GATE ME 2021 Official Paper: Shift 2 (Previous Year Paper)

Five persons P, Q, R, S and T.

1) R is to be seated at the second position from the left end of the row.

2) P and T cannot be seated at either end of the row.

Two cases are possible for this condition.

3) P should not be seated adjacent to S.

There are two cases in case - 1.

Hence, total of 3 cases are possible.

The correct answer is '​(i) and (ii)'.

Key Points

• 'The no of' is singular and takes singular verb. Thus, 'the number of candidates is staggering' is correct.
• Similarly, 'A no of' is plural and takes plural verb. Thus, 'a number of candidates are staggering' is correct.
• Thus, both the statement i and ii is correct.

Concept:

The Watched will beep together at the time of their Least Common Multiple (LCM).

Calculation:

Given:

Watch X beeps at = 30 sec

Watch Y beeps at = 32 sec

They beep together at = 10.00 AM

LCM of (30 & 32) = 480 sec

And X and Y will beep together after 480 sec or 8 minutes.

So Next beep time = 10.08 AM

Explanation:

Given:

⊕ ÷ ⊙ = 2, Δ × ⊗ = 10, ⊙ + Δ = 5; Δ × ⊗ = 10

∵ \(\frac{⊕}{⊙} = 2\), \(\frac{⊕}{Δ} = 3\)

∴ \(\frac{⊙ }{Δ}=\frac{3}{2}\) 

\(⊙=\frac{3Δ}{2}\)

now, ⊙ + Δ = 5 

\(\frac{3Δ}{2} + Δ = 5 \)

5Δ = 10

∴ Δ = 2

∵ ⊙ + 2 = 5 

∴ ⊙ = 3

Now,  Δ × ⊗ = 10

2 × ⊗ = 10

∴ ⊗ = 5

Now, \(\frac{⊕}{⊙} = 2\)

\(\frac{⊕}{3} = 2\)

∴ ⊕ = 6

Hence (⊗ - ⊕)2 = (5 - 6)² = 1

Explanation:

Tracing path followed by Mr. X,

Hence Mr. X will be in North-West direction from the starting point.

Explanation:

The Venn diagram according to the given statements is as follows:

Conclusions:

All risk seekers are wealthy. → False (All wealthy are risk seekers but can't say vice versa is also true)

Only some entrepreneurs are risk seekers → False (All entrepreneurs are wealthy and All wealthy are risk seekers so all entrepreneurs will be risk seekers)

Hence neither conclusion I nor II is correct.

Concept:

\(P(E) = \frac{Number~of~favorable~outcomes}{Total~number~of~outcomes}\)

Calculation:

Given:

Number of blue balls = 15, Number of black balls = 45

Total number of balls = 15 + 45 = 60 balls

Balls are drawn from the box at random without replacement.

The probability of drawing first blue ball and second black ball is;

\(P(E)= \frac{{{15_{{C_1}}}}}{{{60_{{C_1}}}}}\times \frac{{{45_{{C_1}}}}}{{{59_{{C_1}}}}}\)

\(P(E) = \frac{15\times45}{60\times59}\)

\(P(E) = \frac{45}{236}\)

Explanation:

Let the radius of the circumscribed circle is R, and the radius of the inscribed circle is r.

Then area of circumscribed circle = πR²

Then the area of the inscribed circle = πr2

We know the angle between two adjacent sides in an equilateral triangle is 60°.

ΔOAB, \(Sin 30 =\frac{r}{R}\)

R = 2r

The ratio of the area of the inscribed circle to the area of the circumscribed circle 

\(=\frac{\pi r^2}{\pi R^2}\)

\(=\frac{\pi r^2}{\pi (2r)^2}\)

\(=\frac{1}{4}\)

Explanation:

let us consider a square sheet of side 1 unit.

Now fold the sheet along the main diagonal.

Now fold this along their line of symmetry.

again fold this along their line of symmetry.

The side of the resulting shape = \(\frac{1}{2}\) unit

 Area of resultant shape =\(\frac{1}{2}\times side^2\)

 Area of resultant shape = \(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\)

 Area of resultant shape = \(\frac{1}{8}\)

The correct answer is States an effect of the first sentence.

Key Points

• Statement one i.e. the world is going through pandemic is the main cause. Statement two i.e. crisis faced due to that pandemic tells us about the effect of pandemic on air travel industry.
• Cause and effect is a relationship between events or things, where one is the result of the other or others.
• Thus, the second sentence states an effect of the first sentence.

Concept:

If a matrix 'A', having a characteristic root 'λ', then we have a non-zero vector 'X' which satisfies the equation [A - λI][X] = 0 or AX = λX.

Where the non-zero vector 'X' is called characteristic vector or Eigen-vector.

Properties of Eigen Value:

if λ is the eigen value of A , then eigen values of A^m is λ^m

Calculation: 

Given:

Ap = α²p  ... eq (i)

We know that

AX = λX  .... eq (ii)

On comparing eq (i) and eq (ii)

λ = α² 

So the eigen value of matrix A is λ = α2 

Eigen value of A2 = λ² = α4.

Concept:

Laplace Transforms:

L (e^at) = \(\frac {1}{s \ - \ a}\)

L (e^-at) = \(\frac {1}{s\ +\ a}\)

Calculation:

Given:

L [f(t)] = \(\frac {s \ + \ 3}{{(s\ +\ 1)}{(s \ + \ 2)}}\)

The above equation through partial fractions can be written as:

\(\frac {s\ +\ 3}{{(s\ + \ 1)}{(s \ + \ 2)}}\) = \(\frac {A}{(s \ + \ 1)}\) + \(\frac {B}{(s \ + \ 2)}\)

s + 3 = A(s + 2) + B(s + 1)

s + 3 = (A + B)s + 2A + B

Comparing co-efficients on both sides, we get

A + B = 1 and 2A + B = 3

We get, A = 2, B = -1

So, \(\frac {s \ + \ 3}{{(s \ + \ 1)}{(s \ +\ 2)}}\) = \(\frac {2}{(s \ + \ 1)}\) - \(\frac {1}{(s \ + \ 2)}\)

f(t) = L^-1\(\left( {\frac{2}{{s \ + \ 1}}} \right)\) - L^-1\(\left( {\frac{1}{{s \ +\ 2}}} \right)\)

f(t) = 2e^-t - e^-2t

So, f(0) = 2e^-0 - e^-0 = 2 - 1 = 1

∴ f(0) = 1

Explanation:

The binomial distribution of a random variable X is given by,

P(X = k) = \(\rm ^nC_kp^kq^{n-k}\)

where n is the number of observations, p is the probability of success & q is the probability of failure.p nCkpkqn−knCkpkqn−k

Total probability, p + q = 1

q = 1 - p

Mean = np 

Variance = npq = np(1 - p)

Explanation:

White Cast iron: It is white because carbon is in the form of carbide known as cementite which is the hardest constituent of Iron.

• The carbon content in White Cast Iron is 1.75 to 2.3%.
• It has high tensile strength, low compressive strength, and high brittleness.
• It is used in abrasion-resistant parts where it's brittleness is of minimum concern such as shell liners, slurry pumps, ball mills, lifter bars, extrusion nozzles, cement mixers, pipe fittings, flanges, etc.

Additional Information

Spheroidal Cast Iron: It is also called ductile cast iron or high strength cast iron.

• This type of cast iron is obtained by adding small amounts of magnesium (0.1 to 0.8%) to the molten grey iron.
• The addition of magnesium causes the graphite to take the form of small nodules or spheroids instead of the normal angular flakes.

Grey Cast iron: 

• The carbon content in Grey Cast Iron is 2.5 to 3.5%.
• It is grey because carbon is present in the form of free Graphite.
• High compressive strength, low tensile strength, and easily machinable

Malleable cast iron: It is a heat-treated iron-carbon alloy

• The graphite structure is formed into irregularly shaped spheroidal particles
• Malleable cast iron is produced from white cast iron by annealing.
• During annealing treatment graphite nucleates and grows from the Fe₃C to form nodules.

Explanation:

Particle size usually is controlled by screening that is, by passing the metal powder through screens of various mesh sizes in addition to screening several other methods are available for particle size analysis, which are:

1. Laser scattering from a laser illuminates a sample consisting of particles suspended in a liquid medium. The particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.
2. Sedimentation, which involves measuring the rate at which particles settle in a fluid.
3. Microscopic analysis, which may include the use of transmission and scanning electron microscopy.

Explanation:

In contouring systems:

The machining path is usually constructed from a combination of linear and circular segments. It is only necessary to specify the coordinates of the initial & final points of each segment and the feed rate. The operation of producing the required shape based on this information is termed interpolation & the corresponding unit is the “Interpolator”.

• So, in CNC machine tool, the function of an interpolator is to generate a reference signal prescribing the shape of the part to be machined.
• The interpolator coordinates the motion along the machine axes, which are separately driven, to generate the required machining path.
• An interpolator coordinates these axis motions in such a way that the programmed path is constantly maintained from the beginning to the end of the movement.
• The two most common types of interpolators are linear and circular.

Explanation:

Laser beam machining:

• Laser machining is a technology that uses a laser beam (a narrow beam of intense monochromatic light) to cut required shapes or profiles or patterns in almost all types of materials.
• In this process, the output of a high-power laser beam is directed in a programmed manner towards the material required to be cut.
• A lens system focuses the emitted laser beam falls the work surface, removing a small portion of the material by vaporization and high-speed ablation.
• The process can be used to make precise holes in thin sheets and materials. Laser beam cutting finds its applications in a variety of fields.
• The fields where the laser beam has been successfully used are cloth and plastic cutting, laser marking, laser welding, laser drilling, cleaning and surface treatments.

Advantages:

• The ability to cut almost all materials.
• No limit to cutting paths as the laser point can move in any paths.
• No cutting lubricants are required.
• As there is an absence of direct contact between the tool and workpiece; thus, no forces are induced and as a result, it is not necessary to provide the work-holding system to hold the workpiece.
• The fragile materials are easy to cut on a laser without any support.
• There is no tooling cost or associated wear costs due to it.
• The laser produces high-quality cuts without extra finishing requirements.

Important Points

Ablation: 

A complex phenomenon occurring when an energy threshold is reached which causes measurable or observable damage and material removal by vaporization. 

 

Additional Information

Electrochemical Machining: 

In electrochemical machining, the metal is removed due to electrochemical action i.e. Ion displacement where the workpiece is made anode and the tool is made the cathode. A high current is passed between the tool and workpiece through the electrolyte. Metal is removed by the anodic dissolution and is carried away by the electrolyte.

The tool material used in ECM should have the following property

• It should have high electrical conductivity
• It should be easily machinable and it should have high stiffness
• Its corrosion resistance should be high.

The advantages of ECM include

• Complex shapes can be made accurately
• The surface finish is good due to atomic level dissolution
• Tool wear practically absent
• Its material removal rate is the highest.

 

Chemical Machining:

It is a machining process in which material is dissolved and removed from the workpiece by controlled chemical reaction using reactive chemical solution i.e. strong alkaline or acidic reagents.

Since the chemical solutions used can dissolve all of the workpiece material, the parts which are not to be dissolved to be applied with a mask that resists the chemical action of the solution, so that only the unmasked portion gets removed by the chemical solution.

The steps to be followed in a typical chemical machining operation are:

• Clean the workpiece thoroughly: This step is necessary to ensure that the masking material will adhere to the workpiece well.
• Apply a chemical resistance mask on the workpiece surface where no material is to be removed.
• Dip the workpiece surface into a chemical solution called etchant and leave it for sufficient time to get the necessary depth of etching.
• The etchant is either continuously sprayed on the surface or the part is immersed in a tank of constantly agitating etchant solution.
• Remove the mask and clean the workpiece.

Advantages of chemical machining:

• Tooling cost is very low.
• Complex contour can be easily machined.
• Hard and brittle material can be machined.
• A high surface finish is obtained i.e free from the burr.
• Difficult to machine materials can be easily processed.

Disadvantages of chemical machining:

• The process is slow; therefore the material removal rate is very low.
• The manufacturing cost is high.
• A large floor area is needed.
• Not possible to produce sharp corners.
• Workpiece thickness, which can be machined, is limited.

 

Abrasive jet machining (AJM):

• Abrasive jet machining (AJM) removes material through the action of a focused stream of abrasive-laden gas.
• Micro-abrasive particles are propelled by inert gas at velocities of up to 1000 ft/sec.
• In an abrasive jet, machining material removal takes place due to the impingement of the fine abrasive particles. These particles move with a high-speed air (or gas) stream.
• The abrasive particles are typically are of 0.025 mm or 25 microns in diameter.
• When an abrasive particle impinges on the work surface at a high velocity, the impact causes a tiny brittle fracture and the flowing air carries away the dislodgedged small workpiece particle.
• This process is more suitable when the work material is brittle and fragile.
• One of the most important factors in the AJM is the distance between the work surface and the tip of the nozzle normally called the nozzle tip distance (NTD).
• NTD affects not only the material removal rate (MRR) from the work surface but also the shape and size of the cavity produced.
• When the NTD increases, the velocity of the abrasive particle impinging on the work surface increases due to their acceleration after they leave the nozzle. This in turn increase MRR. With the further increase in NTD, the velocity reduces due to the drag of the atmosphere which initial checks the increase in MRR and finally decreases it.

Applications:

• For drilling holes of intricate shapes in hard and brittle materials
• For machining fragile, brittle and heat-sensitive materials

Limitations:

• MRR is rather low
• Abrasive particles tend to get embedded particularly if the work material is ductile
• Tapering occurs due to flaring of the jet

Concept:

In CPM:

The standard deviation of critical path:

σ_cp = \(\sqrt {Sum;of;variance;along;critical;path} \)

σ_cp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)

Where, σ₁, σ₂, ...., σ₈, σ₉ are the standard deviation of each activity on the critical path   

Calculation:

Given:

σ1, σ2, ...., σ8, σ9 = 3

σcp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)

σ_cp = \(\sqrt {3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2} \)

σ_cp = \(\sqrt {9 \times 9} \) = 9

∴ the standard deviation of the critical path is 9.

Explanation:

Allowance:

• Allowance is minimum clearance or maximum interference.
• It can be negative or positive.
• It is independent of tolerance.
• It is the difference between the maximum material limits i.e. it is the difference between the upper limit of the shaft and the lower limit of the hole.

Explanation:

In forced convective heat transfer, Stanton number (St), Nusselt number (Nu), Reynolds number (Re) and Prandtl number (Pr) are related as

\(St = \frac{{Nu}}{{R{e};Pr}};=\frac{{Nusselt~Number}}{{Reynold's~Number;\times;Prandtl~Number}} \)

Nusselt Number:

Nusselt number is defined as:

\(Nu = \frac{{h{L_c}}}{k}\)

where k is the thermal conductivity of the fluid and Lc is the characteristic length.

\({N_u} = \frac{{{Q_{conv}}}}{{{Q_{cond}}}} = \frac{{h{\rm{\Delta }}T}}{{k\frac{{{\rm{\Delta }}T}}{L}}} = \frac{{hL}}{k}\)

Therefore the Nusselt number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer.

Reynold Number:

Reynolds number is a dimensionless formula that is used to differentiate laminar flow from turbulent flow.

Reynolds number is given by

\(Re = \frac{{\rho ;\times; V; \times ;D}}{\mu }\)

Where,

ρ = Density of fluid, V = velocity of fluid, D = Diameter of pipe, μ = Dynamic viscosity of the fluid

Prandtl Number:

It is defined as the ratio of momentum diffusivity to thermal diffusivity.

\(Pr = \frac{{\mu; {C_p}}}{K} = \frac{{\left( {\frac{\mu }{\rho }} \right)}}{{\left( {\frac{K}{{\rho ;{C_p}}}} \right)}}\)

\(Pr = \frac{\nu }{\alpha } = \frac{{momentum;diffusivity}}{{thermal;diffusivity}}\)

\(\frac{δ }{{{δ _T}}} = {\left( {Pr} \right)^{1/3}};\)

where δ is hydrodynamic boundary layer thickness and δT is thermal boundary layer thickness.

If Prandtl number is greater than one, momentum diffusivity dominates and hydrodynamic boundary layer thickness is more than thermal boundary layer thickness.

Heat transfer coefficient (h) is not a property of the fluid, it is an experimentally determined coefficient value that is complex in nature.

Because h depends on various factors such as Properties of fluid, Type of flow, Type of surface, Geometry of surface, Orientation of surface, etc...

Concept:

The continuity equation for a two-dimensional flow of an incompressible fluid is

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)

Calculation:

Given:

\(\frac{\delta(u^2)}{\delta x} + \frac{\delta(uv)}{\delta y}\)

After differentiating we get,

\(2u\frac{\delta u}{\delta x} + u\frac{\delta v}{\delta y}+v\frac{\delta u}{\delta y}\)

\(u\frac{\delta u}{\delta x} +u\left[\frac{\delta u}{\delta x}+ \frac{\delta v}{\delta y}\right]+v\frac{\delta u}{\delta y}\)

According to continuity equation for 2D incompressible flow \(\frac{\delta u}{\delta x}+ \frac{\delta v}{\delta y}= 0\)

∴ \(u \frac{\delta u}{\delta x} + v \frac{ \delta u}{\delta y}\)

Explanation:

Reynolds Expression for Turbulent Shear stress:

Reynold's through his experiment developed an expression between two layers of a fluid at a small distance apart, which is given as:

\(τ_t=\rho{u'}{v'}\)

where u', v' = fluctuating component of velocity in the direction of x and y respectively.

u' and v' are both varying and hence τ will also vary. To calculate τ, time average is done on both sides of the equation.

\(τ_t=\bar{\rho}\bar{u}'\bar{v}'\)

Concept: 

A streamline is an imaginary curve drawn in space such that tangent drawn to it at any point will give the velocity of fluid-particle at a given instant of time. 

The equation of streamline for 2-D flow is given as:

\( \frac{{{\rm{dx}}}}{{\rm{u}}} = \frac{{{\rm{dy}}}}{{\rm{v}}}\)

where u and v are the x and y components of the velocity.

Calculation:

Given:

u = 2xyt, v = -y²t

\(\frac{{dx}}{{2xyt}} = \frac{{dy}}{{-y^2t}} \Rightarrow \smallint \frac{{dx}}{x} = - 2\smallint \frac{{dy}}{y}\)

In x = -2In y + ln C

ln x + ln y² = ln C

ln xy² = ln C

xy² = C

When equation for streamline passing through x = 1, y = 1 then C = 1

∴ xy2 = 1 is the equation of streamline passing through x = 1, y = 1.

Explanation:

Taking moment about point P

F ×  L - T _RS × L = 0

T RS = F

Force in member RS  = F (compressive)

Now FBD of point R

Taking summation of force in Y direction 

-T_PR sin45° + T_RS = 0

-TPR sin45° + F = 0

T_PR = F √ 2

Forces in the members PR  F √ 2 (tensile)

Explanation:

Kennedy’s theorem states that if three bodies have plane motion relative to one another, then there I-centres i.e. instantaneous centre must lie on a straight line.

There are 4 links, so the total number of instantaneous centers will be 

\(\frac{n(n - 1)}{2}\) = \(\frac{4 \times3}{2}\) = 6

I - centre:

I₁₃ is the point of intersection of the lines joining I₁₂, I₂₃ and I₁₄, I₃₄, These lines will meet at U, Hence U is an I centre.

Similarly, I₂₄ is the point of intersection of the lines joining I_12, I₂₄ and I₂₃, I₃₄, These lines will meet at R, Hence R is an I centre.

I12 = P, I34 = S, I23 = Q, I14 = T, I13 = U, I24 = R

So, All the points P, Q, R, S, T, and U are instantaneous centres of mechanism.

Explanation:

Controlling force curve: 

When the graph between the controlling force as ordinate and radius of rotation of the dead weight balls as abscissa is drawn, then the graph obtained is called the controlling force diagram.

Controlling force, Fc = mω2r.

The relation between the controlling force (FC) and the radius of rotation (r) for the spring controlled governors is given by the following equation.

FC = a.r – b 

when 

FC = ar – b  (governor is stable) i.e. B

FC = ar        (governor is Isochronous) i.e D

FC = ar + b  (governor is unstable) i.e F

Explanation:

The Strain Energy of Distortion or Distortion energy per unit volume is given by**:**

Distortion energy per unit volume = (Total Strain energy) - (Energy of dilation)

Total Strain Energy is given by:

\({\rm{U}} = \frac{1}{{2{\rm{E}}}}\left{ {{\rm{\sigma }}_1^2 + {\rm{;\sigma }}_2^2 + {\rm{\sigma }}_3^2 - 2{\rm{\mu }}\left( {{{\rm{\sigma }}_1}{{\rm{\sigma }}_2} + {{\rm{\sigma }}_2}{{\rm{\sigma }}_3} + {{\rm{\sigma }}_3}{{\rm{\sigma }}_1}} \right)} \right}\)

Energy of dilation:

\(U_v=\frac{(1-2\mu)}{6E}\times \frac {(\sigma_1+\sigma_2+\sigma_3)^2}{3}\)

Distortion energy per unit volume = (Total Strain energy) - (Energy of dilation)

Ud = \(\frac{1\ +\ \mu}{6E}[(σ_1-σ_2)^2+(σ_2-σ_3)^2+(σ_3-σ_1)^2]\)..... eq (i)

In simple tension test:

U_d = \(\left(\frac{1\ +\ \mu}{6E}\right)2σ_{ym}^2\) .... eq (ii)

Maximum distortion energy theory (Von mises theory)

• According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy per unit volume reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from a simple tension test.
• von misses stress under triaxial condition is given by:

\({σ _{vm}} = \sqrt{{\frac{1}{2}\left{ {{{\left( {{σ _1} - {σ _2}} \right)}^2} + {{\left( {{σ _2} - {σ _3}} \right)}^2} + {{\left( {{σ _3} - {σ _1}} \right)}^2} } \right}} }\)

Now if we compare von misses stress and distortion energy per unit volume equation then,

Ud =\(\frac{1\ +\ \mu}{3E};\times {σ_{vm}^2}\)

U_d ∝ σ²_vm

σ_vm ∝  \(\sqrt{U_d}\)

The von Mises stress at a point in a body subjected to forces is proportional to the square root of the distortional strain energy per unit volume.

Concept:

Simpson's one-third rule:

\(I = \frac{h}{3}\left[ {{y_0} + {y_n} + 4\left( {{y_1} + {y_3} + {y_5} \ldots } \right) + 2\left( {{y_2} + {y_4} \ldots } \right)} \right]\)

where n = number of intervals, interval size, \(h = \frac{b\ -\ a}{n}\)

Calculation:

Given:

f(x) = \(\int^{5.2}_4\) ln x dx, h = 0.3

b = 5.2, a = 4

\(n = \frac{b\ -\ a}{h} = \frac{5.2\ -\ 4}{0.3}\) = 4

\(I = \frac{h}{3}\left[ {{y_0} + {y_4} + 4\left( {{y_1} + {y_3} } \right) + 2\left( {{y_2} } \right)} \right]\)

\(I = \frac{0.3}{3}\left[ ({{\ln4} + {\ln5.2})+ 4\left( {{\ln4.3} + {\ln4.9} } \right) + 2\left( {{\ln 4.6} } \right)} \right]\)

∴ I = 1.8278 ≈ 1.83

Concept: 

Cartesian form:

Z = x + iy 

Polar form:

Z = r(cos θ + isin θ) 

Exponential form:

Z = re^iθ

Where, r = \(\sqrt {{x^2} + {y^2}} \) and θ = tan^-1(y/x).

Combining all of them

Z = x + iy = r(cos θ + isin θ) = reiθ

According to De-movier's theorem:

Zⁿ = (reiθ)ⁿ = rⁿe^niθ

Calculation:

Given:

Let Z = 1 + i

Comparing with cartesian form x = 1 and y = 1

θ = tan-1(y/x) = tan-1(1) = \(\frac{π}{4}\), 

∴ r = \(\sqrt {{x^2} + {y^2}}\) = \(\sqrt{{1^2} + {1^2}}\) = \(\sqrt{2}\)

Z = reiθ = \( {\sqrt 2 {e^{i\frac{π }{4}}}} \)

Zⁿ = (reiθ)n = rneniθ

(1 + i)⁸ = \({\left( {\sqrt 2 {e^{i\frac{π }{4}}}} \right)^8}\) = 16 × e^i2π = 16 (cos 2π + i sin 2π)

(1 + i)8 = 16 × 1 = 16.

Concept:

It is the ratio of inertia force to the elastic force of flowing fluid.

\(M = \sqrt \frac{Inertia force}{Elastic force} = \frac{V}{C}\)

where C is the velocity of sound in the fluid, V is the velocity of the body moving in the fluid.

For adiabatic flow:

\(C = \sqrt {\frac{\gamma P}{ρ }} =\sqrt {\frac{\gamma \times\rho\times R\times T}{ρ }} = \sqrt {\gamma RT}\)

The relation of stagnation temperature T_o and Mach number is:

\( \frac{{{T_0}}}{T} = 1 + \left( {\frac{{γ - 1}}{2}} \right){M^2}\)

Calculation:

Given:

V = 300 m/s, T = 330 K, γ = 1.4, R = 287 J/kg-K.

\(M = \frac{V}{\sqrt{\gamma RT}} = \frac{300}{\sqrt{1.4\times 287\times 330 }}\) = 0.823

We know that,

\( \frac{{{T_0}}}{T} = 1 + \left( {\frac{{γ - 1}}{2}} \right){M^2}\)

\(T_0 = T\left{1 + \left( {\frac{{1.4 - 1}}{2}} \right){0.823^2}\right}\)

T_o =  330 (1 + 0.1355) = 374.71 K

∴ Stagnation temperature T_o = 374.71 K

Concept:

The coefficient of performance is the ratio of heat extracted in the refrigerator to the work done on the refrigerant.

COP = \(\frac{Q_1}{W_R}\)

Q1 = Heat rejected from the cold reservoir

WR = Work required to pump the heat out

Calculation:

Given:

Refrigeration effect Q₁ = 150 kJ/kg, COP = 10

COP = \(\frac{Q_1}{W_R}\)

W_R = \(\frac{Q_1}{COP} = \frac{150}{10}\) = 15 kJ/kg

Heat rejected by the refrigerant in the condenser is

Q₂ = Q₁ + W_R

Q2 = 150 + 15 = 165 kJ/kg

Concept:

Let total mass = m, mass of liquid mf, mass of vapour m_g 

∴ m = m_f + m_g

Net volume = volume of liquid + volume of vapour

V =  vf × mf  +  vg × mg

Calculation:

Given:

mf  = 0.2m, mg = 0.8m,  vf = 0.001084 m3/kg, v_g = 0.46242 m3/kg, V = 50 m3

V =  vf × mf  +  v_g × m_g

50 =  0.001084 × 0.2m + 0.46242 × 0.8m

m = 135.08 kg

∴ mass of liquid vapour mixture in the tank is 135.08 kg.

Concept:

Mach number

Mach number is defined as the ratio of inertia force to elastic force.

\(M = \sqrt {\frac{{Inertia;force}}{{Elastic;force}}} = \sqrt {\frac{{\rho A{V^2}}}{{KA}}} \)

\(\therefore M=\sqrt {\frac{{{V^2}}}{{\frac{K}{\rho }}}} = \frac{V}{{\sqrt{\frac{K}{\rho }} }} = \frac{V}{C}\)

where V = velocity of a body moving in fluid and C = velocity of sound in the fluid.

\(C=\sqrt{\gamma RT}\)

where γ = ratio of specific heats, R =  gas constant, T = temperature in Kelvin.

Calculation:

Given:

M = 0.6, T = 350 K, R = 320 J/kgK, γ = 1.3

\(C=\sqrt{\gamma RT}\)

\(C=\sqrt{1.3 \times 320 \times 350}=381.575;m/s\)

\(M = \frac{V}{C};;;;\)

\(0.6 = \frac{V}{381.575};;;;\)

V = 228.94 ≈ 229 m/s.

Additional Information

For the compressible fluid flow, the Mach number is an important dimensionless parameter.

On the basis of the Mach number, the flow is defined.

Concept:

Euler's buckling load \(P_e = \frac{{{\pi ^2}EI_{min}}}{{{L_e^2}}}\)

where L_e = effective length of a column

1. If one end of the column is fixed and the other end free L_e = 2L
2. If one end fixed and the other end is pinned or hinged L_e = \(\frac{L}{\sqrt2}\)
3. If both ends of the column are fixed Le = \(\frac{L}{2}\)
4. If both ends of a column are hinged Le = L

Calculation:

Given:

For the column whose one end fixed and another end free:

Buckling load P₁ = 100 N, L₁ = 2L

If the free end is replaced by pinned joint L₂ = \(\frac{L}{\sqrt2}\)

Since the column is the same \(P_e \propto \frac{1}{L_e^2}\)

\(\frac{P_2}{P_1} = \frac{L_1^2}{L_2^2}\)

P₂ = \(100 \times\frac{4L^2}{\frac{L^2}{2}}\)  = 800 N

Concept:

P = Hydrostatic pressure, a = side of the cubical block 

Under the hydrostatic condition, hydrostatic pressure will be the same in all direction i.e.,

P_x = P_y = P_z = -P 

Negative because Hydrostatic pressure is compressive in nature

Also, the strain will be the same for all side i.e.,

ϵ_x = ϵ_y = ϵ_z = ϵ

\(\epsilon=\frac{σ_x}{E}-\mu\frac{σ_y}{E}-\mu\frac{σ_z}{E}\)

Calculation:

Given:

P = 250 N/mm², a = 200 mm, Elastic modulus (E) = 2 × 10⁵ N/mm², Poisson's ratio (μ) = 0.3

σ_x = σ_y = σ_z = -P

\(\epsilon=\frac{σ_x}{E}-\mu\frac{σ_y}{E}-\mu\frac{σ_z}{E}\)

\(\epsilon=\frac{{\delta a}}{a} = \frac{1}{E}\left[ {{P_x} - μ \left( {{P_y} + {P_z}} \right)} \right]\)

\(\frac{{\delta a}}{a} = \left( {\frac{{ - P }}{E}} \right)\left( {1 - 2μ } \right)\)

\(\delta a = - \frac{{ {250}\times200 \times\left{ {1 - (2\times0.3)} \right}}}{{2 \times {{10}^5}}}\) = -0.10 mm

The reduction in the side of a cube is 0.10 mm.

Explanation:

\(I=\int^{π/2}_0\int^{\cosθ}_0r\sin θ ;drdθ\)

The order of 'drdθ' decides the priority of integration.

Here the limit of r varies from 0 to cos θ and, the limit of θ varies from 0 to π/2.

\(I=\int^{π/2}_0\int^{\cosθ}_0(r;dr)\sin θ ;dθ\)

\(I=\int^{π/2}_0\left[\frac{r^2}{2}\right]_0^{cos\theta}\sin θ ;dθ\)

\(I=\int^{π/2}_0\frac{1}{2}\cos^2 \theta .\sin θ ;dθ\)

\(I=\frac{1}{2}\int^{π/2}_0\cos^2 \theta .\sin θ ;dθ\) ....eq (1)

We know that;

\(I=\int^{π/2}_0\sin^m θ \cos^n \theta ;dθ=\frac{\Gamma\left(\frac{m\ +\ 1}{2}\right)\Gamma\left(\frac{n\ +\ 1}{2}\right)}{2\Gamma\left(\frac{m\ +\ n\ +\ 2}{2}\right)}\)

On comparing with eq (1)... m = 1 and n = 2.

\(I=\frac{1}{2}\int^{π/2}_0\cos^2 \theta .\sin θ ;dθ=\frac{\Gamma\left(\frac{1\ +\ 1}{2}\right)\Gamma\left(\frac{2\ +\ 1}{2}\right)}{2\Gamma\left(\frac{1\ +\ 2\ +\ 2}{2}\right)}=\frac{\Gamma\left(1\right)\Gamma\left(\frac{3}{2}\right)}{2\Gamma\left(\frac{5}{2}\right)}\)

\(\frac{\Gamma\left(1\right)\Gamma\left(\frac{3}{2}\right)}{2\Gamma\left(\frac{5}{2}\right)}=\frac{\Gamma\left(1\right)\ \times\ \frac{1}{2}\Gamma\left(\frac{1}{2}\right)}{2\ \times\ \frac{3}{2}\ \times\ \frac{1}{2}\Gamma\left(\frac{1}{2}\right)}=\frac{1}{3}\)

\(I=\frac{1}{2}\int^{π/2}_0\cos^2 \theta .\sin θ ;dθ=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\)

Concept:

Stationary point is the point at which the derivative of the function f(x) is zero.

Calculation:

Given:

x and r are matrices of order n × 1

Q is a symmetric matrix of order n × n

Let us take n = 2 

x = \(\begin{vmatrix} x_1\ x_2 \end{vmatrix}_{2\ \times\ 1}\)r = \(\begin{vmatrix} r_1\ r_2 \end{vmatrix}_{2\ \times\ 1}\)Q = \(\begin{vmatrix} a &b \ b &c \end{vmatrix}_{2\ \times\ 2}\)

\(x^T = \begin{vmatrix} x_1 & x_2 \end{vmatrix}\) , \(r^T = \begin{vmatrix} r_1 & r_2 \end{vmatrix}\)

\(f(x) = \frac{1}{2} x^T Qx-r^Tx\)

f(x) = \(\frac{1}{2} \begin{vmatrix} x_1 & x_2 \end{vmatrix} \begin{vmatrix} a &b \ b &c \end{vmatrix} \begin{vmatrix} x_1\ x_2 \end{vmatrix}\) - \(\begin{vmatrix} r_1 & r_2 \end{vmatrix} \begin{vmatrix} x_1\ x_2 \end{vmatrix}\)

f(x) =\(\frac{1}{2}\)([ax_1^2 + cx_2^2 + 2bx_1x_2]\) - [r₁x₁ + r₂x₂]

For stationary point derivative of f(x) is equal to 0 such that:

\(\frac{\partial f}{\partial x_1} = 0\), \(\frac{\partial f}{\partial x_2} = 0\)

\(\frac{\partial f}{\partial x_1} =\) ax₁ + bx₂ - r₁ = 0

\(\frac{\partial f}{\partial x_2} =\) cx₂ + bx₁ - r₂ = 0

writing above equation in matrix form we get,

⇒ \(\begin{vmatrix} a &b \ b &c \end{vmatrix} \begin{vmatrix} x_1\ x_2 \end{vmatrix} = \begin{vmatrix} r_1\ r_2 \end{vmatrix}\)

⇒ Qx = r

⇒ x = Q^-1r

Concept:

This is the variable separable form of differential equation.

Calculation:

Given:

(1 + y)\(\frac{dy}{dx}\) = y

\(\frac{(1 + y)}{y}\)dy = dx

\(\frac{1}{y}\)dy + 1dy = dx

Integrating both side,

\(\smallint \frac{1}{y}dy + ;\smallint 1dy = ;\smallint dx + C\)

lny + y = x + C

at x = 1, y = 1

ln1 + 1 = 1 + C

C = 0

So, lny = (x - y)

y = \({e^{\left( {x \ - \ y} \right)}}\) = \(\frac {e^x}{e^y}\)

yey = ex

Explanation:

Product (m) → i ... m

Workstation (n) → j ... n

a_ij = time, r_i = selling price, h_i = holding cost

T → t = 1, 2, ... T

d_it = demand of the product in time t, c_jt = capacity of workstation in time t, S_it = Number of products sold in time t, x_it = Number of product produced in time t,

I_it = Number of product i hold in inventory at end of period t

Capacity constraint

a_ijx_it ≤ c_jt

Inventory constraint

I_it = I_{i, t - 1} + x_it - S_it

Concept:

Let ‘P’ be the atmospheric pressure in (kPa)

‘ϕ’ be the relative humidity, ‘Pv’ be the partial pressure of the air in (kPa), ‘Pvs’ be the saturation pressure of water in (kPa), ‘ω’ be the specific humidity of air in (kg/kg of dry air).

\(ϕ = \frac{{{P_v}}}{{{P_{vs}}}} \)

\(ω = 0.622;\times\frac{{{P_v}}}{{\left( {P\ - \ {P_v}} \right)}} \)

Given:

Calculation:

P = 101 kPa, ϕ = 0.6, P_vs = 3.6 kPa

\(ϕ = \frac{{{P_{v}}}}{{{P_{vs}}}} \Rightarrow 0.6 = \frac{{P_v}}{{{3.6}}}\)

P_v = 2.16 kPa

\(ω = 0.622 \times\frac{{{P_v}}}{{\left( {P\ - \ {P_v}} \right)}} \)

\(\omega = 0.622\times\frac{{2.16}}{{\left( {101 \ - \ 2.16} \right)}} \)

∴ ω =  0.0135928 kg of water vapor/kg of dry air.

∴ ω = 13.59 gm of water vapor/kg of dry air.

Concept:

To find slope and deflection of unsymmetric load Castigliano's theorem is used and the two theorems of Castigliano's for structural analysis are as follows:

Castigliano’s 1^st theorem: 

The partial derivative of total strain energy of the system with respect to any particular deflection at a point is equal to the force applied at that point in the same direction as that of the deflection.

\(\frac{{\partial {\rm{U}}}}{{\partial {\rm{\delta }}}} = {\rm{w}}\)

Castigliano’s 2^nd theorem: 

The partial derivative of strain energy of the system with respect to load at any point is equal to deflection at that point.

\(\frac{{\partial {\rm{U}}}}{{\partial {\rm{w}}}} = {\rm{\delta }}\)

Also, the partial derivative of strain energy of the system with respect to the couple at any point is equal to slope at that point.

\(\frac{{\partial {\rm{U}}}}{{\partial {\rm{M}}}} = {\rm{\theta }}\)

Where U is the strain energy of the system.

Calculation:

Given:

First, we find strain energy of the frame,

U = U_PQ + U_QR

\({U_{PQ}} = \frac{{{M^2}{L}}}{{2EI}}= \frac{{{(FL)^2}{L}}}{{2EI}}= \frac{{{F^2L^3}}}{{2EI}}\)

To find strain energy of member QR, consider a small element in member QR and Energy stored in the small element:

\(dU = \frac{{M_x^2}}{{2EI}}dx\)

For member QR: Mx = Fx

\({U_{QR}} = \frac{1}{{2EI}}\mathop \smallint \limits_0^{{L}} {\left( { {M_x}} \right)^2}dx = \frac{1}{{2EI}}\mathop \smallint \limits_0^{L} {(Fx)^2}dx\)

\({U_{QR}} = \frac{{{F^2}{L^3}}}{{6EI}}\)

\(\begin{array}{l} U = {U_{PQ}} + {U_{QR}} = \frac{{{F^2L^3}}}{{2EI}} + \frac{{{F^2L^3}}}{{6EI}} = \frac{{{2F^2}{L^3}}}{{3EI}} \end{array}\)

∴ By Castigliano’s theorem horizontal deflection at free end R is

\( {(\delta _H)_R} = \frac{{\partial U}}{{\partial F}} = \frac{{4F{L^3}}}{{3EI}}\)

Concept:

Velocity Ratio of a belt drive:

The velocity ratio is the ratio of the speed of the driven pulley to that of the driving pulley. The speed is inversely related to the diameter of the pulley.

Velocity ratio = \(\frac {N_2}{N_1}\) = \(\frac {D_1}{D_2}\)

where, N1 = Speed of driver in rpm, N2 = Speed of follower in rpm, D1 = Diameter of the driver, D2 = Diameter of the follower.

Velocity ratio of gear:

In gear drive, the speed is inversely related to the diameter as well as teeth in gear.

Velocity ratio = \(\frac {N_2}{N_1}\) = \(\frac {T_1}{T_2}\)

where, N1 = Speed of the driver, N2 = Speed of the driven, T1 = Number of teeth on the driver, T2 = Number of teeth on the driven.

Note:

• Two externally mating gear rotates in opposite direction.
• In open belt drive, the direction of the driven pulley is the same, whereas in crossed belt drive the direction of the driven pulley is opposite.

Calculation:

Given:

D₁ = 150 mm, D₂ = 250 mm, N₁ = 2500 rpm.

T₂ = 18, T₃ = 44, T₄ = 15, T₅ = 33, T₆ = 36, T₇ = 16

Let, the Clockwise (CW) direction of motion is taken as positive and Counter Clockwise (CCW) as negative.

Between pulley (1) and (2). 

\(\frac {N_2}{N_1}\) =  \(\frac {D_1}{D_2}\)

\(\frac {N_2}{2500}\) = \(\frac {150}{250}\)

N2 = 1500 rpm (CCW)  .... [∵ N₁ = CCW]

Between gear 2 and gear 3:

Pulley 2 and gear 2 mounted on the same shaft, ∴ gear 2 has N₂ = 1500 rpm CCW.

\(\frac{{{N_3} ;= ;{N_4}}}{{{N_2}}}\) = \(\frac {T_2}{T_3}\)    [∵ N₃ = N₄ compound gear]

N4 = \(\frac {18}{44}\) × 1500 = 613.64 rpm (CW)

Between gear 4 and gear 5:

\(\frac {N_5}{N_4}\) = \(\frac {T_4}{T_5}\)

N5 = \(\frac {15}{33}\) × 613.64 = 278.93 rpm (CCW)

Between gear 5 and gear 6:

\(\frac{{{N_6}; = ;{N_7}}}{{{N_5}}}\) = \(\frac {T_5}{T_6}\)

N7 = \(\frac {33}{36}\) × 278.93 = 255.68 rpm (CW)

∴ N7 = 255.68 rpm (CW)

Concept:

Transmissibility is defined as the ratio of force transmitted to the foundation (FT) to the disturbing force (F).

\(T=\frac{{{F}_{T}}}{{{F}_{un}}}=\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}~}}\)

\({{\omega }_{n}}=\sqrt{\frac{{{K}_{eq}}}{m}}\)

• When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ωn = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ωn < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ωn > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > √2. Here the force transmitted to the foundation increases as the damping is increased.

Calculation:

Given:

Forced frequency ω = 40 rad/s, m = 100 kg

We know force transmitted is directly proportional to the transmissibility ratio, more is the transmissibility ratio more is force transmitted.

The transmissibility ratio is proportional to the frequency ratio so we find the frequency ratio of each isolator.

1. For k = 640 kN/m, ζ = 0.70

\( {\omega _n} = \sqrt {\frac{k}{m}} =\sqrt {\frac{640\times 10^3}{100}}= 80; rad/sec\)

\(\frac{ \omega}{\omega _n} = \frac{40}{80} = 0.5\)

2. For k = 640 kN/m, ζ = 0.07

\( {\omega _n} = \sqrt {\frac{k}{m}} =\sqrt {\frac{640\times 10^3}{100}}= 80; rad/sec\)

\(\frac{ \omega}{\omega _n} = \frac{40}{80} = 0.5\)

3. For k = 22.5 kN/m, ζ = 0.70

\( {\omega _n} = \sqrt {\frac{k}{m}} =\sqrt {\frac{22.5\times 10^3}{100}}= 15; rad/sec\)

\(\frac{ \omega}{\omega _n} = \frac{40}{15} = 2.66\)

4) For k = 22.5 kN/m, ζ = 0.070

\( {\omega _n} = \sqrt {\frac{k}{m}} =\sqrt {\frac{22.5\times 10^3}{100}}= 15; rad/sec\)

\(\frac{ \omega}{\omega _n} = \frac{40}{15} = 2.66\)

For the different values of ζ  and frequency ratios, the force transmitted graph has been plotted.

The isolators in the ascending order of the force transmitted to the foundation are 4-3-1-2.

Explanation:

Apply energy method:

Let system is displaced by a small angle 'θ' 

Since the system is in static equilibrium position

So net energy E, and \( \frac{dE}{dt} =0\)

Energy of spring = \(\Rightarrow U = \frac{1}{2}kx^2 \)

Energy of pulley during rotation \(\Rightarrow U = \frac{1}{2}Jω^2 \)

Energy of bucket mass m \(\Rightarrow U = \frac{1}{2}mv^2 \)

Net energy  

\(E= \frac{1}{2}kx^2 +\frac{1}{2}Jω^2 +\frac{1}{2}mv^2\)

x = rθ , w = \(\dot{\theta}\), v = rω   

\(E= \frac{1}{2}kr^2θ^2 +\frac{1}{2}J \dot{θ} ^2 +\frac{1}{2}mr^2w^2\)

\(E= \frac{1}{2}kr^2θ^2 +\frac{1}{2}J \dot{θ} ^2 +\frac{1}{2}mr^2\dot{θ}^2\)

\(\frac{dE}{dt}= \frac{1}{2}kr^2(2θ)\dot{\theta} +\frac{1}{2}J (2\dot{θ})\ddot\theta +\frac{1}{2}mr^2(2\dot{θ})\ddot\theta = 0\)

\(\frac{dE}{dt}= kr^2θ \dot{\theta} +J \dot{\theta}\ddot\theta +mr^2\dot{\theta}\ddot\theta =0\)

\(\frac{dE}{dt}= kr^2θ +J \ddot\theta +mr^2\ddot\theta =0\)

\(\frac{dE}{dt}= kr^2θ +(J +mr^2)\ddot\theta =0\)

\( kr^2θ +(J +mr^2)\ddot\theta =0\)

\(\ddot\theta+ \frac{kr^2θ} {(J +mr^2)} =0\)

Comparing \(\ddot\theta+ {\omega^2_n\theta} {} =0\)

\(\omega^2_n = \frac{kr^2} {(J +mr^2)} \)

\(\omega_n=\sqrt{\frac{kr^2}{J+ mr^2}}\)

Concept:

According to the Newton-Raphson method:

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

where x_n is starting guess.

Calculation:

Given:

f(x) = x3 - x - 3 

\(f'\left( x \right) = 3{x^2} -1\)

Starting guess (x0 = 2)

Now first iterations

\(x_1=x_0-\frac{(x_0^3;-;x_0;-;3)}{3x_0^2;-;1}\)

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}} = 2 - \frac{{{{\left( { 2} \right)}^3} -2-3 }}{{3{{\left( { 2} \right)}^2} - 1}} = 1.7273\)

Second iterations

\(x_2=x_1-\frac{(x_1^3;-;x_1;-;3)}{3x_1^2;-;1}\)

x₁ = 1.7273

\({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}} = 1.7273 - \frac{{{{\left( { 1.7273} \right)}^3} -1.7273-3 }}{{3{{\left( { 1.7273} \right)}^2} - 1}} = 1.67369\)

x₂ = 1.67

Concept:

Economic Order Quantity (EOQ):

There are mainly two costs associated with inventory. Ordering cost and inventory carrying cost. Above two cost is opposite in nature so order quantity must be balanced quantity, to obtain a minimum total cost. This quantity is known as Economic Order Quantity.

• At EOQ, the total cost is minimum.
• At EOQ, Ordering cost = Holding cost

The expression for EOQ in the production build-up model is given by:

\(Q = \sqrt {\frac{{2D{C_o}}}{{{C_H}}};\left( {\frac{P}{{P - D}}} \right)} \)

Where, D = Annual or Yearly demand of Inventory (units/yr), C_o = Setup cost (Rs/setup), C_H = Cost of holding one unit in inventory for one complete year (Rs/unit/yr), d = Demand or Consumption rate (unit/day), P = Production or build up rate (unit/day)

Calculation:

Given:

p = 30000 unit/day, d = 15000 unit/day, C_H = 20 Rs/unit/yr, C_o = 1800 Rs/setup.

Number of working days in a year = 300

∴ D = d × (Number of working days in a year) = 300 × 15000 units/yr

P = p × (Number of working days in a year) = 300 × 30000 units/yr

So, Economic batch quantity,

\(Q = \sqrt {\frac{{2D{C_o}}}{{{C_H}}};\left( {\frac{P}{{P - D}}} \right)} \)

\(Q = \sqrt {\frac{{2 × 15000 × 300 × 1800}}{{20}} × \left( {\frac{{30000 \times300}}{{(30000\times 300) - (15000\times 300)}}} \right)} \)

Q = 40249.22 ≈ 40250 units.

Concept:

Chvorinov’s law for solidification time.

\(Time = k{\left( {\frac{{Volume}}{{Area}}} \right)^2}\)

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

For same material 'k' is same.

∴ Time ∝ (Modulus)2

Modulus for cylinder shape \(= \frac{{\frac{{\pi {D^2}}}{4} \times D}}{{\left( {2 \times \frac{{\pi {D^2}}}{4}} \right) ;+ ;(\pi D \times D)}} = \frac{D}{6}\)  (∵ H = D)

Calculation:

Given:

Steel rectangular plate with dimensions = 75 mm × 125 mm × 20 mm, t = 2 minutes

Modulus of steel plate \(= \frac{{75; \times ;125 ;\times ;20}}{{2\left[ {(75 ;\times ;125) ;+ ;(20 ;\times ;125); + ;(75 ;\times ;20)} \right]}} = 7;mm\) 

Cylinder H = 50 mm, D = 50 mm.

Modulus for cylinder shape \(= \frac{D}{6}=\frac{50}{6}=8.33;mm\) 

Time ∝ (Modulus)2

\(\Rightarrow \frac{{Time;of;cylinder}}{{Time;of;plate}} = {\left( {\frac{{Module;of;cylinder}}{{Module;of;plate}}} \right)^2}\)

\(\Rightarrow \frac{t}{2} = {\left( {\frac{{8.33}}{{{{7}}{{}}}}} \right)^2}\)

∴ t = 2.83 minutes

Concept:

Melting efficiency is given by:

\({η _m} = ;\frac{{{H_m}}}{{{H_i ×η_h}}}\)

where, Hm = Heat required to melt the metal, Hi = Heat input required, η_m = melting efficiency, η_h = heat conversion efficiency.

Calculation: 

Given:

Heat required to melt (Hm) = 20 J/mm3

Heat transfer efficiency, (ηth) = 10 % = 0.1

Diameter d = 5 mm and thickness h = 1 mm, time =  0.1 s, ηm = 100 % = 1, η_h = 10 % = 0.1

Volume of nugget = \(V = \frac{{\pi}}{{{{4}}{{}}}} ×\ d^2 × h\)

\(V = \frac{{\pi}}{{{{ 4;}}{{}}}} ×\ 5^2 × \ (1)\)

V = 19.63 mm3

Hm = 20 J/mm3

H_m = 20 × 19.63 = 392.7 J.

Power required to melt \(= \frac{{392.7}}{Time}= \frac{{392.7}}{{{{ 0.1;}}{{}}}} =3927;W\)

\({η _m} = ;\frac{{{H_m}}}{{{H_i ×η_h}}}\)

\({1} = ;\frac{{{3927}}}{{{H_i×0.1}}}\)

Hi = 39270 Watt = 39.27 kW

Concept:

For grinding operation of metal

\(Specific; Energy(e) =\frac{Power}{MRR}\)

\({\rm{Power;}}\left( {\rm{P}} \right) = {F_c} × V = {F_c} × \frac{{\pi DN}}{{60}}\)

Material removal rate (MRR) = Table speed × width × depth of cut

Where, Fc = Tangential force, V = Grinding speed, N = Speed of a wheel in rpm, e = Specific energy consumption, MRR = Material removal rate, D = Grinding wheel diameter

Calculation:

Given:

Length = 300 mm, width = 10 mm, height = 50 mm, D = 150 mm, V = 40 m/sec, Table speed = 5 m/min = \(\frac{5000}{60};mm/sec \),

Depth of cut = 50 μm = 50 × 10^-3 mm = 0.050 mm, F_c = 40 N

Power = F_c × V = 40 × 40 = 1600 W

Material removal rate (MRR) = Table speed × width × depth of cut = \(\frac{5000}{60}\) × 10 × 0.05 = 41.666 mm³/sec

\(Specific; Energy(e) =\frac{Power}{MRR} =\frac{1600}{41.666} = 38.4; J/mm^3\)

The value of the specific grinding energy under the aforesaid grinding conditions is 38.4 J/mm³.

Concept:

Shear angle (ϕ) = tan^-1\(\left( {\frac{{r\cosα }}{{1 - r\sinα }}} \right)\)

where r = cutting ratio and α = orthogonal rake angle.

Cutting ratio (r) = \(\frac {t}{t_c}=\frac {V_f}{V_c}\)

where t = uncut chip thickness, t_c = chip thickness, V_f = chip velocity, V_c = cutting velocity. 

Shear stress (τ_s) = \(\frac{F_s}{A_s}=\frac{{{F_s}}}{{\frac{{bt}}{{\sinϕ }}}}\)

where, \(\frac {bt}{sinϕ}\) = shear area, F_s = shear force.

Pure orthogonal cutting:

feed (mm/rev) = t (uncut chipthickness in mm).

depth of cut (d) = width (b)

∴ bt = fd

∴ \({\tau _s} = \frac{{{F_s}}}{{\frac{{bt}}{{\sinϕ }}}}=\frac{{{F_s} \times \sinϕ }}{{f d}}\)

Calculation:

Given:

Rake angle (α) = 0°, Shear force (F_s) = 400 N, Cutting velocity (V_c) = 100 m/min, Chip velocity (V_f) = 20 m/min

r = \(\frac {t}{t_c}\) = \(\frac {V_f}{V_c}\) = \(\frac {20}{100}\) = 0.2

ϕ = tan-1\(\left( {\frac{{r\cosα }}{{1 - r\sinα }}} \right)\) = tan^-1 (r)   [∵ α = 0°]

ϕ = tan^-1 (0.2) = 11.3099°

ϕ = 11.3099°

\({\tau _s} = \frac{{{F_s} \times \sinϕ }}{{f d}}\) = \( \frac{{400 \times \sin \left( {11.309} \right)}}{{2 \times 0.1}}\) = 392.23 MPa

∴ the shear strength, τs of the material will be 392.23 MPa.

Concept:

Rolling Operation:

The assumptions used while deriving equations of rolling is

Initial volume = Final volume

L_i × W_i × t_i = L_f × W_f × t_f

where, L_i, W_i, t_i are the initial length, width and thickness, and L_f, W_f, t_f are the final length, width and thickness

Calculation:

Given:

L_i = 3600 mm, W_i = 250 mm, t_i = 50 mm.

Thickness reduces by 10 % and Width increases by 3 %

∴ t_f = 0.90 × 50 mm, W_f = 1.03 × 250 mm

Now, Initial volume = Final volume

Li × Wi × ti = Lf × Wf × tf

3600 × 250 × 50 = L_f × 1.03 × 250 × 0.90 × 50

∴ L_f = 3883.49 mm ≈ 3883.5 mm.

Concept:

∠OAB, sin θ = \(\frac {h_1 - h_2}{l}\)    [∵ The diameter of both rollers are the same.]

Calculation:

Given:

Let take h₂ = 76.2 mm, l = 254 mm, θ = 30° 

sin θ = \(\frac {h_1 - h_2}{l}\)

sin 30° = \(\frac {h_1 - 76.2}{254}\)

h₁ = 203.2 mm.

Let's take h₁ = 76.2 mm

sin θ = \(\frac {h_1 - h_2}{l}\)

sin 30° = \(\frac {76.2 - h_2}{254}\)

∴ h₂ = -50.8 mm (which is not possible).

Concept:

Mean Absolute Percentage Error (MAPE):

Mean absolute percentage error (MAPE)

\(MAPE= \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {\frac{{D_i - F_i}}{{D_i}}} \right|}}{n} \times 100 = \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {\frac{{E_i}}{{D_i}}} \right|}}{n} \times100 \)

Calculation:

Given:

 

Mean Absolute Percentage Error (MAPE):

\(MAPE = \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {\frac{{D_i - F_i}}{{D_i}}} \right|}}{n} \times 100 \)

\( \therefore \frac{{\left| {\frac{{ 25}}{{225}}} \right| + \left| {\frac{{ - 20}}{{220}}} \right| + \left| {\frac{{ - 15}}{{285}}} \right| + \left| {\frac{{ 20}}{{290}}} \right| + \left| {\frac{{20}}{{250}}} \right|}}{5} \times 100 =8.072; % \)

Concept:

Effectiveness for parallel flow is given by:

\(\epsilon_p = \frac{{1 - {e^{ - NTU\left( {1 + R} \right)}}}}{{1 + R}}\)

Effectiveness for counter flow is given by:

\(\epsilon_p = \frac{{1 - {e^{ - NTU\left( {1 - R} \right)}}}}{{1 - R{e^{ - NTU\left( {1 - R} \right)}}}}\)

where \(R = \frac{{{C_{min}}}}{{{C_{max}}}} \)

When there is a phase change (condenser or evaporator) C_max = infinity

So R = 0

Then the effectiveness is given by:

\(\epsilon = {{1 - {e^{ - NTU }}}}{{}}\)

Effectiveness is the ratio of actual heat transfer rate to maximum possible heat transfer rate.

\(\varepsilon = \frac{{{{\dot q}_{act}}}}{{{{\dot q}_{max}}}}\)

\(\varepsilon = \frac{{{C_c}\left( {{T_{co}} ;- ;{T_{ci}}} \right)}}{{{C_{min}}{\rm{\Delta }}{T_{max}}}}\)

\(\varepsilon = \frac{{{T_{hi}} - {T_{ho}}}}{{{\rm{\Delta }}{T_{max}}}} = \frac{{{T_{co}} - {T_{ci}}}}{{{\rm{\Delta }}{T_{max}}}}\)

ΔTmax = T_s - T_ci

Calculation:

Given:

T_s = 350K, T_ci = 300K, U = 1500 W/m2K, C = 4000 J/kgK, ṁ = 100 kg/s, A = 400 m2

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

C_min = 100 × 4000 W/K.

\(NTU = \frac{{1500× 400}}{{{100× 4000_{}}}} \)

NTU = 1.5

\(\epsilon = {{1 - {e^{ - NTU }}}}{{}}\)

\(\epsilon = {{1 - {e^{ - 1.5 }}}}{{}}\)

\(\epsilon = {{ 0.777 }}{{}}\)

\(\varepsilon = \frac{{{T_{hi}} - {T_{ho}}}}{{{\rm{\Delta }}{T_{max}}}} = \frac{{{T_{co}} - {T_{ci}}}}{{{\rm{\Delta }}{T_{max}}}}\)

ΔTmax = T_s - T_ci

ΔTmax = 350 - 300 = 50 K

\(\varepsilon = \frac{{{T_{co}} - {T_{ci}}}}{{{\rm{\Delta }}{T_{max}}}}\)

\(\varepsilon =0.777 = \frac{{{T_{co}} - {300_{}}}}{{{\rm{}}{50_{}}}}\)

T_co = 338.85 K ≈ 339 K.

Concept:

Thermal Boundary layer (TBL):

The thermal boundary layer is a thin region inside which temperature gradients are present in the normal direction to the plate. These temperature gradients are present due to heat transfer between the hot plate and free stream fluid.

In the layer just touching the slab, the heat flux leaving by the plate is equal to the heat flux gained by air.

\(\left.\begin{matrix} -k_{slab}\frac{dT}{dy} \end{matrix}\right|_{y = 0, slab}\) = \(\left.\begin{matrix} -k_{air}\frac{dT}{dy} \end{matrix}\right|_{y = 0, air}\) ...

(i) Calculation:

Given:

Temperature variation of air (T) = 300 + 200 e^-5y

k_air = 1 W/m-K, k_slab = 100 W/m-K, \(\left.\begin{matrix} \frac{dT}{dy} \end{matrix}\right|_{y = 0, slab}\) = ??

\(\left.\begin{matrix} -k_{air}\frac{dT}{dy} \end{matrix}\right|_{y = 0, air}\) = -1 × {0 + 200 × (-5) × e^-5y}

\(\left.\begin{matrix} -k_{air}\frac{dT}{dy} \end{matrix}\right|_{y = 0, air}\) = -1 × {200 × (-5) × 1} = 1000 K/m

Using Equation (I)

\(\left.\begin{matrix} -k_{slab}\frac{dT}{dy} \end{matrix}\right|_{y = 0, slab}\) = \(\left.\begin{matrix} -k_{air}\frac{dT}{dy} \end{matrix}\right|_{y = 0, air}\)

\(\left.\begin{matrix} -100 ×\frac{dT}{dy} \end{matrix}\right|_{y = 0, slab}\) = 1000 

\(\left.\begin{matrix} \frac{dT}{dy} \end{matrix}\right|_{y = 0, slab}\) = -10 K/m.

∴ Magnitude of \(\left.\begin{matrix} \frac{dT}{dy} \end{matrix}\right|_{y = 0, slab}\) = 10 K/m.

Concept:

The difference between the mass flow rate at inlet and outlet is equal to the rate of change of water in the tank i.e.,

\(\dot{m}_{i} - \dot{m}_{o} \) = \(\frac{{d{m_T}}}{{dt}}\)

\(\dot{m}_{i} = 0\)

∴ \(- \dot{m}_{o} \) = \(\frac{{d{m_T}}}{{dt}}\)

mass = density × Volume = density × Arrea × Velocity = density × Area × \(\sqrt {2gh}\)

-ρAo\(\sqrt{2gH}\) = \(\frac{d}{{dt}}\left( {A \times H \times \rho } \right)\)

\(dt =- \frac{A}{{{A_o}}} \times \frac{1}{{\sqrt {2g} }} \times \frac{{dH}}{{\sqrt H }}\)

\(t = - \frac{A}{{{A_o}}} \times \frac{1}{{\sqrt {2g} }} \times ;\mathop \smallint \limits_i^f \frac{{dH}}{{\sqrt H }}\)

\(t = - \frac{A}{{{A_o}}} \times \frac{1}{{\sqrt {2g} }} \times 2 \times \left[ {\sqrt {{H_f}} - \sqrt {{H_i}} } \right]\)...(i)

Calculation:

Given:

A = 1 m², A_o = 1 cm² = 10^-4 m², g = 9.8m/s², H_i = 1 m, H_f = 0.25

Using equation (i)

\(t = - \frac{A}{{{A_o}}} \times \frac{1}{{\sqrt {2g} }} \times 2 \times \left[ {\sqrt {{H_f}} - \sqrt {{H_i}} } \right]\)

\(t = - \frac{1}{{{{10}^{ - 4}}}} \times \frac{1}{{\sqrt {2 \times 9.8} }} \times \left( {1 - 2} \right)\)

t = 2258.8 s

Explanation:

Let mass flow rate at location 2 is m kg/s

Applying energy balance for open feed water heater

m₂ h₂ + m₅ h₅ = m₆h₆

Mass balance for open feed water heater

m2 + m5 = m6 

So

m2 h2 + m5 h5 = (m2 + m₅) h₆

Calculation:

Given:

h2  = 2624 kJ/kg, h₅ = 226.7 kJ/kg, h₆ = 708.6 kJ/kg, m₅ = 100 kg/s

m2 h2 + m5 h5 = (m2 + m5) h6

m2 × 2624 + 100 × 226.7 = (m2 + 100) × 708.6

m2 = 25.16  kg/s

Concept:

Conservation of mass:

According to this principle, the mass at the inlet and outlet should be equal.

∴ Using conservation of mass we have

Q̇ + A_iV_i = A_oV_o

Where, A_i = Area of the cross-section of the jet, A_o = Area of pipe, V_i = Velocity of the jet at the inlet to pipe, V_o = Outlet velocity, Q = Flow rate of entrained water.

Calculation:

Given:

Diameter of pipe (d_o) = 0.32 m, V_o = 6 m/s, A_i = 0.01 m², V_i = 35 m/s

Q̇  + AiVi = AoVo

Q̇ = \(\left(\frac{{\pi d_o^2}}{4}\times V_o\right)\) - A_iV_i

Q̇ = \(\left{\frac{{\pi}}{4} × {(0.32)^2} × 6 \right}\) - (0.01 × 35) = 0.13255 m³/s

Q̇ = 132.55 ltr/s     [∵ 1 m³ = 10³ ltr]

∴ The total water that leaves the pipe is 132.55 ltr/s.

Concept:

H_inlet = H_turbine + H_{L, stator + Guide} + H_rotor + H_{draft tube} + \(\frac{{V_3^2}}{{2g}}\) ...

(i) \({η _H} = \frac{{{H_{turbine}}}}{{{H_{inlet}}}}\)

Where, η_H is the hydraulic efficiency, HL, stator + Guide = Head loss in stator and guide vanes, V₃ = Velocity at the exit of the draft tube, g = acceleration due to gravity

Calculation:

Given:

Hinlet = 200 m, HL, stator + Guide = 5 m, Hrotor = 10 m, Hdraft tube = 2 m, V₃ = 3.5 m/s, g = 9.8 m/s2

From equation (i)

200 = Hturbine + 5 + 10 + 2 + \(\frac{{3.5^2}}{{2 \times 9.8}}\)

Hturbine = 182.375 m

∴ \({η _H} = \frac{{{H_{turbine}}}}{{{H_{inlet}}}}\) = \(\frac {182.375}{200}\) = 0.9118 = 91.87 %

Concept:

Entropy:

Entropy is a measure of the disorder of a system.

Mathematically entropy is calculated by ΔS:

\(ΔS= {Q \over T}\)

where Q is the heat transferred energy to the system during a process, and T is the temperature in K at which the process takes place.

When the process is irreversible then entropy generation in the system (δs)gen is always positive, the heat transfer will decide whether the entropy will increase or decrease.

Now,

\(\dot{S}_{gen}\) = \(\dot{S}_{e}\) - \(\dot{S}_{i}\)

\(\dot{S}_{gen}\) = \((\dot{m}_{2}s_{2} + \dot{m}_{3}s_{3})_{e} - (\dot{m}_{1}s_{1})_{i}\)

Where \(\dot{S}_{gen}\) = Entropy generation, \(\dot{S}_{e}\) = Entropy at exit, \(\dot{S}_{i}\) = Entropy at inlet, \(\dot{m}_{2}, \dot{m}_{3}\) = mass flow rate at end 2 and 3, \(\dot{m}_{1}\) = mass flow rate at inlet, s1, s2, s3 = entropy per unit mass at their respective ends.

\(\dot{S}_{gen}\) = \(\dot{m}_{2}\)s2 + \(\dot{m}_{3}\)s3 - \((\dot{m}_2 + \dot{m}_{3})\)s1 =  \(\dot{m}_{2} (s_{2} - s_{1}) + \dot {m}_{3} (s_{3} - s_{1})\)...(i)

We know that,

sf - si = \({c_p}\ln \frac{{{T_f}}}{{{T_i}}} - R\ln \frac{{{P_f}}}{{{P_i}}}\) ...(ii)

Where subscript 'i' & 'f' stands for initial and final condition.

Calculation:

Given:

P₁ = 500 kPa, P₂ = P₃ = 100 kPa, T₁ = 300 K, T₂ = 340 K, T₃ = 240 K, \(\dot{m}_{1}\) = 5 kg/s, \(\dot{m}_{2}\) = 3 kg/s, \(\dot{m}_{3}\) = 2 kg/s

Using equation (i) and (ii)

\(\dot{S}_{gen}\) = \(\dot{m}_{2} (s_{2} - s_{1}) + \dot {m}_{3} (s_{3} - s_{1})\) = \(\dot{m}_{2}\) \(\left[ {{c_p}\ln \frac{{{T_2}}}{{{T_1}}} - R\ln \frac{{{P_2}}}{{{P_1}}}} \right]\) + \(\dot{m}_{3}\) \(\left[ {{c_p}\ln \frac{{{T_3}}}{{{T_1}}} - R\ln \frac{{{P_3}}}{{{P_1}}}} \right]\)

∴ \(\dot{S}_{gen}\) = \(3 \times \left[ {1.005\ln \left( {\frac{{340}}{{300}}} \right) - 0.287\ln \left( {\frac{{100}}{{500}}} \right)} \right] + 2 \times \left[ {1.005\ln \left( {\frac{{240}}{{300}}} \right) - 0.287\ln \left( {\frac{{100}}{{500}}} \right)} \right]\)

\(\dot{S}_{gen}\) = 2.24 kW/K

Concept:

When a body is just about to slide then it is in an equilibrium state and at this state, ∑FH = 0 and ∑Fv = 0.

Calculation:

Given:

We draw the FBD of the block when it is about to move upward.

∑F_H = 0

μ_sN cos 30 + N sin 30 = 750    (1)    

∑F_v = 0

 - μsN sin 30 + N cos 30 = 900      (2)

Divide eqution 1 and 2

\( \frac{{{μ_sN cos 30 + N sin 30 }}}{{ - μ_sN sin 30 + N cos 30 }} =\frac{750}{900}\)

\( \frac{{{μ_s\sqrt{3}; +; 1 }}}{{ - μ_s; +; \sqrt{3} }} =\frac{5}{6}\)

μ_s = 0.1728

 

Why the block moves upward?

Because if we consider it is moving in a downward direction then the value of the coefficient of static friction comes negative which is practically not possible in that's why we have to considered block is about to move in an upward direction and we draw FBD of this condition and solve the problem.

Concept:

Applying Newton 2^nd Law for the system:

ΣFnet = msystema 

Net torque = Iα

Calculation:

Given:

For wheel: diameter D = 0.8 m, radius R = 0.4 m, mass M = 1 kg.

For axle: diameter d = 0.2 m, radius r = 0.1 m, mass M = 1.5 kg.

Net mass of the system = (2 × wheel) + axle = (2 × 1) + 1.5 = 3.5 kg.

∴ mass of system = 3.5 kg.

The 10 N of force is applied on the axle and axle is carrying two wheel.

Applying Newton 2^nd Law for the system:

ΣF_net = m_systema 

10 - 2F = 3.5a ..... eq (1)

For wheel:

Wheel is rotating Net torque = Iα 

2F × 0.4 - 10 × 0.1 = I_wheel × α 

2F × 0.4 - 10 × 0.1 = 2MR² × \(\frac{a}{R}\)  [∵ I_wheel = 2MR² & \(\alpha=\frac{a}{R}\)]

2F × 0.4 - 1 = 0.8a ....eq (2)

Solving eq (1) and (2)

a = 1.3637 m/s²

Concept:

The slope of the beam = ratio of the area of the Bending moment diagram and Flexural rigidity.

\(Slope= \frac{{{\rm{A_{BM\ diagram}}}}}{{\rm{EI}}} ​​{{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Given:

A_{BM diagram}= 10000 Nm2 , EI = 200 × 106 Nm2

By using equation (1),

\(⇒ Slope= \frac{{{\rm{10000}}}}{{\rm{200 × 10 ^6}}} \)

⇒ Slope = 50 × 10 ^-6  radian

⇒ Slope = 50 micro radian.

Concept:

The fluctuation of energy in Cycle ΔE = \(∫ T_{mean} × dθ \)

where θ = angle turned in one revolution in radian, T_mean = mean torque 

The maximum fluctuation of energy ΔE = IC_s(ω_mean)²

where I = Moment of inertia of flywheel about the axis of rotation

C_s = Coefficient of fluctuation of speed = \(\frac{N_{max}\ -\ N_{min}}{N_{mean}}\)

Calculation:

Given:

T(θ) = 12000 + 2500 sin (2θ) Nm , N_mean = 200 rpm

ωmean =\(\frac{2\pi N}{60}\) = 20.94 rad/sec.

Variation of the speed from the mean speed not to exceed ±0.5 %

⇒ Cs = + 0.5 - (-0.5) = 1 % = 0.01

The period of sin 2θ is π, hence the turning moment diagram will complete one cycle in π period.

T_mean = \(\frac{1}{\pi}\)∫T dθ = \(\frac{1}{\pi}\)∫(12000 + 2500 sin 2θ) dθ = 12000 Nm.

The torque exerted on the crankshaft in one stroke is equal to the mean resisting torque on the flywheel:

T = T_mean

The fluctuation of energy in one stroke is  ΔE = \(∫_{0}^{\frac{\pi}{2}} (T\ -\ T_{mean})dθ\) 

ΔE = \(∫_{0}^{\frac{\pi}{2}} (T\ -\ T_{mean})dθ\) = \(∫_{0}^{\frac{\pi}{2}} 2500\sin\ (2θ);dθ\) = 2500 J

ΔE = ICs(ωmean)2 

I = \(\frac{\Delta E}{C_s(\omega_{mean})^2}\) = \(\frac{2500}{0.01;\times;(20.94)^2}\) = 569.94 kgm² \(\approx\)  570 kgm².

Concept:

In Δ APC

\( \tan\theta =\frac{{AP}}{{PC}} \)

In Δ BQC

\( \tan\theta =\frac{{BQ}}{{QC}} \)

\(\tan \theta=\frac{{AP}}{{PC}} = \frac{{BQ}}{{QC}}\)

Calculation:

Given:

S₁ = 450 MPa, N₁ = 1000 cycles, S₂ = 150 MPa, N₂ = 10⁶ cycles.

\(\tan \theta=\frac{{AP}}{{PC}} = \frac{{BQ}}{{QC}}\)

\(\frac{{\log 450 ;-; \log150}}{{6;- ;3}} = \frac{{\log200;- ;\log150}}{{6; -;\log N}}\)

\(\frac{{\log 3}}{{3}} = \frac{{\log{(20/15)}}}{{6 -\log N}}\)

6 - log N = 0.785578

log N = 5.214422

Number of cycle (N) equals 10^5.214422 = 163842 cycles.