The sentence is stating the highest peak in the world. Since it is a specific thing, we need to use the definite article ‘the’ before it. Also the sentence is using the superlative degree and so we say ‘the highest peak in the world’ making option A the correct answer. There cannot be many highest peaks in the world and so options C and D are incorrect.

The context of the sentence is asking a person who has been deprived of something because of a theft. The word to be used to fill the blank is ‘lose’ which means to be deprived of something.‘Loose’ means something that is not fitted. ‘Louse’ is the singular form of the word ‘lice’ that is a parasite that lives in the skin of mammals and birds.‘Loss’ is a noun that means the process of losing someone or something. Eg: He suffered tremendous loss in his business.

The sentence is looking for a contrast as it is joined by the conjunction ‘despite’. The best pair of words that can fit the context of the sentence is ‘effectiveness…prescribed’. Though the medicine is ‘effective’ in treating diabetes, it is not being ‘prescribed’ widely. A new medicine cannot have a ‘prescription’ or ‘availability’ for treating a disease. ‘Proscribed’ means forbidden by law. In case we use ‘acceptance…proscribed’ the sentence will not make any sense because it will mean that though the medicine is accepted widely, it is not forbidden by law. 

Total no of unripe fruits = 5692000 × (15/100) = 853800

Total no of ripe fruits = 5692000 – 853800 = 4838200

No of unripe apples = 853800 × (45/100) = 384210

No of ripe apples = 4838200 × (34/100) = 1644988   → as 66% are oranges

So, total no of apples = 384210 + 1644988 = 2029198

If it is supposed, all are staying in straight line, we get the following possible arrangement,

Now here as Arun is farther away than Ahmed but closer than Susan from where I live, he has to be between 5 km to 7 km. Here the only possible option is 6.02 km.

Probability that a person is infected = 0.5

Out of the infected people,

Probability of developing disease = 0.3

Probability of not developing disease = 0.7

So probability that person infected but not showing symptoms = 0.5 × 0.7 = 0.35

So the percentage is 35%.

The paragraph states that the subject was very confident about his good friends helping him in his times of need because he had always helped them before in their time. Thus, inference III follows.

Since the events of the last week proved him wrong, this means that his confidence was broken and his friends had not helped him. Thus inference 4 also follows.

Hence option B is the correct answer.

1. Leela is older than her cousin Pavithra.

Leela > Pavithra

2. Pavithra’s brother Shiva is older than Leela.

Shiva > Leela > Pavithra

3. When Pavithra and Shiva are visiting Leela, all three like to play chess.

4. Pavithra wins more often than Leela does.

Now let’s check each option,

A) When Shiva plays chess with Leela and Pavithra, he often loses → any information about Shiva’s play is not given.

B) Leela is the oldest of the three → this is false as Shiva is the oldest.

C) Shiva is a better chess player than Pavithra → any information about Shiva’s play is not given.

D) Pavithra is the youngest of the three → This is true.

\(\begin{array}{l} {q^{ - a}} = {r^{ - 1}};;;{r^{ - b}} = {s^{ - 1}};;;{s^{ - c}} = {q^{ - 1}}\ \Rightarrow {q^a} = {r^1};,;{r^b} = {s^1};,;{s^c} = {q^1}\ \Rightarrow {q^a} = {s^{ac}} = {\left( {{r^b}} \right)^{ac}} = {r^{abc}} = {r^1}\ \Rightarrow {r^1} = {r^{abc}} \end{array}\)

So, abc = 1.

The work done by Q in 1 hour = 1/(25 × 12) = 1/300

The work done by R in 1 hour = 1/(50 × 12) = 1/600

Q takes a sick leave of 2 days so, the work done by Q in 5 days (i.e. total 7 days)

= (5 × 12)/(25 × 12)

R worked for 18 hours daily for 7 days, so work done by R in 7 days = (7 × 18)/(50 × 12)

So the ratio of work done by Q and R in 7 days \(= \frac{{\left( {\frac{{5 \times 12}}{{25 \times 12}}} \right)}}{{\frac{{7 \times 18}}{{;50 \times 12}}}} = \frac{{20}}{{21}}\)

So the ratio will be 20:21.

Concept:

It is case of equal matrix

∴ L. H. S = R. H. S

Calculation:

\(\left[ {\begin{array}{*{20}{c}} 2&5\ { - 4}&3 \end{array}} \right]\left{ {\begin{array}{*{20}{c}} x\ y \end{array}} \right} = \left{ {\begin{array}{*{20}{c}} 2\ { - 30} \end{array}} \right}\)

2x + 5y = 2     ----(I)

-4x + 3y = -30   ----(II)

Multiply eq (1) by 4 and (II) by 2

∴8x + 20y = 8       ---(III)

-8x + 6y = -60       ---(IV)

Adding eq (I) and (II)

26y = -52 ⇒ y = -2

Putting the value of y is eq (I)

2x + 5 (-2) = 2 ⇒ x = 6

Concept:

This is a definition of Laplace transform

\(L\left{ {f\left( t \right)} \right} = ;\mathop \smallint \nolimits_0^\infty {e^{ - st}}f\left( t \right)dt = f\left( s \right)\)

Some important Laplace transforms are:

\(L\left( {{t^n}} \right) = \frac{{n!}}{{{s^{n + 1}}}}\)

\(L\left( {{t^n}} \right) = \frac{{n!}}{{{s^{n + 1}}}}\)

\(L\left( {{t^n}{e^{at}}} \right) = \frac{{n!}}{{{{\left( {s - a} \right)}^{n + 1}}}}\)

Concept:

if f(z) is an analytic function then Cauchy-Riemann condition will be satisfied.

i.e., \(\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}{\rm{}} & \frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)

Calculation:

Given:

u = 2xy​

\(\frac{{\partial u}}{{\partial x}} = 2y \Rightarrow \frac{{\partial v}}{{\partial y}} = 2y\)

\(\frac{{\partial u}}{{\partial y}} = 2x \Rightarrow \frac{{\partial v}}{{\partial x}} = ; - 2x\)

Now,

\(dv = \frac{{\partial v}}{{\partial x}}dx + \frac{{\partial v}}{{\partial y}}dy\)

dv = -2x dx + 2y dy

Integrating both sides

\(\smallint dv = \smallint \left( { - 2x} \right)dx + \smallint 2ydy\)

v = -x² + y² + constant

Alternate solution

f(z) = u + iv

If u(x,y) is given, then

\(f\left( z \right) = \smallint {u_x}\left( {z,0} \right)dz - i\smallint {u_y}\left( {z,0} \right)dz + Constant\)

\({u_x} = \frac{{\partial u}}{{\partial x}} = 2y;\)and \({u_y} = \frac{{\partial u}}{{\partial y}} = 2x\)

u_x(z,0) = 0, u_y(z,0) = 2z

\(f\left( z \right) = - i\smallint \left( {2z} \right)dz + Constant\)

F(z) = -i(z²) + constant

F(z) = -i(x + iy)² + constant

F(z) = -i{(x² – y²) + 2ixy)} + constant

F(z) = 2xy + i(-x² + y²) + constant

v(x,y) = -x² + y²

If v(x,y) is given, then

\(f\left( z \right) = \smallint {v_y}\left( {z,0} \right)dz + i\smallint {v_x}\left( {z,0} \right)dz + Constant\)

Explanation:

Poisson distribution formula,

\(;P\left( x \right) = \frac{{{e^{ - λ }}{λ ^x}}}{{x!}}\)

where λ = mean value of occurrence within an interval

P(x) = probability of x occurrence within an interval

For Poisson Distribution we have

Mean = Variance = (Standard Deviation)²

∴ Standard Deviation = \(\sqrt {Mean} = \sqrt \mu\)

Concept:

Newton-Raphson method:

It has order of convergence 2 and number of guesses required is 1.

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

Calculation:

x = 10 cos x

⇒ f(x) = x – 10 cos x ⇒ f'(x) = 1 + 10 sin x

\(\therefore At;n = 0 \Rightarrow {x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)

\({x_0} = \frac{\pi }{4}\)

\(f\left( {\frac{\pi }{4}} \right) = \frac{\pi }{4} - 10\cos \frac{\pi }{4} = ; - 6.285\)

\(f'\left( {\frac{\pi }{4}} \right) = 1 + 10\sin \frac{\pi }{4} = 8.071\)

\(\therefore {x_1} = \frac{\pi }{4} + \frac{{6.285}}{{8.071}} = 1.564\)

∴ Value of Iteration after 1^st iteration = 1.56

Mistake Point: While calculation be aware that mod of virtual calculation is in radian, if you are taking angle is π-terms.

Concept:

Lami's Theorem:

If a body is in equilibrium under the effect of three forces, then each force is proportional to the sine of the angle between the other two.

\(i.e\frac{P}{{\sin \propto }} = \frac{Q}{{\sin \beta }} = \frac{R}{{\sin\gamma }}\)

Also,

\(\frac {T_{QR}}{\sin (180\ -\ \alpha)} = \frac {{W}}{\sin (\beta \ +\ \alpha)} = \frac {T_{PR}}{\sin(180\ -\ \beta)} \)

sin(180° - θ) = sin θ 

Calculation:

Let T be tension is spring.

Lami’s Theorem

\(\frac{T}{{\sin 90^\circ }} = \frac{{100}}{{\sin 120^\circ }} = \frac{F}{{\sin 150^\circ }}\)

\(F = \frac{{100\sin 150}}{{\sin 120}}\)

F = 57.73 N

Concept:

Conservation of energy principle i.e. the sum of its mechanical energy remains constant.

Calculation:

A body of mass m at any height H will possess potential energy (mgH) and kinetic energy \(\left( {\frac{1}{2};m{v^2}} \right)\) if it is in motion.

∵ There is no friction on the wall. Therefore no energy deteriorates.

\(mg{H_i}; + \frac{1}{2}mV_i^2 = mg{H_f} + \frac{1}{2}mV_f^2\)

H_i = H, H_f = 0, V_i = 0

\({V_f} = \sqrt {2gH}\)

Conclusion:

The conservation of energy principle does not have any effect due to the curvature of the path if the motion is frictionless or on the smooth wall.

Concept:

Torsional Rigidity: It is defined as the product of modulus of rigidity (G) and polar moment of Inertia.

⇒ For the same material modulus of rigidity remains the same.

\(\frac{{{J_2}}}{{{J_1}}} = \frac{{G{I_{p2}}}}{{G{I_{p1}}}} = \frac{{{I_{P2}}}}{{{I_{P1}}}} = \frac{{\frac{\pi }{2}\left( {r_4^4 - r_3^4} \right)}}{{\frac{\pi }{2}\left( {r_2^4 - r_1^4} \right)}}\)

\(\frac{{{J_2}}}{{{J_1}}} = \frac{{r_4^4 - r_3^4}}{{r_2^4 - r_1^4}}{{;;;;;}} \ldots \left( 1 \right)\).

Calculation:

Given:

The area of correction is the same.

\(\Rightarrow \pi \left( {r_2^2 - r_1^2} \right) = \pi \left( {r_4^2 - r_3^2} \right){{;;;;;}} \ldots \left( 2 \right)\) 

\(\Rightarrow r_2^2 - r_1^2 = r_4^2 - r_3^2\)

From equation (I) and (2),

\(\Rightarrow\frac{{{J_2}}}{{{J_1}}} = \frac{{\left( {r_4^2 - r_3^2} \right)\left( {r_4^2 + r_3^2} \right)}}{{\left( {r_2^2 - r_1^2} \right)\left( {r_2^2 + r_1^2} \right)}} = \frac{{r_4^2 + r_3^2}}{{r_2^2 + r_1^2}}\)

\( \because {a^4} - {b^4} = {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2} = \left( {{a^2} + {b^2}} \right)\left( {{a^2} - {b^2}} \right)\)

\(\because {r_4} > {r_2};;& ;;{r_3} > {r_1}\) i.e both numerator terms are greater than denomination term and there is addition is numerator & denominator both

\(\therefore r_4^2 + r_3^2 > r_2^2 + r_1^2\)

\(\Rightarrow\frac{{{J_2}}}{{{J_1}}} > 1\)

Concept:

\(I = \frac{{{a^4}}}{{12}}\)

\({\delta _{free;end}} = \frac{{P{L^3}}}{{3EI}},;\delta \propto \frac{1}{I} \Rightarrow \delta \propto \frac{1}{{{a^4}}}\)

∵ Initial side = a

Final side = 1.19 a

\(\therefore {\delta _i} = \frac{k}{{{a^4}}},{\delta _f} = \frac{k}{{{{\left( {1.19a} \right)}^4}}} = \frac{k}{{2.005;{a^4}}} = 0.498\frac{k}{{{a^4}}}\)

Where k → constant ⇒ δ_f = 0.498 δ_i

\(\frac{{{\delta _f} - {\delta _i}}}{{{\delta _i}}} = \left( {\frac{{{\delta _f}}}{{{\delta _i}}} - 1} \right) \times 100% = \left( {0.498 - 1} \right) \times 100% = \left( { - 0.501} \right) \times 100\)

= 50% decreases.

Concept:

Gyroscopic couple:

If the axis of a spinning or rotating body is gives an angular motion about an axis perpendicular to the axis of spin, an angular acceleration acts on the body about the third perpendicular axis. The couple required to produce this acceleration is known as gyroscopic couple or torque.

τ = Iωω_p = I α

I = polar moment of inertia, ω = angular velocity of spin axis, ω_p = angular velocity of precision axis

∝ = Angular acceleration

Calculation:

I = 10 kg-m², ω = 100 rad/s

\({\omega _p} = \frac{V}{r} = \frac{{20}}{{100}} = 0.2;rad/sec\)

τ = Iωωp 

τ = 10 × 100 × 0.2 = 200 N-m

Concept:

Steady state Amplitude:

\(A = \frac{{{f_0}/k}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\varepsilon \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

At resonance ω = ω_n i.e. damped frequency is equal to natural frequency

\(A = \frac{{{f_0}/k}}{{2\varepsilon }}\)

Calculation:

\({\left( {Amplitude} \right)_{Resonance}} = \frac{{{F_0}}}{{2\varepsilon k}} = \frac{{100}}{{2 \times 0.25 \times 10 \times 1000}}\)

x = 0.02 m = 20 mm

Concept:

When force is applied, spring as a whole is compressed or elongated but this overall deformation comes about due to torsional/bending deformation of the spring wire.

Load per unit defection is known as stiffness or spring Constance (k)  

\(k = \frac{{G{d^4}}}{{8{D^3}n}}\)

where G = Shear modulus of the spring material, d = Wire diameter, D = Mean coil diameter, n = number of active turns.

So, for a given helical spring, the spring constant depends on coil diameter, wire diameter and the number of active turns.

Note: This question is under ambiguity as modulus of rigidity defines material strength under torsion but according to option spring constant of helical compression spring does not depend on material strength.

Concept:

For turbulent flow

\(u = \bar u + u'\)

\(\bar{u}=\frac{1}{T} \int_0^T u d t\)

where,

u = velocity in x-direction at any instant.

\(\bar u\) = Average velocity in x-direction.

u’ = Fluctuating component of velocity in x-direction.

Time average of fluctuating component of velocity,

\(\overline{u^{\prime}}=\frac{1}{T} \int_0^T u^{\prime} d t=\frac{1}{T} \int_0^T(u-\bar{u}) d t\)

\(\overline{u^{\prime}}=\frac{1}{T} \int_0^T u d t-\frac{1}{T} \int_0^T \bar{u} d t=\bar{u}-\left[\frac{1}{T} \times(\bar{u} T)\right]=0\)

Important Points

Interaction effect between two fluctuating components over a long period is non-zero.

\(\mathop \smallint \limits_{ - T}^T u'v'dt \ne 0\)

where v’ → fluctuating component of velocity in y-direction

Explanation:

When a body is either wholly or partially immersed in a fluid, a lift is generated due to the net vertical component of hydrostatic pressure forces experienced by the body. This lift is called the buoyant force and the phenomenon is called buoyancy.

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of the liquid displaced by the body and acts vertically upward through the centroid of the displaced volume.

Thus, the net weight of the submerged body, (the net vertical downward force experienced by it) is reduced from its actual weight by an amount that equals the buoyant force.

FB = ρghA = ρgV

FB = f(Vdisplaced, ρ)

Concept:

The insulating materials are added to decrease the heat loss through the cross-section, however, this addition of insulation does not guarantee that there will be a decrease in the heat transfer since the addition of insulation has the following effects on the resistance to the heat flow

• The conduction resistance increases due to increases in the outer radius
• The convection resistance decreases due to an increase in the surface area of the heat flow,

Therefore the critical thickness of the insulation at which the heat flow through the cross-section is maximum but beyond this thickness, there is always a reduction in the heat transfer.

The critical radius cylindrical cross-section​ is given as

\({r_c} = \frac{k}{h}\)

And for spherical cross-section

\({r_c} = \frac{2k}{h}\)

Where k is the thermal conductivity of the insulating material and h is the outside heat transfer coefficient.

Calculation:

Given:

\( \Rightarrow {{\rm{r}}_{\rm{c}}} = \frac{{{{\rm{K}}_{{\rm{ins}}}}}}{{\rm{h}}} = \frac{{0.15 \times 1000}}{{25}}{\rm{;mm}}\)

r_c = 6 mm

∴ r_c > r₀

∴ Adding insulation increases heat transfer up to 6 mm beyond it decreases heat transfer.

Concept:

Both Heat and Work are boundary phenomena. Both are observed only at the boundaries of the system and both represent energy crossing the boundary of the system.

Point function:

• The thermodynamic properties which depend on the end state only (independent of the path followed) are known as point function like temperature, pressure, density, volume, enthalpy, entropy etc.

Path function: 

• The thermodynamic properties which depend on the end states, as well as the path followed, are known as path function like heat and work.

Only thermodynamic properties are exact differential and properties are point function. As both heat and work are path functions so, they are not thermodynamic properties. So, they are inexact differentials.

Stoichiometric Requirement:

\({C_3}{H_8} + a;\left( {{O_2} + 3.76;{N_2}} \right) \to bC{O_2} + c{H_2}O + 3.76a;{N_2}\)

On comparing both side

H: 2c = 8 ⇒ c = 4

C: b = 3

O: 2a = 2b + c ⇒ a = 5

\({C_3}{H_8} + 5;\left( {{O_2} + 3.76;{N_2}} \right) \to 3C{O_2} + 4{H_2}O + 18.8;{N_2}\)

Assuming no hydrocarbons in the products:

\({C_3}{H_8} + 5;{O_2} \to 3C{O_2} + 4{H_2}O\)

∵ Now 10% deficit oxygen is used:

Oxygen = 0.9 × 5 = 4.5

\({C_3}{H_8} + 4.5;{O_2} \to aC{O_2} + bCO + c{H_2}O\)

On comparing

H: 2c = 8 ⇒ c = 4

O: 2a + b + c = 9 ⇒ 2a + b = 5    ---(I)

C: a + b = 3     ---(II)

Eq. (I) and (II): a = 2, b = 1

\(\therefore {C_3}{H_8} + 4.5;{O_2} \to 2C{O_2} + 1;CO + 4{H_2}O\)

Volume % of CO in product:

\({V_{CO}} = \frac{b}{{a + b + c}} = \frac{1}{{2; + ;1; + ;4}} \times 100;% = 14.28;% \)

Concept:

Specific speed of turbine: The speed required to produce unit power working under a unit heat is called specific speed of turbine.

\({N_S} = \frac{{N\sqrt P }}{{{H^{5/4}}}}\)

Calculation:

N_S1 = N_S2, H₁ = H₂

\(\frac{{{N_1}}}{{{N_2}}} = 2\)

\(\therefore {N_1};\sqrt {{P_1}} ; = {N_2};\sqrt {{P_2}}\)

\(\frac{N_2}{N_1}=\sqrt\frac{P_1}{P_2}\)

\(\Rightarrow \frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{N_2}}}{{{N_1}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4} = 0.25\)

Explanation:

Regeneration in vapour power cycle (i.e. Rankine cycle)

Regeneration means to take some part of heat from the expanding steam in the turbine to decrease the overall heat supplied in the process.

The thermal efficiency of the Rankine cycle can be increased by the use of regenerative heat exchange.

In the regenerative cycle, a portion of the partially expanded steam is drawn off between the high - and low - pressure turbines. The steam is used to preheat the condensed liquid before it returned to the boiler. In this way, the amount of heat added at the low temperatures is reduced. Therefore, the mean effective temperature of heat addition is increased, and cycle efficiency is increased.

⇒ At point 2 some amount of steam is taken out for heating purpose of water.

Ideal cycle is approached in real practice by heating the feed water by bleeding steam from the turbine at certain points after doing work. The bled steam is mixed with the condensate in a separate heater.

Effect of Regeneration:

→ Decrease in turbine work due to decrease in mass flow rate of steam.

→ Decrease in heat rejection in condenser due to reduction in mass flow rate.

→ Increased in mean temperature of heat addition.

So increases the efficiency of engine.

Important Point:

Efficiency of Ideal Regeneration Rankine cycle = Efficiency of Carnot cycle

\({\eta _{IRRC}} = {\eta _C} = {\eta _{Stirling;cycle}} = 1 - \frac{{{T_L}}}{{{T_H}}}\)

Feed water are used as heat exchanger in regenerative cycle.

Explanation:

Jominy test:-

→ Distance from tip (bottom) to the location where there is 50% pearlite & 50% martensite is called Jominy distance

→ Jominy test is used for finding hardenability i.e. case of martensite formation

Hardenability is measured by the Jominy end quench test performed for standard cylindrical specimen, standard austenitization conditions, and standard quenching conditions (jet of water at a specific flow rate and temperature)

Explanation:

→ Welding always requires a very high temperature, since it involves melting down of metals. It is a well-known fact that metals have very high melting points.

→ For soldering filler melting point temperature is less than 427°C. Lead + Zn is used as a filler material. ZnCl₂ & NH₄Cl is used as a flux.

→ For brazing, filler melting point temperature is greater than 427°C but less than work piece melting point temperature. Cu + Zn i.e. brass is used as a filler material. Borax or Boric acid is used as a flux.

→ So T_welding > T_brazing > T_soldering

Pouring Basin →

→ Pouring Basin act as are reservoir from which molten metal moves smoothly into the sprue.

→ Pouring basin stop slag from entering the mould cavity by means of a skimmer or skim bob.

→ The depth of the pouring basin is 2.5 times the sprue entrance diameter for smooth metal flow and to prevent vortex formation:

Runner:

→ It is generally located in a horizontal plane (parting plane) which connects the sprue to its ingates, thus, allowing the metal to enter the mould cavity.

Ingate (Gate):

→ These are the opening through which molten metal enters the mould cavity.

→ The shape and cross-section of the ingate should be such that it can readily be broken off after casting solidification and also to allow the metal to enter quickly into other mould cavities.

Choke:

Also called sprue base well.

→ Choke is that part of the gating system which has the smallest cross-sectional area and helps in controlling the rate of metal flow, thus ensuring lower metal flow velocity in runners and minimizing sand erosion in runners.

Explanation:

Electron Beam Machining (EBM): In this machining workpiece placed in the vacuum chamber and High‐voltage electron beam directed toward the workpiece. The energy of electron beam melts/ vaporizes selected region of the workpiece. The electron beam is moved by deflection coils.

The operation is performed in the vacuum to prevent the reduction of electron velocity.

Important Points

α = 0, F_c = 500 N, F_T = 250 N

μ = ?

Calculation:

\(\mu = \frac{F}{N} = \frac{{{F_c}\sin \alpha + {F_T}\cos \alpha }}{{{F_c}\cos \alpha - {F_T}\sin \alpha }}\)

sin 0 = 0, cos 0 = 1

\(\therefore \mu = \frac{F}{N} = \frac{{{F_T}}}{{{F_C}}} = \frac{{250}}{{500}} = \frac{1}{2}\)

Merchant circle diagram

For α ≠ 0

ϕ → shear plane angle

For α = 0

Merchant circle diagram is a rectangle i.e. F_t = F and F_c = N

Alternate method

\(\because\mu = \tan \beta ; = \frac{f}{N} = \frac{{{F_T}}}{{{F_c}}}\)     [∵ from merchant circle diagram for α = 0]

\(\mu = \frac{{250}}{{500}} = \frac{1}{2}\)

Plug Gauges are mainly used for measuring hole dimension. The Go-gauge checks the lower limit of a straight hole and the NO-Go gauge checks the upper limit.

Snap Gauges: Used as a quick means for checking sizes within certain limits by comparing the size of the parts with the opening of the gauge.

Feeler Gauge is used to measure gap width or clearance between the mating parts.

Fillet Gauge: It is also known as Radius gauge. Components are machined to have a curved formation on the edges or at the junction of two parts, which are known as a fillet. The gauges used to check the radius of fillet are fillet gauges.

f(x) = 2x³ – 3x² [-1, 2]

Global Minimum t f(x) = ?

f (-1) = 2(-1)³ – 3(-1)² = -2 – 3 ⇒ f (-1) = -5

f (2) = 2(2)³ – 3(2)² = 16 – 12 ⇒ f (2) = 4

f’(x) = 6x² – 6x

f’(x) = 0 ⇒ 6x² – 6x = 0 ⇒ 6x (x - 1) = 0 ⇒ x = 0, 1

So, f (0) = 0

f (1) = 2(1)³ – 3(1)² ⇒ f (1) = -1

⇒ Key concept: In this question since it is asked for global minimum no. you have to check for all the value which are in the range of [-1, 2].

∴ Global Minimum of f(x) ⇒ f (-1) = -5

y’’ + 9y = 0

y’’ + 9y = 0

Differential Eq. is of 2^nd order and Homogeneous.

Differential Eq. → D² + 9 = 0

Auxiliary Eq. →

∴ m² + 9 = 0 ⇒ m² = -9 = 9 i²

m = ± 3i

On comparing α ± iβ on root of above eqn.

i.e. Root of above eq. is imaginary α = 0, β = 3

So, Solution above eq.

y = C.F = e^αx [C₁ cos β_x + C₂ sin β_x]

∴ y = C₁ cos 3x + C₂ sin 3x

From given boundary condition: y (0) = 0

At x = 0, y = 0

⇒ C₁ = 0

y(π/2) = √2

\(\sqrt 2 = {C_1}\cos 3\left( {\frac{\pi }{2}} \right) + {C_2}\sin \frac{{3\pi }}{2} = {C_2}\sin \frac{{3\pi }}{2} = - {C_2}\)

⇒ C₂ = -√2

∴ y = -√2 sin 3x

y(π/4) = ?

\(y = - \sqrt 2 \sin \frac{{3\pi }}{4} = - \sqrt 2 \times \frac{1}{{\sqrt 2 }} = - 1\)

\(y\left( {\frac{\pi }{4}} \right) = - 1\)

\(\mathop \smallint \limits_{ - \infty }^\infty \frac{{\sin x}}{{{x^2} + 2x + 2}}dx\)

Concept:

Cauchy Integral Theorem:

\(\oint \left[ {\frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}} \right]dz = 2\pi i\frac{{{f^n}\left( a \right)}}{{n!}}\)

Where z = a be any point inside the close region.

Cauchy’s Residue Theorem:

\(\frac{1}{{2\pi i}};\oint f\left( z \right)dz = \) Sum of Residue at Pole or singularity with in the region

Res at z = a

\(= \frac{1}{{\left( {n - 1} \right)!}};\mathop {\lim }\limits_{z \to a} {\left( {\frac{d}{{dz}}} \right)^{n - 1}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]\)

Calculation:

We know that \({e^{ix}} = \cos x + i\sin x\)

Let x replace by z.

\(\therefore I = \mathop \smallint \limits_{ - \infty }^\infty \frac{{Imaginary;\left[ {{e^{iz}}} \right]}}{{{z^2} + 2z + 2}}dz\)

Now, z² + 2z + 2 = 0 then roots of z are:

\(\therefore z = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 8} }}{2} = \frac{{ - 2 \pm \sqrt {4{i^2}} }}{2} = - 1 \pm i\)

z = -1 ± i is the only pole lying in f(z) > 0

Here n = 1

\(Re{s_{z = a}} = \frac{1}{{\left( {n - 1} \right)!}};\mathop {\lim }\limits_{z \to a} {\left( {\frac{d}{{dz}}} \right)^{n - 1}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]\)

\(\therefore Re{s_{at;z = - 1 \pm i}} = \frac{1}{{\left( {1 - 1} \right)!}}\mathop {\lim }\limits_{z \to - 1 \pm i} {\left( {\frac{d}{{dz}}} \right)^{1 - 1}}\left[ {\left( {z - {{\left( {1 \pm i} \right)}^1}} \right).\frac{{{e^{iz}}}}{{\left[ {z - \left( { - 1 + i} \right)} \right]\left[ {z - \left( { - 1 - i} \right)} \right]}}} \right]\)

\(\therefore Re{s_{at;z = - 1 + i}} = \mathop {\lim }\limits_{z \to - 1 + i} \left[ {z - \left( { - 1 + i} \right)} \right] \cdot \frac{{{e^{iz}}}}{{(z - \left( { - 1 + i} \right)\left[ {z - \left( { - 1 - i} \right)} \right]}}\)

\(= \mathop {\lim }\limits_{z \to - 1 + i} \frac{{{e^{iz}}}}{{z - \left( { - 1 - i} \right)}}\)

\(= \frac{{{e^{i\left( { - 1 + i} \right)}}}}{{ - 1 + i - \left( { - 1 - i} \right)}} = \frac{{{e^{ - i + {i^2}}}}}{{ - 1 + i + 1 + i}} = \frac{{{e^{ - i - 1}}}}{{2i}}\)

\(\therefore \mathop \oint \nolimits_c^{} \frac{{Imaginary;\left( {{e^{iz}}} \right)}}{{{z^2} + 2z + 2}}dx = Imaginary;\left[ {2\pi i \cdot \frac{{{e^{ - i - 1}}}}{{2i}}} \right]\)

\(= Imaginary;\left[ {\pi {e^{ - i}} \cdot {e^{ - 1}}} \right]\)

\(= Imaginary;\left[ {\frac{{\pi {e^{ - i}}}}{e}} \right];;;\left[ {{e^{ix}} = \cos x + \sin x} \right]\)

\(= \frac{\pi }{e};Imaginary;\left[ {\cos 1 - i\sin 1} \right]\)

Taking only imaginary part

\(\oint \frac{{Imaginary;\left( {{e^{iz}}} \right)}}{{{z^2} + 2z + 2}}dz = - \frac{{\pi \sin 1}}{e}\)

Concept:

Gauss Seidel Method:

In Gauss Seidel method, the value of x calculated is used in next calculation putting other variable as 0.

x₁ + 2x₂ + 3x₃ = 5

Putting x₂ = 0, x₃ = 0 ⇒ x₁ = 5

2x₁ + 3x₂ + x₃ = 1

Putting x₁ = 5, x₃ = 0 ⇒ x₂ = -3

3x₁ + 2x₂ + x₃ = 3

Putting x₁ = 5, x₂ = -3 ⇒ x₃ = 3 – 3(5) – 2 (-3)

x₃ = 3 – 15 + 6

x₃ = -6

Mistake Point → Don’t arrange them diagonally because It is given in question solve as per given order.

Given: T = 20 N, μ = 0.25, cos θ = 0.8, sin θ = 0.6, g = 10 m/s²

Friction always acts opposite to the direction of motion.

N = mg cos θ = m × 10 × 0.8 = 8m

∵ It is given static friction. Therefore there will be no motion i.e. a = 0

mg sin θ = m × 10 × 0.6 = 6m

mg sin θ = μ_sN + T

6m = 0.25 × 8m + 20

6m = 2m + 20 ⇒ m = 5 kg

Concept:

Truss is a member of structure such that it does not change it shape under load except deformation of one of member of truss.

Its degree of freedom is 0.

P & R is hinged joint

\(\tan 30^\circ = \frac{{PR}}{{PQ}} \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{PR}}{L} \Rightarrow PR = \frac{L}{{\sqrt 3 }}\)

\(\cos 30^\circ = \frac{{PQ}}{{QR}} \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{L}{{QR}} \Rightarrow QR = \frac{{2L}}{{\sqrt 3 }}\)

Solving by joint method:

∑F_x = 0 ⇒ R_x + P_x = 0

∑F_y = 0 ⇒ R_y = W + P_y

At point Q.

∑F_y = 0

F_QR sin 30 – W = 0 ⇒ F_QR/2 = W

F_QR = 2W

Force towards point is taken compressive (+ve)

F_QR = 2W (compressive)

∑F_x = 0 ⇒ F_PQ + F_QR cos 30° = 0 ⇒ F_PQ = -√3/2 F_QR

F_PQ = -√3 W

F_PQ = √3 W (tensile)

Alternate Method

Applying sin rule

\(\frac{{{F_{QR}}}}{{\sin 90^\circ }} = \frac{{{F_{PQ}}}}{{\sin 60^\circ }} = \frac{W}{{\sin 30^\circ }}\)

⇒ F_QR = 2W compressive

⇒ F_PQ = √3 W tensile

Explanation:

\({\delta _{AB}} = - \frac{{RL}}{{AE}} = \frac{{ - \left( {P - F} \right)L}}{{A\left( {3E} \right)}}\)

\({\delta _{BC}} = \frac{{FL}}{{AE}}\)

Total Deflection = 0

δ_AB + δ_BC = 0

\(\frac{{ - \left( {P - F} \right)L}}{{3AE}} + \frac{{FL}}{{AE}} = 0 \Rightarrow \frac{{FL}}{{AE}} = \frac{{\left( {P - F} \right)L}}{{A\left( {3E} \right)}}\)

\(\Rightarrow 3F = P - F \Rightarrow \frac{P}{F} = 4\)

MOI of a rectangle about the centre of gravity

\(I = \frac{{b{d^3}}}{{12}}\)

But since in question, MOI is asked about its base so. use parallel axis theorem i.e. I_x-x = I_G + Ay²

∴ I = (I_xx)_Rectangle – (I_x-x)_Circle

\(I = \left[ {\frac{{b{d^3}}}{{12}} + bd;{{\left( {10} \right)}^2}} \right] - \left[ {\frac{\pi }{{64}}{{\left( d \right)}^4} + \frac{\pi }{4}{{\left( d \right)}^2} \times {{\left( {10} \right)}^2}} \right]\)

\(I = \left[ {\frac{{10{{\left( {20} \right)}^3}}}{{12}} + 10 \times 20;{{\left( {10} \right)}^2}} \right] - \left[ {\frac{\pi }{{64}}{{\left( 8 \right)}^4} + \frac{\pi }{4}{{\left( 8 \right)}^2} \times {{\left( {10} \right)}^2}} \right]\)

I_x-x= 26666.66 – 5227.61

I_x-x = 21439.05

Calculation:

Given:

E = 200 GPa, P = 1 N, L = 1 m

(σ_b)_A = ?

R_Bx = 0

R_By + R_Cy + 1N = 1N

R_By = -R_Cy

∑M_B = 0

⇒ -R_Cy × 3 + P × 2 – P × 1 = 0 ⇒ -3R_Cy + P = 0

\({R_{Cy}} = \frac{P}{3}\therefore {R_{By}} = \frac{{ - P}}{3}\)

Now Taking Moment along A

∑ M_A = R_By × 1.5 + P × 0.5

 \({M_A} = \frac{{ - P}}{3} \times 1.5 + 1 \times 0.5 = 0 \Rightarrow {M_A} = 0\)

\(\therefore {\left( {{\sigma _b}} \right)_A} = \frac{{My}}{I} = 0\)

∵ In question it is given that crank is perpendicular to stroke line of piston

∴ Torque = Force × Crank length

\(= 5000;N \times \frac{{200}}{{1000}}m\)

= 1000 N-m

T = 1 kN-m

Concept:

A gear train having a relative motion of axis is called a planetary or an epicyclic gear train. In an epicyclic gear train, the axis of at least one of the gears also moves relative to the frame.

Calculation:

T₂ = 60, T₃ = 20, T₄ = 100

N₂ → fixed = 0, N₄ = 100 rpm ccw i.e. -100 rpm.

N₅ = ? i.e. arm.

\(\frac{{{N_3}}}{{{N_2}}} = - \frac{{{T_2}}}{{{T_3}}}\) [External Mesh]

Let gear 2 is given x revolutions.

\({N_3} = - \frac{{{T_2}}}{{{T_3}}}x = - \frac{{60}}{{20}}x = - 3x\)

\(\frac{{{N_4}}}{{{N_3}}} = \frac{{{T_3}}}{{{T_4}}} \Rightarrow {N_4} = \frac{{{T_3}}}{{{T_4}}}{N_3} = \frac{{20}}{{100}} \times - 3x = - \frac{3}{5}x\)

∵ N₂ fixed ∴ y = -x as x + y = 0

N₄ = -100

\( - \frac{{3x}}{5} + y = - 100 \Rightarrow \frac{3}{5}y + y = - 100 \Rightarrow \frac{8}{5}y = - 100\) 

\(y = - \frac{{500}}{8} \Rightarrow y = - 62.5;rpm\) 

Arm revolution = 62.5 rpm ccw

Total Energy of system, TE = K.E due to translator motion + K.E due to rotary motion + P.E of spring

Displacement for mass: x = aθ

\(\begin{array}{l} KE = \frac{1}{2}M{{\dot x}^2} + \frac{1}{2}I{{\dot \theta }^2} = \frac{1}{2}M{a^2}{{\dot \theta }^2} + \frac{1}{2}.\frac{1}{2}M{a^2}{{\dot \theta }^2}\ = \frac{1}{2}M{a^2}{{\dot \theta }^2} + \frac{1}{4}M{a^2}{{\dot \theta }^2} = \frac{3}{4}M{a^2}{{\dot \theta }^2}\ PE = \frac{1}{2}K{{x'}^2} \end{array}\)

x' = (a + d)θ

⇒ PE = K {(a + d)θ}²

\(\begin{array}{l} T.E = \frac{3}{4}M{a^2}{{\dot \theta }^2} + \frac{1}{2}K{\left( {a + d} \right)^2}{\theta ^2}\ \frac{d}{{dt}}KE + \frac{d}{{dt}}PE = 0\ \frac{3}{2}M{a^2}\ddot \theta + K{\left( {a + d} \right)^2}\theta = 0 \Rightarrow \ddot \theta + \frac{{K{{\left( {a + d} \right)}^2}}}{{\frac{3}{2}M{a^2}}}\theta = 0 \Rightarrow \ddot \theta + \frac{{2K{{\left( {a + d} \right)}^2}}}{{3M{a^2}}}\theta = 0 \end{array}\)

Natural frequency: \({\omega _n} = \sqrt {\frac{{2K{{\left( {a + d} \right)}^2}}}{{3M{a^2}}}} \)

Concept:

According to Tresca Theory,

 \(Max \left{ {\left( {{\sigma _1} - {\sigma _2}} \right),\left( {{\sigma _2} - {\sigma _3}} \right),\left( {{\sigma _3} - {\sigma _1}} \right)} \right} \le \frac{{{\sigma _{yt}}}}{{{n_T}}}{{;;;;;}} \ldots \left( 1 \right)\)

According to Von–Mises

\({\left( {{\sigma _1} - {\sigma _2}} \right)^2} + {\left( {{\sigma _2} - {\sigma _3}} \right)^2} + {\left( {{\sigma _3} - {\sigma _1}} \right)^2} + 6\left( {\tau _{xy}^2 + \tau _{yz}^2 + \tau _{zx}^2} \right) \le 2{\left( {\frac{{{\sigma _{yt}}}}{{{n_v}}}} \right)^2}{{;;;;;}} \ldots \left( 2 \right)\)

Where,

Maximum shear stress (Tresca) theory,  FOS = nT, Von-Mises (Maximum distortion energy) Theory, FOS = nV

Calculation:

Given:

σ₁ = 100 MPa, σ₂ = 100 MPa, σ₃ = 0, σ_yt = 200 MPa

By using equation (1),

\(Max\left{ {0,100, - 100} \right} \le \frac{{200}}{{{n _T}}} ⇒ {n_T} = 2\)

By using equation (2),

\({\left( 0 \right)^2} + {\left( {100} \right)^2} + {\left( {0 - 100} \right)^2} \le 2{\left( {\frac{{200}}{{{n _V}}}} \right)^2}\)

\(⇒10000 + 10000 \le 2{\left( {\frac{{200}}{{{n _V}}}} \right)^2}\)

_{⇒ nv} = 2

∴ n_T = n_V

Concept:

Hydrostatic Law \(\frac{{\partial P}}{{\partial Z}} = - pg;\left[ {Going;Upward} \right]\)

\(\frac{{\partial P}}{{\partial Z}} = + pg;\left[ {Going;Downward} \right]\)

P = ρgh

Calculation:

P_A – ρ₀gh₀ – ρ_airgh_a + ρ_wg × h_w – P_B = 0

\({P_A} - 800 \times 10 \times \frac{{200}}{{1000}} - 1.16 \times 10 \times \frac{{80}}{{1000}} + 1000 \times 10 \times \frac{{380}}{{1000}} - {P_B} = 0\)

P_A – 1600 – 0.928 + 3800 – P_B = 0

P_A + 2199.072 – P_B = 0

P_A – P_B = -2.199 kPa

|P_A – P_B|= 2.19 kPa

The magnitude of pressure difference between pipes A and B is 2.19 kPa

Concept:

For pipe flow, the dynamic similarity will be obtained if the Reynolds number in the model and prototype are equal.

(Re)_Model = (Re)_Prototype

\(\frac{{{\rho _m}{V_m}{D_m}}}{{{\mu _m}}} = \frac{{{\rho _p}{V_p}{D_p}}}{{{\mu _p}}}\)

\(\frac{{{V_m}{D_m}}}{{{\nu _m}}} = \frac{{{V_p}{D_p}}}{{{\nu _p}}}\)

Calculation:

Given: ν_oil = 1.0 × 10^-5 m²/s, v_w = 0.89 × 10^-6 m²/s, D_oil = 0.5 m, D_w = 20 mm = 0.02 m, V_oil = 10 m/s, V_w = ?

\({\left( {\frac{{VD}}{v}} \right)_P} = {\left( {\frac{{VD}}{v}} \right)_m}\)

\(\frac{{10 \times 0.5}}{{1.0 \times {{10}^{ - 5}}}} = \frac{{{V_w} \times 0.02}}{{0.89 \times {{10}^{ - 6}}}}\)

V_Water = 22.25 m/s

\(U = {U_0}\left[ {2\left( {\frac{y}{\delta }} \right) - {{\left( {\frac{y}{\delta }} \right)}^2}} \right]\)

\(\frac{{{{\dot m}_{bd}}}}{{{{\dot m}_{a - b}}}} = ?\)

Concept:

ṁ = ρAV [Continuity equation for compressible fluid]

Let take width of plate be w

Calculation:

∴ ṁ_a-b = ρ × A × U₀ = ρ × w × δ × U₀      …1)

\({\dot m_{c - d}} = \mathop \smallint \nolimits_0^\delta \rho w \times dy \times U\)

\({\dot m_{c - d}} = \mathop \smallint \nolimits_0^\delta \rho w \times {U_0}\left[ {2\left( {\frac{y}{\delta }} \right) - {{\left( {\frac{y}{\delta }} \right)}^2}} \right]dy\)

\( = \rho w{U_0}\left[ {\frac{{2.{y^2}}}{{2\delta }} - \frac{{{y^3}}}{{3{\delta ^2}}}} \right]_0^\delta = \rho w{U_0}\left[ {\frac{{{\delta ^2}}}{\delta } - \frac{{{\delta ^3}}}{{3{\delta ^2}}}} \right] = \rho w{U_0}\left[ {\delta - \frac{\delta }{3}} \right] = \frac{{2\rho w{U_0}\delta }}{3}\)

\({\dot m_{b - d}} = {\dot m_{a - b}} - {\dot m_{c - d}} = \rho w{U_0}\delta - \frac{2}{3}\rho w{U_0}\delta \)

\({\dot m_{b - d}} = \frac{1}{3}\rho w{U_0}\delta \)     …3)

From equation 1) & 3)

\(\frac{{{{\dot m}_{b - d}}}}{{{{\dot m}_{a - b}}}} = \frac{1}{3} = 0.33\)

Mistake: Be careful while taking area of flow as the profile is continuously changing.

Concept:

It is unsteady flow as the fluid parameter are changing with respect to time.

Therefore, lumped unsteady analysis is valid here.

Calculation:

\(\frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = exp\left( { - \frac{{hA}}{{\rho V{C_p}}} \times t} \right)\) 

Where A → surface area

D = 10 mm, T = 350 K, T_∞ = 300 K, T₀ = 1000 K

Time constant for a Thermocouple \(\tau = \frac{{\rho V{C_p}}}{{hA}} = 16;sec\)

\(\therefore \frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = \exp \left( { - \frac{t}{\tau }} \right) = \exp \left( { - \frac{t}{{16}}} \right)\) 

\( = \frac{{350 - 300}}{{1000 - 300}} = \exp \left( { - \frac{t}{{16}}} \right)\) 

\( + \frac{{50}}{{700}} = \exp \left( { - \frac{t}{{16}}} \right)\) 

Taking logarithm both side

\(\ln \left( { + \frac{5}{{70}}} \right) = \ln \left{ {\exp \left( { - \frac{t}{{16}}} \right)} \right} = - \frac{t}{{16}}\) 

\( + 2.639 = + \frac{t}{{16}}\) 

t = 42.22 sec

Concept:

For similar object the view factor remains the same i.e. F₁₂ = F₁₄

→ Sum of all view factor for an enclosure = 1

Calculation:

F₁₂ = F₁₄ = 0.26

F₁₂ + F₁₄ + F₁₃ = 1

0.26 + 0.26 + F₁₃ = 1

F₁₃ = 0.48

Net radiation heat loss or gain on side 1 = Q₁₂ + Q₁₄ + Q₁₃

Q₁₂ = Q₁₄ [∵ Temperature & view factor are same]

∴ Q_Net,1 = 2Q₁₂ + Q₁₃

\(=\frac{2\sigma \left( T_{1}^{4}-T_{2}^{4} \right)}{\frac{1-{{\epsilon }_{1}}}{{{A}_{1}}{{\epsilon }_{1}}}+\frac{1}{{{A}_{1}}{{F}_{12}}}+\frac{1-{{\epsilon }_{2}}}{{{A}_{2}}{{\epsilon }_{2}}}}+\frac{T\left( T_{1}^{4}-T_{3}^{4} \right)}{\frac{1-{{\epsilon }_{1}}}{{{A}_{1}}{{\epsilon }_{1}}}+\frac{1}{{{A}_{1}}{{F}_{13}}}+\frac{1-{{\epsilon }_{3}}}{{{A}_{3}}{{\epsilon }_{3}}}}\)

∵ T₁ = T₃ ⇒ Q₁₃ = 0

∵ All surfaces are black surface ⇒ ϵ = 1

∴ Net radiation heat loss or gain on side 1 is

Q_Net,1 = 2Q₁₂ = 2 σ × A₁F₁₂ (T₁⁴ – T₂⁴)

Q_Net,1 = 2 × 5.67 × 10^-8 × 0.5 × L × 0.26 (1200⁴ - 800⁴)

Q_net/L = 24530.68 W/m

∵ Temperature of side 1 is greater than side 2

So there will be a loss of 24530.68 W/m of heat from side 1.

Concept:

Since in question not any velocity profile is given so we will use Blasius equation.

Calculation:

 Pr = 1, T = 500 K, L = 1.5 m, T_∞ = 300 K

\({\left( {{P_r}} \right)^{\frac{1}{3}}} = \frac{{{\delta _{hy}}}}{{{\delta _{th}}}} \Rightarrow \frac{{{\delta _{hy}}}}{{{\delta _{th}}}} = 1\)

∴ δ_hy = δ_th

\(Re = \frac{{\rho VD}}{\mu } = \frac{{VD}}{v} = \frac{{10 \times 0.5}}{{30 \times {{10}^{ - 6}}}} = 1.66 \times {10^5}\) 

∵ Re ≤ 5 × 10⁵

∴ Flow is laminar flow

∴ Blasius equation for laminar flow

\({\delta _{hy}} = \frac{{5x}}{{\sqrt {{R_{ex}}} }} = \frac{{5 \times 0.5}}{{\sqrt {1.66 \times {{10}^5}} }} \times 1000;mm\) 

δ_hy = 6.12 mm

δ_hy = δ_th = 6.12 mm

Mistake Point: While calculating Reynold number. Calculate the Reynold no. at a point at which hydrodynamic boundary length is required.

Concept:

According to Clausius Clapeyron equation

\(\frac{{dP}}{{dT}} = \frac{{Latent;heat \times {P_{sat}}}}{{T_{sat}^2 \times R}}\left[ {When;{V_g} \gg > {V_f}} \right]\) 

\(\frac{{dP}}{{dT}} = \frac{{Latent;heat}}{{{T_{sat}}\left( {{V_g} - {V_f}} \right)}}\)

Calculation:

\(0.189 = \frac{{LH}}{{298\left( {43.38 - 0} \right)}}\)

LH = 2443.24 kJ/kg

Given: P = a + bV [∵ Pressure varies linearly with volume]

P₁ = 100 kPa, V₁ = 0.2 m³, P₂ = 200 kPa, V₂ = 0.1 m³, R = 0.275 kJ/kJ-k

100 = a + 0.2 b     ---(1)

200 = a + 0.1 b    ----(2)

From eq. (1) & (2)

a = 300, b = - 1000

W = ∫ Pdv = \(\smallint _1^2;\left( {a + bV} \right)dV = \left[ {aV + b\frac{{{V^2}}}{2}} \right]_1^2\)

\(W = a;\left( {{V_2} - {V_1}} \right) + \frac{b}{2}\left( {V_2^2 - V_1^2} \right)\)

\(W = 300;\left( {0.1 - 0.2} \right) + \left( {\frac{{ - 1000}}{2}} \right);\left( {{{0.1}^2} - {{0.2}^2}} \right)\)

= - 30 – 500 (– 0.03) = - 30 + 15 = - 15 kJ

∴ In question it has asked only magnitude

Therefore W = 15 kJ

Alternative Method:

∵ Work done = Area enclosed in P – V diagram = Area of trapezium

= ½ (Sum of Parallel distance) × (Distance b/w them)

\(W = \frac{1}{2}\left( {200 + 100} \right) \times \left( {0.2 - 0.1} \right) = 15;kJ\)

Ideal Rankine cycle

S₁ = S₂

6.7428 = s_f + x₂ s_fg

6.7428 = 1.213 + x₂ (6.2434)

x₂ = 0.8857

h₂ = h_f + x.h_fg = 384.39 + (.8857) 2278.6 = 2402.546 kJ/kg

Turbine work (W_T) = h₁ – h­₂

W_T = 3115.3 – 2402.54 = 712.76 kJ/kg

Pump Work (W_p) = V_fdp [V_f at 75 kPa]

W_p = 0.001037 (3000 - 75) kJ/kg = 3.03 kJ/kg

Net work done = W_T - W_P = 709.73 kJ/kg

Heat supplied (Q_s) = h₁ – h₄

W_P = h₄ – h₃ = 3.03 ⇒ h₄ = 387.42 kJ/kg

h₃ = h_f at 75 kPa = 384.39 kJ/kg

\({\eta _{th}} = \frac{{{W_{net}}}}{{{Q_s}}} = \frac{{709.73}}{{{h_1} - {h_4}}} = \frac{{709.73}}{{3115.3 - 384.39}}\)

η_th = 26.01 %

Toughness: The ability of the material to absorb energy up-to-the fracture point is known as toughness.

Toughness = Area enclosed under the stress-strain curve up to fracture limit

Area of trapezoid = ½ × Sum of parallel sides × Distance between them

\(U = \frac{1}{2} \times \frac{{0.2}}{{100}} \times 100 + \frac{1}{2}\left( {100 + 140} \right)\frac{{\left( {0.6 - 0.2} \right)}}{{100}} + \frac{1}{2}\left( {140 + 130} \right)\frac{{\left( {.8 - 0.6} \right)}}{{100}};\)

Please take care that strain is given in % so you have to divide with 100.

Toughness = 0.1 + 0.48 + 0.27

Toughness = 0.85 MPa

\(\frac{{MJ}}{{{m^3}}} = \frac{{MNm}}{{{m^3}}} = \frac{{MN}}{{{m^2}}} = MPa\) [Be careful of unit]

Concept:

Specific Latent Heat (L):

It is the heat required or liberated to change the phase of a substance at constant temperature. Unit - kJ/kg

Example- Latent heat of fusion or melting, Latent heat of Vaporization

Q = Mass × Specific Latent heat

Q = mL

Calculation:

Given:

Q = 10 kW

It is given that 10 kW heat is removed till the molten metal is completely solidified i.e. it is the latent heat which is required for its phase change i.e. temperature remains constant.

\(Q = \frac{{mL}}{t} \Rightarrow 10 = \frac{{2 × L}}{{\left( {20 - 10} \right)}}\)

L → latent heat of fusion L = 50 kJ/kg.

Note: Here Q is given into kW, so to convert kW into J, we need to divide it by time which is already given into the graph.

VT^0.14 f^0.7 d^0.4 = Constant = K

45 (30)^0.14 (0.35)^0.7 (2.0)^0.4 = K

K = 45.842

∵ Speed, feed and depth of cut are increased individually by 25 %

i.e.

f₂ = 1.25 f₁ = 1.25 (0.35) = 0.4375 min/rev

V₂ = 1.25 V₁ = 1.25 (45) = 56.25 m/min

d₂ = 1.25 d₁ = 1.25 (45) = 2.5 mm

VT^0.14 f^0.7 d^0.4 = 45.82

∴ 56.25 × T^0.14 × (0.4375)^0.7 × (0.5)^0.4 = 45.82

T^0.14 = 1.007

\(T = {\left( {1.007} \right)^{\frac{1}{{0.14}}}}\)

T = 1.051 min

Concept:

According to modulus method of riser design:

Modulus of Riser = 1.2 × Modulus of Casting

Modulus \(= \frac{{Volume}}{{Surface;Area}}\)

∵ Diameter of Riser = Height of Riser (d = H)

# Bottom surface of cylindrical riser does not contribute as cooling surface.

∴ Surface area of Riser \(= \pi dH + \frac{\pi }{4}{d^2}\)

\({\left( {{A_s}} \right)_{Riser}} = \pi {d^2} + \frac{\pi }{4}{d^2} = \frac{5}{4}\pi {d^2}\)

Volume of Riser:

\({\left( V \right)_{Riser}} = \frac{\pi }{4}{d^2}h = \frac{\pi }{4}{d^3}\)

\({\left( {\frac{V}{{{A_S}}}} \right)_{Riser}} = \frac{{\frac{\pi }{4}{d^3}}}{{\frac{5}{4}\pi {d^2}}} = \frac{d}{5}\)

Surface area of casting:

\({\left( {{A_s}} \right)_{casting}} = \pi dh + 2 \times \frac{\pi }{4}{d^2}\)

Volume of Riser:

\({\left( V \right)_{Riser}} = \frac{\pi }{4}{d^2}h\)

\({\left( {\frac{V}{{{A_S}}}} \right)_{Casting}} = \frac{{\frac{\pi }{4}{d^2}h}}{{\pi dh + 2 \times \frac{\pi }{4}{d^2}}} = \frac{{\frac{1}{4}dh}}{{h + \frac{1}{2}d}} = \frac{{\frac{1}{4} \times 200 \times 100}}{{100 + \frac{1}{2} \times 200}} = 25\)

∵ M_R = 1.2 M_C

\(\Rightarrow {\left( {\frac{V}{{{A_S}}}} \right)_{Riser}} = 1.2{\left( {\frac{V}{{{A_S}}}} \right)_{Casting}}\)

\(\frac{d}{5} = 1.2;\left[ {25} \right] \Rightarrow d = 150;mm\)

d = h_R = 150 mm

Concept:

Maximum Possible reduction in Rolling (Δh) = μ²R

Where μ is coefficient of friction, R is Roll radius

Calculation:

(Δh) = μ²R

Δh = (0.08)² × 300 = 1.92 mm

Explanation:

Here point P is not at origin:

\(\left[ {\begin{array}{*{20}{c}} {x' - {x_p}}\ {y' - {y_p}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\cos \theta ;;;;;;\sin \theta }\ { - \sin \theta ;;;;;\cos \theta } \end{array};;;} \right]\left[ {\begin{array}{*{20}{c}} {x - {x_p}}\ {y - {y_p}} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {x' - 1}\ {y' - 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.8;;;;;;0.6}\ { - 0.6;;;;;;0.8} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {4 - 1}\ {5 - 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {0.8;;;;;0.6}\ { - 0.6;;;;;;0.8} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\ 2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3.6}\ { - 0.2} \end{array}} \right]\)

X’ = 3.6 + 1 = 4.6

Y’ = -0.2 + 3 = 2.8

Concept:

The economic order quantity is given by:

\(EOQ = \sqrt {\frac{{2 \times Demand \times Purchasing;cost}}{{Holding;cost}}} \)

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

C_o = ordering cost, C_h = holding constant, D = demand

The cycle time is related to EOQ as:

EOQ = D × T

Calculation:

D = 10,000, C_o = Rs. 2000, C_h = 14.4% of Rs 100

\({Q^*} = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}} = \sqrt {\frac{{2 \times 10,000 \times 2000}}{{\frac{{14.4}}{{100}} \times 100}}} = 1666.66\)

Time between two consecutive Order (T)

\(T = \frac{{{Q^*}}}{{Demand}}\)

\(T = \frac{{1666.66}}{{10000}}year\)   [1 year = 12 month]

\(T = \frac{{1666.66}}{{10000}} \times 12;month\)

T = 2 months

Concept:

Z_max = 15x₁ + 20x₂

Constraint:

12x₁ + 4x₂ ≥ 36       ---(1)

12x₁ – 6x₂ ≤ 24      ---(2)  

x₁, x₂ ≥ 0

From eq (1): 12x₁ + 4x₂ = 36

At x₁ = 0, x₂ = 9

x₂ = 0, x₁ = 3

(0, 9) and (3, 0)

From eq (2): 12x₁ – 6x₂ = 24

At x₁ = 0, x₂ = -4

x₂ = 0, x₁ = 2

(0, -4) and (2, 0)

Draw these points on graph.

∵ Feasible region is not bounded here so above linear programming problem has unbounded solution.