Your Answer : (Not Answered)

Correct Answer : A

The Christiansen Uniformity Coefficient is 93.26%.

Christiansen's coefficient is

\[ CU = 100\left(1 - \frac{\sum|x_i - m|}{n\,m}\right) \]

where m is the mean depth. The 12 depths run evenly from 50 to 39 mm, so \(m = (50 + 39)/2 = 44.5\) mm. The absolute deviations are 5.5, 4.5, 3.5, 2.5, 1.5, 0.5 on each side, totalling \(2 \times 18 = 36\) mm:

\[ \begin{aligned} CU &= 100\left(1 - \frac{36}{12 \times 44.5}\right) \\ &= 100(1 - 0.0674) = 93.26\% \end{aligned} \]

91.66 and 89.70 do not follow from Christiansen's formula with these depths, and 36.28 is far too low for depths that vary by only about ±12% around the mean.

Your Answer : (Not Answered)

Correct Answer : D

Conjunctive use means the combined use of surface and ground water resources.

Conjunctive use plans the use of canal (surface) water and well (ground) water together, so that each makes up for the other's weaknesses. Groundwater can be pumped in dry years or at times when canal supplies fail, and surface water can recharge aquifers in wet years. It also helps control waterlogging in canal commands by pumping out the rising groundwater.

Combining irrigation with hydropower is multipurpose use, use by cooperative farmers is a form of management, and use in both kharif and rabi is a cropping matter. None is conjunctive use.

Your Answer : (Not Answered)

Correct Answer : C

The answer keyed is that sub-irrigation is useful where there is capillary movement in the root zone.

In sub-irrigation, water is applied below the surface, by raising the water table or through buried pipes, and reaches the roots by capillary rise. It works where the soil has good capillary movement into the root zone, a fairly permeable surface layer and an impervious layer below to hold the water table up.

Saline water is unsuitable, because water rising by capillarity and evaporating leaves salt at the surface. A heavy clay transmits water too slowly. And the absence of a hard pan is a drawback, since a hard pan below the root zone is what keeps the raised water table from draining away.

Your Answer : (Not Answered)

Correct Answer : B

The correct matching is 1 - c, 2 - d, 3 - a, 4 - e, 5 - b.

1. A tensiometer measures soil water tension, so it is used to monitor soil moisture.
2. A piezometer is an open tube showing the hydrostatic pressure head at a point.
3. A lysimeter is a soil-filled tank whose water balance gives evapotranspiration, the consumptive use of a crop.
4. An elbow meter measures the pressure difference across a pipe bend, from which the volumetric flow rate is found.
5. A Pitot tube converts velocity head to pressure, and is analysed with Bernoulli's equation.

Option A swaps the piezometer and Pitot tube, option C mismatches the first three instruments, and option D swaps the elbow meter and Pitot tube.

Your Answer : (Not Answered)

Correct Answer : C

The keyed answer is that overlap increases with an increase in the wind velocity.

Wind distorts and shortens the wetted pattern of a sprinkler, carrying spray downwind and leaving the upwind side dry. To keep the application uniform, the sprinklers are set closer together in windy conditions; design guides reduce spacing from about 65% of the wetted diameter in calm air to 50% or less in strong wind, which means more overlap.

Raising or lowering the sprinkler head changes the wetted diameter but is not the condition the overlap is designed for, and calm air allows the widest spacing, the least overlap.

Your Answer : (Not Answered)

Correct Answer : B

Trickle irrigation is also known as drip irrigation.

In drip (trickle) irrigation, water is delivered at low pressure through a network of pipes and emitters that release it drop by drop, or as a slow trickle, directly at the root zone of each plant. It gives water use efficiencies of about 90% and is widely used for orchards, vegetables and sugarcane.

Sprinkler irrigation sprays water over the crop like rain. Surge irrigation is a surface method in which water is released into furrows in intermittent surges. Neither is another name for trickle irrigation.

Your Answer : (Not Answered)

Correct Answer : A

The emitter is an orifice type.

Emitter discharge follows \(q = k P^{x}\), where k is the emitter constant and x the emitter exponent that identifies the flow regime. Solve for x:

\[ \begin{aligned} 4 &= 0.3 \times 110^{x} \\ 110^{x} &= 13.33 \\ x &= \frac{\ln 13.33}{\ln 110} = \frac{2.590}{4.700} \approx 0.55 \end{aligned} \]

An exponent close to 0.5 means fully turbulent flow, which is the behaviour of orifice emitters. Long-path emitters run partly laminar, with x of about 0.7 (1.0 for fully laminar flow). Pressure-compensating emitters are designed for x near 0, so that discharge hardly changes with pressure. In the standard classification by flow regime, an exponent of 0.5 points to the orifice class.

Your Answer : (Not Answered)

Correct Answer : D

The lateral spacing is 20.0 m.

A sprinkler's application rate is its discharge spread over the area it serves, sprinkler spacing times lateral spacing:

\[ \begin{aligned} q &= 1.3\ \text{L/s} = 4.68\ \text{m}^{3}/\text{h} \\ S_{l} S_{m} &= \frac{q}{I} = \frac{4.68}{0.0117} = 400\ \text{m}^{2} \\ S_{m} &= \frac{400}{20} = 20\ \text{m} \end{aligned} \]

The discharge given is already the actual sprinkler discharge, so the key does not apply the coefficient of discharge or the uniformity coefficient again. The distractors come from doing so: multiplying the area by 0.9 gives 18.0 m, by 0.8 gives 16.0 m, and by both (0.72) gives 14.4 m.

Your Answer : (Not Answered)

Correct Answer : A

In Lacey's theory the velocity does not depend on the depth of water as such.

Lacey's regime equations express the velocity through the silt factor f and the hydraulic mean depth R:

\[ \begin{aligned} V &= \sqrt{\tfrac{2}{5} f R} \\ V &= 10.8 R^{2/3} S^{1/3} \end{aligned} \]

The second, the regime flow equation, brings in the bed slope S. So hydraulic mean depth, slope and silt factor all enter the velocity. The depth of water itself does not appear: R is the area divided by the wetted perimeter, not the flow depth.

Your Answer : (Not Answered)

Correct Answer : A

The critical depth is about 1.09 m.

At critical flow \(Q^{2}/g = A^{3}/T\), where for this section \(A = (5 + y)y\) and the top width is \(T = 5 + 2y\). The left side is \(20^{2}/9.81 = 40.77\). Trying y = 1.09 m:

\[ \begin{aligned} A &= (6.09)(1.09) = 6.638\ \text{m}^{2} \\ T &= 7.18\ \text{m} \\ A^{3}/T &= 292.5/7.18 = 40.74 \end{aligned} \]

That matches 40.77, so y = 1.09 m. 1.18 m is the critical depth of a 5 m wide rectangular channel, \((q^{2}/g)^{1/3}\) with q = 4 m²/s, which ignores the extra area of the side slopes; at 1.18 m the trapezoidal expression is already about 52.6. 2.12 and 2.62 m are far deeper than either.

Your Answer : (Not Answered)

Correct Answer : B

The keyed pair is 20.26 m bottom width and 1.56 m depth.

Lacey's regime equations with silt factor f = 1 give the velocity, area and wetted perimeter:

\[ \begin{aligned} V &= \left(\frac{Qf^{2}}{140}\right)^{1/6} = 0.750\ \text{m/s} \\ A &= Q/V = 33.3\ \text{m}^{2} \\ P &= 4.75\sqrt{Q} = 23.75\ \text{m} \end{aligned} \]

For a side slope of ½ H : 1 V, \(P = b + \sqrt{5}\,d\) and \(A = bd + d^{2}/2\). Solving these two equations gives a depth of about 1.6 m and a bottom width of about 20.2 m, which matches the keyed pair to within the rounding of the regime constants.

The pair with 1.38 m depth gives an area of only about 29 m², too small for the regime velocity. 23.75 m is the whole wetted perimeter mistaken for the bottom width, and 32.78 m is wider still.

Your Answer : (Not Answered)

Correct Answer : C

In a semi-modular outlet the discharge depends upon the water level in the distributary only.

A semi-modular (flexible) outlet, such as a pipe outlet with free fall or an open flume outlet, is designed so that flow through it is free or supercritical. As long as the minimum working head is available, the water level in the watercourse below it has no effect, while the discharge still changes with the head in the distributary.

Discharge independent of both levels describes a modular (rigid) outlet. Discharge dependent on both levels describes a non-modular outlet, such as a submerged pipe. No common outlet depends on the watercourse level alone.

Your Answer : (Not Answered)

Correct Answer : B

A super-passage is a cross-drainage work in which the canal is taken below the natural stream: the drainage passes over the canal in a trough, so the canal flows beneath it. The opposite arrangement, where the canal is carried over the drainage, is an aqueduct. Crossing at nearly the same level is done by a level crossing, and diverting water from a river into a canal is the function of a headworks or a weir, not of a super-passage.

Your Answer : (Not Answered)

Correct Answer : C

A marginal bund is the river training work built along the banks on the upstream side of a diversion site (a weir or barrage) to prevent the land from being flooded when the high flood level rises because of the structure. Bed pitching protects the bed against scour, a spur is built out from a bank to deflect the current away from it, and the river proper is simply the natural river channel and not a training work.

Your Answer : (Not Answered)

Correct Answer : D

The main canal is the type classified by discharge and relative importance in the network. The main canal takes water from the river or reservoir and carries the largest discharge; it feeds branch canals, which feed distributaries and minor canals. Irrigation, navigation and power canals are classified by the purpose they serve (supplying crops, carrying boats, or feeding a hydropower plant), not by size and rank within a network.

Your Answer : (Not Answered)

Correct Answer : C

The incorrect statement is that waterlogging increases the capillary water available to plants. In a waterlogged soil the pores are filled with water and air is shut out, so roots cannot get oxygen, and plants do not benefit from the extra water. The other effects occur: the soil becomes colder, since the water has a high heat capacity and evaporates; salts are brought to the surface by capillary rise and deposited as water evaporates; and weeds and other wild plants grow in the wet soil, which cuts crop yield.

Your Answer : (Not Answered)

Correct Answer : C

The net depth that the canal has to supply is the crop requirement less the effective rainfall: 60 - 20 = 40 cm = 0.40 m over a base period of 20 days. The duty at the field for this crop is

With 20 percent losses on the way, only 80 percent of the water at the head reaches the field, so the duty at the head is \(432 \times 0.8 = 345.6\) ha/cumec. The intensity of irrigation and the command area are not needed for the duty, which is stated per cumec on the crop area.

Your Answer : (Not Answered)

Correct Answer : A

Duty on the field is the area irrigated by one cumec over the base period, and for a delta Δ and base period B it is \(D = 8.64\,B/\Delta\). Here B = 12 days and Δ = 0.40 m, so

Since 20 percent of the water is lost in the water course, only 80 percent of the water at the head reaches the field, so the duty at the head of the water course is lower: \(259.2 \times 0.8 = 207.36\) ha/cumec. The value 259.2 is the duty at the field without the loss. The area of 6 km\(^2\) is not needed for the duty.

Your Answer : (Not Answered)

Correct Answer : C

The Kor depth is the depth of water to be supplied during the Kor period, the first and most critical watering of a crop. The volume needed is the area times the depth: 2200 ha = 22 x 10\(^6\) m\(^2\), so the volume is \(22 \times 10^6 \times 0.14 = 3.08 \times 10^6\) m\(^3\). The Kor period of 20 days is 1.728 x 10\(^6\) s, so the discharge is

The options 1.56, 1.63 and 1.91 m\(^3\)/s would come from taking the depth or the period slightly wrong; they do not correspond to the data given.

Your Answer : (Not Answered)

Correct Answer : A

Duty D (ha/cumec), base period B (days) and delta Δ (m) are related by \(\Delta = 8.64\,B/D\). The constant 8.64 comes from converting 1 cumec flowing for one day (86,400 m\(^3\)) into hectare-metres. Therefore

The values 0.423, 0.542 and 0.611 m result from using the wrong base period or the wrong constant in the relation, for example forgetting that the duty is expressed for the whole base period of 60 days.