All questions of Engineering Hydrology for Civil Engineering (CE) Exam

The percentage of total quantity of water in the world that is saline is about
  • a)
    71%
  • b)
    33%
  • c)
    67%
  • d)
    97%
Correct answer is option 'D'. Can you explain this answer?

Anand Mehta answered
Explanation:

The world's total quantity of water is about 1.386 billion cubic kilometers. Out of this, only about 2.5% is freshwater, and the remaining 97.5% is saline water.

The percentage of total quantity of water in the world that is saline can be calculated as follows:

Saline water quantity = 97.5% of 1.386 billion cubic kilometers
= 1.3515 billion cubic kilometers

Total water quantity = 1.386 billion cubic kilometers

Percentage of saline water = (Saline water quantity / Total water quantity) x 100
= (1.3515 / 1.386) x 100
= 97.48%

Therefore, the percentage of total quantity of water in the world that is saline is about 97%.

Conclusion:

The world's water resources are mainly composed of saline water, which is not suitable for human consumption or agricultural purposes. The limited freshwater resources need to be conserved and managed sustainably to meet the growing demand for water worldwide.

The mass curve of rainfall of a storm is a plot of
  • a)
    rainfall depths for various equal durations plotted in decreasing order
  • b)
    rainfall intensity vs time in chronological order
  • c)
    accumulated rainfall intensity vs time
  • d)
    accumulated precipitation vs time in chronological order
Correct answer is option 'D'. Can you explain this answer?

Avinash Mehta answered
The mass curve of rainfall of a storm is a plot of accumulated precipitation against time in chronological order. It shows how the total rainfall changes over time during a storm. It is useful for understanding how much and how quickly rainfall will accumulate during a storm, which can help with flood forecasting and stormwater management. The mass curve can also help to identify how much rainfall is concentrated in a short period of time ( peak rainfall rate) and how much rainfall is spread over the duration of the storm.

Orographic precipitation occurs due to air masses being lifted to higher altitudes by
  • a)
    the density difference of air masses
  • b)
    a frontal action
  • c)
    the presence of mountain barriers
  • d)
    extratropical cyclones
Correct answer is option 'C'. Can you explain this answer?

Lavanya Menon answered
The moist air masses may get lifted-up to higher altitudes due to the presence of mountain barriers and consequently undergo cooling, condensation and precipitation, such a precipitation is known as orographic precipitation.

The shape of the recession limb of a hydrograph depends on
  • a)
    basin as well as storm characteristics
  • b)
    storm characteristics only
  • c)
    basin characteristics only
  • d)
    base flow only
Correct answer is option 'C'. Can you explain this answer?

Zoya Sharma answered
The climatic factors control the rising iimb and the recession limb is independent of storm characteristics, being determined by catchment characteristics only.

The flow-mass curve is an integral curve of
  • a)
    the hydrograph
  • b)
    the hyetograph
  • c)
    the flow duration curve
  • d)
    the S-curve
Correct answer is option 'A'. Can you explain this answer?

Maulik Joshi answered
The flow mass curve is a plot of the cumulative discharge volume against time plotted in chronological order. The ordinate of the mass curve, V at any time t is,

Since the hydrograph is a plot of Q vs t, it is easy that the flow mass curve is an integral curve of the hydrograph.

The Muskingum method of flood routing assumes the storage S is related to inflow rate I and outflow rate Q of a reach as S =
  • a)
    K[x l - (1 - x)Q]
  • b)
    K[x Q + (1 - x) I]
  • c)
    K[xI + (1 - x)Q]
  • d)
    Kx [l - (1 - x)Q]
Correct answer is option 'C'. Can you explain this answer?

For a given channel reach by selecting a routing interval Δt and using the Muskingum equation, the change in storage is,
S2 - S1 = k(xI2 - I1) + 1 - x(Q2 - Q1)
Where suffixes 1 and 2 refer to the conditions before and after the time interval Δt.

Interception losses
  • a)
    include evaporation, through flow and stemflow
  • b)
    consists of only evaporation loss
  • c)
    includes evaporation and transpiration losses
  • d)
    consists of only stemflow
Correct answer is option 'B'. Can you explain this answer?

Ameya Roy answered
A part of rain may be caught by vegetation and subsequently evaporated. The volume of water so caught is called interception. Interception loss is solely due to evaporation and does not include transpiration, through fall or stemflow.

The probable maximum flood is
  • a)
    the standard project flood of an extremely large river
  • b)
    a flood adopted in the design of all kinds of spillways
  • c)
    a flood adopted in all hydraulic structures
  • d)
    an extremely large but physically possible flood in the region
Correct answer is option 'D'. Can you explain this answer?

The Probable Maximum Precipitation (PMP) may be defined as the maximum depth of precipitation for a given duration that may possibly occur on a given catchment at any time of the year. Hence such a precipitation would result from the possible severest storm that may result from the worst possible combinations of hydrological conditions in the area.
Probable Maximum Flood (PMF) is the extreme flood that is physically possible in a region as a result of severe most combinations, including rare combinations of meteorological and hydrological factors. The PMF is used in situations where the failure of the structure would result in loss of life and catastrophic damage and as such complete security from potential floods is sought.

Thermohygrograph gives a continuous recording of
  • a)
    discharge on a thermal sensitive paper
  • b)
    temperature and pressure
  • c)
    temperature and humidity
  • d)
    solar radiation and wind velocity
Correct answer is option 'C'. Can you explain this answer?

Gauri Roy answered
Thermohygrograph is a device that records continuous data of temperature and humidity. It is commonly used in industries and laboratories where temperature and humidity control is crucial. Here is a detailed explanation of the answer:

Temperature and Humidity Recording:
Thermohygrograph gives a continuous recording of temperature and humidity. It has two sensors, one for temperature and the other for humidity. These sensors record the respective values and plot them on a chart over a period of time. The chart is usually a circular disc that rotates once a day or a week, depending on the model.

Working Principle:
The working principle of a thermohygrograph is based on the expansion and contraction of a special type of paper that is sensitive to temperature and humidity. The paper is coated with a chemical that changes color with the change in temperature and humidity. The paper is wrapped around the rotating disc, and the sensors record the values and plot them on the paper.

Applications:
Thermohygrograph is widely used in industries that require controlled temperature and humidity, such as pharmaceuticals, food processing, and storage facilities. It is also used in laboratories to monitor the temperature and humidity during experiments. It is an essential tool for climate studies and weather forecasting.

Advantages:
Thermohygrograph provides a continuous recording of temperature and humidity, which is useful for identifying any fluctuations or variations in the environment. It helps in maintaining the quality and safety of products that are sensitive to temperature and humidity. It also helps in predicting weather patterns and climate changes.

Conclusion:
In conclusion, Thermohygrograph is a device that records continuous data of temperature and humidity. It is widely used in industries and laboratories for monitoring and controlling the environment. It provides accurate and reliable data that helps in maintaining the quality and safety of products.

The percentage of total quantity of fresh water in the world available in the liquid form is about .
  • a)
    30%
  • b)
    70%
  • c)
    11%
  • d)
    51%
Correct answer is option 'A'. Can you explain this answer?

About 71 percent of the Earth's surface is water-covered, and the oceans hold about 96.5 percent of all Earth's water. Also, % of total water available in liquid form out of total quantity of freshwater is 30% and rest is in form of ice and glaciers.

The Indian Meteorological department has changed over from Symon’s, gauge to fibreglass raingauges of two sizes. The collector areas of these gauges are:
  • a)
    (1000 cm2 and 500 cm2)
  • b)
    (400 cm2 and 20 cm2)
  • c)
    (200 cm2 and 100 cm2)
  • d)
    (100 cm2 and 50 cm2)
Correct answer is option 'C'. Can you explain this answer?

The Indian Meteorological department (IMD) has changed over to the use of fibreglass reinforced polyester raingauges, which is an improvement over the system gauge. These come in different combinations of collector and bottle. The collector in two sizes having areas of 200 and 100 cm2 respectively.

The term ‘mean annual flood’ denote
  • a)
    mean of floods in partial-duration series
  • b)
    mean annual flow
  • c)
    a flood with a recurrence interval of 2.33 years
  • d)
    a flood with a recurrence interval of N/2 years, where N = number of years of record
Correct answer is option 'C'. Can you explain this answer?

Arnab Saini answered
Gumbel’s distribution has the property which gives T = 2,33 years for the average of the annual series when N is very large. Thus the value of a flood with 7 = 2.33 years is called the mean annual flood.

The following is not a direct stream flow determination technique:
  • a)
    Dilution method
  • b)
    Ultrasonic method
  • c)
    Area-velocity method
  • d)
    Slope-area method
Correct answer is option 'D'. Can you explain this answer?

Aniket Mehta answered
Explanation:

The correct answer is option 'D' - Slope-area method. Let's understand why the slope-area method is not a direct stream flow determination technique.

Slope-Area Method:
The slope-area method is a commonly used indirect method for determining stream flow in open channels. It involves measuring the cross-sectional area of the channel and the slope of the water surface. By applying the principles of conservation of mass and energy, the flow rate can be calculated. However, it is not considered a direct stream flow determination technique because it relies on measurements and calculations rather than direct measurement of the flow.

Direct Stream Flow Determination Techniques:
Direct stream flow determination techniques involve directly measuring the flow rate of water in a stream or channel. These techniques provide more accurate and precise measurements compared to indirect methods. Some of the commonly used direct stream flow determination techniques include:

1. Dilution Method:
The dilution method involves adding a known quantity of a tracer (such as a dye or salt) to the stream and measuring the change in concentration downstream. By knowing the initial concentration of the tracer and the rate of dilution, the flow rate can be calculated.

2. Ultrasonic Method:
The ultrasonic method utilizes ultrasonic sensors to measure the velocity of the water flow. By combining the velocity measurements with cross-sectional area measurements, the flow rate can be determined.

3. Area-Velocity Method:
The area-velocity method involves measuring the cross-sectional area of the channel and the velocity of the water flow. By multiplying the area and velocity, the flow rate can be calculated.

These direct stream flow determination techniques provide accurate and reliable measurements, making them suitable for various applications in water resources engineering, hydrology, and environmental studies.

Conclusion:
The slope-area method is not considered a direct stream flow determination technique because it relies on measurements and calculations rather than direct measurement of the flow. The dilution method, ultrasonic method, and area-velocity method are examples of direct stream flow determination techniques that provide more accurate and precise measurements.

When specific information about the density of snowfall is not available, the water equivalent -of snowfall is taken as
  • a)
    50%
  • b)
    30%
  • c)
    10%
  • d)
    90%
Correct answer is option 'C'. Can you explain this answer?

Vandana Desai answered
-
The 
density
 of water is 1000 kg/m^3 and 
snow density
 is usually measured as a ratio to this. So 
snow
 which is 100 kg/m^3 is specified as 100/1000, or 10 percent (of the 
density
 of water). The water content divided by the 
snow
 depth also gives the 
density
 of the 
snow
.
-
When specific information about the density of snowfall is not available, the water equivalent of snowfall is taken as 10%

In the hydrological cycle the average residence time of water in the global
  • a)
    atmospheric moisture is larger than that in the global rivers
  • b)
    oceans is smaller than that of the global groundwater
  • c)
    rivers is larger than that of the global groundwater
  • d)
    oceans is larger than that of the global groundwater
Correct answer is option 'D'. Can you explain this answer?

Pallabi Tiwari answered
Explanation:
The hydrological cycle is the continuous movement of water on, above, and below the surface of the Earth. It involves the processes of evaporation, transpiration, precipitation, infiltration, and runoff. Water moves through different reservoirs such as the atmosphere, oceans, rivers, lakes, groundwater, and glaciers. The residence time of water refers to the average time that a water molecule spends in a particular reservoir.

Residence time of water in oceans:
- The oceans are the largest reservoir of water on the planet, covering about 70% of the Earth's surface.
- The average residence time of water in the oceans is about 3,000 years.
- This means that a water molecule in the ocean will stay there for an average of 3,000 years before it evaporates into the atmosphere or is transported to another reservoir.

Residence time of water in groundwater:
- Groundwater is the water that is stored in the pores and fractures of rocks beneath the Earth's surface.
- The average residence time of water in groundwater is about 1,400 years.
- This means that a water molecule in groundwater will stay there for an average of 1,400 years before it is discharged into a stream or river, or extracted by a well.

Residence time of water in rivers:
- Rivers are the channels through which water flows from one place to another, eventually reaching the oceans.
- The average residence time of water in rivers is about 16 days.
- This means that a water molecule in a river will stay there for an average of 16 days before it is discharged into the ocean.

Residence time of water in atmospheric moisture:
- Atmospheric moisture refers to the water vapor in the Earth's atmosphere.
- The average residence time of water in atmospheric moisture is about 9 days.
- This means that a water molecule in atmospheric moisture will stay there for an average of 9 days before it either condenses into clouds and precipitates, or is transported to another location by winds.

Conclusion:
From the above explanation, it is clear that the average residence time of water in oceans is larger than that of the global groundwater. Hence, option D is the correct answer.

The standard Symons' type raingauge has a collecting area of diameter
  • a)
    12.7 cm
  • b)
    10 cm
  • c)
    5.08 cm
  • d)
    25.4 cm
Correct answer is option 'A'. Can you explain this answer?

Collecting Area of Standard Symons Type Raingauge

The standard Symons type raingauge is an instrument used to measure rainfall. It has a cylindrical container with a funnel-shaped top that collects rainwater. The collected water is then measured to determine the amount of rainfall. The collecting area of the standard Symons type raingauge is given as:

a) 12.7 cm

Explanation:

The collecting area of a raingauge is the surface area of the funnel-shaped top that collects rainwater. The standard Symons type raingauge has a diameter of 10 inches (25.4 cm) and a height of 10 inches (25.4 cm). The funnel-shaped top of the raingauge has a diameter of 8 inches (20.32 cm) at the top and a diameter of 5 inches (12.7 cm) at the bottom. Therefore, the collecting area of the raingauge is the area of the circular base with a diameter of 5 inches (12.7 cm).

Hence, the correct answer is option A, i.e., 12.7 cm.

The standard project flood is
  • a)
    smaller than the probable maximum flood
  • b)
    the same as the probable maximum flood
  • c)
    the same as the design flood
  • d)
    larger than the probable maximum flood by a factor implying safety factor
Correct answer is option 'A'. Can you explain this answer?

The flood that would result from a sever combination of meteorological and hydrological factors that are reasonably applicable to the region. Extremely rare combinations of factors are excluded.

If a 4-hour unit hydrograph of a certain basin has a peak ordinate of 80 m3/s, the peak ordinate of a 2-hour unit hydrograph for the same basin will be
  • a)
    equal to 80 m3/s
  • b)
    greater than 80 m3/s
  • c)
    less than 80 m3/s
  • d)
    between 40 m3/s to 80 m3/s
Correct answer is option 'B'. Can you explain this answer?

With the reduction of unit hydrograph duration, 1 cm excess rainfall will occur in reduced period. So peak ordinate of UH will increase and time base will decrease. Thus the peak ordinate of 2 hr UH will be greater than 80 m3/s.

A hydraulic structure has been designed for a 50 year flood. The probability that exactly one flood of the design capacity will occur in the 75 year life of the structure is
  • a)
    0.02
  • b)
    0.220
  • c)
    0.336
  • d)
    0.780
Correct answer is option 'C'. Can you explain this answer?

Ashwin Gupta answered
Solution:

Given: Probability of design capacity flood in 50 years = 1/50

We need to find the probability of exactly one flood of design capacity in the 75-year life of the structure.

To solve the problem, we can use the Poisson distribution formula, which gives the probability of a certain number of events occurring in a fixed interval of time, given the average rate of occurrence.

The formula is:

P(X=k) = (e^(-λ) * λ^k) / k!

where X is the random variable representing the number of events, k is the number of events we want to find the probability for, λ is the average rate of occurrence, and e is the base of the natural logarithm.

Let's calculate the value of λ first:

The expected number of design capacity floods in 75 years is:

λ = (75 / 50) * 1 = 1.5

Now, let's use the Poisson distribution formula to find the probability of exactly one flood of design capacity in 75 years:

P(X=1) = (e^(-1.5) * 1.5^1) / 1! = 0.336

Therefore, the answer is (c) 0.336, which is option (c).

Consider the following chemical emulsions:
1. Methyl alcohol
2. Cetyl alcohol
3. Stearyl alcohol
4. Kerosene
Which of the above chemical emuisions is/are used to minimize the loss of water through the process of evaporation?
  • a)
    1 only
  • b)
    1 and 4
  • c)
    2 and 4
  • d)
    2 and 3
Correct answer is option 'D'. Can you explain this answer?

Shilpa Pillai answered
Certain chemicals such as cetyl alcohol (hexadecanol) and stearyl alcohol (octadecanol) form monomoiecular layers on a water surface. These layers, act as evaporation inhibitors by preventing .the water molecules to escape past them.

The % of earth covered by oceans is about
  • a)
    31%
  • b)
    51%
  • c)
    71% '
  • d)
    97%
Correct answer is option 'C'. Can you explain this answer?

The % of earth covered by oceans is about 71%.

Explanation:

Coverage of the Earth's Surface

The Earth's surface is covered by different types of landmasses such as mountains, valleys, deserts, forests, and water bodies such as oceans, seas, lakes, and rivers.

Percentage of the Earth's Surface Covered by Oceans

The oceans are the largest water bodies on Earth and cover about 71% of the Earth's surface. The remaining 29% of the Earth's surface is covered by landmasses such as continents and islands.

Importance of Oceans

Oceans are important for various reasons such as:

- They regulate the Earth's climate and weather patterns
- They provide a habitat for diverse marine life
- They serve as a source of food and livelihood for millions of people
- They facilitate trade and transportation
- They offer recreational activities such as swimming, surfing, and diving.

Conclusion

In summary, the % of Earth covered by oceans is about 71%, making them a crucial part of the Earth's ecosystem and human activity.

Under identical conditions evaporation from sea water is
  • a)
    about 2-3% less than that from fresh water
  • b)
    the same as that from fresh water
  • c)
    about 2-3% more than that from fresh water
  • d)
    8-9% more than that from fresh water
Correct answer is option 'A'. Can you explain this answer?

Avinash Mehta answered
Under identical conditions, evaporation from seawater is typically about 2-3% less than evaporation from fresh water. This is because the salt in the seawater raises the boiling point of the water, making it more difficult for the water to turn into vapor. Additionally, the surface tension of seawater is higher than that of freshwater, which also reduces evaporation.

An instantaneous unit hydrograph is a direct runoff hydrograph
  • a)
    of 1 cm magnitude due to a rainfall excess of 1-h duration
  • b)
    that occurs instantaneously due to a unit rainfall excess of duration-D h
  • c)
    of unit rainfall excess precipitating instantaneously over the catchment
  • d)
    occurring at any instant in a long storm
Correct answer is option 'C'. Can you explain this answer?

Explanation:

An instantaneous unit hydrograph is a hydrograph that represents the direct runoff response of a catchment to a unit rainfall excess that occurs instantaneously over the catchment. It is a fundamental concept in hydrology and is widely used in the analysis and design of stormwater management systems.

Key Points:

- Direct Runoff: An instantaneous unit hydrograph represents the direct runoff component of the total streamflow. It does not include the baseflow component, which is the groundwater contribution to streamflow.

- Unit Rainfall Excess: The unit rainfall excess refers to a rainfall excess of a unit magnitude, such as 1 cm or 1 inch. It is the portion of rainfall that exceeds the infiltration capacity of the catchment and directly contributes to streamflow.

- Instantaneous Precipitation: The unit rainfall excess is assumed to occur instantaneously over the catchment. This means that the entire unit excess rainfall is assumed to be applied to the catchment at a single point in time.

- Catchment Response: The instantaneous unit hydrograph represents the response of the catchment to the unit rainfall excess. It shows how the streamflow hydrograph changes over time in response to the input of rainfall.

- Duration: The instantaneous unit hydrograph does not have a specific duration. It is a theoretical concept that represents the response of a catchment to a unit rainfall excess regardless of the duration of the excess rainfall.

Conclusion:

In summary, an instantaneous unit hydrograph is a hydrograph that represents the direct runoff response of a catchment to a unit rainfall excess that occurs instantaneously over the catchment. It is a theoretical concept that helps in understanding the relationship between rainfall and streamflow in a catchment.

A flood-hydrograph due to an isolated storm consists of
  • a)
    surface runoff, interflow, and base flow phases
  • b)
    surface runoff only
  • c)
    base flow only
  • d)
    surface runoff and base flow phases only
Correct answer is option 'A'. Can you explain this answer?

Bhaskar Rane answered
Hydrographs and Flood Events
A hydrograph is a graphical representation of the flow rate or discharge of a river or channel over a specific period of time. It depicts the rise and fall of water levels in response to precipitation, and is commonly used in hydrology and hydraulic engineering to analyze flood events.

Isolated Storm and Flood Hydrograph
An isolated storm refers to a singular rainfall event that occurs over a relatively short period of time, typically within a few hours. When this storm occurs, it can produce a flood hydrograph that illustrates the changes in river discharge over time. The flood hydrograph resulting from an isolated storm typically consists of three distinct phases: surface runoff, interflow, and base flow.

Surface Runoff
Surface runoff is the portion of precipitation that flows over the ground surface and eventually enters streams or rivers. It occurs when the rate of rainfall exceeds the infiltration rate of the soil, causing excess water to flow overland. Surface runoff is a major contributor to flood events, as it can rapidly increase river discharge. In the flood hydrograph, the surface runoff phase is characterized by a steep rising limb, indicating a rapid increase in river discharge.

Interflow
Interflow refers to the lateral movement of water within the soil, just below the ground surface. It occurs when the soil becomes saturated and excess water starts to move horizontally through the soil layers. Interflow contributes to the overall flow of water towards streams or rivers, but its contribution to flood events is relatively smaller compared to surface runoff. In the flood hydrograph, the interflow phase is represented by a slower rising limb, indicating a gradual increase in river discharge.

Base Flow
Base flow is the portion of streamflow that is sustained by groundwater seepage into a river or channel. It represents the long-term average flow of a river under normal weather conditions, and it is typically lower than the peak flow associated with flood events. However, during a flood event, the base flow can contribute to the overall river discharge. In the flood hydrograph, the base flow phase is depicted by a relatively flat portion after the peak flow, indicating a gradual recession of river discharge.

Conclusion
In summary, the flood hydrograph resulting from an isolated storm consists of three phases: surface runoff, interflow, and base flow. Surface runoff is the primary contributor to flood events, as it represents the rapid increase in river discharge during heavy rainfall. Interflow and base flow play smaller roles in flood events but contribute to the overall hydrograph shape. Understanding the different phases of a flood hydrograph is crucial for analyzing and managing flood risk in hydraulic engineering and hydrology.

The Muskingum method of flood routing gives Q2 = C0I2 + C1I1 + C2Q1. The coefficients in this equation will have values such that
  • a)
    C0 + C1 = C2
  • b)
    C- C1 - C2 =1
  • c)
    C0 + C1 + C2 = 0
  • d)
    C0 + C1 + C2 = 1
Correct answer is option 'D'. Can you explain this answer?

Muskingum Method for Flood Routing

The Muskingum method is a hydrological technique used for flood routing in rivers and streams. It is a linear reservoir model that predicts the outflow hydrograph from an upstream river reach to a downstream river reach. The method is based on the principle of conservation of mass and the assumption that the flow in a river can be represented by a single channel with uniform characteristics.

Equation for Flood Routing

The Muskingum method of flood routing gives the following equation:

Q2 = C0I2 + C1I1 + C2Q1

where Q2 is the outflow at the downstream reach, Q1 is the inflow at the upstream reach, I1 and I2 are the inflows at the current and previous time steps, and C0, C1, and C2 are the coefficients that determine the routing characteristics.

Coefficients in the Equation

The coefficients in the Muskingum equation are determined by the channel characteristics and the routing parameters. The values of the coefficients depend on the channel geometry, roughness, and slope, as well as the time step and the length of the reach.

The coefficients have the following properties:

- C0 + C1 + C2 = 1, which ensures that the outflow is a linear combination of the inflows.
- C0, C1, and C2 are all positive and less than or equal to 0.5, which ensures stability and accuracy of the method.
- The values of the coefficients depend on the ratio of the time step to the reach length, which affects the storage and attenuation characteristics of the reach.

Conclusion

In summary, the Muskingum method is a widely used technique for flood routing in rivers and streams. The method is based on a linear reservoir model and uses the Muskingum equation to predict the outflow hydrograph at the downstream reach. The coefficients in the equation are determined by the channel characteristics and the routing parameters, and have specific properties that ensure stability and accuracy of the method.

If the average annual rainfall and evaporation over land masses and oceans of the earth are considered it would be found that
  • a)
    over the land mass the annual evaporation is the same as the annual precipitation
  • b)
    about 9% more, water evaporates from the oceans than what falls back on them as precipitation
  • c)
    over the oceans about 19% more rain falls than what is evaporated
  • d)
    over the oceans about 19% more water evaporates than what falls back on them as precipitation
Correct answer is option 'B'. Can you explain this answer?

Sravya Tiwari answered
Whenever precipitation occurs, a part of rainfall is infiltrated into the soil and other part is surface runoff meeting in the rivers. 
Moreover, not all the water evaporated from the sea is transported towards land and most of it fall back as precipitation in oceans itself.
Therefore, it is found that about 9% more, water evaporates from the ocean than what falls back them as precipitation.

The Double mass curve technique is adopted to
  • a)
    check the consistency of raingauge records
  • b)
    to find the average rainfall over a number of years
  • c)
    to find the number of rainguages required
  • d)
    to estimate the missing rainfall data
Correct answer is option 'A'. Can you explain this answer?

Rahul Chauhan answered
If the conditions relevant to the recording of a raingauge station have undergone a significant change during the period of record, inconsistency would arise in the rainfall data of that station. The checking for inconsistency of record is done by the double mass curve technique. This technique is based on the principal that when each recorded data comes from the same parent population, they are consistent.

The St Venant equations for unsteady open-channel flow are
  • a)
    continuity and momentum equations
  • b)
    momentum equation in two different forms
  • c)
    momentum and energy equations
  • d)
    energy equation
Correct answer is option 'A'. Can you explain this answer?

The St Venant equations for unsteady open-channel flow are the continuity and momentum equations. The continuity equation describes the conservation of mass in an open-channel flow, while the momentum equation describes the conservation of momentum in an open-channel flow.
The St Venant equations are named after the French engineer Jean Baptiste Marie Charles St Venant, who developed them in the 19th century to describe the flow of water in open channels. They are widely used in the field of hydraulic engineering to analyze and predict the behavior of open-channel flows, such as those found in rivers, canals, and other watercourses.

The St Venant equations are based on the principles of fluid mechanics, and they can be used to predict the flow rate, velocity, and other properties of a fluid in an open channel under various conditions. They are typically used in conjunction with other mathematical and computational models to predict the behavior of open-channel flows in complex systems, such as those found in natural watercourses or engineered water management systems.

The hydrologic cycle states that
  • a)
    total inflow - total outflow = constant
  • b)
    subsurface inflow = subsurface outflow
  • c)
    inflow into basin = outflow from the basin .
  • d)
    mass inflow - mass outflow = change in mass storage
Correct answer is option 'D'. Can you explain this answer?

The Hydrologic Cycle

The hydrologic cycle is the continuous circulation of water between the earth's surface and the atmosphere. It involves various processes such as evaporation, precipitation, infiltration, and runoff. The balance between the inflow and outflow of water is crucial for the sustainability of water resources.

Mass Inflow and Outflow

The mass inflow and outflow refer to the amount of water that enters and leaves a particular system. The mass inflow includes the amount of water that enters a basin through precipitation, surface runoff, and subsurface flow. The mass outflow includes the amount of water that leaves the basin through evaporation, transpiration, surface runoff, and subsurface flow.

Change in Mass Storage

The change in mass storage refers to the difference between the inflow and outflow of water in a particular system. It is the amount of water that is stored in the basin at a particular time. The change in mass storage can be positive or negative depending on the inflow and outflow of water.

The Correct Answer

The hydrologic cycle states that the mass inflow and outflow are equal, and the change in mass storage is the difference between the inflow and outflow of water. Therefore, the correct answer is option D, "mass inflow - mass outflow = change in mass storage." This equation is important in hydrology as it helps to understand the water balance of a particular system and to manage water resources effectively.

In the moving-boat method of stream-flow measurement, the essential measurements are
  • a)
    the velocity recorded by the current meter, the depths and the speed of the boat.
  • b)
    the velocity and direction of the current meter, the depths and the time interval between depth readings
  • c)
    the depth, time interval between readings, speed of the boat and velocity of the stream
  • d)
    the velocity and direction of the current meter and the speed of the boat
Correct answer is option 'B'. Can you explain this answer?

The moving-boat technique measures the velocity over the width of a segment by suspending the current meter at a constant depth during the transverse of the boat across the stream.
The measured velocity and the additional information of the depth gives the discharge.
Thus, velocity of the river is determined by whether the current meter measures along or opposite of the stream and the velocity it measures.
The depth multiplied with velocity gives volume per unit width of water flown. The volume divided with time gives discharge flown.

As floodwave passess a given section of a river, life lime of occurrence of the maximum stage and that maximum discharge will be such that
  • a)
    the maximum discharge passes down before the maximum stage is attained
  • b)
    the maximum stage is attained before the maximum discharge passes down
  • c)
    the two events occur simultaneously
  • d)
    no specific sequence would be universally assignable
Correct answer is option 'C'. Can you explain this answer?

Nitya Nambiar answered
Explanation:

When a floodwave passes through a given section of a river, the maximum stage and maximum discharge will occur at some point during the flood event. The question asks about the relationship between the occurrence of these two events.

Option A: The maximum discharge passes down before the maximum stage is attained

This option suggests that the peak flow of water passes downstream before the water level reaches its highest point. However, in a flood event, the water level usually rises before the peak flow occurs. This option is not correct.

Option B: The maximum stage is attained before the maximum discharge passes down

This option suggests that the water level reaches its highest point before the peak flow passes downstream. In many cases, this is indeed the situation during a flood event. As the water level rises, it eventually reaches its maximum, and then the peak flow of water follows. This option is plausible.

Option C: The two events occur simultaneously

This option suggests that the maximum stage and maximum discharge occur at the same time. This is a possibility during a flood event, especially if there are no significant delays in the propagation of the floodwave. The water level and flow rate could both reach their peaks simultaneously. This option is plausible.

Option D: No specific sequence would be universally assignable

This option suggests that there is no consistent relationship between the occurrence of the maximum stage and maximum discharge. However, in reality, there is often a relationship between the rise in water level and the subsequent peak flow. This option is not correct.

Conclusion:

Based on the explanations above, the correct answer is option C: the maximum stage and maximum discharge occur simultaneously.

Evapotranspiration is confined to
  • a)
    daylight hours
  • b)
    night-time only
  • c)
    land surfaces only
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Transpiration is essentially confined to day light hours and rate of transpiration depends upon the growth period of the plant. Evaporation on the other hand, continues all through the day and night although the rates are different.

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