QUESTIONS
- Describe the various types of recording rain gages. When is the use of telemetry necessary?
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Recording rain gages can be of the following types: (1) tipping bucket,
(2) weighing mechanism, and (3) float chamber.
The tipping bucket features a two-compartment bucket pivoted on a knife's edge. The device
is calibrated so that when one of the compartments is full (with a fixed
amount of rain) and the other is empty, the bucket overbalances and tips.
A weighing gage has a device that weighs the rain or snow collected in a
bucket. As it fills with precipitation, the bucket moves downward and
its movement is transmitted to a pen on a strip-chart recorder. Float
gages are essentially water level gages.
A float located inside a chamber is connected to a pen on a strip-chart recorder. The float rises
as the collected rainwater enters the chamber, and the rise of the float
is recorded on the chart.
The application of telemetry in hydrologic data collection is usually
found in connection with operational hydrology and real-time flood
forecasting.
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- Explain the physical basis for radar measurements of precipitation. What radar wavelength should be used to sense heavy rains?
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The physical basis for radar measurement of precipitation is the
relationship between rainfall intensity and the scattering of
electromagnetic radiation. A radar emits electromagnetic radiation and
monitors the amount of radiation that is received after being subjected
to atmospheric scattering. A longer-wavelength radar (10-cm) should be
used to sense heavy rains.
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- What is the water equivalent of the snowpack? What is a snow course?
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The water equivalent is the depth of water that is obtained after melting
a certain depth of snowpack. Water equivalent is a measure of the amount
of water remaining in storage in the snowpack.
A snow course is
an established course or path for the purpose of performing snow
measurements, particularly water equivalent.
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- Explain the procedure to determine a catchment's water equivalent using a snow chart.
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A snow chart is used together with the catchment's hypsometric curve to
determine the catchment's water equivalent. The catchment's elevation
difference is divided into several equal increments. For each elevation
increment, a subarea is obtained from the hypsometric curve, and a
corresponding water equivalent is obtained from the snow chart. The
catchment water equivalent is obtained by weighing the individual
water equivalents in proportion to their respective subareas.
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- Why is the pan-evaporation measurement likely to be greater than the actual lake evaporation?
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The difference between pan evaporation and actual lake evaporation is
attributed to the pan installation and exposure. For example, a pan
installed on supports above the ground surface is subject to extra
radiation on the sides. On the other hand, a buried pan is subject to
appreciable heat exchange between it and the surrounding soil. These
and other environmental factors cause the pan evaporation measurement to
be different from the actual lake evaporation.
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- What is an evapotranspirometer? On what principle is it based? What is a lysimeter?
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An evapotranspirometer is an instrument or apparatus designed to measure
potential evapotranspiration.
It is based on the principle of conservation of mass (in this case, moisture). Water enters an
evapotranspirometer tank only from above, either as natural or artificial
precipitation, and it leaves it only through the bottom pipes to be
collected in measuring cans. During one time period, the difference
between the amount of water entering each soil tank and the amount of
water accumulated in the respective collecting can is the water lost to
evapotranspiration, provided the proper allowance is made for changes in
moisture storage in the soil tank.
A lysimeter is an instrument or apparatus designed to measure actual evapotranspiration.
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- Describe the two types of equipment to measure infiltration rates in the field.
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Two types of equipment are used to measure infiltration rates in the
field: (1) flooding type infiltrometer, and (2) sprinkler type
infiltrometer.
The flooding infiltrometer consists of two concentric metal rings that are inserted a distance of 2 to 5 an into the ground.
The rate at which water must be applied to the inner ring to maintain a
constant head of 0.5 an is taken as a measure of infiltration rate.
In the sprinkler infiltrometer, a simulated rainfall condition is applied
over a small plot by using sprinklers. The simulated rainfall is
continued for as long as necessary to attain an equilibrium runoff
condition at the plot outlet. Average infiltration rate is calculated as
the difference between the constant rainfall rate and the constant (i.e.,
equilibrium) runoff rate.
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- How does the catchment size affect the analysis of infiltration rates from rainfall runoff
data?
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For very large basins, say in excess of 20,000 km2, the time elapsed
between rainfall and runoff may be so great that it may be practically
Impossible to determine the amount of runoff produced by a storm event
within a reasonable length of time. Therefore, infiltration analysis
using rainfall-runoff data is limited to basins with negligible long-term
storage.
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- How is soil moisture determined by the water balance method?
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The water balance method is based on the assumption that soil moisture
can be represented as the difference between precipitation (input) and
evapotranspiration (output). Thus, moisture content can be evaluated
directly from readily available precipitation and evapotranspiration
data. When rainfall exceeds evapotranspiration, the soil moisture
increases; conversely, when evapotranspiration exceeds rainfall, the soil
moisture decreases.
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- What are the two properties of a good rating curve? Describe the two types of control in open channel flow. How does control affect the rating? Explain.
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The two properties of a good rating are its stability and permanence. A
stable rating remains constant in time, i.e., the effects of flow
nonuniformity, unsteadiness, or erosion and sedimentation are negligible.
A permanent rating is one that is not likely to be disturbed by human
activities.
The two types of control in open channel flow are: (1)
section control and (2) channel control.
A control is characterized by a single-valued rating, i.e., a one-to-one correspondence between discharge
and stage. Section control forces the rating to be single-valued (at or
near critical flow) at a given cross section. Channel control typically
occurs in a long and relatively prismatic channel, far from features that
can cause backwater. In this case, the flow will readily reach
equilibrium, with a single-valued discharge-stage rating. The absence or
loss of control renders a rating unstable, with the loss of the
one-to-one correspondence between discharge and stage.
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- Describe two types of recording gages to measure stage. When is a telemetric gage needed?
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A recording gage to measure stage can be of two types: (1) float-actuated or (2)
pressure-actuated. In the float-actuated recorder, a pen recording the
water level on a strip chart is actuated by a float on the surface of the
water. The recorder and float are housed in a suitable enclosure on top
of a stilling well connected to the stream by two intake pipes. The
pressure-actuated recorder eliminates the need for the stilling well.
The sensing element of the recorder is a diaphragm, which is submerged in
the stream. The changing water level produces a change in pressure on
the diaphragm, which is transmitted to the recorder.
A telemetric gage is ideally suited for applications where speed of
processing is of the utmost importance, e.g., for operational hydrology
or real-time flood forecasting.
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- Describe the various means to carry out stream velocity measurements using current meters.
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Techniques for measuring stream velocity with a current meter vary with
stream size. If the stream is wadable, the meter is affixed to a
graduated depth rod. If the stream is too deep to wade, the meter is
suspended on a cable and is held in the water with a sounding weight.
Measurements using cable suspension are made from bridges, cableways, or
boats. For the heavier sounding weights or when using a boat, a sounding
reel may be required.
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- When are chemical and physical methods to measure stream velocity applicable? What is the crucial assumption in the dilution method to measure stream velocity?
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Chemical and physical methods to measure stream velocity are applicable
when it is impractical to use current meters. Such is the case of very
shallow streams, very large rivers, or tidal flaw.
The crucial assumption in the dilution method to measure stream velocity
is the assumption of complete mixing of the substance at the downstream
sampling point.
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- What is the slope-area method? When is it used? What is the recommended minimum fall to preserve accuracy?
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The slope-area method is an approximate method to estimate flood-peak
discharge. It is based on a steady nonuniform flow evaluation of the
carrying capacity of a stream, using high water marks to estimate the
water surface slope. It is used in situations where it is impractical or
impossible to carry out stream gaging, particularly during and
immediately following a large flood event. The recommended minimum fall
to preserve accuracy is 0.15 m.
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PROBLEMS
A snow sample 20 cm high melted into 3 cm of water. What was the density of the snow sample?
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The density of the snow sample (in percentage) is:
Snow density = ( 3 cm / 20 cm ) ×100 = 15 percent. ANSWER.
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- What is the water equivalent of a snow accumulation measuring 9 in. with a density of 8%?
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The water equivalent W of the snow accumulation is:
W = 9 in. × (8 / 100) = 0.72 in. ANSWER.
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- The following elevation-area-snow-water-equivalents have been measured in a certain catchment:
| Elevation (m) |
2000 |
2500 |
3000 |
3500 |
4000 |
| Cumulative area (km2) |
0 |
255 |
432 |
519 |
605 |
| Snow-water equivalent (mm) |
0 |
0 |
8 |
22 |
30 |
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Determine the catchment's overall snow-water equivalent.
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| Elevation (m) |
2000 |
2500 |
3000 |
3500 |
4000 |
| Cumulative area (km2) |
0 |
255 |
432 |
519 |
605 |
| Snow-water equivalent (mm) |
0 |
0 |
8 |
22 |
30 |
| Incremental Area (km2) |
|
255 |
177 |
87 |
86 |
Average water equivalent
per elevation increment (mm) |
|
0 |
4 |
15 |
26 |
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The catchment's overall snow-water equivalent W is the average water
equivalent weighted in terms of the incremental area:
W = [(0 × 255) +
(4 × 177) + (15 × 87) + (26 × 86)] / 605 = 7.02 mm. ANSWER.
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- The following snow-chart and hypsometric data have been measured in a certain catchment:
| Elevation (m) |
0 |
10 |
20 |
30 |
40 |
50 |
60 |
70 |
80 |
90 |
100 |
| Cumulative area (km2) |
0 |
22 |
39 |
54 |
64 |
76 |
81 |
88 |
94 |
97 |
100 |
| Snow-water equivalent (mm) |
3 |
3 |
4 |
4 |
5 |
5 |
5 |
7 |
7 |
8 |
8 |
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Determine the catchment's overall snow-water equivalent.
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| Elevation (m) |
0 |
10 |
20 |
30 |
40 |
50 |
60 |
70 |
80 |
90 |
100 |
| Cumulative area (km2) |
0 |
22 |
39 |
54 |
64 |
76 |
81 |
88 |
94 |
97 |
100 |
| Snow-water equivalent (mm) |
3 |
3 |
4 |
4 |
5 |
5 |
5 |
7 |
7 |
8 |
8 |
Incremental area (percent) |
- |
22 |
17 |
15 |
10 |
12 |
5 |
7 |
6 |
3 |
3 |
Average water equivalent
per elevation increment (mm) (mm) |
- |
3.0 |
3.5 |
4.0 |
4.5 |
5.0 |
5.0 |
6.0 |
7.0 |
7.5 |
8.0 |
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The catchment's overall snow-water equivalent W is the average water
equivalent weighted in terms of the incremental area:
W = [(3.0 × 22) +
(3.5 × 17) + (4.0 × 15) + (4.5 × 10) + (5.0 × 12) + (5.0 × 5) + (6.0 ×
7) + (7.0 × 6) + (7.5 × 3) + (8.0 × 3)] / 100 = 4.46 mm. ANSWER.
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- Given the following stream gaging data, calculate the dischare.
| Vertical no. |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
| Distance to reference point (m) |
15 |
20 |
25 |
30 |
35 |
40 |
45 |
50 |
55 |
60 |
65 |
| Sounding depth (m) |
0.0 |
0.5 |
0.8 |
1.2 |
1.5 |
2.5 |
3.0 |
2.0 |
1.2 |
0.8 |
0.0 |
| Velocity at 0.2 depth (m/s) |
0.0 |
0.5 |
0.7 |
0.9 |
1.2 |
1.4 |
1.7 |
1.3 |
0.9 |
0.7 |
0.0 |
| Velocity at 0.8 depth (m/s) |
0.0 |
0.4 |
0.6 |
0.7 |
0.8 |
1.1 |
1.3 |
1.0 |
0.7 |
0.6 |
0.0 |
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| Vertical no. |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
| Width (m) |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
5 |
| Depth (m) |
0.0 |
0.5 |
0.8 |
1.2 |
1.5 |
2.5 |
3.0 |
2.0 |
1.2 |
0.8 |
0.0 |
| Average velocity (m/s) |
0.00 |
0.45 |
0.65 |
0.80 |
1.00 |
1.25 |
1.50 |
1.15 |
0.80 |
0.65 |
0.00 |
| Unit discharge (m3/s) |
0.00 |
1.12 |
2.60 |
4.80 |
7.50 |
15.62 |
22.50 |
11.50 |
4.80 |
2.60 |
0.00 |
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The discharge is the sum of the unit discharges: Q = 73.04 m3/s. ANSWER.
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A certain substance is introduced at point A of a stream at the rate of 50 L/s with a concentration of
12,000 ppm. At a downstream point B, after complete mixing, the concentration of the substance is measured to be 15 ppm. Calculate the stream discharge.
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Using Eq. 3-6:
Q = [(12,000/15) - 1] × 50 L/s / (1000 L/m3) = 39.95 m3/s. ANSWER.
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Calculate the flood discharge of a certain stream by the slope-area method, given the following data: upstream
flow area Au = 402 m2, upstream wetted perimeter Pu = 98 m,
upstream αu = 1.11, downstream flow area Ad = 453 m2,
downstream wetted perimeter Pd = 105 m, downstream αd = 1.13,
fall F = 0.5 m, reach length L = 870 m, and reach Manning n = 0.04.
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The upstream conveyance (Eq. 3-7a) is 25,753.
The downstream conveyance(Eq. 3 -7b) is 30,013.
The reach conveyance (Eq. 3-8) is 27,802.
The first approximation to the energy slope (Eq. 3-9) is 0.0005747.
The first approximation to the peak discharge (Eq. 3-10) is 666.49 m3/s.
Since the reach is expanding, the energy loss coefficient k = 0.5. The
result of the iteration is summarized below.
| Iteration No. |
hvu
(m) |
hvd
(m) |
Energy slope
(m/m) |
Peak discharge
(m3/s) |
| 1 |
0.156 |
0.125 |
0.0005925 |
676.74 |
| 2 |
0.160 |
0.129 |
0.0005925 |
676.74 |
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The flood discharge is: Q = 677 m3/s. ANSWER.
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Calculate the flood discharge of a certain stream by the slope-area method, given the following data: upstream
flow area Au = 3522 m2, upstream wetted perimeter Pu = 650 m,
upstream αu = 1.17, downstream flow area Ad = 3259 m2,
downstream wetted perimeter Pd = 621 m, downstream αd = 1.21,
fall F = 0.35 m, reach length L = 1,250 m, and reach Manning n = 0.028
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The upstream conveyance (Eq. 3-7a) is 388,045.
The downstream conveyance (Eq. 3 -7b) is 351,497.
The reach conveyance (Eq. 3-8) is 369,319.
The first approximation to the energy slope (Eq. 3-9) is 0.00028.
The first approximation to the peak discharge (Eq. 3-10) is 6180 m3/s.
Since the reach is contracting, the energy loss coefficient k = 1. The
result of the iteration is summarized below.
| Iteration No. |
hvu
(m) |
hvd
(m) |
Energy slope
(m/m) |
Peak discharge
(m3/s) |
| 1 |
0.184 |
0.222 |
0.0002496 |
5835.0 |
| 2 |
0.164 |
0.198 |
0.0002528 |
5872.0 |
| 3 |
0.166 |
0.200 |
0.0002528 |
5872.0 |
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The flood discharge is: Q = 5872 m3/s. ANSWER.
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150902 10:15 |
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