BPSC AE Paper 6 (Civil) 14 Oct 2022 Official Paper (Previous Year Paper)

Concept:

Consumptive use of water by a crop is the depth of water consumed by the plant in the process of transpiration and evaporation, during crop growth.

These two processes (i.e. transpiration and evaporation) collectively termed as evapo-transpiration.

Consumptive use of water by a crop is expressed as the depth of water per unit area for specified period, such as days, months or seasons. The value of consumptive use of water is needed to calculate the irrigation requirement of the crop

Concept:

The crop ratio is calculated as the ratio of the area under cultivation for the Rabi season to the area under cultivation for the Kharif season. It helps in understanding the distribution of water and cultivation patterns between different crop seasons.

Calculation:

Given:

• Gross Command Area (GCA) = 50,000 hectares
• Culturable Command Area (CCA) = 80% of GCA
• Intensity of irrigation for Rabi season = 60%
• Intensity of irrigation for Kharif season = 20%

1. Calculate the Culturable Command Area (CCA):

$\text{CCA} = 0.8 \times \text{GCA}$

$\text{CCA} = 0.8 \times 50,000 , \text{hectares}$

$\text{CCA} = 40,000 , \text{hectares}$

2. Calculate the area irrigated during the Rabi season:

$\text{Area for Rabi} = \text{Intensity of irrigation for Rabi} \times \text{CCA}$

$\text{Area for Rabi} = 0.6 \times 40,000 , \text{hectares}$

$\text{Area for Rabi} = 24,000 , \text{hectares}$

3. Calculate the area irrigated during the Kharif season:

$\text{Area for Kharif} = \text{Intensity of irrigation for Kharif} \times \text{CCA}$

$\text{Area for Kharif} = 0.2 \times 40,000 , \text{hectares}$

$\text{Area for Kharif} = 8,000 , \text{hectares}$

4. Calculate the Crop Ratio:

$\text{Crop Ratio} = \frac{\text{Area for Kharif}}{\text{Area for Rabi}}$

$\text{Crop Ratio} = \frac{8,000}{24,000}$

$\text{Crop Ratio} = 0.333$

Concept:

Storage efficiency of irrigation is a measure of how effectively water is stored in the root zone during irrigation compared to the amount of water that was required in the root zone before irrigation. It is calculated using the ratio of the water stored to the water required, expressed as a percentage.

Calculation:

Given:

• Water pumped into a canal: 120 cu m (not directly needed for this calculation)
• Water supplied to the field: 80 cu m (not directly needed for this calculation)
• Water stored in the root zone: 60 cu m
• Water required in the root zone prior to irrigation: 80 cu m

Formula for Storage Efficiency:

$\eta_s = \frac{\text{Water stored in the root zone}}{\text{Water required in the root zone prior to irrigation}} \times 100$

Substituting the given values:

$\eta_s = \frac{60 , \text{cu m}}{80 , \text{cu m}} \times 100$

$\eta_s = \frac{60}{80} \times 100$

$\eta_s = 0.75 \times 100$

$\eta_s = 75\%$

Concept:

For an irrigation land:

SC = Saturation capacity, FC = Field capacity, OMC = Optimum moisture content, PWP = Permanent welting point and UWP = Ultimate welting point

1. Equivalent depth of water held at field capacity (x) = S × d × Fc
2. Equivalent depth of water held at PWP (x’) = S × d × (PWP)
3. Available moisture/storage capacity of soil (y) = S × d × (Fc - PWP)
4. Readily available moisture content (dw) = S × d × (Fc - OMC)

Given:

Field capacity (F. C) = 20 % = 0.20

Permanent willing point (PWP) = 10 % = 0.10

Depth of the root zone (d) = 100 cm

Dry Unit weight of soil (γd) = 2.0 g/cc

Calculations:

The storage capacity $= \frac{{{\gamma _d} \times d \times \left( {F.C - PWP} \right)}}{{{\gamma _w}}}$

The storage capacity $= \frac{{2 \times 100 \times \left( {0.20 - 0.10} \right)}}{1} = 20;cm$

Hence, the storage capacity of the soil is 20 cm.

Explanation:

Duty: It is the number of hectares of land irrigated for full growth of a given crop by a supply of 1 cumec of water continuously during the entire base period of that crop.

Duty is the area that can be irrigated by the discharge of 1 cumec of water.

At the head of the canal, there are numerous losses to occur later which requires more amount of water to irrigate a particular field. However, if considered on the field, all losses have already occurred and a lesser amount of water is required to irrigate the same considered area.

∴ Duty of water changes from place to place, it will be maximum at the field and minimum at the head of the canal.

Concept:

Aquifer: 

It is an underground geological formation that contains water and a sufficient amount of water can be extracted economically using water wells. It generally comprises layers of sand and gravel and fractured bedrock.

• Unconfined Aquifer: When the water table serves as the upper boundary of the aquifer, the aquifer is known as an unconfined aquifer. The unconfined aquifer is also known as the water table aquifer and phreatic aquifer. An impervious layer is generally served as the bottom boundary of an unconfined aquifer and the top layer is not confined.

• Confined Aquifer: An aquifer that is bounded by two impervious layers at the top and bottom of the aquifer is called a confined aquifer. The confined aquifer is also known as a pressure aquifer or artesian aquifer. The top and bottom layers of a confined aquifer are generally impervious.

Aquiclude:

• It contains a large amount of water in pores, but Impermeable to the flow of water (extraction of water is very difficult). It may be considered as close to water movement.
• A good example of aquiclude is clay.

Aquitard:

• Aquitard form by that material through which the only seepage is possible but extraction of water is not so easy (Insignificant yield) as in aquifer.
• A good example of aquitard is silty clay material.

Concept

As per Lacey’s formula, the relation between discharge, Q, and the wetted perimeter is given as:

$P = 4.75;\sqrt Q$

For a rectangular channel, wetted perimeter P is given as:

P = B + 2 × y

Where y is the depth of flow.

If the channel is very wide the P ≈ B (as y ≪ B)

Calculation:

Given: Q = 100 cumecs; y = 5 m, B =?

Now,

$P = 4.75 \times \sqrt {100} = 47.5;m$

Further,

P = B + 2 × y

47.5 = B + 2 × 5

⇒ B = 37.5 m

Concept:

Specific yield (Sy) is the ratio of the volume of water that drains from a saturated rock (due to gravity) to the total volume of the rock.

The Specific Retention (Sr) of a rock or soil is the ratio of the volume of water a rock can retain against gravity to the total volume of the rock.

Therefore, the total porosity is equal to the volume or water that a rock will yield by gravity drainage (Sy) and the volume held by surface tension (Sr) and is given as:

${S_y} + {S_r} = \eta$

Calculations:

Given:

Sy = 30%

η = 50%

${S_y} + {S_r} = \eta$

$30\% + {S_r} = 50\%$

${S_r} = 50\% - 30\%$

${S_r} = 20\%$

Concept:

Lacey followed Lindley’s hypothesis, which states that “dimensions and slope of a channel to carry a given discharge and silt load in easily erodible soil are uniquely determined by nature”.

According to Lacey:

“Silt is kept in suspension by the vertical component of eddies generated at all points of forces normal to the wetted perimeter”.

So the eddies are generated from the bottom as well as the side of the channel.

Regime Channel: “A channel is said to in regime if there is neither silting nor scouring”.

According to Lacey, there may be three regime conditions:

(i) True regime; (ii) Initial regime; and (iii) Final regime.

Important Points:

According to Kennedy’s theory Silt carried by the flowing water is kept in suspension by the vertical

component of eddies which are generated from the bed of the channel.

Explanation:

Garret's diagram is a graphical representation of the design of canal dimensions based on Kennedy's theory. The diagram is discharge plotted on the abscissa (X-axis), and slope of the channel on the left side of the ordinate (Primary Y-axis), and the depth of the channel and critical velocity (v∘) on the right side of the ordinate (secondary Y-axis). Garret diagram is applicable only for the side slope of the channel is 0.5 : 1 (H : V).

Kennedy's Theory:

Kennedy proposed his theory based on the charge and grade of silt present in the Upper Doab Canal (Punjab) and its distributaries. He observed that there is always one velocity for a given condition, called "critical velocity", v∘, which characterizes a non-silting and non-scouring channel in a steady regime. This velocity depends on the depth of water in the channel.

v∘ = 0.546 × y0.64

Where y = depth of water, m.

This relation was applicable to the grade of silt present only in the Upper Doab Canal. He introduced a factor called the "critical velocity ratio, (CVR), m" in his equation to account for the varying grades of silt in other regions. Thus, he propounded :

v∘ = 0.546 m y0.64

Where m = v/v∘ = critical velocity ratio.

Concept: 

Lacey's scour depth is given by:

$R; = ;1.35{\left( {\frac{{{q^2}}}{f}} \right)^{\frac{1}{3}}};$

Where, q = Discharge per meter width, f = silt factor

Calculations:

Given, q = 4 cumec, f = 2

$R; = ;1.35{\left( {\frac{{{4^2}}}{2}} \right)^{\frac{1}{3}}};=;2.7;m$

Important Points 

Various formulas deduced by lacey is given below:

1) silt factor ⇒ $\rm{f = 1.76\sqrt {{d_{mm}}}}$

2. velocity of flow ⇒$V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}$

3) hydraulic mean depth ⇒$R = \frac{{5{V^2}}}{{2f}}$

4. wetted perimeter ⇒ $P = 4.75\sqrt Q$

Explanation:

Trap Efficiency:

• Trap efficiency of a reservoir is the ratio of total sediments retained by the reservoir to the total sediments entering into the reservoir (inflow sediments).
• Trap efficiency of the reservoir is the function of the ratio of reservoir capacity and total inflow.
• The reservoir capacity is inversely related to the number of sediments deposited. As time passes, the rate of silting or the rate of deposition of sediments reduces but the amount of sediments deposited will be increased which in turn reduces the capacity of the reservoir. Hence, trap efficiency is also gets reduced.

Other factors affecting trap efficiency:

• Size of reservoir and stream: A larger reservoir on a small stream has less capacity inflow ratio and hence has higher trap efficiency.
• The velocity of flow of stream: Higher velocity less deposition of sediments, lower is the trap efficiency.

Concept:

The maximum compressive stresses occur at the heel (mostly during reservoir empty condition) or at the toe (at reservoir full condition) and on planes normal to the face of the dam.

∴ For reservoir empty condition maximum compressive force will be at the heel.

∴ For reservoir full condition maximum compressive force will be at the toe of the dam.

The figure below is the pressure distribution diagram for reservoir full condition:

Explanation:

The elementary profile of a dam

• An elementary profile of a gravity dam is the theoretical shape of its cross-section when it is subjected to only three main forces i.e.Self-weight, Water pressure, and Uplift pressure.
• Moreover, the elementary profile has zero top width and no freeboard.
• The right-angle triangle is the most suitable section for the theoretical profile.
• The elementary profile is hypothetical because an actual gravity dam has some minimum top width and freeboard, and it will also be subjected to forces other than the three main forces considered in the elementary profile.

Explanation:

River Training Works

River training works are constructed for several reasons but the fundamental objective is to restrict the horizontal movement of the river channel along a certain alignment.

The measures are adopted to train the river because rivers in alluvial flood plains are prone to meandering and hence cause a lot of damage to life and property.

Additional InformationMeandering

• Meandering is considered a wave phenomenon.
• A meander is formed when moving water in a stream erodes the outer banks and widens its valley and the inner part of the river has less energy and deposits what it is carrying.
• A meandering channel tends to be deeper towards the concave bank and progressively gets shallower towards the convex bank.

Explanation:

An elementary profile of a gravity dam is shown:

Let the resultant pass through any point at a distance e from the centre of CB. Hence ‘e’ is the eccentricity.

The components of R are shown as above.

\(\therefore {{\rm{\sigma }}{\rm{c}}} = \frac{{{{\rm{R}}{\rm{v}}}}}{{{\rm{b}} \times 1}} - \frac{{\left( {{{\rm{R}}_{\rm{v}}} \times {\rm{e}}} \right) \times \left( {\frac{{\rm{b}}}{2}} \right)}}{{1 \times \frac{{{{\rm{b}}^3}}}{{12}}}};\)

For no tension at base,

σc ≥ 0

\( \Rightarrow \frac{{{{\rm{R}}{\rm{v}}}}}{{{\rm{b}} \times 1}} \ge \frac{{\left( {{{\rm{R}}{\rm{v}}} \times {\rm{e}}} \right)\left( {{\rm{b}}/2} \right)}}{{\left( {{{\rm{b}}^3}/12} \right)}}\)

⇒ e ≤ b/6

Hence when the resultant force lies in the middle-one-third region, there is no tension anywhere at the base of the elementary profile of a gravity dam.

Important Points

• Gravity dam is made of concrete or masonry and resists the water force using the weight of the dam only.
• An elementary profile of a gravity dam is a triangular section having zero width at water level and maximum base width where hydrostatic pressure is maximum
• The resultant force R mainly comprises the water force and weight of the dam.

Explanation:

Rainfall hyetograph shows Intensity of rainfall v/s time. For example,

Note:

Explanation:

The following table shows the meaning of different types of curves in Hydrology.

Calculation:

Chezy's Formula:

$V = C\sqrt{RS}$  ............................ (i)

Manning's Formula:

$V = \frac{1}{n}R^{2/3}S^{1/2}$  .......................... (ii)

where C = Chezy's Constant

R = Hydraulic Radius

S = Hydraulic Gradient Line

 n = Manning's Roughness Coefficient

From equations (i) and (ii)

$V = C\sqrt{RS} = \frac{1}{n}R^{2/3}S^{1/2}$

$C(R^{1/2}) = \frac{1}{n}(R^{2/3})$

$C = \frac{1}{n} (R^{1/6})$

Concept:

The specific energy head in a rectangular channel with given alternate depths can be calculated using the formula:

$E = \frac{y_1^2 + y_2^2 + y_1 \times y_2}{y_1 + y_2}$

Calculation:

Given:

• Alternate depth $y_1 = 1.5$ m
• Alternate depth $y_2 = 2.5$ m

Substitute the given values into the formula:

$E = \frac{(1.5)^2 + (2.5)^2 + (1.5 \times 2.5)}{1.5 + 2.5}$

Calculate the numerator:

• $y_1^2 = (1.5)^2 = 2.25 , \text{m}^2$
• $y_2^2 = (2.5)^2 = 6.25 , \text{m}^2$
• $y_1 \times y_2 = 1.5 \times 2.5 = 3.75 , \text{m}^2$
• $\text{Numerator} = 2.25 + 6.25 + 3.75 = 12.25 , \text{m}^2$

Calculate the denominator:

$y_1 + y_2 = 1.5 + 2.5 = 4 , \text{m}$

Calculate the specific energy head $E$:

$E = \frac{12.25 , \text{m}^2}{4 , \text{m}}$

$E = 3.06 , \text{m}$

Concept:

The average shear stress on the channel boundary in open channel flow can be determined using the formula:

$\tau_0 = \gamma R S$

Where:

• $\tau_0$ = Shear stress on the channel boundary (N/m²)
• $\gamma$ = Unit weight of water (10 kN/m³)
• $R$ = Hydraulic radius (m)
• $S$ = Slope of the channel

Calculation:

Given:

• Bottom width of the channel ($b$) = 2 m
• Uniform flow depth ($y$) = 1.25 m
• Longitudinal slope of the channel ($S$) = 0.0003
• Unit weight of water ($\gamma$) = 10 kN/m³

1. Calculate the Cross-Sectional Area ($A$) of the Flow:

$A = b \times y$

$A = 2 \times 1.25$

$A = 2.5 , \text{m}^2$

2. Calculate the Wetted Perimeter ($P$) of the Channel:

$P = b + 2y$

$P = 2 + 2 \times 1.25$

$P = 4.5 , \text{m}$

3. Calculate the Hydraulic Radius ($R$):

$R = \frac{A}{P}$

$R = \frac{2.5}{4.5}$

$R = 0.5556 , \text{m}$

4. Calculate the Shear Stress ($\tau_0$):

$\tau_0 = \gamma R S$

$\tau_0 = 10 \times 0.5556 \times 0.0003$

$\tau_0 = 0.0016668 , \text{kN/m}^2$

$\tau_0 = 1.6668 , \text{N/m}^2 \approx 1.67 , \text{N/m}^2$

Explanation:

Froude's Number is given by

Fr= $\frac{{\rm{v}}}{{\sqrt {{\rm{g}} \times \left( {\frac{{\rm{A}}}{{\rm{T}}}} \right)} }}$

Where,

A = Area of cross-section, and T = Top width

For a triangular channel as stated above

${{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {\frac{{{\rm{gA}}}}{{\rm{T}}}} }}$

T = 2y + 2y = 4y

${\rm{A}} = \frac{1}{2} \times 4{\rm{y}} \times {\rm{y}} = 2{y^2}$

$\therefore {{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {{\rm{g}} \times \frac{{2{{\rm{y}}^2}}}{{4{\rm{y}}}}} }}$

${{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {{\rm{g}}\left( {\frac{{\rm{y}}}{2}} \right)} }}$

Explanation

Specific energy is the total head with respect to the channel bottom. It is a summation of the velocity head and the depth of flow at a given cross-section. Specific energy has proven to be very useful in the analysis of open-channel flow.

For rectangular channel 

E= 1.5yc or (3/2)y_c

or

y_c = (2/3)E

Where E=  Specific energy , yC = Critical depth 

For triangular section 

E= 1.25yc

Concept:

To determine the type of bed slope of a rectangular channel, we need to compare the normal depth ($y_n$) with the critical depth ($y_c$). The type of slope is categorized based on whether the normal depth is greater than, less than, or equal to the critical depth.

Calculation:

Given:

• Width of the rectangular channel ($b$) = 3.0 m
• Normal depth ($y_n$) = 0.8 m
• Discharge ($Q$) = 9.5 cumec (cubic meters per second)

1. Calculate the Cross-Sectional Area ($A$) of Flow:

$A = b \times y_n$

$A = 3.0 \times 0.8$

$A = 2.4 , \text{m}^2$

2. Calculate the Hydraulic Radius ($R$):

The hydraulic radius $R$ is given by:

$R = \frac{A}{P}$, where $P$ is the wetted perimeter.

$P = b + 2y_n$

$P = 3.0 + 2 \times 0.8$

$P = 4.6 , \text{m}$

$R = \frac{2.4}{4.6}$

$R \approx 0.5217 , \text{m}$

3. Calculate the Critical Depth ($y_c$):

The critical depth for a rectangular channel is calculated using the formula:

$y_c = \left( \frac{Q^2}{g \times b^2} \right)^{1/3}$

$y_c = \left( \frac{9.5^2}{9.81 \times 3^2} \right)^{1/3}$

$y_c = \left( \frac{90.25}{88.29} \right)^{1/3} \approx 1 , \text{m}$

4. Determine the Slope Type:

• Mild Slope: $y_n > y_c$
• Steep Slope: $y_n < y_c$
• Critical Slope: $y_n = y_c$

Given:

• $y_n = 0.8 , \text{m}$
• $y_c = 1 , \text{m}$

Since $y_n < y_c$, the slope of the channel is considered steep.

Explanation:

Continuity Equation: This equation is always used to describe the conservation of mass in fluid flow. For a hydraulic jump in a rectangular channel, the continuity equation ensures that the flow rate before the jump equals the flow rate after the jump. This is mathematically expressed as:

Q=A_1​V_1​=A_2​V

Momentum Equation: This is crucial for analyzing the forces involved in a hydraulic jump, as it relates to the change in momentum across the jump. The momentum equation in terms of depth and velocity is used to relate the initial and sequent depths of the hydraulic jump. The specific force or momentum flux function is given by:

These two equations are the fundamental tools needed to derive the relationship between the initial and sequent depths of a hydraulic jump and the initial Froude number in a rectangular channel.

Concept:

To determine the initial Froude number and energy loss in a hydraulic jump, we use specific formulas based on the depths before and after the jump.

Given Data:

• Initial depth ($y_1$) = 0.40 m
• Sequent depth ($y_2$) = 1.60 m

Calculation of Initial Froude Number ($F_1$):

The Froude number ($F_1$) can be calculated using the relationship between initial and sequent depths:

$\frac{y_2}{y_1} = \frac{1}{2} \left(\sqrt{1 + 8F_1^2} - 1\right)$

Substituting the given values:

$\frac{1.60}{0.40} = \frac{1}{2} \left(\sqrt{1 + 8F_1^2} - 1\right)$

Simplifying:

$4 = \frac{1}{2} \left(\sqrt{1 + 8F_1^2} - 1\right)$

$8 = \sqrt{1 + 8F_1^2} - 1$

$\sqrt{1 + 8F_1^2} = 9$

$1 + 8F_1^2 = 81$

$8F_1^2 = 80$

$F_1^2 = 10$

$F_1 = \sqrt{10} \approx 3.16$

Calculation of Energy Loss ($\Delta E$):

The energy loss in a hydraulic jump is given by:

$\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2}$

Substituting the given values:

$\Delta E = \frac{(1.60 - 0.40)^3}{4 \times 0.40 \times 1.60}$

$\Delta E = \frac{1.20^3}{4 \times 0.40 \times 1.60}$

$\Delta E = \frac{1.728}{2.56}$

$\Delta E = 0.675 , \text{m}$ or 27/40

Hence, the most appropriate answer is option 1.

Concept:

IRC has fixed the maximum limit of superelevation in plain and rolling terrains and in snow-bound areas as 7.0% or 0.07 ( 1 in 15) taking such mixed traffic into consideration. On hilly roads not bound by snow, a maximum superelevation of up to 10% has been recommended. On urban road stretches with frequent intersections, it may be necessary to limit the maximum superelevation to 4.0 percent, keeping in view the convenience in construction and that of turning movements of vehicles.

Additional Information

Superelevation:

Superelevation is the transverse slope provided to counteract the effect of centrifugal force and reduce the tendency of vehicles to overturn and to skid laterally outwards by raising the pavement's outer edge with respect to the inner edge. It is represented by e.

$e+f={v^2\over gR}$

Where v = Speed of the vehicle,

R = Radius of the horizontal curve,

f = Lateral friction,

g = Acceleration due to gravity.

Explanation:

Transition curve is a curve in plan which is provided to change the horizontal alignment from straight to circular curve gradually means the radius of transition curve varies between infinity to R or R to infinity.

IRC recommends Spiral or clothoid as the ideal transition curve due to following reasons:

i) It satisfies that rate of change of centrifugal acceleration is constant i.e., Ls.R = constant.

Where Ls = length of transition curve R = radius of curve.

ii) The calculation and field implementation of spiral curve is simple and easy.

iii) It enhances aesthetics also.

Concept:

The maximum values of the ruling, limiting, and exceptional gradients recommended by the IRC in different terrains are given below

Explanation:

As per IRC, the Centrifugal force corresponding to 75 % of the design speed is counteracted by providing the super elevation on horizontal curves. IRC recommends following formula for calculation of Super elevation :

$e = \frac{{{V^2}}}{{225R}}$

Where

e is the super elevation

R is the radius of the horizontal circular curve in ‘m’.

V is the design speed in ‘km/hr’.

So, The speed gets reduced by 100 - 75 = 25 %.

Note:

1. This has been done because India has mixed traffic conditions i.e. slow-moving and fast-moving vehicles.
2. The Superelevation calculated using the above formula should be less than the maximum permissible value and shall be more than the minimum value which is the camber of the road.
3. The remaining centrifugal force is balanced by friction on road.

Explanation:

30th Highest Hourly Volume: It is the hourly volume that will be reached only thirty times or exceeded only 29 times in a year and all other hourly volume of the year will be less than this value.

30th highest hourly traffic value is found to be satisfactory from the consideration of the facility as well as the cost. This is because the cost will be much lesser when compared to the peak hourly volume and there will be congestion only for 29 hours in the year and this is considered reasonable.

Important Points:

Other Important Terms:

Average Annual Daily Traffic (AADT): The average 24-hour traffic volume at a given location over a full 365-day year, i.e. the total number of vehicles passing the site in a year divided by 365.

Average Annual Weekday Traffic (AAWT): The average 24-hour traffic volume occurring on weekdays over a full year. It is computed by dividing the total weekday traffic volume for the year by 260.

Average Daily Traffic (ADT): An average 24-hour traffic volume at a given location for some period of time less than a year. It may be measured for six months, a season, a month, a week, or as little as two days. An ADT is a valid number only for the period over which it was measured.

Average Weekday Traffic (AWT): An average 24-hour traffic volume occurring on weekdays for some period of time less than one year, such as for a month or a season.

Concept:

Spot speed:

Spot speed is the instantaneous speed of the vehicle at the specified location.

The different methods to find spot speed is:

• Radar speedometer which automatically records instantaneous speed
• Photographic method
• Pressure contacts tubes
• Enoscope
• Direct timing procedure

Explanation:

According to IRC,

Explanation:

Origin and Destination studies:

(i) These studies are essential in planning new highway facilities or improving existing road system.

(ii) It is also used for planning mass rapid transit system.

Method:

(i) Vehicle or licence plate method

(ii) Road side interview method

(iii) Tag on car method

(iv) Home interview method

Representation of Origin and Destination data:

Desire line: 

• It shows the desire direction of flow of traffic. It is a straight line connecting origin and destination.
• Width of the desire line is directly proportional to the number of trip in both direction. The road having maximum thickness should be given more importance

Explanation:

Basic capacity: The basic capacity is the maximum numbers of passenger cars that can pass a given point on a lane or a roadway during one hour under the most ideal roadway and traffic conditions that can possibly be attained.

Possible capacity: It is the maximum number of vehicles that can pass a given point on a lane or roadway during one hour under the prevailing roadway and traffic conditions.

Practical Capacity: It is the maximum number of vehicles that can pass a given point on a lane or roadway during one hour with creating unreasonable delay under the prevailing roadway and traffic conditions.

Design capacity: it is the practical capacity or a smaller value determined for use in designing the highway to accommodate the design hourly volume.

Theoretical Capacity: It is the number of vehicles passing any point in one hour per lane

Explanation:

Abrasion Test: 

• Abrasion test is carried out to test the hardness property of aggregates.

Los Angeles abrasion test:

• The principle of Los Angeles abrasion test is to find the percentage wear due to relative rubbing action between the aggregate and steel balls used as abrasive charge.

​Los Angeles testing machine:

• The Los Angeles abrasion testing machine shall consist of a hollow steel cylinder arranged for rotating about its axis in a horizontal position.
• The hollow cylinder shall be made of steel with its ends closed, the cylindrical shaft mounted on stub shafts attached to both ends but not entering it.
• The principle of the Los Angeles Abrasion Test is to find the percentage of wear due to relative rubbing action between the aggregate and steel balls used as an abrasive charge.
• Los Angeles machine consists of a circular drum of internal diameter 700 mm and length 520 mm mounted on a horizontal axis enabling it to be rotated. An abrasive charge consisting of cast iron spherical balls of 48 mm diameters and weight 340-445g is placed in the cylinder along with the aggregates.
• The cylinder is then locked and rotated at the speed of 30-33 rpm for a total of 500-1000 revolutions depending upon the gradation of aggregates.
• After specified revolutions, the material is sieved through a 1.7 mm sieve, and the passed fraction is expressed as a percentage total Weight of the sample.

Explanation:

Penetration test:

• Hardness of bitumen is obtained by the penetration test.
• It measures the distance a standard blunt pointed needle will vertically penetrate a sample of material at 25°C, the load being 100 gram and time of application of load being 5 seconds.
• The unit of penetration is 1/10 mm. Thus 80/10 grade bitumen means 8 - 10 mm penetration.
• 30/40 and 80/100 grade bitumen are more commonly used, depending upon bitumen construction type and climate conditions.

Important Points

• Grade of bitumen is decided on the basis of the penetration test. If the grade of bitumen is A-30 then it means this bitumen is manufacturing from Assam petroleum and penetration value is 3 mm. If the grade is S-30 then this bitumen is manufactured from other than Assam petroleum.
• In hot climate, a lower penetration grade bitumen like 30/40 bitumen is preferred.
• Tar is soft, hence penetration test is not used in the case of tar.

Explanation:

Bitumen:

• Bitumen is a petroleum product obtained by the distillation of petroleum crude oil.
• Bitumen is hydrocarbon material of either natural of pyrogenous origin, found in gaseous, liquid semisolid or solid form.
• Bituminous material is widely used in road construction and maintenance.

Bitumen emulsion:

• The bituminous emulsion is widely used in road construction, especially for maintenance and repair works.
• It is formed by mixing together molten bitumen with hot water in presence of an emulsifying agent in such a manner that the bitumen is mechanically broken into fine globules.
• When it is applied on the road, water is removed by evaporation or absorption into a medium as a result the emulsion breaks up and bitumen is left in place.
• It is a condition in which bitumen is suspended in an aqueous medium. It is used for Patch-up work.
• The biggest advantage of bitumen emulsion is that it can be used in the rainy season also.
• Also emulsions have been used in soil stabilization, particularly for stabilization of sands in desert areas.

Cutback bitumen:

• Cutback bitumen is a type of bitumen that is obtained when the viscosity of bitumen is reduced by volatile diluent.
• It is generally used in colder regions.

Coal tar:

• Tar is also a bituminous material obtained by the destructive distillation of coal or wood.
• It has the same appearance as bitumen but both have different characteristics.

Explanation:

Water Borne Diseases:

• The term waterborne disease is reserved largely for infections that predominantly are transmitted through contact with or consumption of microbially polluted water.
• Many infections may be transmitted by microbes or parasites that accidentally, possibly as a result of exceptional circumstances, have entered the water.

Examples of water-borne diseases:

• Typhoid Fever
• Cholera
• Giardia
• Amoebic Dysentery
• E-coli
• Salmonella
• Amoebic dysentery:
• Amebiasis, or amoebic dysentery, is a gastrointestinal illness that develops when an organism called a Protozoan parasite Entamoeba histolytica enters your intestines.
• The illness may cause diarrhea, nausea, stomach cramps, and fever.
• Healthcare providers usually treat it with antibiotics.

Malaria is caused by protozoa called Plasmodium vivax. Malaria is an acute febrile illness spread through the bites of infected mosquitoes. It is a mosquito-borne disease​

Additional Information 

Infectious hepatitis:

• Hepatitis is inflammation of the liver tissue.
• Some people or animals with hepatitis have no symptoms, whereas others develop yellow discoloration of the skin and whites of the eyes (jaundice), poor appetite, vomiting, tiredness, abdominal pain, and diarrhea.
• Hepatitis is most commonly caused by the virus Hepatoviral  A, B, C, D, and E.

Infectious jaundice:

• Jaundice is the yellowing of the skin, mucous membranes, and the whites of the eyes that occurs when the body does not process bilirubin as it should.
• Jaundice typically occurs due to an underlying disorder that either causes the production of too much bilirubin or prevents the liver from eliminating it.

Poliomyelitis:

• The words polio (grey) and myelon (marrow, indicating the spinal cord) are derived from the Greek.
• It is the effect of poliomyelitis virus on the spinal cord that leads to the classic manifestation of paralysis.
• Poliomyelitis or polio is a crippling and potentially deadly infectious disease caused by poliovirus that spreads from person to person invading the brain and spinal cord resulting into paralysis.

Explanation:

Water quality for different compounds As per l.S.10500 are as follows:

Explanation:

Turbidity can be measured using Turbidity rod method that consists of a platinum needle having

Diameter = 1 mm and

Length = 25 mm

at the tip of aluminium rod.

This rod is inserted in the water sample to be tested and the height at which this needle just become invisible is noted and represented the turbidity of water sample.

This representation is done in terms of standard units which is obtained by the addition of Silica as Silica Dioxide (SiO2) in milligrams in one litres of pure water.

The standard unit is expressed as Silica Turbidity Unit (STU)

Hence, turbidity is measured on Standard silica scale.

Explanation:

BOD removal efficiency of various Biological Units

Additional Information

Oxidation ditch

• An oxidation ditch is a modified activated sludge biological treatment process that utilizes long solids retention times (SRTs) to remove biodegradable organics.
• Oxidation ditches are typically complete mix systems, but they can be modified to approach plug flow conditions.
• Retention time is much longer in an oxidation ditch usually two days.
• BOD removal Efficiency is 97% and and bacteria removal efficiency is 98%.

Explanation:

Initially, landfill is one of the rubbers and plastic waste disposal methods commonly used all the time but now, it is not preferred as it will take up to 450 years for plastic to break down, and rubber will take a much longer time. So, a landfill is not the best method for the disposal of plastic and rubber waste.

The rubber and plastic waste is converted into fuel oil through pyrolysis and it has become the best method to dispose of rubber and waste plastic.

Note:

Pyrolysis is the thermal decomposition of biomass that takes place in the absence of oxygen with temperature ranges from 350 to 550 °C that could reach up to 700 °C. This process decomposes organic materials into solid, liquid, and gas mixtures.

Explanation:

Trickling filters are designed primarily to remove dissolved organic matter from wastewater. This is achieved through the biological activity of microorganisms that form a biofilm on the filter media, effectively breaking down and oxidizing the organic pollutants.

Trickling Filters Work:

1. Structure and Design: A trickling filter consists of a bed of media (such as rocks, gravel, or specially designed plastic media) over which wastewater is distributed. The media provides a large surface area for microorganisms to attach and grow.
2. Microbial Biofilm: As wastewater trickles over the surface of the media, a thin layer of microbial biofilm develops on the media. This biofilm is composed of bacteria, fungi, and other microorganisms that are effective in degrading organic matter.
3. Flow of Wastewater: Wastewater is distributed over the surface of the media using a rotating arm or another distribution system, ensuring even coverage. The wastewater then trickles down through the media bed, contacting the microbial biofilm.
4. Biological Oxidation: The microorganisms in the biofilm consume the organic matter present in the wastewater as a food source. The organic matter is oxidized to carbon dioxide, water, and new microbial cells. This process reduces the biochemical oxygen demand (BOD) of the wastewater, which is a measure of the amount of organic matter present.
5. Effluent Collection: After passing through the media, the treated water is collected at the bottom of the filter and directed to a secondary clarifier. In the clarifier, any remaining solids (including sloughed-off biofilm) are settled out before the water is discharged or subjected to further treatment.

Concept:

The population equivalent of a town based on its Biochemical Oxygen Demand (BOD) is calculated using the total BOD of the town per day and the BOD per capita per day. The formula is:

$\text{Population Equivalent} = \frac{\text{Total BOD of the town per day}}{\text{BOD per capita per day}}$

Given Data:

• Total BOD of the town per day = 25000 kg/day
• BOD per capita per day = 50 grams

Calculation:

1. Convert BOD per capita per day to the same unit as total BOD:

Since the total BOD is given in kilograms, convert 50 grams to kilograms:

$50 , \text{grams} = \frac{50}{1000} , \text{kg} = 0.05 , \text{kg}$

2. Calculate the Population Equivalent:

$\text{Population Equivalent} = \frac{25000 , \text{kg/day}}{0.05 , \text{kg/person/day}}$

$\text{Population Equivalent} = \frac{25000}{0.05}$

$\text{Population Equivalent} = 500000$

Explanation:

To prevent the entry of foul-smelling gases from the sewers into the house, the most effective methods involve using water seals and vent pipes. Let's analyze the options provided:

1. Exhaust fan: While an exhaust fan can help ventilate a space and remove odors, it does not specifically prevent the entry of gases from the sewer system into the house. It may help mitigate odors once they are inside but is not a preventative measure for sewer gases.
2. Providing water seals for all the fixtures: Water seals, such as those found in P-traps and S-traps, act as barriers by holding a small amount of water, which prevents sewer gases from traveling back up through the plumbing into the house. While effective, relying solely on water seals might not be sufficient in cases where water evaporates or the traps dry out.
3. Providing water seals for all the fixtures and vent pipe in the plumbing system: This option is the most effective and is considered the standard practice in plumbing design. Water seals prevent gases from entering through fixtures, while vent pipes ensure that sewer gases are safely vented outside the house. The vent pipes also help maintain atmospheric pressure in the plumbing system, preventing siphoning of water from the traps.
4. Providing sufficient vent pipes in the plumbing system: Vent pipes are essential in allowing gases to escape and preventing the build-up of pressure in the plumbing system. While vent pipes are necessary, providing them without water seals would not prevent gas entry through fixtures if water traps dry out.

Explanation:

Traps: As the name suggests, traps simply trap the water and sewer gas. It is a plumbing device used to prevent smell, bacteria, also insects from entering your home. Different types of trap mention bellow,

• Gully trap is provided outside the building before connecting it to the external sewerage line. It also collects wastewater from the kitchen sink, washbasin, bath, and wash area.
• Intercepting trap is provided to prevent the foul gases from the public sewer from entering into the building sewer.
• Anti-siphon trap is used to prevent siphoning action.
• p trap is used with an Indian water closet. The traps are made from UPVC or cast-iron sheets. This trap also has a water seal and prevents entry of foul gases to the house.

Explanation:

Imhoff tank:

(i) Imhoff tank is used for the clarification of sewage by simple settling and sedimentation, along with anaerobic digestion of the extracted sludge. It consists of an upper chamber in which sedimentation takes place, from which collected solids slide down inclined bottom slopes to an entrance into a lower chamber in which the sludge is collected and digested.

(ii) The two chambers are otherwise unconnected, with sewage flowing only through the upper sedimentation chamber and no flow of sewage in the lower digestion chamber.

(iii) It is a robust and effective settler that causes a suspended solids reduction of 50 to 70%, COD reduction of 25 to 50%, and leads to potentially good sludge stabilization.

It consists of two compartments:

Upper Compartment:

• This compartment has a circular or rectangular shape with V-shaped walls and a slot at the bottom, allowing solids to settle into the digestion compartment, while preventing foul gas from rising up and disturbing the settling process. It is exposed to atmosphere, and thus working condition is aerobic.

Lower Compartment: 

• In this compartment, sludge accumulates and is compacted and partially stabilized through anaerobic digestion. Thus, working condition is anaerobic

Explanation:

COD (Chemical oxygen demand):

It is the amount of oxygen required for the decomposition of both biodegradable & non-biodegradable organic matter present in the system.

BOD (Biochemical oxygen demand):

• It is the amount of oxygen required for the decomposition of biodegradable organic matter present in the system.
• BOD during 5 days at 20°c is taken as standard BOD and is approximately 68% of the ultimate BOD.

​ThOD (Theoretical oxygen demand):

If the type of organic matter and its concentration present in the system is known then the amount of oxygen required for its decomposition can be computed theoritacly and is termed as ThOD.

TOC (Total organic carbon):

TOC is just another way to express the concentration of organic matter present in the system in terms of its carbon content.

BOD to COD ratio:

• It signifies the quality of wastewater with respect to the type of organic matter present in it that is either biodegradable or non-biodegradable.
• If the value of this ratio high that means the biodegradable organic matter is high and vice-versa.

Explanation:

Water is considered acidic when its pH is below 7. The level of acidity can vary depending on the specific pH value.

Less than 6.5 – Water with a pH less than 6.5 is commonly considered mildly acidic.

Less than 6.0 – This refers to more acidic water, but it is still classified as acidic.

Less than 5.5 – This indicates even stronger acidity.