GATE CE 2014 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is option 4 i.e. ​experiences

Key Points    

• As we know 'Present Indefinite Tense' is used to express habits, general truth, unchanging situations, emotions, etc
• Thus, the given sentence expresses a general truth or experience; hence 'Present Indefinite tense' will be used in the sentence.
• Structure of present indefinite tense:
• Subject+ the first form of verb(s/es)
• Example: He plays cricket.
• As the subject is 'a person' is singular; hence, the verb should be used in singular form only according to subject-verb agreement.
• By adding s/es with First form of verb we make the verb in singular form**; i.e. experience + s = experiences​**

​Correct Sentence- A person suffering from Alzheimer’s disease experiences short-term memory loss. 

Additional Information

• Subject-verb agreement: It states that the verb corresponding to the singular subject will be singular and the verb corresponding to the plural subject will be plural.
• Example:
• He plays cricket.
• They play cricket.

The correct answer is option 1 i.e. ​Contentment

Key Points  

• Contentment denotes a state of happiness and satisfaction; ease of mind/ happiness/ fulfilment,etc
• Let's see the meaning of other given options-

 

• Ambition, perseverance and hunger does not add a meaningful sense to the given context
• As per the meaning of all given options only 'Contentment' goes with the blank appropriately to give a contextually and grammatically correct sentence

Correct Sentence- Contentment is the key to their happiness; they are satisfied with what they have.

The correct answer is Option 3 i.e Women's solidarity knows no national boundaries.

Key Points

• The statement "As a woman, I have no country" means that:
• Women don't consider women from other countries and places different.
• Women identify with other women and the ideas of national boundaries are non-existent establishing a sense of solidarity.
• Now let's explore the given options:
• Option 1 is incorrect because nowhere does it say that women don't have a country. The speaker clearly states that "As a woman" she has no country. Here 'no country expresses' the idea of considering all women the same without national segregation.
• Option 2 is incorrect because the statement does not say anything regarding citizens or citizenship of any country.
• Option 3 is correct because it conveys the meaning of the sentence. It clearly puts forward the idea that women's solidarity knows no national boundaries which is exactly what the author is saying by expressing the fact that 'I have no country,' which means there are no segregating national boundaries.
• Option 4 is incorrect because the statement does not touch upon the topic of rights nor does it convey or express anything related to it.
• From the above-given explanation, we find Option 3 to be the correct answer as it expresses the meaning of the statement very well.

Concept:

According to the Binomial distribution 

The mean is given by,

λ = np

n= Number of times event has occurred or repeated

p = Probability of success

Since the average time between successive occurrences has been asked.

Therefore, we need to find a recurrence interval.

The probability that a particular rainfall is equal or exceeded (n=1) is a given by

${\rm{P = }}\frac{{\rm{1}}}{{\rm{T}}}$

where T = recurrence interval

Therefore, recurrence interval

${\rm{T = }}\frac{{\rm{1}}}{{\rm{P}}}$

Calculation:

Given,

Probability of an earthquake greater than Magnitude 6 occurring in the Garhwal Himalayas = 0.04

i.e P = 0.04

T = $1\over 0.04$

T = 25 years

Hence the average time between successive occurrences of such earthquakes is 25 years.

Concept:

The population of new city will be doubled when the population will be = 10 million

Calculation:

Given,

The population of new city = 5 million

It grows by 20% annually.

1) The increase in population for the 1^st year.

${\rm{Increase;in;population}} = \frac{{20}}{{100}} \times 5 = 1{\rm{;million}}$

Hence the population after 1 year = 6 million.

2) The increase in population for the 2^nd year.

${\rm{Increase;in;population}} = \frac{{20}}{{100}} \times 6 = 1.2{\rm{;million}}$

Hence the population after 2 year = 7.2 million.

3) The increase in population for the 3rd year.

${\rm{Increase;in;population}} = \frac{{20}}{{100}} \times 7.2 = 1.44{\rm{;million}}$

Hence the population after 3 year = 8.64 million.

4) The increase in population for the 4th year.

${\rm{Increase;in;population}} = \frac{{20}}{{100}} \times 8.64 = 1.728{\rm{;million}}$

Hence the population after 4 year = 10.368 million.

Hence it can be seen that the population of new city doubles in 3 to 4 years

Given,

​Som is younger to Riaz.

Riaz > Som

Shiv is elder to Ansu.

Shiv > Ansu

Ansu is the youngest in the group.

From the given statements,

Statement 1: Shiv is younger to Riaz.

It implies that Riaz > Shiv

We know that Riaz > Som and Shiv > Ansu

So the eldest child in the group is Riaz, which we can determine by only using statement 1.

Statement 2: Shiv is elder to Som. 

Shiv > Som

Than;

Riaz/Shiv > Som > Ansu (As Ansu is the youngest in the group)

But; we cannot find who is the eldest either Riaz or Shiv.

So, statement 2 cannot determine the answer.

Hence the correct answer is option 1.

The correct answer is Option 3 i.e Exactitude is not critical in dealing with big data.

Key Points

• Let's look at the key points from the above paragraph:
• Moving into a world of big data will demand us to change our conditioned attitude of being exact.
• Exactness is a product of the old, an artefact of the information deprived analogue age.
• In a world filled with big data being obsessed with exactness is unwanted.
• Big data age is no longer deprived of data and thus there is no need to gather every crucial point and be exact.
• Now looking at the points we find that:
• Option 1 is incorrect as it is not the main point; the main point is about Big Data and the need to change Exactitude.
• Option 2 is incorrect, the paragraph does not mention anywhere about Big Data's obsession with Exactness instead it states the opposite.
• Option 3 is correct as it clearly puts forward the main point of the paragraph. The fact that Exactitude is not relevant nor critical in the Big Data age.
• Option 4 is incorrect because the emphasis and of the paragraph is on Big Data and not Sparse data and bais.
• Therefore from the above points, we find that Option 3 is the correct answer as it contains the main point of the paragraph.

Additional Information

• Exactitude: the quality or an instance of being exact

Given,

The total quantity of exports of all the items = 5 lakh tonnes = 5 × 10⁸ kg

The total revenues of all items = 250 crore rupees = 250 × 10⁷ rupees

Now for item 1,

Item 1 exported in kg = (11/100) × (5 × 108) = 55 × 10⁶ kg

Revenue of item 1 exported = (12/100) × (250 × 10⁷) = 3 × 10⁸ rupees

${\rm{Revenue;of;item;}}1{\rm{;exported;per;kg}} = \frac{{3 \times {{10}^8}}}{{55 \times {{10}^6}}} = \frac{{300}}{{55}}$

Now for item 4,

Item 4 exported in kg = (22/100) × (5 × 108) = 11 × 107 kg

Revenue of item 4 exported = (6/100) × (250 × 107) = 15 × 107 rupees

${\rm{Revenue;of;item;}}4{\rm{;exported;per;kg}} = \frac{{15 \times {{10}^7}}}{{11 \times {{10}^7}}} = \frac{{15}}{{11}}$

Now ratio of the revenue generated through export of Item 1 per kilogram to the revenue generated through export of Item 4 per kilogram

$\frac{{{\rm{Revenue;of;item;}}1{\rm{;exported;per;kg}}}}{{{\rm{Revenue;of;item;}}4{\rm{;exported;per;kg}}}} = \frac{{300}}{{55}} \times \frac{{11}}{{15}} = \frac{4}{1}$

Hence the ratio is 4 : 1

Given, 

1. X is 1 km northeast of Y.

2. Y is 1 km southeast of Z.

3. W is 1 km west of Z.

4. P is 1 km south of W.

5. Q is 1 km east of P

We can draw the following diagram according to the given conditions,

From the figure,

ZQ = 1 km

Now,

∠ZYX = 90° 

∴ By Pythagoras theorem in triangle ZYX

${\rm{ZX}} = \sqrt {{{\left( {{\rm{ZY}}} \right)}^2} + {{\left( {{\rm{YX}}} \right)}^2}}$

${\rm{ZX}} = {\rm{;}}\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}}$

${\rm{ZX}} = \sqrt 2$

In right-angle triangle ZQX,

${\rm{QX}} = \sqrt {{{\left( {{\rm{ZQ}}} \right)}^2} + {{\left( {{\rm{ZX}}} \right)}^2}}$

${\rm{QX}} = {\rm{;}}\sqrt {{{\left( 1 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}$

${\rm{ZX}} = \sqrt 3$

Hence, "√3" is the correct answer.

Probability of getting (H.H.H.H) = ½ × ½ × ½ × ½ = 1/16

Probability of getting (H.H.H.H.T) = ½ × ½ × ½ × ½ × ½ = 1/32

Given condition is that (H.H.H.H) is already realized conditional probability of getting next H after

Probability of getting (H,H,H,H,T) when (H,H,H,H) is already happend (It will be the conditional probability of getting (H,H,H,H,T) when (H,H,H,H) has already occured)

P(H,H,H,H,T / H,H,H,H) $= \frac{{\frac{1}{{32}}}}{{\frac{1}{16}}} = 0.5$

Given that,

${\rm{Z}} = \frac{{2 - 3{\rm{i}}}}{{ - 5 + {\rm{i}}}}$

Multiplying and dividing by (-5 - i)

${\rm{Z}} = \frac{{\left( {2 - 3{\rm{i}}} \right)}}{{\left( { - 5 + {\rm{i}}} \right)}} \times \frac{{\left( { - 5 - {\rm{i}}} \right)}}{{\left( { - 5 - {\rm{i}}} \right)}}$

${\rm{Z}} = \frac{{ - 10 - 2i + 15i + 3{i^2}}}{{25 + 5i - 5i - {i^2}}}$

Here, i² = -1

${\rm{Z}} = \frac{{ - 10 - 13i + 3\left( { - 1} \right)}}{{25 - \left( { - 1} \right)}}$

${\rm{Z}} = \frac{{ - 13 - 13i}}{{26}}$

${\rm{Z}} = - \frac{{13}}{{26}} + \frac{{13i}}{{26}};$

${\rm{Z}} = - \frac{1}{2} + \frac{i}{2};$

Z = -0.5 + 0.5i

Concept: 

The standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation, is

${\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{ + P}}{\rm{.y = Q}}}$

Where, P and Q are the functions of 'X'

The integrating factor is given as,

${\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{\smallint P}}{\rm{dx}}}}$

The solution of the above equation is given as,

\(\begin{array}{*{20}{l}} {\left( {{\rm{I}}{\rm{.F}}} \right){\rm{y}} = {\rm{ \smallint}}\left( {{\rm{I}}{\rm{.F}}} \right){\rm{Q}}\ {\rm{dx}}}+C \end{array}\)

Calculation:

Given equation,

$\frac{{{\rm{dP}}}}{{{\rm{dt}}}} + {{\rm{K}}_2}{\rm{P}} = {{\rm{K}}_1}{{\rm{L}}_0}{{\rm{e}}^{ - {{\rm{K}}_1}{\rm{t}}}}$

This equation is a linear equation of first order, therefore comparing it with the general equation

${\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{ + P}}{\rm{.y = Q}}}$

Hence P = K₂

∴ ${\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{\smallint P}}{\rm{dt}}}}$

∴ ${\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{\smallint K_2}}{\rm{dt}}}}$

∴ ${\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{ K_2}}{\rm{t}}}}$

Concept:

Normal/Gaussian/Bell distribution:

Probability distribution function (PDF) for a normal distribution is:

${\rm{PDF = f}}\left( {\rm{x}} \right){\rm{ = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}};} }}{{\rm{e}}^{{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}{{\left( {\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}} \right)}^{\rm{2}}}}}$

where,

x = normal random variable

μ = mean = mode = median

σ = standard deviation. σ2 = variance

Calculation:

\(\mathop \smallint \limits_{ - \infty }^\infty {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 1\)

\(\mathop \smallint \limits_{ - \infty }^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} + \mathop \smallint \limits_{\rm{a}}^{ - \infty } {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 1\)

\(\mathop \smallint \limits_{ - \infty }^a {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 0.5\)

Stress strain for cement-paste:

The stress-strain curve for hardened cement paste is almost linear as shown in the figure.

Stress strain for aggregate:

The aggregate is more rigid than the cement paste and will therefore deform less (i.e. have a lower strain) under the same applied stress.

Stress strain for concrete:

The stress strain curve of concrete lies between those of the aggregate and the cement paste. However this relationship is non-linear over the most of the range.

Section modulus:

The section modulus (Z) of the cross-sectional shape is significant in designing beams. It is a direct measure of the strength of the beam.

A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.

To calculate Z, the distance (y) to the extreme fibres from the centroid (or neutral axis) must be found as that is where the maximum stress could cause failure.

\(\begin{array}{*{20}{l}} {\frac{{\rm{M}}}{{\rm{I}}}{\rm{ = }}\frac{{\rm{\sigma }}}{{\rm{y}}}{\rm{ = }}\frac{{\rm{E}}}{{\rm{R}}}}\ {{\rm{Z = }}\frac{{\rm{I}}}{{{{\rm{Y}}_{{\rm{max}}}}}}} \end{array}\)

where,

I = Second moment of inertia

Y_max = Distance of centroidal axis from the bottom

\({\rm{Z}} = \frac{{\rm{I}}}{{{{\rm{Y}}{{\rm{max}}}}}} = \frac{{{\rm{AY}}{{\rm{max}}}^2}}{{{{\rm{Y}}_{{\rm{max}}}}}}\)

Z = AY_max = First moment of inertia

Hence the first moment of area about the axis of bending for a beam cross-section is section modulus

Explanation

The target mean strength (f­m) is taken as

fm = fck + 1.65 σ

where,

fck = characteristic compressive strength, and σ = standard deviation

The standard deviation for different grade of concrete as per Clause 9.2.4.2 IS:456-2000

CONCEPT:

As per IS 456: 2000,

The Short term Static Modulus of elasticity of concrete Ec is given as-

\({{\rm{E}}{\rm{c}}} = 5000{\rm{;}}\sqrt {{{\rm{f}}{{\rm{ck}}}}} \)

Where,

fck = characteristic compressive strength of concrete,

fck and Ec are measured in MPa.

 

Initial tangent modulus:

• It is a slope to curve from the origin.
• The initial tangent modulus is applicable when material behaves linearly elastic

Secant Modulus

• It is defined as the slope of a line connecting the point at the start of a stress-strain diagram to a point with stress 1/3 rd of ultimate stress.
• It is also known as the static modulus of elasticity.
• In concrete, the stress-strain curve is nonlinear (assumed to be parabolic till a strain of 0.002) so an average increase in stress with respect to an average increase in strain to find elasticity modulus
• So we use the Secant modulus as the modulus of elasticity of concrete.

Important Points

As per IS 456: 2000,

Fine grained soil are classified on the basis of plastic chart. Clay is found to exist above A-line whereas silt & organic soil found to exist below A-line.

The equation for A-line is,

I_P = 0.73 (W_L - 20) 

Suffix and prefix used in soil classification:

Black cotton soil:

Black cotton soils have montmorillonite clay mineral which shows swelling and shrinkage behavior on wetting and drying.

Black Cotton Soil also consists of the excess of Montmorillonite mineral.

Bentonite clay shows swelling and shrinkage characteristics due to moisture content variation due to the presence of Montmorillonite mineral.

Bentonite clay is also called Montmorillonite clay.

Important point:

As black cotton soils are expansive soils i.e having swelling and shrinkage behavior and due to this Lightly loaded structures are more susceptible to damage as a result of volume changes in the soil. Therefore these soils need to be stabilized first before loading.

Investigation says that lime is a good stabilizing agent for the construction of a road in the black cotton soil.

Lime forms the bonding between the soil particles which is missing in black cotton soil and hence stabilizes.

For flexible footing:  Contact pressure is uniform.

For Rigid footing:  Settlement is uniform

 

Hence it can be seen that the contact pressure for a rigid footing resting on clay at the centre and the edges is minimum and maximum respectively

Concept: 

Continutiy equation in Three-Dimension

$\frac{{\partial {\rm{\rho }}}}{{\partial {\rm{t}}}} + \frac{{\partial \left( {{\rm{\rho U}}} \right)}}{{\partial {\rm{X}}}} + \frac{{\partial \left( {{\rm{\rho V}}} \right)}}{{\partial {\rm{Y}}}} + \frac{{\partial \left( {{\rm{\rho W}}} \right)}}{{\partial {\rm{Z}}}} = 0{\rm{;}}$

Where,

U, V, and W are components of velocity in X, Y and Z direction respectively

When the flow is steady$,;;\frac{{\partial {\bf{\rho }}}}{{\partial {\bf{t}}}} = 0$

$∴ \frac{{{\rm{\partial }}\left( {{\rm{\rho U}}} \right)}}{{{\rm{\partial X}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho V}}} \right)}}{{{\rm{\partial Y}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho W}}} \right)}}{{{\rm{\partial Z}}}} = 0$

When flow is steady and incompressible, ρ = constant

$∴ \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} + \frac{{\partial {\rm{W}}}}{{\partial {\rm{Z}}}} = 0{\rm{;}}$

When the flow is steady, incompressible and 2-D, $\frac{{\partial {\bf{W}}}}{{\partial {\bf{Z}}}} = 0$

$∴ \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}$

Calculation:

Given,

${\rm{u}} = \frac{{\rm{x}}}{{{{\rm{T}}_1}}}$, ${\rm{v}} = - \frac{{\rm{y}}}{{{{\rm{T}}_2}}}$ and w = 0

As per continuity equation, for two dimensional incompressible flow,

$∴ \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}$

$∴ \frac{{\partial {\rm{(\frac{{\rm{x}}}{{{{\rm{T}}_1}}})}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{(- \frac{{\rm{y}}}{{{{\rm{T}}_2}}})}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}$

$\frac{1}{{{{\rm{T}}_1}}} - \frac{1}{{{{\rm{T}}_2}}} = 0$

∴T₁ = T₂

Hence The given flow is incompressible if T1 = T2

Anemometer:

Anemometer is a device that is used to measure wind speed in the atmosphere. It is commonly used at weather stations.

Hygrometer:

Hygrometer is a device which is used to measure relative humidity (water vapour content).

Pitot Tube:

Pitot Tube is a device used calculating the velocity of flow at any point in a pipe or a channel. It is based on the principle that if the velocity of flow at a point becomes zero, the pressure there is increased due to the conversion of the kinetic energy into pressure energy. 

Tensiometer:

Tensiometers are used for the direct measurement of negative pore water pressures (rise of water due to capillarity) in the soils by establishing a continuous connection between soil pore water and the measuring system.

Concept:

Time of concentration: 

It is the time required for surface runoff to travel from hydraulically most distant point of the basin to the outlet.

Time of concentration depends upon the slope, the catchment characteristics and the flow path.

For a hydrograph analysis, time of concentration is defined as the time duration from the end of excess rainfall to the point of inflection.

For small catchment areas, lag time (tp) is equal to time of concentration (t_c)

tc = Hydrograph peak time - Half of rainfall duration

where,

t_p = Lag time, t_b = Base time

Q_p = Peak ordinate, T = Storm duration

OA = Approach segment

AB = Rising limb

BC = Crest segment

DE = Recession limb

Calculation:

T hr rainfall = 3hr

Hydrograph peak at 7 hours

Point of inflection at 8.5 hours

Time of concentration is given by,

∴ t_c = Hydrograph peak time - Half of rainfall duration

∴ t_c = 7 - (3/2) = 5.5 hrs

Hence, the time of concentration (in hours) for this catchment is 5.5 hrs.

Concept:

A common method of Hydrologic channel routing is Muskingum method, in which the channel storage in a reach is expressed as a function of both inflow and outflow discharge.

S = k [x Im – (1 - x)Qm]

Where, s = Storage in a channel reach

I = Inflow Discharge

Q = Outflow Discharge

k → Coefficient known as storage time coefficient. It was unit of time, it is approx. equal to time of travel of flood wave through the channel reach.

x → Weightage factor. Its range series from 0 to 0.5

• For naturally occurring channels m = 1, Hence the equation becomes

S = k [x I – (1 - x)Q]

• For x = 0, storage is only a function of outflow discharge. Such reservoirs are also known as Linear Reservoir.

Muskingum model of routing a flood through a stream reach is expressed as

O₂ = K₀I₂ + K₁I₁ + K₂O₁

where K₀, K₁and K₂ = Routing coefficients for the concerned reach

I₁ and I₂ = Inflows to the reach 1 and 2

O₁ and O₂ = Outflows from the reach 1 and 2

K₀ + K₁ + K₂ = 1.0

Concept:

Aerobic bacteria:

Bacteria that utilize oxygen for growth and oxygen-based metabolism is called aerobic bacteria.

Heterotrophs: 

It cannot produce organic compounds from inorganic sources and therefore rely on consuming other organisms in the food chain.

Anaerobic bacteria:

Bacteria that do not utilize oxygen for growth is called Anaerobic bacteria.

Autotrophs:

They store chemical energy in carbohydrate food molecules which they build themselves.

Explanation:

In activated sludge process, in the presence of food mass and oxygen heterotrophic bacteria converts food mass into biomass.

Hence the dominating microorganisms in an activated sludge process reactor are aerobic heterotrophs.

Important point:

Followings are the classification of secondary treatment units:

Concept:

1. Electrostatic Precipitators:-

• In electrostatic precipitators, the emitted gas flue gas is passed through a highly ionized atmosphere high-voltage field; and in that zone particulates get electrically charged and get separated from the gaseous stream with the help of electrostatic forces. Fine particles up to 1 μm wet or dry can be removed easily in an electrostatic pressure.
• Four basic steps required in the operation of a high-voltage single-stage electrostatic precipitator:

a) electrical charging of the particulates

b) collection of charged particles on a grounded surface

c) neutralization of the charge at the collector

d) removal of the particulate for disposal.

2. Gravitational Settling Chambers:

• Settling chambers in air-pollution control systems provide enlarged areas to minimize horizontal velocities and allow time for the vertical velocity to carry the particle to the floor.
• The emitted smokes, when made to pass through a settling chamber, drop some of their larger sized particles in the chamber as per Stoke’s Law. The largest size particle (d) that can be removed with 100% efficiency in such a chamber of length L and height H is given by

​\(d=C.\sqrt{\frac{18\mu .{{v}{h}}.H}{g.L.{{\rho }{p}}}}\)

Where, vh = Horizontal velocity of gas passing through the chamber

• Simple to design and maintain, and low-pressure flow.
• It requires larger space for installation and has low collection efficiency for small-sized particles.
• Only larger sized particles are separated.
• Although theoretically they should be able to remove particulates down to 5 or 10 μm, in actual, they are not practical for the removal of particles much less than 50 μm in size.

3. Fabric Filters

• In fabric filters, the gas stream laden with particles is passed through a woven or felted fabric that filters out particulate matter, allowing the gaseous matter to flow on.
• Small, particles are initially retained on the fabric by direct interception, inertial impaction, diffusion, electrostatic attraction, and gravitational settling.
• The collection of sub-micron particles is accomplished by sieving after a mat of dust gets found on the fabric.
• Filter bags, are capable of removing most particles as small as 0.5 μm and will remove substantial quantities of particles as small as 0.1 μm with an efficiency greater than 99%.

 

4. Wet Scrubbers:

• Wet scrubbers remove the particulates from the incoming gaseous stream by allowing the flue gases to flow up against a falling water stream.
• When aqueous chemical solutions, other than water, like lime, oil etc. it is used for the removal of gaseous pollutants also from the flue gases.
• Generally, collectors operating at very low pressure loss remove only medium- to coarse-size particles, while collectors operating at higher pressure losses are highly efficient at removing fine particles.
• Most commonly used wet collectors are— the spray tower, the wet cyclone scrubber, and the venturi scrubber.​

Explanation:

Hence the two air pollution control devices that are usually used to remove very fine particles from the flue gas are Electrostatic Precipitator and Fabric Filter

Concept:

$\text{e}+\text{f}=\frac{{{\text{V}}^{2}}}{127\text{R}}$

Where

V = Speed of the vehicle, e = super elevation, f = coefficient of lateral friction and R = radius of curve

Calculation:

Given:

V = 110 kmph, e = 8% = 0.08 and f = 0.10

$\text{e}+\text{f}=\frac{{{\text{V}}^{2}}}{127\text{R}}$

$0.08+0.10=\frac{{{110}^{2}}}{127\times \text{R}}$

R = 529.3 m

∴ The ruling design radius of the curve is 529.3 m

Concept:

Land Surveying: 

To determine the boundaries and areas of parcels of land, also known as property survey, boundary survey, or cadastral survey.

Engineering Survey: 

To collect requisite data for planning, design, and execution of engineering projects.

Topographic Surveying:  

It is used to identify and map the contours of the ground and existing features on the surface of the earth or slightly above or below the earth's surface (i.e. trees, buildings, streets, walkways, manholes, utility poles, retaining walls, etc.

Concept:

L - Hospital's Rule:

${\rm{If;}}\mathop {\lim }\limits_{{\rm{X}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}{\rm{;is;of;the;form}}\frac{0}{0}\left( {{\rm{or}}} \right)\frac{\infty }{\infty }$

${\rm{then;}}\mathop {\lim }\limits_{{\rm{X}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \mathop {\lim }\limits_{{\rm{X}} \to {\rm{a}}} \frac{{{\rm{f'}}\left( {\rm{X}} \right)}}{{{\rm{g'}}\left( {\rm{X}} \right)}} = \frac{{{\rm{f'}}\left( {\rm{a}} \right)}}{{{\rm{g'}}\left( {\rm{a}} \right)}}$

When f’(a) = g’(a) ≠ 0

Calculation:

Given,

$\mathop {\lim }\limits_{{\rm{\alpha }} \to {\rm{0}}} \frac{{{\rm{x^{\alpha}-1}} }}{{\ {\rm{\alpha}} }}$

It is of $\frac{0}{0}$ form

∴ By L – Hospital's rule,

${\rm{L}} = \frac{{\frac{{{\rm{d}}\left( {{{\rm{x}}^{{\rm{\alpha }}} - 1}} \right)}}{{{\rm{d\alpha }}}}}}{{\frac{{{\rm{d}}\left( {\rm{\alpha }} \right)}}{{{\rm{d\alpha }}}}}}$

${\rm{L}} = \frac{{{{\rm{x}}^{\rm{\alpha }}}\log {\rm{x}}}}{1}$

L = log x

Concept:

Poisson Distribution

The Poisson distribution can be used to model the vehicle arrival

${\rm{P}}\left( {{\rm{n}},{\rm{t}}} \right) = \frac{{{{\rm{e}}^{ - {\rm{λ t}}}}{{\left( {{\rm{λ t}}} \right)}^{\rm{n}}}}}{{{\rm{n}}!}}$

Where,

λ = Number of vehicles per second

t = Time interval

n = Number of vehicles

Calculation:

Given,

λ = 240 veh/hr = 240/3600 veh/sec = 0.0666 veh/sec

n = 1, t = 30sec

${\rm{P}}\left( {{\rm{n}},{\rm{t}}} \right) = \frac{{{{\rm{e}}^{ - {\rm{λ t}}}}{{\left( {{\rm{λ t}}} \right)}^{\rm{n}}}}}{{{\rm{n}}!}}$

${\rm{P}}\left( {{\rm{1}},{\rm{30}}} \right) = \frac{{{{\rm{e}}^{ - {\rm{0.0666× 30}}}}{{\left( {{\rm{0.0666× 30}}} \right)}^{\rm{1}}}}}{{{\rm{1}}!}}$

P (1, 30) = e^-2 × 2 = 0.27

Loading after removing the cable

${\rm{Axial;stress}} = \frac{{{\rm{Axial;force}}}}{{{\rm{Area;of;the;beam}}}}$

${\rm{Axial;stress}} = {\rm{σ }} = \frac{{50}}{{0.2 × 0.2}} = 1250{\rm{;kN}}/{{\rm{m}}^2}$

Shear force = V = 50 kN

Bending moment = M = 50 × 3 = 150 kN

∴ Axial stress = σ = 1250 kN/m², Bending moment = M = 150 kN, Shear force = V = 50 kN

Calculation:

Given:

Distribution Factor:

Fixed End Moments:

M_AB = M_BA = M_PE = M_EP = M_CD = M_DC = 0

${{\rm{M}}_{{\rm{BP}}}} = - \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = - \frac{{24 \times {8^2}}}{{12}} = - 128{\rm{;kN}}.{\rm{m}}$

${{\rm{M}}_{{\rm{PB}}}} = \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = \frac{{24 \times {8^2}}}{{12}} = 128{\rm{;kN}}.{\rm{m}}$

${{\rm{M}}_{{\rm{PC}}}} = - \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = - \frac{{24 \times {8^2}}}{{12}} = - 128{\rm{;kN}}.{\rm{m}}$

${{\rm{M}}_{{\rm{CP}}}} = \frac{{{\rm{W}}{{\rm{L}}^2}}}{{12}} = \frac{{24 \times {8^2}}}{{12}} = 128{\rm{;kN}}.{\rm{m}}$

Distribution Table:

Hence the magnitude of bending moment at P is 176 kNm

Degree of static indeterminancy = D_S = 0

∴ Number of plastic hinges = D_S + 1 = 1

 

Hence from the principal of virtual work

\(\left( { - {{\rm{M}}{\rm{P}}} \times {\rm{\theta }}} \right) - \left( {{{\rm{M}}{\rm{P}}} \times {\rm{\theta }}} \right) + \left( {{\rm{P}} \times \frac{{\rm{L}}}{2} \times {\rm{\theta }}} \right) - \left( {\frac{{\rm{P}}}{2} \times \frac{{\rm{L}}}{3} \times {\rm{\theta }}} \right) = 0\)

$- 2{{\rm{M}}_{\rm{P}}}{\rm{\theta }} + \frac{{{\rm{PL\theta }}}}{3} = 0$

${\rm{P}} = \frac{{6{{\rm{M}}_{\rm{P}}}}}{{\rm{L}}}$

Given,

Length of the cable = 10 m

After resolving the forces in the cable, we get the following figure

ΣF_Y = 0

2T cos θ = 120

Where Cos θ = 4/5

2 × T × (4/5) = 120  

T = 75 kN

Concept:

Principle stress equation is given by,

${{\rm{σ }}_1},;{{\rm{σ }}_2} = \frac{{{{\rm{σ }}_x} + ;{{\rm{σ }}_y}}}{2} \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_x} - ;{{\rm{σ }}_y}}}{2}} \right)}^2} + {{\left( {{τ _{xy}}} \right)}^2}}$

Where,

σ1 = Maximum principal stress

σ2 = Minimum principal stress

σx = Stress in x direction

σy = Stress in Y direction

τxy = Shear stress in xy plane

Maximum shear stress is given by

\({{\rm{τ }}{{\rm{max}}}} = \sqrt {{{\left( {\frac{{{{\rm{σ }}{\bf{x}}} - ;{{\rm{σ }}{\bf{y}}}}}{2}} \right)}^2} + {{\left( {{{\bf{τ }}{{\bf{xy}}}}} \right)}^2}} \)

Calculation:

Given,

σ_x = -2 MPa, σ_y = 4 MPa, τ_xy = 4 Mpa

\({{\rm{τ }}{{\rm{max}}}} = \sqrt {{{\left( {\frac{{{{\rm{σ }}{\bf{x}}} - ;{{\rm{σ }}{\bf{y}}}}}{2}} \right)}^2} + {{\left( {{{\bf{τ }}{{\bf{xy}}}}} \right)}^2}} \)

\({{\rm{τ }}{{\rm{max}}}} = \sqrt {{{\left( {\frac{{{{\rm{-2 }}{\ }} - ;{{\rm{4 }}_{}}}}{2}} \right)}^2} + {{\left( {{{4}}} \right)}^2}} \)

τ_max = 5 MPa

Concept:

Stability analysis of an infinite slope of cohesive soils

Shear strength of cohesive soils is given by,

S = C + (γ_d H Cos² i) tan ϕ

Shear stress of cohesive soil is given by,

τ = γ_d H cos i sin i

As the normal stress σ depends upon the height H of the slope, an expression for the height can be found by equating the shear stress and shear strength

C + (γ_d H Cos² i) tan ϕ = γ H cos i sin i

${\rm{\gamma_d H}}{\cos ^2}{\rm{i}}\left( {\frac{{\sin {\rm{i}}}}{{\cos {\rm{i}}}} - \tan ϕ } \right) = {\rm{C}}$

${{\rm{H}}_{\rm{C}}} = \frac{{\rm{C}}}{{{\rm{\gamma_d }}\left( {\tan {\rm{i }} - \tan ϕ } \right){{\cos }^2}{\rm{i}}}}$

This the height at which the slope is just stable and is known as critical height (H_C)

Calculation:

Given,

C = 20 kPa, γ_d = 16 kN/m³

i = 40°, H = 5 m

${{\rm{H}}_{\rm{C}}} = \frac{{\rm{C}}}{{{\rm{\gamma_d }}\left( {\tan {\rm{i }} - \tan ϕ } \right){{\cos }^2}{\rm{i}}}}$

${{\rm{5}}} = \frac{{\rm{20}}}{{{\rm{16 }}\left( {\tan {\rm{40 }} - \tan ϕ } \right){{\cos }^2}{\rm{40}}}}$

tan ϕ = 0.413

ϕ = 22.44°

Concept:

Following are some of the methods for in-situ site exploration

1) Pressuremeter Test (PMT):

It is developed by Menard.

It is used to determine stress-deformation characteristics of the soil in natural condition.

2) Static Cone Penetration Test (SCPT):

Cone resistance (f_C) is determined

The relation between resistance of cone (f_C) and standard penetration number (N) is as follows

For Gravel f_C = 800 N to 1000 N

For Sands f_C = 500 N to 600 N

For silty sands f_C = 300 N to 400 N

For silts and Clayey silts f_C = 200 N

3) Standard Penetration Test (SPT):

It is specially used for cohesionless soils. It is used to determine the relative density and the angle of shearing resistance of cohesionless soils.

It can also be used to determine the unconfined compressive strength of cohesive soils.

The standard penetration number is equal to the number of blows required for 300 mm of penetration beyond a seating drive of 150 mm.

If the number of blows for 150 mm drive exceeds 50, it is taken as refusal and the test is discontinued.

4) Vane shear test (VST):

It is used to determine shear strength of a cohesive soil in its natural condition.

The undrained shear strength (C_U) of the soil is determined.

The Vane-Shear test is extremely useful for determining the in-situ shear strength of very soft and sensitive clays. The method cannot be used for sandy soils.

Explanation:

Concept:

Ultimate bearing capacity of a pile is given by,

Q_U = Q_b + Q_S

Where,

Q_b = End bearing resistance = A_b f_b

Q_S = Skin frictional resistance or shaft resistance = A_S f_S

A_b = Base area of pile = π/4 d²

f_b = Bearing capacity at the end of the pile

A_s = Surface area of pile = π d L

f_S = Shear resistance between pile and soil

For sandy soils:

Q_U = A_b σ’_V N_q + A_S K σ’_Vavg tan δ

σ’_V = Effective vertical stress at pile base level

N_q = Depends on ϕ values

K = coefficient of lateral earth pressure

σ’_Vavg = Effective vertical average stress along the pile length

δ = Angle of friction between pile and soil

Calculation:

Given,

d = 500 mm, L = 20 m, ϕ = 30°

γ_d = 20 kN/m³, δ = 2/3ϕ, K = 2.7

N_q = 25

For friction pile for sandy soils

Q_U = A_S f_S

\({{\rm{f}}{\rm{S}}} = \frac{1}{2}{\rm{;}}{{\rm{\bar \sigma }}{\rm{V}}}{\rm{K}}\tan {\rm{\delta }}\)

σ_V = γ_d × L = 20 × 20 = 400 kN/m²

$\tan {\rm{\delta }} = \tan \left( {\frac{2}{3}\phi } \right) = 0.364$

${{\rm{f}}_{\rm{S}}} = \frac{1}{2} \times{\rm{;}}400 \times 2.7 \times 0.364 = 196.56{\rm{;kN}}/{{\rm{m}}^2}$

 ∴ Q_U = 196.56 × π d L

∴ Q_U = 196.56 × π × 0.5 × 20 = 6175 kN

The ultimate bearing of capacity = 6175 kN

Concept:

Immediate settlement of cohesive soils

The linear theory of elasticity is used to determine the elastic settelment of the footings on saturated clay.

\({{\rm{S}}{\rm{i}}} = {\rm{qB}}\left( {\frac{{1 - {{\rm{μ }}^2}}}{{{{\rm{E}}{\rm{S}}}}}} \right){\rm{I}}\)

Where,

q = Uniformly distributed load (kPa), B = Characteristic length of the loaded area

E_S = Modulus of elasticity of the soil (MPa), μ = Poisson’s ratio

I = Influence factor

I = Influence factor for rigid footing = 0.8

Calculation:

Given,

q = 110 kPa, E_S = 30 MPa

μ = 0.3, B = 20 m

I = For rigid footing = 0.8 

\({{\rm{S}}{\rm{i}}} = {\rm{qB}}\left( {\frac{{1 - {{\rm{μ }}^2}}}{{{{\rm{E}}{\rm{S}}}}}} \right){\rm{I}}\)

${{\rm{S}}_{\rm{i}}} = {\rm{110\times 20}}\left( {\frac{{1 - {{\rm{0.3 }}^2}}}{{{{\rm{30}}}}}} \right){\rm{0.8}}$

S_i = 53.38 mm

Concept:

The force exerted by a horizontal jet normally on a vertical fixed plate is given by

In X-direction

F_X = ρ A V², where A and V is area and velocity of jet respectively.

In Y-direction

F_Y = 0

The force exerted by a horizontal jet normally on a vertical moving plate is given by

In X-direction

F_X = ρ A (V – u)², where A and V is area and velocity of jet respectively, and u is the velocity of the plate.

In Y-direction

F_Y = 0

Calculation:

Given,

Q = 15 litres/sec, ρ = 1000 kg/m³

d = dia of nozzle = dia of jet = 30 mm = 0.03 m

${\rm{Velocity;of;jet;}} = \frac{{\rm{Q}}}{{\rm{A}}} = \frac{{15 \times {{10}^{ - 3}}}}{{\frac{{\rm{\pi }}}{4} \times {{\left( {0.03} \right)}^2}}} = 21.22{\rm{;m}}/{\rm{sec}}$

Force on a fixed plate

F_X = ρ A V²

F_X = 1000 × (π/4 × 0.03 × 0.03) × (21.22)²

F_X = 318.29 N

Concept:

Venturimeter:

A venturimeter is the device used for measuring the rate of a flow of a fluid flowing through a pipe. It consists of three parts

(a) A short converging part, (b) Throat, and (c)  Diverging part

It is based on the principle of Bernoulli’s equation

The expression for the rate of flow throw venturimeter

\({{\rm{Q}}{{\rm{act}}}} = {{\rm{C}}{\rm{d}}} \times \frac{{{{\rm{a}}_1}{{\rm{a}}_2}}}{{\sqrt {{\rm{a}}_1^2 - {\rm{a}}_2^2} }} \times \sqrt {2{\rm{gh}}} \)

Where,

a₁ and a₂ = Area at section 1 and 2 respectively

C_d = Co-efficient of venturimeter and its value is less than 1

h = Difference of pressure head

Calculation:

Given,

d₁ = Diameter at section 1 = 0.2 m

d₂ = Diameter at section 2 = 0.1 m

h = 2 m, C_d = 1, g = 9.81 m/sec²

${{\rm{a}}_1} = \frac{{\rm{\pi }}}{4} \times {\left( {0.2} \right)^2} = 0.0314{\rm{;}}{{\rm{m}}^2}$

${{\rm{a}}_2} = \frac{{\rm{\pi }}}{4} \times {\left( {0.1} \right)^2} = 0.0078{\rm{;}}{{\rm{m}}^2}$

\({{\rm{Q}}{{\rm{act}}}} = {{\rm{C}}{\rm{d}}} \times \frac{{{{\rm{a}}_1}{{\rm{a}}_2}}}{{\sqrt {{\rm{a}}_1^2 - {\rm{a}}_2^2} }} \times \sqrt {2{\rm{gh}}} \)

${{\rm{Q}}_{{\rm{act}}}} = 1 \times \frac{{0.0314 \times 0.0078}}{{\sqrt {{{\left( {0.0314} \right)}^2} - {{\left( {0.0078} \right)}^2}} }} \times \sqrt {2 \times 9.81 \times 2}$

Q_act = 0.050 m³/sec

Important point:

The expression for the rate of flow throw orifice meter

\({{\rm{Q}}{{\rm{act}}}} = {{\rm{C}}{\rm{d}}} \times \frac{{{{\rm{a}}_0}{{\rm{a}}_1}}}{{\sqrt {{\rm{a}}_1^2 - {\rm{a}}_0^2} }} \times \sqrt {2{\rm{gh}}} \)

Where,

C_d = Coeffecient of discharge for orifice meter

The C_d for orifice meter is much smaller than the venturimeter

Concept:

Critical flow analysis:

1. Specific energy is minimum for a given discharge.

2. Specific force is minimum for a given discharge.

3. Discharge is maximum for a given specific energy.

4. Discharge is maximum for a given specific force.

5. Froude number = F_r = 1

6. Velocity head is equal to half of hydraulic depth, $\frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} = \frac{{\rm{D}}}{2}$

Relations for critical flow:

General equation valid for critical flow of any shape of channel

$\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}} = {\rm{F}}_{\rm{r}}^2$

For critical flow in reactangular channel

${{\rm{Y}}_{\rm{c}}} = {\rm{critical;depth}} = {\left( {\frac{{{{\rm{Q}}^2}}}{{{\rm{g}}{{\rm{b}}^2}}}} \right)^{\frac{1}{3}}}$

\({{\rm{E}}{\rm{c}}} = {\rm{Specific;energy}} = \frac{3}{2} × {{\rm{Y}}{\rm{c}}}\)

Calculation:

Given,

b = Width of channel = 2.5 m

Q = Discharge = 4 m³/sec

g = 9.81 m/sec²

${{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{Q}}^2}}}{{{\rm{g}}{{\rm{b}}^2}}}} \right)^{\frac{1}{3}}}$

${{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{4}}^2}}}{{{\rm{9.81× }}{{\rm{2.5}}^2}}}} \right)^{\frac{1}{3}}}$

Y_c = 0.64 m

Now,

Q = A × V_c

${\rm{V_c}} = \frac{{\rm{Q}}}{{\rm{A}}} = \frac{4}{{{\rm{B}} \times {{\rm{Y}}_{\rm{c}}}}} = \frac{4}{{2.5 \times 0.64}} = 2.5{\rm{;m}}/{\rm{sec}}$

Hence the velocity of flow corresponding to critical depth = 2.5 m/sec

Important point:

For critical flow in parabolic channel

${Y_c} = {\left( {\frac{{27{Q^2}}}{{8gB}}} \right)^{1/4}}$

\({{\rm{E}}{\rm{c}}} = {\rm{Specific;energy}} = \frac{4}{3} × {{\rm{Y}}{\rm{c}}}\)

For critical flow in triangular channel

${{\rm{Y}}_{\rm{c}}} = {\rm{critical;depth}} = {\left( {\frac{{{{\rm{2Q}}^2}}}{{{\rm{g}}{{\rm{m}}^2}}}} \right)^{\frac{1}{5}}}$

\({{\rm{E}}{\rm{c}}} = {\rm{Specific;energy}} = \frac{5}{4} × {{\rm{Y}}{\rm{c}}}\)

Concept:

Field Capacity (F.C):

It is the amount of water held in the soil after excess water gets drained off to gravity and rate of downward movement has decreased.

In medium textured soil, F.C is about 50% of pore volume.

Soil moisture tension at F.C ranges between one-tenth to one-third atmosphere.

Permanent Wilting Point (P.W.P):

Water content at which plant is no longer able to extract water from the soil for its growth.

Soil moisture tension at P.W.P ranges between 7 to 32 atmosphere.

Depth of irrigation water (d_w):

d_W = G × d × (F.C – P.W.P)

where,

d = Effective root zone of the crop

\({\rm{G}} = {\rm{Specific;gravity;of;soil}} = \frac{{{{\rm{\gamma }}{\rm{d}}}}}{{{{\rm{\gamma }}{\rm{w}}}}}\)

Calculation:

Given,

F.C = 28% = 0.28, P.W.P = 18% = 0.18, d = 70 cm

γ_d = 1.3 g/cm³, γ_w = 1.0 g/cm³

∴ G = 1.3/1 = 1.3

d_W = G × d × (F.C – P.W.P)

d_W = 1.3 × 70 × (0.28 – 0.18)

d_W = 9.1 cm 

We got d_w = 9.1 cm, but we want answer in mm, therefore d_w = 91 mm

But we have no such option, therefore answer is none of the above

Given

${\rm{U}} = {\rm{;}} - \frac{{{{\rm{h}}^2}}}{{8{\rm{\mu }}}}\frac{{{\rm{dp}}}}{{{\rm{dx}}}}\left[ {1 - 4{{\left( {\frac{{\rm{y}}}{{\rm{h}}}} \right)}^2}} \right]$

The maximum velocity occurs at y = 0

$\therefore {{\rm{U}}_{{\rm{max}}}} = {\rm{;}} - \frac{{{{\rm{h}}^2}}}{{8{\rm{\mu }}}}\frac{{{\rm{dp}}}}{{{\rm{dx}}}}\left[ {1 - 4{{\left( {\frac{0}{{\rm{h}}}} \right)}^2}} \right]$

$\therefore {{\rm{U}}_{{\rm{max}}}} = {\rm{;}} - \frac{{{{\rm{h}}^2}}}{{8{\rm{\mu }}}}\frac{{{\rm{dp}}}}{{{\rm{dx}}}}$

The average velocity is given by,

\({{\rm{U}}{{\rm{avg}}}} = \frac{{\rm{Q}}}{{\rm{A}}} = \frac{{\smallint {\rm{U}}.{\rm{dA}}}}{{\rm{A}}} = \frac{{\smallint {{\rm{U}}{{\rm{max}}}}\left( {1 - 4\frac{{{{\rm{y}}^2}}}{{{{\rm{h}}^2}}}} \right){\rm{dA}}}}{{\rm{A}}}\)

\({{\rm{U}}{{\rm{avg}}}} = \frac{{2\mathop \smallint \nolimits_0^{\frac{h}{2}} {{\rm{U}}{{\rm{max}}}}\left( {1 - 4\frac{{{{\rm{y}}^2}}}{{{{\rm{h}}^2}}}} \right){\rm{dy}} \times 1}}{{{\rm{h}} \times 1}}\)

\({{\rm{U}}{{\rm{avg}}}} = \frac{{2{{\rm{U}}{{\rm{max}}}}}}{{\rm{h}}}\left[ {{\rm{y}} - \frac{{4{{\rm{y}}^3}}}{{3{{\rm{h}}^2}}}} \right]_0^{\frac{{\rm{h}}}{2}}\)

\({{\rm{U}}{{\rm{avg}}}} = \frac{{2{{\rm{U}}{{\rm{max}}}}}}{{\rm{h}}}\left[ {\frac{{\rm{h}}}{2} - \frac{{4{{\rm{h}}^3}}}{{24{{\rm{h}}^2}}}} \right]\)

\({{\rm{U}}{{\rm{avg}}}} = \frac{2}{3}{{\rm{U}}{{\rm{max}}}}\)

Concept:

Settling velocity:

The settling velocity of discrete particles can be determined using appropriate Stokes equation depending on Reynolds number

By Stokes law,

${{\rm{V}}_{\rm{s}}} = \frac{{{\rm{g}}\left( {{\rm{G}} - 1} \right){{\rm{d}}^2}}}{{18{\rm{\nu }}}}$

Where,

G = Specific gravity of particles, d = Diameter of particle

ν = Kinematic viscosity

Stokes law holds good for Reynolds number, R_e < 1

Calculation:

Given,

d = 0.06 mm, g = 9.81 m/s², Q = 0.1 m³/sec

G = 2.65, ν = 1.0105 × 10^-2 cm²/sec = 1.0105 × 10^-6 m²/sec

${{\rm{V}}_{\rm{s}}} = \frac{{{\rm{g}}\left( {{\rm{G}} - 1} \right){{\rm{d}}^2}}}{{18{\rm{\nu }}}}$

${{\rm{V}}_{\rm{s}}} = \frac{{9.81 \times \left( {2.65 - 1} \right){{\left( {0.06 \times {{10}^{ - 3}}} \right)}^2}}}{{18 \times 1.0105 \times {{10}^{ - 6}}}}$

V_S = 3.20 × 10^-3 m/sec

${\rm{Surface;area}} = {\rm{A}} = \frac{{\rm{Q}}}{{{{\rm{V}}_{\rm{s}}}}} = \frac{{0.1}}{{3.2 \times {{10}^{ - 3}}}} = 31.25{\rm{;}}{{\rm{m}}^2}$

Hence the minimum surface area required to remove particles of diameter 0.06 mm = 31.25 m²

Concept:

Aluminium sulphate (Alum):

1. Alum is a universal coagulant. It is a very economical and cheap chemical compound.

2. Alum produces effective floc in the water when the pH of water is between 6.5 to 8.5.

3. Besides acting as a coagulant it also removes colour, odour and improves the taste of water.

4. But it requires sufficient amount of alkalinity in water to produce floc and it also imparts non- carbonaceous hardness in water and thus water becomes corrosive.

Calculating any quantity in kg:

${\rm{Quantity;}}\left( {{\rm{kg}}} \right) = {\rm{Flow;rate;}}\left( {\frac{{{\rm{litre}}}}{{{\rm{day}}}}} \right) \times {\rm{No;of;days}} \times {\rm{Dosage;of;compound}}\left( {\frac{{{\rm{kg}}}}{{{\rm{litre}}}}} \right)$

Calculation:

Given,

Q = Flow rate = 35 m³/min

∴ Q (litre/day) = 35 × 10³ × 60 × 24

Alum dosage = 25 mg/litre

∴ Dosage (kg/litre) = 25 × 10^-6 kg/litre

${\rm{Quantity;}}\left( {{\rm{kg}}} \right) = {\rm{Flow;rate;}}\left( {\frac{{{\rm{litre}}}}{{{\rm{day}}}}} \right) \times {\rm{No;of;days}} \times {\rm{Dosage;of;compound}}\left( {\frac{{{\rm{kg}}}}{{{\rm{litre}}}}} \right)$

${\rm{Quantity;}}\left( {{\rm{kg}}} \right) = 35 \times {10^3} \times 60 \times 24 \times 30 \times 25 \times {10^{ - 6}}$

Quantity in kg = 37800 kg

∴ The quantity of alum required for 30 days operation of plant = 37800 kg

Concept:

Chick’s law for disinfection:

\({{\rm{N}}{\rm{t}}} = {{\rm{N}}{\rm{o}}}{{\rm{e}}^{ - {\rm{Kt}}}}\)

Where,

N_t = Number of organisms remained at any time t

N_o = Number of initial organisms at time t = 0

K = Disinfection rate constant

t = Contact time or detention time or time of exposure

Calculation:

Given,

To achieve a 98% kill of micro-organisms,

Effluent flow rate = Q = 2670 m³/day

Chlorine dosage = 15 mg/litre

\({{\rm{N}}{\rm{t}}} = {{\rm{N}}{\rm{o}}}{{\rm{e}}^{ - 0.145{\rm{t}}}}\)

Now, to achieve a kill of 98% micro-organisms that means 2% organisms are alive and at time t = 0 there will be 100% micro-orgainsms alive.

∴ N_t = 2% = 0.02 and N_o = 100% = 1

$0.02 = 1 \times {{\rm{e}}^{ - 0.145{\rm{t}}}}$

${\log _{\rm{e}}}0.02 = - 0.145 \times {\rm{t}} \times {\log _{\rm{e}}}{\rm{e}}$

∴ t = 26.98 min

Now we know that

${\rm{Q}} = \frac{{{\rm{Volume}}}}{{{\rm{time}}}}{\rm{;}}$

∴ Volume = Q × t

 ${\rm{Volume}} = 2670 \times \frac{{26.98}}{{24 \times 60}} = 50.02{\rm{;}}{{\rm{m}}^3} \approx 50{\rm{;}}{{\rm{m}}^3}$

The volume of the disinfection unit required to achieve a 98% kill of micro-organisms = 50.02 m³

Important point:

The percentage removal of micro-organisms

\({\rm{% ;Removal}} = \frac{{{{\rm{N}}{\rm{o}}} - {{\rm{N}}{\rm{t}}}}}{{{{\rm{N}}_{\rm{o}}}}} \times 100\)

The relation between the dose of disinfectant ‘C’ and contact time ‘t’

Cⁿt = constant

Where n = Dilution factor

Concept:

Biochemical oxygen demand (BOD):

BOD is defined as amount of oxygen demanded by the micro-organisms C₅H₇NO₂ (Bacteria) present in it to decompose biodegradable organic matter in wastewater under aerobic conditions.

BOD is measure of strength of waste water

5 day BOD @ 20°c = [(DO) _I - (DO) _F] × D.F

(DO) _I = Initial dissolved oxygen

(DO) _F = Final dissolved oxygen

D.F = Dilution factor

${\rm{D}}.{\rm{F}} = \frac{{300}}{{{{\rm{V}}_{\rm{s}}}}} = \frac{{100}}{{{\rm{\% dilution}}}}$

\({\rm{D}}{{\rm{O}}{{\rm{initial}}}} = \frac{{{{\rm{Q}}{\rm{S}}}{{\left( {{\rm{DO}}} \right)}{\rm{S}}} + {{\rm{Q}}{\rm{R}}}{{\left( {{\rm{DO}}} \right)}{\rm{R}}}}}{{{{\rm{Q}}{\rm{S}}} + {{\rm{Q}}_{\rm{R}}}}}\)

Q_S and Q_R is flow of water in stream and river respectively

(DO)_S and (D)_R is dissolved oxygen in stream and river respectively

Expression for BOD

L_o = Initial organic matter in waste water at time t =0

L_t = Organic matter remained in wastewater at any time t

L_o – L_t = Organic matter removed in time t

\({{\rm{L}}{\rm{t}}} = {{\rm{L}}{\rm{o}}}{{\rm{e}}^{ - {\rm{Kt}}}}\)

Where

K = BOD rate constant

Calculation:

Given,

Q_S = 2 m³/sec, (BOD)_s = 90 mg/litre

Q_R = 12 m³/sec, (BOD)_R = 5 mg/litre

Area = 50 m², K = 0.25/day

Distance of downstream = 10 km

\({\rm{BO}}{{\rm{D}}{{\rm{mix}}}} = \frac{{{{\rm{Q}}{\rm{S}}}{{\left( {{\rm{BOD}}} \right)}{\rm{S}}} + {{\rm{Q}}{\rm{R}}}{{\left( {{\rm{BOD}}} \right)}{\rm{R}}}}}{{{{\rm{Q}}{\rm{S}}} + {{\rm{Q}}_{\rm{R}}}}}\)

${\rm{BO}}{{\rm{D}}_{{\rm{mix}}}} = \frac{{\left( {2 \times 90} \right) + \left( {12 \times 5} \right)}}{{2 + 12}} = 17.14{\rm{;mg}}/{\rm{lit}}$

Now we know that

${\rm{Velocity}} = \frac{{{{\rm{Q}}_{{\rm{mix}}}}}}{{\rm{A}}} = \frac{{2 + 12}}{{50}} = 0.28{\rm{;}}{{\rm{m}}^3}$

${\rm{Time;taken;to;travel;}}10{\rm{;km;distance;}} = \frac{{{\rm{Distance}}}}{{{\rm{velocity}}}} = \frac{{10000}}{{0.28}} = 35714.28{\rm{;sec}} = 0.41{\rm{;days}}$

\({{\rm{L}}{\rm{t}}} = {{\rm{L}}{\rm{o}}}{{\rm{e}}^{ - {\rm{Kt}}}}\)

${{\rm{L}}_{\rm{t}}} = 17.14 \times {{\rm{e}}^{ - 0.25 \times 0.41}} = 15.46{\rm{;mg}}/{\rm{litre}}$

Hence the BOD of river water, 10 km downstream of the mixing point = 15.46 mg/litre

Concept:

Moving observer method:

Moving car or moving observer method of traffic stream measurement has been developed to provide simultaneous measurement of traffic stream variables. It has the advantage of obtaining the complete state with just three observers and a vehicle.

The mean speed by moving the observer method is given as follows

\({{\rm{V}}{\rm{s}}} = \frac{{\rm{L}}}{{{{\rm{t}}{\rm{w}}} - \frac{{{{\rm{m}}_{\rm{w}}}}}{{\rm{q}}}}}\)

t_w = Observation time when the observer is moving with the stream

m_w = Net vehicles overtake the observer when it is moving with the stream

q = Flow density when the observer is at rest in vehicles/min

Calculation:

Total time taken by the student to reach home = 40 min

Student’s stop time = 15 min

∴ t_w = 40 – 15 = 25 min = 0.416 hr

L = 5 km

${\rm{q}} = \frac{{45{\rm{;vehicles}}}}{{15{\rm{;min}}}} = 3\frac{{{\rm{veh}}}}{{{\rm{minute}}}} = 180{\rm{;veh}}/{\rm{hr}}$

m_w = 60 vehicles

\({{\rm{V}}{\rm{s}}} = \frac{{\rm{L}}}{{{{\rm{t}}{\rm{w}}} - \frac{{{{\rm{m}}_{\rm{w}}}}}{{\rm{q}}}}}\)

${{\rm{V}}_{\rm{s}}} = \frac{5}{{0.416 - \frac{{60}}{{180}}}} = 60.48{\rm{;km}}/{\rm{hr}}$

Hence the speed of vehicle stream = 60 km/hr

Concept:

Traffic density (K):

The number of vehicles occupying a unit length of road at a given instant.

It is expressed in vehicles/km

Traffic volume (q):

The number of vehicles occupying moving in a specified direction on a given road during specified.

It is expressed in vehicles/hr or vehicle/day

Traffic capacity (C):

It is the ability of road to accommodate traffic volume

${\rm{C}} = \frac{{1000 × {\rm{V}}}}{{\rm{S}}}$

where S = Average center to center spacing of a vehicle

It is expressed in vehicles/hr/road

The relation between traffic volume, traffic density and speed is given by

q = K × V

Capacity flow or maximum flow (q_max):

\({{\rm{q}}{{\rm{max}}}} = \frac{{{{\rm{V}}{\rm{f}}} × {{\rm{K}}_{\rm{j}}}}}{4}\)

Where,

V_f = Free mean speed i.e maximum speed at zero density

K_j = Jam density i.e maximum density at zero speed

Calculation:

Given,

${\rm{V}} = \left[ {80 - \frac{2}{3}{\rm{K}}} \right]$

K_j is at V = 0

∴ ${\rm{0}} = \left[ {80 - \frac{2}{3}{\rm{K_j}}} \right]$

∴ K_j = 120 veh/km

V_f is at K = 0

∴ ${{\rm{V}}_{\rm{f}}} = \left[ {80 - \frac{2}{3} \times {\rm{0}}} \right]$

∴ V_f = 80 km/hr

${{\rm{q}}_{{\rm{max}}}} = \frac{{{\rm{80}} × {{\rm{120}}}}}{4} =2400$

∴ Capacity on this section of highway would be 2400 veh/hr

Concept:

Webster’s method for an optimum cycle time of a signal is given by

${{\rm{C}}_{\rm{o}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.5L\ +\ 5}}}}{{{\rm{1\ -\ Y}}}}$

Effective green time on ith lane

\({{\rm{G}}{\rm{i}}}{\rm{ = }}\frac{{{{\rm{Y}}{\rm{i}}}}}{{\rm{Y}}}{\rm{ × }}\left( {{{\rm{C}}_{\rm{o}}}{\rm{ - L}}} \right)\)

L = Lost time per cycle = 2N + R

N = Number of phase = Number of independent movements

R = All red time

Y = Sum of the ratios of normal to saturated flows

Y = Y₁ + Y₂ + Y₃

Where,

${{\rm{Y}}_1} = \frac{{{{\rm{q}}_1}}}{{{{\rm{s}}_1}}}$, ${{\rm{Y}}_2} = \frac{{{{\rm{q}}_2}}}{{{{\rm{s}}_2}}}$ and ${{\rm{Y}}_3} = \frac{{{{\rm{q}}_3}}}{{{{\rm{s}}_3}}}$

q₁, q₂ and q₃ is normal flow in phase 1, 2 and 3 respectively

s₁, s₂ and s₃ is saturated flow in phase 1, 2 and 3 respectively

Calculation:

Given,

q₁ = 200 veh/hr, q₂ = 187 veh/hr and q₃ = 210 veh/hr

s = s1 = s2 = s₃ = 1800 veh/hr/lane 

L = Total lost time = 4 × 4 = 16 seconds

C_o = 60 seconds

Y = Y1 + Y2 + Y₃ + Y₄

${\rm{Y}} = \frac{{{{\rm{q}}_1}}}{{{{\rm{s}}_1}}} + \frac{{{{\rm{q}}_2}}}{{{{\rm{s}}_2}}} + \frac{{{{\rm{q}}_3}}}{{{{\rm{s}}_3}}} + {Y_4}$

${\rm{Y}} = \frac{{200}}{{1800}} + \frac{{187}}{{1800}} + \frac{{210}}{{1800}} + {Y_4}$

Y = 0.332 + Y₄

Now

${{\rm{C}}_{\rm{o}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.5L\ +\ 5}}}}{{{\rm{1\ -\ Y}}}}$

${{\rm{60}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.5× 16\ +\ 5}}}}{{{\rm{1\ -\ (0.332+Y_4)}}}}$

Y₄ = 0.185

∴ Y = 0.332 + 0.185 = 0.517

Effective green time for 4^th phase

\({{\rm{G}}{\rm{4}}}{\rm{ = }}\frac{{{{\rm{Y}}{\rm{4}}}}}{{\rm{Y}}}{\rm{ × }}\left( {{{\rm{C}}_{\rm{o}}}{\rm{ - L}}} \right)\)

${{\rm{G}}_{\rm{4}}}{\rm{ = }}\frac{{{{\rm{0.185}}}}}{{\rm{0.517}}}{\rm{ × }}\left( {{{\rm{60}}}{\rm{ - 16}}} \right) = 15.74$

Hence the effective green time for 4^th phase = 15.74 seconds