k1.k2)^.5
2. Hf = h1 -h2 - (velocity head2- velocity head1) *(1- contraction_loss_coeff)
3. V1.A1 = K* sqrt( Hf / L)
4. Q =V1 * A1

Question 3

different type of precepetation

Answer 3

-Rain, 0.5mm < d < 6mm, den>2.5mm/hr
-snowfall, combined flakes of ice crystals
average density = 0.1 g /cm^2
-drizzle, d < 0.5mm, density < 1mm/hr
-glaze, sheet on ground
-sleet, frozen rain drops at sub freezing-temp
-Hail, large lumps of size > 8mm

Question 4

first and last section average width in river discharge calculation when dept and distance to bank is given

Answer 4

W= (W1 +.5*W2)^2 / 2W1

Question 5

finding discharge without sediment movement

Answer 5

According to SHEILDS equation
-critical tractive stress at bottom 
tau.c = 0.056* γw *d* (S-1) 
=  γw.d/11      for S=2.5
-equate with applied stress
tau = γw *R*S
-find R

-critical tractive stess on side slopes
 Tauc' = α *tau.c
α = sqrt [ 1- (sinθ /sinΦ)^2 ] 
-equate with reduced applied stress
tau' = 0.75*γw *R*S

Question 6

canal outlets
non modular
semi modular
rigid moduler

Answer 6

non modular - discharge depends upon dH btween distributary and water course

semi modular- indipendent on water course level, but depends on distributary level

rigid moduler- constant discharge, irrespective of fluctuations either side

Question 7

flexibility of outlet

Answer 7

flexibility = dQ(of outlet) / dQ (of distributing channel)

Question 8

proportionality of outlet

Answer 8

when flexibility ==1

Question 9

senitivity of outlet

Answer 9

sensitivity = dQ / dH (in distributing channel)

Question 10

efficiency of outlet

Answer 10

ratio = head recovered / Head put in

Question 11

Setting of outlet

Answer 11

= depth below FSL / depth of FSL

Question 12

adjustability of outlet

Answer 12

capacity for modification to take into account changed conditions such as silting and scouring of distributing channel

Question 13

type of floods

Answer 13

standard project flood(SPF) = flood of severe conditions
maximum probable flood(MPF) = catastrophic flood
design flood = flood of desired re-occurrence interval
Maximum flood = peak flow obtained from data

Question 14

triangular hydrograph depth- discharge relation shortcut

Answer 14

0.18 T*Q = A*d
Q= cu.m
T = base period in hours
A = area Km^2
d = depth in cm

Question 15

Height of wave depending on Fetch and wind speed

Pressure force generated by wind

Answer 15

hw = 0.032 sqrt(VF) + 0.673 - 0.271 f^0.75

hw = 0.032 sqrt(VF) for F> 32 km

Pw = 2 (water unit weight) .hw^2

Question 16

Groynes length and spacing guideline

Answer 16

L < 0.2W
L > 1-2 times depth

Spacing
S= (2-2.5) times L for convex
S= L for concave

S for wide river > S for narrow
S for permeable groynes > impermeable

Question 17

Design of stilling basin

Answer 17

HL =total head loss
Yc =critical depth
Z= HL/Yc
Y=1+0.93556*Z^0.240 
{0.240 if Z>1,   0.386 if Z<1}

y2=Y*Yc
Basin invert level= water level - y2

Question 18

Disadvantages of drip irrigation

Answer 18

Disadvantages of drip irrigation
 Does not offer frost protection (as sprinklers do)
 Plastic drip-lines and sub-mains may be attacked by rodents and small animals.
 Requires regular flushing (to clear off the dirt collected near the ends of the drip lines) and supervision.
 High skill is required in the design, installation, operation and maintenance

Question 19

Factor of safety of dam overturning, sliding and shear

Answer 19

Foverturning >1.5
Fslide>1
Fshear>1.5

Question 20

True regime

Answer 20

Hence, an artificially constructed channel having a certain fixed section and a certain fixed slope can
behave in regime only if the following conditions are satisfied.
 Discharge is constant;
 Flow is uniform;
 Silt charge is constant (constant amount of silt )
 Silt grade is constant (the type and size of silt is always the same)
 Channel is flowing through a material which can be scoured as easily as it can be deposited and is of
the same grade as is transported.
Hence, a designed channel shall be in ‘true regime’ if the above conditions are satisfied. But in practice,
all these conditions can never be satisfied. And, therefore, artificial channels can never be in ‘true regime’;
they can either be in initial regime or final regime

Question 21

Initial Regime

Answer 21

 It is the first stage of regime attained by a channel after it is in service.
 If a channel is excavated with smaller width and flatter bed slope, then as the flow takes place in the
channel, bed slope of the channel is increased due to deposition of silt on the bed of the channel to
develop an increased flow velocity. Hence, the given discharge is allowed to flow through the channel of
smaller width.
 With increase in the bed slope , the depth may also vary but the width of the channel does’nt change
because the sides of the channel are usually cohesive and hence they resist erosion.
So, keeping the discharge, silt grade, silt charge and width fixed and only by varying bed slope
and depth, the channel attains stability. This condition is known as initial regime.

Question 22

Final Regime

Answer 22

It is the ultimate state of regime attained by a channel when in addition to bed slope and depth, the width
of the channel is also adjusted as per requirement.
 In this condition, the resistance of the sides of the channel is ultimately over come due to continuous
action of water.
 So, the channel adjust its width, depth and bedslope in order to obtain a stable channel. This condition
is known as final regime.
 Such a channel in which all variables are equally free to vary, has a tendency to assume a semi-elliptical
section. The coarser the silt, the flatter is the semi-ellipse, i.e. greater is the width of the water-surface.
The finer the silt, the more nearly the section attains a semicircle

Question 23

Following methods of river training are commonly used:

Answer 23

1. Embankments: The floods may be prevented from submerging the country by constructing earth embankments. They are designed and constructed in the same way as an earth dam. The embankments are generally constructed parallel to the river channel.
2. Guide banks or Bell’s bunds: Rivers in flood plains submerge very large areas during flood periods. Naturally when some structure is to be constructed across such a river, it is very expensive to construct the work spanning whole width of the river. To economise this, some training work may be constructed to confine the flow of water within a reasonable waterway.
   Guide banks are meant for guiding and confining the flow in a reasonable
   waterway at the site of the structure.
3. Spurs or Groynes: They are the structures constructed transverse to the river flow.
   They extend from the bank into the river.
   Groynes serve following purposes:
   (i) They protect the river banks by keeping the flow away from it.
4. Bed pitching and bank revetment: Sometimes to protect the bed and bank against action of water, protection is provided by laying a closely packed stone blocks or boulders or even concrete blocks. This permanent revetment and pitching counteracts the general tendency of the water to notch away the material from bed and banks.
5. Dredging of river: To improve navigability of the river channel, the river section may need to be excavated. This excavation is carried out to increase the depth of flow. The process of underwater excavation is termed as dredging. The machinery used for the purpose is called as dredger.

Question 24

Exit gradient by khosla theory

Answer 24

ic = (H/d) / (π.sqrt(λ))

λ= (1+ sqrt(1+α^2)) /2
α= B/d