GATE ME 2015 Official Paper: Shift 3 (Previous Year Paper)

In order to accurately solve this question, we must attempt to understand the context of the sentence.

We can see that the verb in the blank defines the connection between 'apparent lifelessness' - which means an appearance of being dead - and 'dormant life' - which means hidden life.

However, the word 'dormant' is also used to refer to describe an animal in the state of hibernation wherein the animal has its normal physical functions suspended or slowed down for a period of time.

Hence, an appearance of being dead would help keep such as a living being alive. Out of the given options, the word 'harbours' which means 'provides a safe shelter or place to hide' fits the context perfectly.

Therefore, option 1 is correct.

The idiom 'fish out of water' is used to refer to a person who feels uncomfortable in a specific unfamiliar situation. This could be used to describe the boy in the given sentence.

The other options contain phrases that are nonsensical and carry no meaning.

Hence, option 3 is correct.

The only sentence in which the underlined word has been used correctly is option 2. The word 'elusive' means 'difficult to find, catch, or achieve' and this fits the context of the sentence accurately.

The correct word for options 3 and 4 would have also been elusive. Option 1 forms a nonsensical sentence.

Hence, option 2 is correct.

Tanya is older than Eric. 

T > E

Cliff is older than Tanya. 

C > T

Eric is older than Cliff. 

E > C

If T > E and C > T then C > T > E ⇒ E < C

So if the first two statements are true, then the third statement is false.

Also

If T > E and E > C then T > E > C ⇒ C < T

If C > T and E > C then E > C > T ⇒ T < E

For a match to be played, we need 2 teams.

No of matches = no. of ways of selections 2 teams out of 5 \(= {5_{{{\rm{C}}_2}}} = 10\)

Therefore the concluding diagram can be: Manager can be executive also. Some executives are also leaders that are not a manager.

The correct option to replace the given phrase is option 3. 

Option 1 and option 2 end with the adjective 'necessary' which, when followed by the verb 'reflects' does not make sense.

Also, in option 2, the nouns 'increase' and 'productivity' immediately follow each other, making no logical sense.

Option 3 solves these problems by placing a preposition (in) to show the relationship between the nouns and ends with the adverb necessarily which describes the verb 'reflects'.

Hence, option 3 is correct.

Let area of triangle PQR be ‘A’

\(\begin{array}{l} \frac{{{\rm{SQ}}}}{{{\rm{PQ}}}} = \frac{1}{{1 + 3}} = \frac{1}{4}\ \frac{{{\rm{QT}}}}{{{\rm{QR}}}} = \frac{2}{{2 + 5}} = \frac{2}{7} \end{array}\)

\(\therefore {\rm{Area\ of\ }}{{\rm{\Delta }}^{{\rm{le}}}}{\rm{SQT}} = \frac{1}{2} \times {\rm{SQ}} \times {\rm{QT}}\)

\(\begin{array}{l} = \frac{1}{2} \times \left( {\frac{1}{4}{\rm{PQ}}} \right) \times \left( {\frac{2}{7}{\rm{QR}}} \right)\ = \frac{1}{4} \times \frac{2}{7} \times \left( {\frac{1}{2} \times {\rm{PQ}} \times {\rm{QR}}} \right)\ = \frac{1}{{14}} \times {\rm{Area\ of\ }}{{\rm{\Delta }}^{{\rm{le}}}}{\rm{PQR}}\ {\rm{given\ }}20{\rm{c}}{{\rm{m}}^2} = \frac{1}{{14}} \times {\rm{A}} \end{array}\)

∴ A = 14 × 20 = 280 cm²

Given: –4 ≤ x ≤ 5 and 6 ≤ y ≤ 16

X coordinate has 10 possible values between -4 and 5 

Y coordinate has 11 possible values between 6 and 16 

Right triangle PQR is to be constructed in the xy – plane so that the right angle is at P and line PR is parallel to the-axis

1. Fix P first, P can have 10 values across X-axis and 11 across Y-axis. So, that's  11 × 10 = 110  

Point P: 110 ways

2. Since PR is parallel to X-axis, R is on the same line as P. So, R has  9 ( = 10 − 1 ) values left along the X-axis.

Point R: 9 ways 

3. Since PQR is a right-angle triangle and the right angle is at P, PQ is parallel to Y-axis. So, Q has 10 ( = 11 − 1 ) values left.

Point Q: 10 ways 

So PQR will have: 110 × 9 × 10 = 9900 ways

X be the event that head occurs in each of the first two tosses:

X = {HHT, HHH}

Y be the event that a tail occurs on the third toss:

Y = {HHT, HTT, TTT, THT}

Z be the event that two tails occur in three tosses:

Z = {TTH, THT, HTT}

 ∴ both y and z are dependent.

X is the event that the head occurs in each of the first two tosses and Y is the event that a tail occurs on the third toss. So X and Y are independent.

Z is the event that two tails occur in three tosses and Y is the event that a tail occurs on the third toss. Event {TTH} is not possible. So Y and Z are dependent.

X is the event that the head occurs in each of the first two tosses and Z is the event that two tails occur in three tosses. So if X occurs Z can not occur hence X and Z are dependent.

Concept:

Stress Amplitude: It is the amount of stress which deviates from mean stress.

Stress Amplitude = (σ­­_max – σ_min)/2 

Stress Ratio: It is the ratio of the minimum stress experienced during a cycle to the maximum stress experienced during a cycle.

Stress Ratio = σ_min / σ_max

σ_max = Maximum stress experienced in a cycle

σ_min = Minimum stress experienced in a cycle

Calculation:

σ_max = 250 MPa σ_min = 50 MPa

\(\begin{array}{l} {\rm{Stress\ Amplitude}} = \frac{{{{\rm{σ }}_{{\rm{max}}}} - {{\rm{σ }}_{{\rm{min}}}}}}{2} = \frac{{250 - 50}}{2} = 100\ {\rm{MPa}}\ {\rm{Stress\ Ratio}} = \frac{{{{\rm{σ }}_{{\rm{min}}}}}}{{{{\rm{σ }}_{{\rm{max}}}}}} = \frac{{50}}{{250}} = 0.2 \end{array}\)

Concept:

Mass moment of inertia: It is the resistance offered by a body to its angular acceleration. 

Mathematically,

 \(I = \smallint {r^2}dm\)

Explanation:

As it can be observed from the above expression that the mass moment of inertia depends strongly on the distribution of the mass from the axis of rotation. 

As the distance of the mass from the axis of rotation increases, the mass moment of inertia will also increase. 

From all the given options, the above wheel has maximum mass located away from the axis of rotation Thus will have a maximum moment of inertia.

Explanation:

Gear Train:

Two or more gears are made to mesh with each other to transmit power from one shaft to another. Such a combination is called a gear train.

When there is only one gear on each shaft, it is known as a simple gear train.

Since the gear 1 drives the gear 2, therefore gear 1 is called the driver and the gear 2 is called the driven or follower.

Idler Gear:

The purpose of an idler gear used in gear trains is to have the required direction of rotation at the output shaft.

• It is used for changing the direction.
• It does not change the velocity ratio or torque.
• It may be noted that when the number of intermediate gears (idler gear) is odd, the motion of both the gears (i.e. driver and driven or follower) is like.
• But if the number of intermediate gears is even, the motion of the driven or follower will be in the opposite direction of the driver.

The ratio for the given gear train is given by the following formula

\(\frac{{{{\rm{\omega }}_2}}}{{{{\rm{\omega }}_6}}} = \frac{{{{\rm{\omega }}_2}}}{{{{\rm{\omega }}_3}}}\frac{{{{\rm{\omega }}_3}}}{{{{\rm{\omega }}_5}}}\frac{{{{\rm{\omega }}_5}}}{{{{\rm{\omega }}_6}}} = \frac{{{{\rm{N}}_3}}}{{{{\rm{N}}_2}}}\frac{{{{\rm{N}}_5}}}{{{{\rm{N}}_4}}}\frac{{{{\rm{N}}_6}}}{{{{\rm{N}}_5}}} = \frac{{{{\rm{N}}_3}{{\rm{N}}_6}}}{{{{\rm{N}}_2}{{\rm{N}}_4}}}\)

∴ Only Gear(5) is the Idler gear, as the number of teeth on the wheel '5' does not appear in the velocity ratio.

Concept:

Coriolis acceleration: Coriolis component of acceleration exists when there is a sliding motion of a slider which is sliding on a link which itself is rotating. The magnitude of the Coriolis component is given by 2Vω in the direction obtained by rotating the velocity vector by 90° in the direction of rotation of the link.

a_c = 2Vω  

a_c = Coriolis Acceleration

V = Component of velocity which causes sliding or velocity of the slider i.e. V_Q/P 

ω = Angular velocity of the link on which slider is present i.e. ω₂

Calculation:

In the given question,

Link 2 is rotating in the clockwise direction, So the slider velocity will be rotated by 90° in the clockwise direction.

Explanation:

Formula σ = K εⁿ, where, σ represents the applied stress on the material, ε is the strain, K is the strength coefficient.

The value of the strain hardening exponent lies between 0 and 1.

If n = 0 means that a material is a perfectly plastic

n = 1 means perfectly elastic solid

For stainless steel SS 304 the value of n lies between 0.10 and 0.50.

Explanation:

Mechanism and shapes formation by cutting:

• The motion responsible for the cutting action is known as the primary motion or cutting motion.
• The motion responsible for gradual feeding the uncut portion is termed as the secondary motion or feed motion.
• The line generated by the cutting motion is called generatrix and the line generated from the feed motion is called the directrix.
• Various geometries can be obtained depending on the shapes of the generatrix and the directrix and their relative directions.

Generation of various surfaces:

Full Mould Casting:

• The process "Full - mould casting" also called as "Cavity Less Casting" or "Evaporative casting" or "Disposable pattern casting" is a variant of sand casting in the sense that the process utilizes an expendable pattern, which instead of being removed from the mould gets burnt and evaporated when the molten metal is poured into the mould.
• The pattern material is expanded polystyrene foam.
• Since the pattern gets destroyed in the process, the method is mainly used for product research in the foundry industry and for one - of - a kind complex casting. The method is also known as the " Lost Foam" process.

Kendall's notation for queues:

(a/b/c) : (d/e/f)

a = Inter-arrival rate of distribution, b = Service time distribution, c = Number of servers, 

d = System capacity (queue discipline), e = Populationn size, f = Service discipline

Arriving and service distributions can of the following type:

M: Exponential, D: Deterministic, E_k: Erlang with parameter k, H_k: Hyperexponential with parameter k, G: General

Couette flow: It is a steady incompressible, laminar flow of a viscous fluid in the space between two surfaces, one of which is moving tangentially relative to a stationary surface. The configuration often takes the form of two parallel plates or the gap between two concentric cylinders.

Explanation:

Bell–Coleman cycle or Reversed Brayton cycle:

where,

Process 1-2: Isentropic compression

Process 2-3: Constant pressure heat rejection

Process 3-4: Isentropic expansion

Process 4-1: Constant pressure heat absorption 

T-S diagram represents a reversed Brayton cycle used in the air conditioning of aeroplanes where the air is used as a refrigerant.

Reversed Carnot cycle:

Vapour Compression Refrigeration cycle:

Prandtl Number: It is the ratio of momentum diffusivity (ν) and thermal diffusivity (α). 

Pr = ν / α 

ν = Momentum diffusivity (Kinematic Viscosity) = μ / ρ ,α = Thermal diffusivity = k / ρc_p, μ = Dynamic viscosity, ρ = Density of fluid

c_p = Specific heat at constant pressure

Important Non-dimensional numbers:

• Biot number  → Ratio of internal thermal resistance to boundary layer thermal resistance
• Grashof number  → Ratio of buoyancy to viscous force
• Prandtl number  → Ratio of momentum to thermal diffusivities
• Reynolds number  → Ratio of inertia force to viscous force

Concept:

The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer.

\(\epsilon = \frac{{{\rm{Actual;heat;transfer}}}}{{{\rm{Maximum;possible;heat;transfer}}}} = \frac{Q}{{{Q_{max}}}}\)

\(NTU = \frac{{UA}}{{{C_{min}}}}\)

NTU is a measure of the effectiveness of the heat exchanger.

The NTU is a measure of the heat transfer size of the exchanger; larger the value of NTU, the closer the heat exchanger approaches its thermodynamic limit.

Capacity Ratio:

\(C_r = \frac{{{C_{min}}}}{{{C_{max}}}}\)

Effectiveness of a parallel flow heat exchanger:

\({\epsilon_{parallel}} = \frac{{1 - \exp \left[ { - NTU\left( {1 + C_r} \right)} \right]}}{{1 + C_r}}\)

Effectiveness of a counter-flow heat exchanger:

\({\epsilon_{counter}} = \frac{{1 - \exp \left[ { - NTU\left( {1 - C_r} \right)} \right]}}{{1 ~-~ C_r\exp \left[ { - NTU\left( {1 - C_r} \right)} \right]}}\)

For condenser and evaporator (C_r = 0)

In condenser and evaporators in which one fluid remains at a constant temperature throughout the exchanger. Here Cmax = ∞ and thus \(C_r = \frac{{{C_{min}}}}{{{C_{max}}}} = 0\)

By using the above case, we arrive at the following common expression for parallel flow as well as counter – flow heat exchangers.

\(\epsilon = 1 - {\rm{exp}}\left( { - NTU} \right)\)

Concept:

Trapezoidal rule states that for a function y = f(x)

xn = x0 + nh, where n = Number of sub-intervals

h = step-size

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]\)     …1)

For a trapezoidal rule, the number of sub-intervals must be a multiple of 1.

Calculation:

x = 1, y (0) = 1×ln1 = 0

x = 2, y (n) = 2ln2

\(\mathop \smallint \limits_1^2 {\rm{x}}\ln {\rm{xdx}} = \frac{1}{2}\left[ {0 + 2\ln 2} \right] = \ln 2 = 0.69\)

Key Points:

Apart from the trapezoidal rule, other numerical integration methods are;

A) Simpson’s one-third rule

B) Simpson’s three-eighth rule

A) Simpson’s one-third rule:

For applying this rule, the number of subintervals must be a multiple of 2.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]\)     ..2)

B) Simpson’s three-eighths rule:

For applying this rule, the number of subintervals must be a multiple of 3.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]\)     …3)

Concept:

Probability of event A given B has occurred:

\(P(A|B) = \frac{{P(A \cap B)}}{{P(B)}}\)

Calculation:

\(P(Y/X) = \frac{{{\rm{P}}\left( {{\rm{X}}\mathop \cap \nolimits {\rm{Y}}} \right)}}{{{\rm{P}}\left( {\rm{X}} \right)}} = \frac{{\frac{1}{{12}}}}{{\frac{1}{4}}} = \frac{1}{3}\)

Concept:

Every square matrix will satisfy its characteristics equation.

Characteristic equation of A is |A - λI| = 0

Calculation:

Given:

\({\rm{Let\ A}} = \left[ {\begin{array}{*{20}{c}} 4&2\ 1&3 \end{array}} \right]\)

Characteristic equation of A is |A - λI| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {4 - {\rm{\lambda }}}&2\ 1&{3 - {\rm{\lambda }}} \end{array}} \right| = 0\)

⇒ λ² – 7λ + 10 = 0

⇒ λ = 2, 5

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{ - \sin {\rm{x}}}}{{2\sin {\rm{x}}~ + ~{\rm{x}}\cos {\rm{x}}}}} \right){\rm{}}\left( {\frac{0}{0}{\rm{form}}} \right)\)

Applying L – Hospital Rule, we get

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{ - \cos {\rm{x}}}}{{2\cos {\rm{x}} ~+~\cos {\rm{x}} ~- ~{\rm{x}}\sin {\rm{x}}}}} \right)\)

= -1/3

Concept:

In the case of a thin cylinder, absolute shear stress is given by:

\(\tau_{max}=\frac{\sigma_1}{2}=\frac{pd}{4t}\)

In the case of a thin cylinder, in-plane of shear stress is given by:

\(\tau_{max}=\frac{\sigma_1 - \sigma_2}{2}=\frac{pd}{8t}\)

Calculation:

Given:

Pressure = 500 kPa, Inner radius = 2 m, Inner diameter = 4 m, Thickness = 10 mm.

In-plane shear stress:

\(\tau_{max}=\frac{\sigma_1 - \sigma_2}{2}=\frac{pd}{8t}\)

\(\tau_{max}=\frac{pd}{8t}\)

\(\tau_{max}=\frac{500;\times;10^3;\times;4;\times;10^3}{8;\times;10;\times;10^6}=25;MPa\)

Concept:

Lami's Theorem: It is an equation that relates the magnitude of the three co-planner, concurrent and non-collinear forces that keeps a body in equilibrium. It states that each force is proportional to the sine of the angle between the other two forces.

Calculation:

\(\frac{{{{\rm{T}}_1}}}{{\sin 120^\circ }} = \frac{{{{\rm{T}}_2}}}{{\sin 150^\circ }} = \frac{{500}}{{\sin 90^\circ }}\)

T₁ = 500 × sin 120° and  T₂ = 500 sin 150°

T₁ = 433 N and T₂ = 250 N

Statement P:

The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is known as damping factor or damping ratio.

\(\zeta = \frac{C}{{{C_c}}}\)

• Overdamped System: ζ > 1
• Underdamped: ζ < 1
• Critical Damping: ζ = 1: The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion.

Statement Q:

Logarithmic decrement:

\(\delta = \ln \frac{{{x_0}}}{{{x_1}}} = \ln \frac{{{x_1}}}{{{x_2}}}\)

If the system executes n cycles:

\(\delta = \frac{1}{n}\ln \frac{{{x_o}}}{{{x_n}}} = \frac{{2\pi \xi }}{{\sqrt {1 - {\xi ^2}} ;}}\)

\(\xi = \frac{\delta }{{\sqrt {4{\pi ^2} + {\delta ^2}} }}\)

\(\delta = \frac{{4\pi \zeta }}{{\sqrt {1 - {\zeta ^2}} }}\)

From the above formula, we can say that the damping factor can be calculated with the help of logarithmic decrement.

Statement R:

In the case of damping due to dry friction between moving surfaces resisting the force of constant magnitude acts opposite to the relative motion.

Statement S:

Viscous drag force is

F_D = Cẋ

Where c = damping coefficient; ẋ = relative velocity

Hence we can say that Drag Force is directly proportional to relative velocity and not it’s square.

So option S is incorrect

Concept:

Two types of co-ordinate systems are used to define and control the position of the tool in relation to the workpiece.

The co-ordinate systems used are:

(1) Absolute co-ordinate system

(2) Incremental co-ordinate system

Absolute System:

In the absolute system (G90), all dimensions or positions are given from a zero or reference point.

Incremental System:

In the incremental system (G91), dimensions or positions are given from a previous known point.

Calculation:

In the incremental system. Point P will become the new origin for point Q.

x-coordinate of point Q = 4

y-coordinate of point Q = 7

Q (x,y) = Q (4,7)

Welding is a process of joining two similar or dissimilar metals by fusion, with or without the application of pressure and with or without the use of filler metals. The fusion of metal takes place by means of heat. The heat may be obtained from electric arc, electric resistance, chemical reaction, friction or radiant energy.

Based on the composition of joints, the joining process may be classified as:

Autogenous Welding: The process of joining similar metals by melting the edges together, without the addition of filler metals. Ex. Friction Welding, Diffusion Welding, Laser Beam Welding, Electron beam welding, Resistance Welding. 

Homogeneous Welding: The process of joining similar metals with the help of a filler rod of the same metal. Ex. Gas Metal Arc Welding, Electroslag Welding, Submerged Arc Welding.

Heterogeneous Welding: The process of joining dissimilar metals using a filler rod. Ex. Plasma Arc Welding, Friction Stir Welding.

Explanation:

Pipes in parallel:

If the main pipe divides into two or more branches and again join together downstream to form a single pipe, then the branched pipes are said to be connected in parallel (compound pipes).

• The rate of flow in the main pipe is equal to the sum of the rate of flow through branch pipes
• The loss of head for each branch pipe is the same

\({\rm{head\ loss\ h}} = {{\rm{h}}_{{{\rm{L}}_1}}} = {{\rm{h}}_{{{\rm{L}}_2}}} = {{\rm{h}}_{{{\rm{L}}_3}}}\)

\({h_f}_L = \frac{{{f_1}{L_1}V_1^2}}{{2g{D_1}}} = \frac{{{f_2}{L_2}V_2^2}}{{2g{D_2}}} = \frac{{{f_3}{L_3}V_3^2}}{{2g{D_3}}}\)

Important Points

Pipes in series:

Pipes in series or compound pipes are defined as the pipes of different lengths and different diameters connected end to end (in series) to form a pipeline.

• Discharge through each pipe is the same
• The total loss of energy or head loss will be the sum of the losses in each pipe

\(v = {v_f} + x;\left( {{v_g} - {v_f}} \right)\)

v = V/m = 0.5 m³/kg

\(0.5 = 0.00106 + x;\left( {0.8908 - 0.00106} \right)\)

\(x = 0.56\)

\(0 \le x \le 1\)

x = 0, Saturated liquid

x = 1, Saturated vapor

As x value lies between 0 and 1, therefore it is a mixture of saturated liquid and saturated vapor.

Let's take an example and check option 3 is correct or not 

Hence

\(\vec \nabla = \hat i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + \hat k\frac{\partial }{{\partial z}}\)

\(\vec V = F_1\hat i + F_2\hat j + F_3\hat k\)

∴ A curl of a vector is

\(\vec \omega = \vec \nabla \times \vec V = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ F_1&{F_2}&F_3 \end{array}} \right|\)

\(\vec \nabla \times \vec V = \hat i\left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}} \right) - \hat j\left( {\frac{{\partial {F_3}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial z}}} \right) + \hat k\left( {\frac{{\partial {F_2}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial y}}} \right)\)

\(\vec \nabla .\left( {\vec \nabla \times \vec V} \right) = \frac{\partial }{{\partial x}}\left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}} \right) - \frac{\partial }{{\partial y}}\left( {\frac{{\partial {F_3}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial z}}} \right) + \frac{\partial }{{\partial z}}\left( {\frac{{\partial {F_2}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial y}}} \right)\)

\(\vec \nabla .\left( {\vec \nabla \times \vec V} \right) = \frac{{{\partial ^2}{F_3}}}{{\partial x\partial y}} - \frac{{{\partial ^2}{F_2}}}{{\partial x\partial z}} - \frac{{{\partial ^2}{F_3}}}{{\partial x\partial y}} + \frac{{{\partial ^2}{F_1}}}{{\partial y\partial z}} + \frac{{{\partial ^2}{F_2}}}{{\partial x\partial z}} - \frac{{{\partial ^2}{F_1}}}{{\partial y\partial z}},=0\)

Hence we can see that option c is verified.

Some of the important identities: (U: scalar, a: vector)

1. curl grad U = ∇ × ∇U = 0
2. div curl a = ∇.∇ × a = 0
3. ∇. (Ua) = U diva + (grad U).a
4. ∇ × (Ua) = U∇ × a + (∇U) × a

Processes involved in compression engine (Diesel cycle) are:

1-2: Reversible adiabatic compression

2-3: Constant pressure heat addition

3-4: Reversible adiabatic expansion

4-1: Constant volume of heat rejection

Heat applied, Q_s = c_p (T₃ – T₂)

Heat rejected, Q_r = c_v (T₄ – T₁)

\(\eta = 1 - \frac{{{Q_r}}}{{{Q_s}}} = 1 - \frac{1}{\gamma }\frac{{\left( {{T_4} - {T_1}} \right)}}{{\left( {{T_3} - {T_2}} \right)}}\)

Where, 

λ = c_p/c_v

Concept:

I_XX = I_Rectangle-XX - 2 × [I_{Square-centroid} + Ah²]

IRectangle-XX = Moment of inertia of rectangle about X-X = BD³/12

ISquare-centroid = Moment of inertia of the cut portion i.e. square portion about its centroid = a⁴/12

A = Area of square

h = Distance between the centroid of square and X-X

Calculation:

Given: B = 60 mm, D = 120 mm, a = 30 mm

Moment of Inertia. \({I_{XX}} = \frac{1}{{12}}\left[ {{{\left( {120} \right)}^3} × 60} \right] - 2\left[ {\frac{1}{{12}} × {{\left( {30} \right)}^4} + 30 × 30 × 30^2} \right]\) 

\(= 6.885 × {10^6}\ m{m^4}\)

Concept:

V = r × ω 

r = Distance of point from axis of rotation

ω = Angular velocity

Calculation:

Given:

V_A = 80 m/s, V_B = 140 m/s

r_B – r_A = 300                _____1)

V_A  = ω × r_A = 80 m/s

V_B = ω × r_B = 140 m/s

\(\therefore ;\frac{{{V_B}}}{{{V_A}}} = \frac{{{r_B}}}{{{r_A}}} = 1.75\)   _______2)  

Solving (1) & (2),

Radius of wheel (r_B) = 700 mm.

∴ Diameter of wheel is 1400 mm.

Using Newton's method and taking moment about point O.

m(2a)²ẍ  + k(a)²x = 0

m4a²ẍ + ka²x = 0

\(\mathop {\ddot x}{+} \frac{{k{a^2}}}{{4m{a^2}}}x = 0\)

\(\mathop {\ddot x}{+} \frac{{k{}}}{{4m{}}}x = 0\)

Comparing the above equation with ẍ + ω_n²x = 0 

∴ \({\omega _n} = \sqrt {\frac{k}{{4m}}} \)

Concept:

Gruebler's equation: This equation is used to find out the degree of freedom of a planner mechanism.

F = 3(n-1) - 2j - h

n = Number of links 

j = Number of revolute pair or binary joint

h = Number of higher pairs

Calculation:

Number of links, N = 6

Total number as binary joints, j = 7

h = 0

F = 3 (N-1) – 2j

= 15-14 = 1.

Concept:

For ball bearing,

FL^1/3 = K

F = Radial Load

L = Life in number of revolutions

Calculation:

Given:

F₁ = 2 kN, L₁ = 540 million revolutions, L₂ = 1 million revolutions

\({F_1}L_1^{\frac{1}{3}} = {F_2}L_2^{\frac{1}{3}}\)

2 × 540^1/3 = F₂(1)^1/3

∴ F₂ = 16.286 kN

Concept:

Δh_max = μ²R

Δhmax = Maximum reduction of the plate

\(μ = \sqrt {\frac{{{{\left( {{\rm{Δ }}h} \right)}_{max}}}}{R}}\)

Calculation:

(Δh)_max = 25 – 20 = 5 mm

R = 250 mm

\(μ = \sqrt {\frac{{{{\left( {{\rm{Δ }}h} \right)}_{max}}}}{R}}\)

\(\therefore μ = \sqrt {\frac{5}{{250}}} =0.1414\)

Concept:

From Chvorinov's Rule,

\({t_s} = k{\left( {\frac{V}{A}} \right)^2}\)

t_s = time of solidification of the casting

V = Volume of casting

A = Area of casting from which heat is transferred

Calculation:

Neglect heat loss of cylinder top and bottom(as mentioned in the question), so considering the curved surface area of the cylinder only.

For the cylindrical cavity (1) 

\({t_1} = k{\left( {\frac{{\frac{\pi }{4}{d^2}h}}{{\pi dh}}} \right)^2} = k{\left( {\frac{d}{4}} \right)^2}\) since d = h.

For cubic casting (2) side of cube a = 2× height of cylinder h = 2d, since d = h.

\(\begin{array}{l} {t_2} = k{\left( {\frac{{{a^3}}}{{6{a^2}}}} \right)^2} = k{\left( {\frac{a}{6}} \right)^2} = k{\left( {\frac{d}{3}} \right)^2}\left( {a = 2d} \right)\ \therefore \frac{{{t_1}}}{{{t_2}}} = \frac{{{{\left( {\frac{d}{4}} \right)}^2}}}{{{{\left( {\frac{d}{3}} \right)}^2}}} = {\left( {\frac{3}{4}} \right)^2} = 0.5625 \end{array}\)

Concept:

\(EOQ = \sqrt {\frac{{2D{C_0}}}{{{C_h}}}} \)

D = Annual demand

C₀ = Ordering cost / order

C_h = Annual holding cost / unit

Calculation:

Given:

D = 2000 kg, C₀ = Rs. 100 / order

C_h = Rs. (0.09 × 25) = Rs. 2.25/unit 

\(\begin{array}{l} EOQ = \sqrt {\frac{{2D{C_0}}}{{{C_h}}}} \ = \sqrt {\frac{{2 × 2000 × 100}}{{0.09 × 25}}} = 421.637 \end{array}\)

No. of orders = 2000/EOQ = (2000 / 421.637) = 4.743

Length of cycle = 365/(No. of orders) = 365 / 4.743 = 76.948 days

Concept:

Chip thickness ratio (r) = t / t_c

In orthogonal turning,

t = feed in mm

t_c = thickness of chip

\(\tanϕ = \frac{{r\cosα }}{{1 - r\sinα }}\)

Shear strain (γ) = cot ϕ + tan(ϕ - α)

ϕ  = shear angle

α = rake angle

Calculation:

Given:

t = 0.14 mm, t_c = 0.28 mm

 r = t / tc = 0.5

\(\tanϕ = \frac{{r\cosα }}{{1 - r\sinα }}\)

\(\tan\phi = \frac{{0.5\cos10^\circ }}{{1 - 0.5\sin10^\circ }}\)

\(\tan\phi = 0.539\)

∴ ϕ = 28.3345° ≈ 28°20’

Shear strain (γ) = cot ϕ + tan(ϕ - α) = 2.1859 ≈ 2.19

Explanation:

Sequence no 5 hence the program will start with N05.

Incremental mode hence the next code will be G91.

As we can see in the diagram we have to go in the counter-clockwise direction to go from (20,20) to (10,10) hence we will use the code G03.

And considering the start point as (0,0) which is actually(20, 20) we have to reach (10,10) hence the co-ordinates will be (X = -10, Y = -10) and the radius R will be 10 so R10.

Concept:

Velocity is usually measured indirectly by measuring the difference between the stagnation and static pressures (pitot tube). The stagnation pressure is found at point 0. To obtain the stagnation pressure the opening of the tube has to be perpendicular to the fluid flow.

The lower end, which is bent through 90° is directed in the upstream direction. The liquid rises up in the tube due to the conversion of kinetic energy into pressure energy. The velocity is determined by measuring the rise of the liquid in the tube. 

\({V_{1}} = {C_v}\sqrt {2gh}\)

\(h = x\left[ {\frac{{{s_g}}}{{{s_0}}} - 1} \right] \)

x = differential manometric reading 

s_g = specific gravity of the manometric fluid

s₀ = specific gravity of water

Calculation:

Given:

sg = 10, s0 = 1, x = 0.01 m

C_v = 1 (Given)

\(h = x\left[ {\frac{{{s_g}}}{{{s_0}}} - 1} \right] = 0.01\left( {10 - 1} \right) = 0.09 m\)

\(\therefore Velocity\ of\ flow = \sqrt {2 \times 9.8 \times 0.09} = 1.328 \ {m}/{s}\)

Concept:

COP_Ref = Refrigeration effect / Work done

COPRef = (h₁ - h₄) / (h₂ - h₁)

Calculation:

Given:

h₁ = 402 kJ/kg, h₂ = 432 kJ/kg, h₄ = 237 kJ/kg

Refrigerating effect = (h1 - h4) = 402-237 = 165 kJ/kg.

Work done = (h2 - h1) = 432-402 = 30 kJ/kg

\(COP = \frac{{165}}{{30}} = 5.5\)

Concept:

Applying SFEE (Steady Flow Energy Equation)

\({h_1} + \frac{{V_1^2}}{2} + g{Z_1} + \frac{{dQ}}{{dm}} = {h_2} + \frac{{V_2^2}}{2} + g{Z_2} + \frac{{dW}}{{dm}}\)

1 – Entry point

2 – Exit point

Calculation:

Given:

h₁ = 2993 kJ/kg, h₂ = 226 kJ/kg

dQ/dm = - 50 kJ/kg

Neglecting the change in potential and velocity head

\({h_1} + \frac{{dQ}}{{dm}} = {h_2} + \frac{{dW}}{{dm}}\)

2993 - 50 = 226 + dW/dm

dW/dm = (2993 - 226 - 50) = 2717 kJ/kg

Work output = 2717 kJ/kg

Concept:

\(\phi = \frac{{{P_w}}}{{{P_s}}} \)

P_w = Pressure of water vapor

P_s = Pressure of saturated air

Humidity Ratio, \(w = 0.622\frac{{{P_w}}}{{{P_a} - {P_w}}}\)

Pa = Atmospheric pressure

Calculation:

\(\phi = \frac{{{P_w}}}{{{P_s}}} = 0.6 = \frac{{{P_w}}}{{5.63}}\)

∴ P_w = 3.378 kPa

 \(w = 0.622\frac{{{P_w}}}{{{P_a} - {P_w}}}\) 

\(w= 0.622 \times \frac{{3.378}}{{100 - 3.378}}\)

w = 0.021745 kg/kg of dry air or 21.745 g/kg of dry air

Concept:

For the surface-1,

\(F_{11} + F_{12} =1\)

And we know that,

\(F_{12}=1\) (as the surface-1 is convex, all the radiation radiated by surface 1 will be absorbed by surface 2)

From the theorem of reciprocity,

\(A_2F_{21} = A_1F_{12}\)

Calculation:

Given:

r₁ = r and r₂ = 4r

A₁ = 4πr²

A₂ = 2π(4r)² + π(4r)² = 48πr²

\(A_2F_{21} = A_1F_{12}\)

48πr² × F₂₁ = 4πr² × 1

F₂₁ = 4/48 = 1/12

F₂₁ = 1/12

Concept:

Green's theorem:

\(\mathop \smallint \nolimits_C^{} \left( {Mdx + Ndy} \right) = \int!!!\int \left( {\frac{{∂ N}}{{∂ x}} - \frac{{∂ M}}{{∂ y}}} \right)dxdy\)

Calculation:

Consider a strip of width dy

y varies from 0 to 1

x varies from 0 to 1-y 

M = 3x - 8y², N = 4y - 6xy

 ∂M/∂y = -16y, ∂N/∂x = -6y

 \(\mathop \smallint_C\left[ {\left( {3x - 8{y^2}} \right)dx + \left( {4y - 6xy} \right)dy} \right]=\mathop \smallint_C\left( {Mdx + Ndy} \right)\) 

\(= \mathop \smallint \nolimits \mathop \smallint \nolimits \left( {\frac{{∂ N}}{{∂ x}} - \frac{{∂ M}}{{∂ y}}} \right)dxdy\)

\(\begin{array}{l} = \mathop \smallint \limits_{y= 0}^1 \mathop \smallint \limits_{x = 0}^{1 - y} \left[ { - 6y - \left( { - 16y} \right)} \right]dxdy\ = \mathop \smallint \limits_{y = 0}^1 \left[ {\mathop \smallint \limits_{x = 0}^{1 - y} 10ydx} \right]dy\ = 10\mathop \smallint \limits_{y = 0}^1 yx\left. \right|_0^{1 - y}dy\ = 10\mathop \smallint \limits_{y = 0}^1 y\left[ {\left( {1 - y} \right) - 0} \right]dy = 10\left. {\left( {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{3}} \right)} \right|_0^1\ = 10\left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{5}{3} = 1.66 \end{array}\)

Concept:

For metrix \(P=\left[ {\begin{array}{*{20}{c}} {a}&b\ { c}&{d} \end{array}} \right]\) then |P| = (a × d) - (b × c)

Inverse of matrix P is given by

\(P^{-1} = {adj P\over det.(P)}\)

\(P^{-1} = {adj P\over |P|}\)

Multipllication of imaginary number i × (-i) = 1

Calculation:

Given:

 Let, \(P=\left[ {\begin{array}{*{20}{c}} {4+ 3i}&-i\ { i}&{4 - 3i} \end{array}} \right]\)

|P| = (4 + 3i) × (4 - 3i) – (i) × (-i) = 16 + 9 – 1 = 24

\(\begin{array}{l} adj P = \left[ {\begin{array}{*{20}{c}} {4 - 3i}&{ i}\ -i&{4 + 3i} \end{array}} \right]\ \therefore {P^{ - 1}} = \frac{1}{{24}}\left[ {\begin{array}{*{20}{c}} {4 - 3i}&{ i}\ -i&{4 + 3i} \end{array}} \right] \end{array}\)

Concept:

By Newton – Raphson Method,

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Calculation:

 f(x) = x3 + 2x2 + 3x – 1

f’(x) = 3x2 + 4x + 3 

1^st iteration,

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)

 f(x₀) = f(1) = 5

 f'(x0) = f'(1) = 10

\(x_1=1 - \frac{{f\left( 1 \right)}}{{f'\left( 1 \right)}} = 1 - \frac{5}{{10}} = \frac{1}{2}\)

2^nd iteration,

\({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\)

f(x₁) =  f(1/2) = f (0.5) = 1.125

f'(x₁) = f'(0.5) = 5.75

\(x_2= 0.5 - \frac{{f\left( {0.5} \right)}}{{f'\left( {0.5} \right)}} = 0.3043\)

\(f\left( t \right) = \begin{array}{*{20}{c}} {2;}&{0 < t < 1}\ {0;}&{otherwise} \end{array}\)

\(\therefore L\left[ {f\left( t \right)} \right] = \mathop \smallint \limits_0^1 2{e^{ - st}}dt = 2\left. {\frac{{{e^{ - st}}}}{{ - s}}} \right|_0^1 = \frac{{2 - 2{e^{ - s}}}}{s}\)

Concept:

Spinning speed of the bullet,

\(T = \frac{S}{V}{{;;;;;}} \ldots \left( 1 \right)\)

The spinning speed of the bullet,

\(ω =\frac{2\pi}{T}{{;;;;;}} \ldots \left( 2 \right)\)

Where,

S = Distance travelled by bullet in one helical rotation, V = Speed of the bullet, T  = time period of helical spin of bullet

Calculation:

Given:

1 turn of the helix has a length of 0.5 m i.e. the bullet rotation in a length of 0.5 m ⇒ S = 0.5 m, V = 20 m/s 

By using equation (1),

\(⇒ T= \frac{{0.5}}{{20}} = 0.025\ sec\)

By using equation (2),

\(⇒ω =\frac{{2\pi }}{{0.025}}\ rad/s\)

⇒ ω = 251.3 rad/s

Concept:

By drawing the bending moment diagram (BMD), we can find out the maximum bending moment.

Calculation:

Finding out the reactions at the supports

\({R_A} + {R_B} = 60;kN\)

Taking moment about point B

\({R_A} \times 4 + 20 \times 2 = 40 \times 2\)

\({R_A} = 10;kN\)

\({R_B} = 50kN\)

The moment at any point on a section at a distance of x 

\({M_x} = {R_A} \times x - w \times \frac{{{x^2}}}{2} = 10x - 5{x^2}\)

x is measured from A

Bending moment diagram (BMD) :

At x = 1, M₁ = 5 kN-m

 x= 4, M_B = -40 kN-m

|M_B| = 40 kN-m

Maximum bending momentum occurs at reaction B.

It has a magnitude of 40 kN-m.

Concept:

Power (P) = T_mean × ω

T_mean = Mean torque

ω = Angular velocity = 2πN/60

P = 2πNT/60

N = RPM of engine 

For a two-stroke engine,

\({T_{mean}} = \frac{1}{π }\mathop \smallint \nolimits_0^π Tdθ \)

For a four-stroke engine,

\({T_{mean}} = \frac{1}{{2π }}\mathop \smallint \nolimits_0^{2π } Tdθ \)

Calculation:

T = (10000 + 1000 sin2θ - 1200 cos2θ) N-m

N = 100 rpm

\(\begin{array}{l} {T_{mean}} = \frac{1}{π }\mathop \smallint \limits_0^π Tdθ = \frac{1}{π }\mathop \smallint \limits_0^π \left( {10000 + 1000sin2θ - 1200cos2θ } \right)dθ \ \Rightarrow {T_{mean}} = \frac{1}{π }\left[ {10000θ - 500cos2θ - 600sin2θ } \right]\begin{array}{*{20}{c}} π \ 0 \end{array} = 10000\ N - m \end{array}\)

and Power,

\(Power = \frac{{2π NT}}{{60}} = \frac{{2 × π × 100 × 10000}}{{60}} = 104719.75\ W\ or\ 104.719\ kW\)

Concept:

The given loading system will be subjected to primary and secondary shear forces.

When the line of action of the load does not pass through the centroid of the rivet system and thus all rivets are not equally loaded, then the joint is said to be an eccentric loaded riveted joint. The eccentric loading results in secondary shear caused by the tendency of force to twist the joint about the centre of gravity in addition to direct shear or primary shear.

Let P = Eccentric load on the joint, and e = Eccentricity of the load i.e. the distance between the line of action of the load and the centroid of the rivet system i.e. G.

The primary shear force on any bolt is given by:

\({P_s} = \frac{P}{n},;where;n;is;the;number;of;bolts\)

The secondary shear force is given by:

\(\frac{{{F_1}}}{{{l_1}}} = \frac{{{F_2}}}{{{l_2}}} = \frac{{{F_3}}}{{{l_3}}} = \frac{{{F_4}}}{{{l_4}}}\)

\({F_i} = \frac{{Pe{l_i}}}{{{l_1}^2 + {l_2}^2 + {l_3}^2 + {l_4}^2}},;where;{l_i};is;the;dis{\rm{tan}}ce;from;the;CG;of;system\)

The primary (or direct) and secondary shear load may be added vectorially to determine the resultant shear load (R) on each rivet.

\(R = \sqrt {{{\left( {{P_s}} \right)}^2} + {F^2} + 2{P_s}F\cos \theta } \)

θ is the angle between the primary or direct shear load (Ps) and secondary shear load.

Calculation:

P = 9kN, e = 250 mm

Primary shear force

\(P_p' = P_q' = P_r' = \frac{p}{3} = \frac{9}{3} = 3kN\)

Secondary shear force:

By symmetry, C.G lies at the center of bolt Q.

∴ r_p = 30 mm, r_r = 30 mm, r_q = 0

\(\begin{array}{l} C = \frac{{Pe}}{{r_q^2 + r_q^2 + r_r^2}} = \frac{{9000 \times 250}}{{{{30}^2} + {0^2} + {{30}^2}}} = 1250\ P''_p = \frac{{Per_p}}{{r_q^2 + r_q^2 + r_r^2}}=C{r_p} = 37.5kN\ P_r' = C{r_r} = 37.5kN\ P_q'' = 0 \end{array}\)

Resultant shear force

Due to symmetry, the stress in P and R will be equal in magnitude

\(\begin{array}{l} {P_p} = {P_r} = \sqrt {{{\left( 3 \right)}^2} + {{\left( {37.5} \right)}^2}} = 37.62 kN\ \tau = \frac{P}{A} = \frac{{38.5648}}{{\frac{\pi }{4}{{\left( {0.012} \right)}^2}}} = 332.633~ MPa \simeq 333~ MPa \end{array}\)

Concept:

\(\tan \alpha = \frac{{D_1 - D_2}}{{2L}}\)

\(\alpha = {\tan ^{ - 1}}\left( {\frac{{D_1 - D_2}}{{2L}}} \right)\)

Calculation:

\(\alpha = {\tan ^{ - 1}}\left( {\frac{{D_1 - D_2}}{{2L}}} \right)\)

\(\alpha = {\tan ^{ - 1}}\left( {\frac{{80 - 65}}{{2\times55}}} \right)\)

α = 7.765°

Taper angle = 2α = 15.53°

Rate of taper, \(T = \frac{{80 - 65}}{{55}} = 0.27\)

\(Set\ over = \frac{{T \times L}}{2} = \frac{{0.27 \times 90}}{2} = 12.15\)

Concept::

Naval Research Laboratory Method:

\({\rm{Shape\ factor}} = \frac{{{\rm{L}} + {\rm{W}}}}{{\rm{H}}} \) 

L denotes the length

W denotes the width

H denotes the thickness

(Note: L > W > H) 

Calculation:

Given:

L = 25 cm, W = 15 cm, H = 5 cm

V_r = Riser Volume

V_c = Casting Volume

Vc  = L × W × H 

\({V_r} = \frac{{\pi {d^2}h}}{4} = \frac{{\pi {d^3}}}{4}\)  [∵ h = d]

\({\rm{Shape\ factor}} = \frac{{{\rm{L}} + {\rm{W}}}}{{\rm{H}}} = \frac{{25 + 15}}{5} = 8\)

Then from the table

\(\therefore \frac{{{V_r}}}{{{V_c}}} = 0.5\)

\({V_r} = 0.5;{V_c}\)

V_r = 0.5 × 25 × 15 × 5 = 937.5

\(\frac{\pi }{4}{d^3} = 937.5\ c{m^3}\)

∴ d = 10.61 cm

Concept:

Unbounded Solution: 

• If the feasible region is not bounded, it is possible that the value of the objective function goes on increasing without leaving the feasible region.
• This is known as an unbounded solution.

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing t (0,0) else the opposite side of (0,0).

Calculation:

Given:

Maximum Z = 3X1 + 2X2

Subject to

• 2X1 + 3X2 ≤ 9

X1 – 5X2 ≥ -20

X1, X2 ≥ 0

• Plotting the graph for the given constraints as shown in the figure. From the figure, we can see that LPP has an unbounded solution.

• So, the correct answer is option 1.

Explanation:

Thoma's cavitation parameter (σ): It is the ratio of Net Positive Suction Head (NPSH) to the total head.

\(\sigma = \frac{{NPSH}}{H} = \frac{{\left( {\frac{{{P_A}}}{{\rho g}} - \frac{{{P_V}}}{{\rho g}} - z - {h_f}} \right)}}{H}\)

NPSH: It is defined as the net head developed at the suction port of the pump, in excess of the head due to the vapour pressure of the liquid at the temperature in the pump. NPSH must be positive for preventing the liquid from boiling. Boiling or cavitations may damage the pump. If NPSH reaches zero then the liquid starts boiling and cavitation starts.

Cavitation: It is defined as the phenomenon of formation of vapour bubbles of a flowing liquid in a region where the pressure of the liquid falls below its vapour pressure and the sudden collapsing of these vapour bubbles in a region of higher pressure. When the vapour pressure collapse, very high pressure is created.

Effects of cavitation:

(i) The metallic surfaces are damaged and cavities are formed on the surfaces.

(ii) Due to the sudden collapse of the vapour bubble, considerable noise and vibrations are produced.

(iii) Due to cavitation, the metal of the runner vanes and the draft tube is gradually eaten away, which results in lowering the efficiency of the turbine.

Concept:

The entropy of a system plus its surroundings (i.e. an isolated system) can never decrease (2^nd law).

The second law states (ΔS)system + (ΔS)surr. = 0, where Δ = final - initial

(ΔS)_system + (ΔS)_surr > 0 irreversible (real world)

(ΔS)_system + (ΔS)_surr = 0 reversible (frictionless, ideal)

In an ideal case if Q is the heat supplied from a source at T, its availability or the maximum work it can deliver is Q (1-T₀ /T) where T₀ is the temperature of the surroundings. Invariably it will be less than this value. The difference is termed as irreversibility.  

Alternatively,

Irreversibility: It is the amount of work to be done to restore the system to the original state.

Mathematically,

I = T₀ ΔS_universe

Calculation: 

Given:

Q = 1000 W, T₀ = 25 + 273 = 298

Irreversibility: 

\(I=T_0Δ S_{universe}\)

\(Δ S_{universe}=Δ S_{system}+Δ S_{gen}\)

\(Δ S_{universe}=Δ S_{system}=\Sigma\left(\frac{Q}{T}\right)\)

\(Δ S_{universe}=Q\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=1000\left(\frac{1}{300}-\frac{1}{400}\right)=0.833; W/K\)

\(I=T_0Δ S_{universe}=298\times0.833=248;W\)

Concept:

Heat transfer (Q) = ΔT / R_th

ΔT = Thermal potential difference

R_th = Total Thermal Resistance = Convective Resistance + Conductive Resistance = R_Conv + R_Cond

Total thermal resistance \(\frac{1}{{{R_{th}}}} = \frac{1}{{\frac{1}{h} + \frac{l}{k}}}\) 

Calculation:

Given:

h = 20 W/m²K, k = 0.9 W/mK

\(\begin{array}{l} \frac{1}{{{R_{th}}}} = \frac{1}{{\frac{1}{{20}} + \frac{{0.18}}{{0.9}}}}=4\frac{W}{{{m^2}K}}\ Q = \frac{{{\rm{Δ }}T}}{{{R_{th}}}} = \left[ {27 - \left( { - 5} \right)} \right] \times 4 = 128\frac{W}{{{m^2}}} \end{array}\)

Concept:

M_T R_m = m₁R₁ + m₂R₂ + m₃R₃ + ..............+ m_nR_n

Rm = (m1R1 + m2R2 + m3R3 + ..............+ mnRn) / M_T   ...................(1)

M_T = m1 + m2 + m₃ +...................+ m_n

R_1, R₂, R₃ are the characteristic gas constant of individual gases.

R_m = Characteristic gas constant of the mixture of gases

R (Characteristic gas constant) = (universal gas constant) / (molecular mass of gas)

R = R_u / M   

Calculation:

Given:

1,2, and 3 represent the gases N2, O2, and CO2 respectively.

m₁ = 0.6M_T, m₂ = 0.3M_T,m₃ = 0.1M_T 

R₁ =  R_u / 28, R₂ =  Ru / 32, R₃ =  Ru / 44

R_u = 8314 J/kmol-k

Using equation 1,

\({R_m} = ({m_1}{R_1} + {m_2}{R_2} + {m_3}{R_3})/{M_T};\)

\({R_m} = \frac{{\left( {\frac{{0.6{M_T}{R_u}}}{{28}} + \frac{{0.3{M_T}{R_u}}}{{32}} + \frac{{0.1{M_T}{R_u}}}{{44}}} \right)}}{{{M_T}}}\)

\({R_m} = \frac{{0.6{R_u}}}{{28}} + \frac{{0.3{R_u}}}{{32}} + \frac{{0.1{R_u}}}{{44}}\)

R_m = 274.9961 J/kg-K