GATE ME 2017 Official Paper: Shift 1 (Previous Year Paper)

Since the person was one among many, the correct word to be used in the first blank is ‘friends’ and the second blank must be filled by the word adverb ‘keenly’ as it describes the verb ‘felt’.

L = √(5² + 12²) = 13 cm = Slant height

Here, 2πr = Lq ….. As cone is rolling over circumference

So, q = 10π/13.

If suppose P is right:

P states that S has at least 3 cars but then R says that S has at least one car which also becomes a true statement. Hence P is not saying right.

If suppose R is right:

R states that S has at least one car, but then Q believes that S has less than 3 cars, which becomes a true statement. Hence R is not saying right.

If suppose Q is right:

Q believes that S has less than 3 cars. Thus P’s statement that S has at least 3 cars and R’s statement that S has at least one car still remains false or indefinite.

But only one of P, Q and R is right. Thus R is definitely false. That means S does not have even 1 car.

Hence S has no car.

Since the speakers became agitated or angry the debate must have become ‘heated’ or ‘intense’.

All the values put either in ascending or descending order first.

Now number of observations equal to 100 [even]

∴ The median of these values = Avg of two middle most observations.

= (50^th observation + 51^thobservation)/2

= (30000 + 30000)/2

= 30,000

We start from unit digit to search for answer.

The only possible digit is 0, which gives 10 – 9 = 1

Similarly for tenth place digit, only possible digit is 0.

Now, since, the ten thousand’s place has 5, but still the answer is only 4 digit number, it means that 1 is being carried over to thousand’s place. Hence thousand’s place has to be 0.

Now, the 100^th place can have either 0 or 1 and accordingly the second number will be 48889 or 48989.

So, the given question can either be 50100 – 48989 = 1111

OR 50000 – 48889 = 1111

In first case: Sum of missing digits = 0 + 1 + 0 + 0 + 9 = 10

In 2^nd case: Sum of missing digits = 0 + 0 + 0 + 0 + 8 = 8

So unique solution cannot be determined.

Explanation:

|x – 1| ≤ 2

⇒ -2 ≤ (x – 1) ≤ 2

⇒ -1 ≤ x ≤ 3

|y + 2| ≤ 3

⇒ -3 ≤ (y + 3) ≤ 3

⇒ -6 ≤ y ≤ 0

x – y ≥ – 2

y ≤ 1

x ≥ -1      y ≥ -5

Total Area = Area of rectangle ABCD + Area of ΔABE = 4 × 6 + ½ (4 × 4) = 32

James tells Mark that he was the best player in his own family. This means Mark was not the best player in his family and so James makes this comment. Thus we can infer that Steve was known to play better than Mark.

Stuart’s demonstration had no effect on his superiors, thus option A can be eliminated. The Europeans were prejudiced and thus option B is eliminated. Since Stuart was reprimanded by the commander in chief, option D can be eliminated. The accurate statement is option C.

From the graph,

Statement (i) is correct,

The time taken for curd formation @25°C = 120 min,

The time taken for curd formation @ 37°C = 80 min, hence (ii) is incorrect.

Concept:

From the property of eigenvalues

Product of eigenvalues = |P| = Determinant of matrix

Sum of Eigenvalues = Sum of Principal elements of Matrix

Calculation:

\(P = \left[ {\begin{array}{*{20}{c}} 2&0&1\ 4&{ - 3}&3\ 0&2&{ - 1} \end{array}} \right]\)

Product of Eigenvalues = |P| = 2(3 - 6) + 1(8 - 0) = 2

Concept:

\(L{t_{x \to 0}}\frac{{\sin x}}{x} = 1\)

Calculation:

\(L{t_{x \to 0}}\left( {{x^2} - \frac{{\sin x}}{x}} \right) = 0 - 1 = - 1;;;;as;L{t_{x \to 0}}\frac{{\sin x}}{x} = 1\)

Alternate solution

If limit x → 0 is put than it is a form of (0/0). Note that in case of (0/0) and (∞/∞) form, apply L-hospital rule.

\(\mathop {\lim }\limits_{x \to 0} \frac{{3{x^2} - \cos x}}{1} = 0 - 1 = - 1\)

Explanation:

One approach is to put options and check if they satisfy the solution.

In this way putting option b

u (x,y) = f(x – cy)

\(\frac{{\partial u}}{{\partial y}} + C\frac{{\partial u}}{{\partial x}} = f'\left( {x - cy} \right)\left( { - c} \right) + cf'\left( {x - cy} \right) = 0\)

The solution of \(\frac{{\partial u}}{{\partial y}} + C\frac{{\partial u}}{{\partial x}} = 0\) is

U(x,y) = f (x – cy)

Alternate solution

Let u = f(ax + by)

\(\therefore \frac{{\partial u}}{{\partial \left( {ax + by} \right)}} = f'\left( {ax + by} \right)\)

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

\(\frac{{\partial u}}{{\partial \left( {ax + by} \right)}} \times \frac{{\partial \left( {ax + by} \right)}}{{\partial y}} + C\frac{{\partial u}}{{\partial \left( {ax + by} \right)}} \times \frac{{\partial \left( {ax + by} \right)}}{{\partial x}} = 0\)

⇒ b + ca = 0

b = - ac,

If a = 1

b = -c

∴ u = f(x - cy)

Concept:

Given equation is

\(\frac{{{d^2}y}}{{d{x^2}}} + 16y = 0\)

This is a homogeneous second order differential equation,

So (D² + 16)y = 0

D² = m²

⇒ m² + 16 = 0     ⇒ m = ± 4i = 0 ± 4i

Solution is given as in this case roots are complex, m = α ± i β

y = (C₁ cos βx + C₂ sin βx) e^αx

= (C₁ cos 4x + C₂ sin 4x) e^ox = C₁ cos 4x + C₂ sin 4x

Now y’ = -4C₁ sin 4x + 4C₂ cos 4x

Applying Boundary condition,

y’ (0) = 1  ⇒ -4C₁ sin (0) + 4C₂ cos(0) = 1

\(4{C_2} = 1 \Rightarrow {C_2} = \frac{1}{4}\)

Putting another boundary condition.

\(y'\left( {\frac{\pi }{2}} \right) = - 1\)

\( - 4{C_1}\sin \left( {\frac{{4\pi }}{2}} \right) + 4{{\rm{C}}_2}\cos \left( {\frac{{4\pi }}{2}} \right) = - 1 \Rightarrow {C_2} = \frac{{ - 1}}{4}\)

So this equation has no solution.

Concept:

Mean = \(\sum\limits_x {x.{f_x}} \)

Calculation:

Mean = \(\sum\limits_x {x.{f_x}} \)

= 1.(1/6) + 2.(1/6) + 3.(1/6) + 4.(1/6) + 5.(1/6) + 6.(1/6)

= 3.5

Explanation:

As the pipe is of uniform diameter and the flow is steady incompressible, therefore the discharge remains constant at every section. As the diameter is constant, the avg. velocity is constant everywhere as shown in figure.

Average Velocity = Flow rate/ Cross-section Area

Concept:

Every fluid flow must satisfy its corresponding mass conservation (continuity) equation.

For a 2 – dimensional incompressible flow

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

\(\vec V = u\hat i + \nu \hat j\)

Calculation:

u = 5 + a₁x + b₁y, ν = 4 + a₂x + b₂y

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial \nu }}{{\partial y}} = 0 \Rightarrow {a_1} + {b_2} = 0\)

Concept:

Power is required to overcome losses, hf

Power = \(ρ Qgh\)

where h = frictional head loss

the frictional head loss will be equal to the pressure difference head

hence \(h =\frac{\Delta P}{ρ g} \)

Calculation:

\(\dot m = ρ Q \Rightarrow Q = \frac{{\dot m}}{ρ } = \frac{1}{{{{10}^3}}} = {10^{ - 3}}\ m^3/s\)

\(h =\frac{\Delta P}{ρ g} \)

\(h = \frac{100 \times10^3}{10^3\times9.81}=10.19 m\)

Power = 9810 × 10^-3 × 10.19 = 100 W

Explanation:

Pump:

• A pump is a device that moves fluids (liquids or gases), or sometimes slurry. By mechanical action, typically converted from electrical energy into hydraulic energy.
• Pumps can be classified into three major groups according to the method they use to move the fluid:

1. Direct lift
2. Displacement
3. Gravity

Centrifugal Pump: It has rotating part eg. impeller, Vane.

Gear Pump: In this pump there is gear mechanism which is rotating part.

Jet Pump: Here pump utilizing ejector principle which have nozzle and diffusor not rotating parts.

Vane Pump: It consist of rotating disc which called as rotor in which number of radial slots are there where sliding vanes is inserted

Additional Information

Concept:

For phase change Heat Exchanger, LMTD of parallel and counter flow will be same. So taking as parallel flow.

\({\rm{LMTD}} = \frac{{\Delta {T_1} - \Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}} \)

Calculation:

\({\rm{\Delta }}{{\rm{T}}_1} = {{\rm{T}}_{\rm{h}}}_1 - {T_c}_1 = 100 - 20 = 80\)

\({\rm{\Delta }}{{\rm{T}}_2} = {{\rm{T}}_{\rm{h}}}_2 - {T_c}_2 = 100 - 50 = 50\)

\({\rm{LMTD}} = \frac{{\Delta {T_1} - \Delta {T_2}}}{{\ln \left( {\frac{{\Delta {T_1}}}{{\Delta {T_2}}}} \right)}} = \frac{{80 - 50}}{{\ln \left( {\frac{{80}}{{50}}} \right)}}\)

LMTD = 63.829°C

Concept:

Specific enthalpy ΔH = C_pΔT 

\({C_p} - {C_v} = \vec R\)

C_p and C_v are molar specific heats

Calculation:

\({C_v} = 2.5;\vec R\)

\({C_p} - {C_v} = \vec R\)

ΔT = 100 k

\( \Rightarrow {C_p} = \vec R + {C_v}\)

\({C_p} = 3.5\vec R\)

\({\rm{\Delta }}H = 3.5\vec R*100\)

\({\rm{\Delta }}H = 3.5 \times \frac{{8.314}}{{mol - k}} \times 100;K\)

ΔH = 2909.9J / mol

Concept:

Heat Pump:

• A heat pump is a device that works on a reversed Carnot cycle and transfers heat from a lower temperature body to a higher temperature body.
• \(CO{P_{HP}} = \frac{{{Q_1}}}{W} \)

Calculation:

Given:

Q₁ (heat rejected from HP) = 10 + 15 = 25 kW, W = 15 kW

\(CO{P_{HP}} = \frac{{{Q_1}}}{W} = \frac{{10 + 15}}{{15}} = 1.67\)

Important Points

COP of heat pump = \(\frac{{{T_H}}}{{{T_H} - {T_L}}}\) is used only for reversible heat pump.

\(CO{P_{HP}} = \frac{{{Q_1}}}{W};\) Can be used for any type of heat pump.

Concept:

Volumetric strain

\({\epsilon_v} = \frac{{{\rm{\Delta }}V}}{V}\)

\({\epsilon_v} = \frac{{\left( {1 - 2\mu } \right)}}{E}\left( {{\sigma _x} + {\sigma _y} + {\sigma _z}} \right)\)

Where μ = Poisson’s ratio

Calculation:

\({\epsilon_v} = \frac{{\left( {1 - 2\mu } \right)}}{E}\left( {{\sigma _x} + {\sigma _y} + {\sigma _z}} \right)\)

For an incompressible material, change in volume (ΔV) = 0 ⇒ ϵ_v = 0

\(0 = \frac{{\left( {1 - 2\mu } \right)}}{E}\left( {{\sigma _x} + {\sigma _y} + {\sigma _z}} \right)\)

\( \Rightarrow 1 - 2\mu = 0 \Rightarrow \mu = \frac{1}{2} = 0.5\)

Concept:

Tangential and radial velocity for the polar co-ordinate is given by,

\({V_t} = rw = {t^2}\frac{{d\theta }}{{dt}} ,{V_r} = \frac{{dr}}{{dt}} \)

Calculation:

Given:

r(t) = t2, θ(t) = t , t = 2 sec

\({V_t} = rw = {t^2}\frac{{d\theta }}{{dt}} = {t^2};;{V_r} = \frac{{dr}}{{dt}} = 2t = 2 \times 2 = 4\)

V_t = 2² × 1 = 4; V_r = 4

\(V = \sqrt {V_r^2 + V_t^2} = \sqrt {{4^2} + {4^2}} = \sqrt {32} \)

\(K.E. = \frac{1}{2}m{V^2} = \frac{1}{2}\times1\times{\left( {\sqrt {32} } \right)^2} = 16\)

Concept:

Power (P)

P = T × W

Torsion equation

\(\frac{T}{J} = \frac{{G\theta }}{L} = \frac{\tau }{r}\)

Calculation:

Given:

P = 40 kW, N = 500 rpm, D = 40 mm

As, P = T × W

\(40 \times {10^3} = {\rm{T}} \times \frac{{2{\rm{\pi }} \times 500}}{{60}} ⇒ {\rm{T}} = 763.9{\rm{;Nm}} = 763.9 \times {10^3}{\rm{Nmm}}\)

Now,

\(\frac{T}{J} = \frac{\tau }{R} ⇒ \tau = \frac{{TR}}{J} = \frac{{763.9 \times {{10}^3} \times 20 \times 32}}{{\pi \times {{40}^4}}}\)

τ_max. = 60.79 N/mm² = 60.79 MPa

Calculation:

Second moment of area (neutral axis), \(I = \frac{{\pi {d^4}}}{{64}}\) 

Section modulus max \( = \frac{I}{{{y_{max}}}}\)

\(ratio = \frac{{\frac{I}{I}}}{{{y_{max}}}} = {y_{max}}\)

For a circular section \({y_{max}} = r = \frac{d}{2}\).

Concept:

The area of the velocity-time graph gives the distance covered.

Calculation:

Distance covered = Area under V-t curve form t = 0 to t = 5 sec.

= Area of Δ A0I + Area of Rectangle ABHI + Area of Trapezoidal BCGH + Area of Trapezoidal CDFG

⇒  Area of Δ A0I + Area of rectangle ABHI + Area of Δ BCJ + Area of rectangle BJHG + Area of Δ KCD + Area of rectangle KDFG

\( ⇒ \frac{1}{2} \times 1 \times 1 + 1 \times 1 + \frac{1}{2} \times 3 \times 1 + 1 \times 1 + \frac{1}{2} \times 2 \times 2 + 2 \times 2\)

⇒ 0.5 + 1 + 1.5 + 1 + 2 + 4 = 10 m

Explanation:

\(\frac{{m{d^2}x}}{{d{t^2}}} + \frac{{dx}}{{dt}} + kx = F\left( t \right)\)

Or \(m\ddot x + c\dot x + kx = 0\)

[By taking the sum of inertia force and external force on a body in a direction in to be zero]

⇒ Solution K = Ae^αt + Be^αt

\(\therefore {\alpha ^2} + \frac{c}{m}{\alpha ^2} + \frac{k}{m} = 0\)

\({\alpha _{1,2}} = \frac{{ - C}}{{2m}} \pm \sqrt {{{\left( {\frac{C}{{2m}}} \right)}^2} - \left( {\frac{k}{m}} \right)} \)

The ration \({\left( {\frac{C}{{2m}}} \right)^2};to\frac{s}{m}\) gives the degree of dampness and the square root of this ratio is called the damping ratio.

\(\xi = \sqrt {\frac{{{{\left( {\frac{C}{{2m}}} \right)}^2}}}{{\left( {\frac{k}{m}} \right)}}} = \frac{C}{{2\sqrt {km} }}\)

Explanation:

Riveted joint:

• The permanent joining of parts with the help of rivets is called a riveted joint.
• Rivets are the cylindrical-shaped metal body used to leakproof the joining of dissimilar metals.
• When two pieces of sheet joint with the help of riveted joint such that one sheet overlaps the other one then this type of joint is called lap joint.
• There are basically three types of failure that occur in lap rivet joints:

where,n = no. of rivets, dr = diameter of rivet, t = thickness of sheet, P = Pitch of sheet, dh = diameter of hole

Calculation:

In the given case:

Net resisting area = (W- 2d)t

Tensile load F is:

F = S_f (W – 2d)t

\(t = \frac{F}{{{S_f}\left( {W - 2d} \right)}}\)

Concept:

\({C_P} = \frac{{USL - LSL}}{{6\sigma }}\)

U = 15.02

L = 14.98

σ = 0.004

\({C_P} = \frac{{15.02 - 14.98}}{{6 \times 0.004}} = 1.667\)

Concept:

According to von Mises theory, for yielding

\({\sigma _{yt}} = \sqrt {\frac{1}{2}\left[ {{{\left( {{\sigma _1} - {\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} - {\sigma _3}} \right)}^2} + {{\left( {{\sigma _3} - {\sigma _1}} \right)}^2}} \right]} \)

Calculation:

Given:

σ1 = +180 MPa, σ2 = -100 MPa, σ3 = 0

Yield stress is:

\({\sigma _{yt}} = \sqrt {\frac{1}{2}\left[ {{{\left( {{\sigma _1} - {\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} - {\sigma _3}} \right)}^2} + {{\left( {{\sigma _3} - {\sigma _1}} \right)}^2}} \right]} \)

\({\sigma _{yt}} = \sqrt {\frac{1}{2}\left[ {{{\left{ {180 - \left( { - 100} \right)} \right}}^2} + \left( { - 100 - {0^2}} \right) + {{\left( {0 - 180} \right)}^2}} \right]} \)

σ_yt = 245.76 MPa

Explanation:

Electric discharge Machining: -

• EDM is an electro thermal non-traditional machining process, where electrical energy is used to generate electrical spark and material removal mainly occurs due to thermal energy of spark.
• EDM is mainly used to machine difficult to machine materials and high strength temperature resistant alloys.
• Work material to be machined by EDM has to be electrically conductive.

Ultrasonic Machining (USM):-

• USM is mainly used for machining brittle materials (which are poor conductors of electricity). It is used for machining alloys, hard and brittle materials, glass, ceramics and carbides etc.

Chemical Machining: -

• Chemical machining is the controlled dissolution of workplace by means of strong chemical reagent. For chemical machining there should not be any residual stress. For eg., in chemical milling, if part to be machined has residual stress from the previous processing, these stresses first should be relieved in order to prevent warping after chemical milling.

Ion Beam Machining: -

• The system consist of an ion source that produces sufficient intense beam of ions, for the removal of atoms from the work surface by impingement of ions. It is used for Nano-machining. IBM is performed in vacuum.

Additional Information

The unconventional machining process and its characteristics and the application areas are discussed in the table below:

Explanation:

Metal deposit rate (MDR) = Area (A) x Weld Speed (V) = Constant

A₁ V₁ = A₂V₂; V₂ = 2V₁ (given)

\({A_1}{V_1} = {A_2}\left( {2{V_1}} \right) \Rightarrow {A_2} = \frac{{{A_1}}}{2}\)

So cross sectional area reduce by half (50%)

Concept:

RPM of lead screw = \(\frac {Pitch~ of Metric ~thread}{Pitch ~of ~lead ~screw}\times Spindle ~RPM\)

Calculation:

Pitch = 0.8 mm

Lead screw pitch = 1.5 mm

Spindle speed = 1500 rpm

In one rotation of lead, it should advance 0.8 mm.

Fraction of revolution in one rotation of spindle \( = \frac{{0.8}}{{1.5}}\)

RPM of lead screw \( = \frac{{0.8}}{{1.5}} \times 1500 = 800;rpm\)

Alternate Solution

Pitch of thread, P_t = 0.8 mm

Speed of spindle, N_t = 1500 rpm

Pitch of screw (P_s) = 1.5 mm

Now N_s × P_s × Z = N_t × P_t × Z_t

N_s × 1.5 × 1 = 1500 × 0.8 × 1

N_s = 800 rpm

Explanation:

The engineering stress-strain curve for mild steel is

Ultimate tensile strength represents the maximum stress that a material can withstand without Fracture.

Concept:

\(P = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}}\ 0&1&0\ {\frac{{ - 1}}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

(i) \(\left| P \right| = \frac{1}{{\sqrt 2 }}\left( {\frac{1}{{\sqrt 2 }}} \right) - 0 + \frac{1}{{\sqrt 2 }}\left( {\frac{1}{{\sqrt 2 }}} \right) = 1\)

(ii) for Matrix to be orthogonal. PP^T = P^TP = I

\({P^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&0&{\frac{{ - 1}}{{\sqrt 2 }}}\ 0&1&0\ {\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

\(P{P^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}}\ 0&1&0\ {\frac{{ - 1}}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&0&{\frac{{ - 1}}{{\sqrt 2 }}}\ 0&1&0\ {\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 1&0&0\ 0&1&0\ 0&0&1 \end{array}} \right] = I\)

As PP^T = I, so P is orthogonal

(iii) \({P^{ - 1}} = \frac{{adj;P}}{{\left| P \right|}}\)

\( = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&0&{\frac{{ - 1}}{{\sqrt 2 }}}\ 0&1&0\ {\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}} \end{array}} \right] = {P^T}\)

As option A, B and C are correct. So option D is wrong.

Concept:

Curl \(\vec V\) = \(∇ \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {2yz}&{3xz}&{4xy} \end{array}} \right| = x\hat i - 2y\hat j + z\hat k\)

∇ = \( \frac{\partial }{{\partial x}}.i + \frac{\partial }{{\partial y}}.j + \frac{\partial }{{\partial z}}.k\)

div \(\vec V\) = ∇.\(\vec V\)= \( \frac{\partial }{{\partial x}}\left( V_x \right) + \frac{\partial }{{\partial y}}\left( { V_y} \right) + \frac{\partial }{{\partial z}}\left( V_z \right)\)

 \(∇ .\left( {∇ \times \vec V } \right) \) = div (Curl \(\vec V\)) = 0

or,

 \(∇ .\left( {∇ \times \vec V } \right) = \frac{\partial }{{\partial x}}\left( x \right) + \frac{\partial }{{\partial y}}\left( { - 2y} \right) + \frac{\partial }{{\partial z}}\left( z \right) \)

= 1 – 2 + 1 = 0

Concept:

\(x = \cos \left( {\frac{{\pi u}}{2}} \right),;y = \sin \left( {\frac{{\pi u}}{2}} \right)\)

\({x^2} + {y^2} = {\cos ^2}\left( {\frac{{\pi u}}{2}} \right) + {\sin ^2}\left( {\frac{{\pi u}}{2}} \right) = 1\)

So it represents an equation of circle in x-y plane.

Given 0 ≤ u ≤ 1

So, 0 ≤ x ≤ 1,  0 ≤ y ≤ 1

i.e., 0 ≤ θ ≤ π/2

So we get a quarter circle in x-y plane and by revolving it 360°, we get a hemisphere.

Area of hemi-sphere = 2π(1)² = 2π

Concept:

Trapezoidal Rule:

\(I= \frac{{\rm{h}}}{2}{\rm{;}}[{{\rm{y}}_0} + 2({{\rm{y}}_1} + {{\rm{y}}_2} + \ldots + {{\rm{y}}_{{\rm{n}} - 1}}) + {{\rm{y}}_{\rm{n}}}]\)

Simpson rule:

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

Calculation:

Given:

Given points on the curve f(x) are

P(0, 3), Q (0.5, 4) & R (1,5)

Trapezoidal rule:

\(I= \frac{{\rm{h}}}{2}{\rm{;}}[{{\rm{y}}_0} + 2({{\rm{y}}_1} + {{\rm{y}}_2} + \ldots + {{\rm{y}}_{{\rm{n}} - 1}}) + {{\rm{y}}_{\rm{n}}}]\)

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

\(∴;\frac{{0.5}}{2}\left[ {\left( {3 + 5} \right) + 2\left( 4 \right)} \right] = 4\)

Simpsons rule:

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

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

\( ∴;\frac{{0.5}}{3}\left[ {\left( {3 + 5} \right) + 4\left( 4 \right)} \right] = 4\)

∴ Difference in the result = 4 – 4 = 0

Concept:

Displacement thickness is given by

\({{\rm{\delta }}^{\rm{*}}} =\int\limits_0^{\rm{\delta }} \left( {1 - \frac{{\rm{u}}}{{{{\rm{U}}_\infty }{\rm{;}}}}} \right){\rm{dy}}\)

Where,

u – velocity of the fluid

U∞ - Free stream velocity

Calculations:

Given:

\(\frac{u}{{{u_\infty }}} = \sin \left( {\frac{{xy}}{{2\delta }}} \right)\)

\(Displacement;Thickness;{\delta ^*} = \mathop \smallint \limits_0^\delta \left( {1 - \frac{u}{{{u_\infty }}}} \right)dy\)

\({\delta ^*} = \mathop \smallint \limits_0^\delta \left( {1 - \sin \frac{{\pi y}}{{2\delta }}} \right)dy\)

\({\delta ^*} = \left[ {y + \frac{{\cos \frac{{\pi y}}{{2\delta }}}}{{\frac{\pi }{{2\delta }}}}} \right]_0^\delta \)

\({\delta ^*} = \delta + \frac{{2\delta }}{\pi }\left( 0 \right) - 0 - \frac{{2\delta }}{\pi }\)

\({\delta ^*} = \delta - \frac{{2\delta }}{\pi } \Rightarrow \frac{{{\delta ^*}}}{\delta } = 1 - \frac{2}{\pi }\)

Explanation

As pipes are in series, the discharge will be the same

Q₁ = Q₂ = Q

l₁ = l₂ = l

For horizontal pipe

\({\rm{\Delta }}{P_1} = {h_{{f_1}}} = \frac{{{f_1}L{Q^2}}}{{12d_1^5}}\)

\({h_{{f_2}}} = \frac{{{f_2}L{Q^2}}}{{12d_2^5}}\)

\(\frac{{{\rm{\Delta }}{P_1}}}{{{\rm{\Delta }}{P_2}}} = \frac{{{f_1}}}{{d_1^5}} \times \frac{{d_2^5}}{{{f_2}}}\)

Now \(f = \frac{K}{{{{\left( {{R_e}} \right)}^n}}} = \frac{K}{{{{\left( {\frac{{Vd}}{v}} \right)}^n}}}\)

As \(Q = \frac{\pi }{4}{d^2} \times V \Rightarrow V \propto \frac{1}{{{d^2}}}\)

\(\therefore f \propto \frac{1}{{{{\left( {\frac{1}{{{d^2}}} \cdot d} \right)}^n}}} \Rightarrow f \propto {d^n}\)

Now \(\frac{{{\rm{\Delta }}{P_1}}}{{{\rm{\Delta }}{P_2}}} = \frac{{{f_1}}}{{d_1^5}} \times \frac{{d_2^5}}{{{f_2}}}\)

\( = \frac{{d_1^n}}{{d_1^5}} \times \frac{{d_2^5}}{{d_2^n}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^{5 - n}}\)

\(\therefore \frac{{{\rm{\Delta }}{P_1}}}{{{\rm{\Delta }}{P_2}}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^{5 - n}}\)

Explanation:

\(\vec V = \left( { - {x^2} + 3y} \right)\hat i + 2xy\hat j\)

u = - x² + 3y; v = 2xy

\({a_x} = u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( { - 2x} \right) + 2xy\left( 3 \right)\)

a_x = [-(1)² + 3 (-1)] (-2(1))+6(1)(-1)

a₁ = 8 – 6

a_x = 2

\({a_y} = u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( {2y} \right) + \left( {2xy} \right)\left( {2x} \right)\)

\({a_y} = \left[ { - {{\left( 1 \right)}^2} + 3\left( { - 1} \right)} \right]\left[ {2\left( { - 1} \right)\left] + \right[4{{\left( 1 \right)}^2}\left( { - 1} \right)} \right]\)

a_y = 8 – 4

a_y = 4

\(a = \sqrt {a_x^2 + a_y^2;} = \sqrt {{2^2} + {4^2}} = \sqrt {20} \)

\(= 2\sqrt 5 \)

Concept:

From first law, Q = du + W

Q = 0 (Adiabatic Process)

du + W = 0 ⇒ W = - du

Stirrer work = work done on the system

Calculation:

• (-100) = du ⇒ du = 100 kJ

mC_v (T₂ – T₁) = 100

1 × 1 × (T₂ – T₁) = 100

(T₂ – 300) = 100  So T₂ = 400 K

\({S_2} - {S_{1;}} = m\left( {{C_v}\ln \frac{{{T_2}}}{{{T_1}}} + R\ln \frac{{{V_2}}}{{{V_1}}};} \right)\)

Rigid container ⇒ V₁ = V₂

\({S_2} - {S_1} = m{C_v}\ln \frac{{{T_2}}}{{{T_1}}} = 1 \times 1000 \times \ln \left( {\frac{{400}}{{300}}} \right)\)

S₂ – S₁ = 287.68 J/k

Concept:

For isentropic process

\(\frac{{{T_1}}}{{{T_{2s}}}} = {\left( {\frac{{{P_1}}}{{{P_{2s}}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

Calculation:

Given:

T₁ = 1200 K

For Process 1 - 2

\(\frac{{{T_1}}}{{{T_{2s}}}} = {\left( {\frac{{{P_1}}}{{{P_{2s}}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

\(\frac{{1200}}{{{T_{2s}}}} = {\left( {10} \right)^{\frac{{1.4 - 1}}{{1.4}}}}\)

T_2s = 621.537 K

\({\eta _{Ts}} = \frac{{{T_1} - {T_2}}}{{{T_1} - {T_{2s}}}} \Rightarrow 0.9 = \frac{{1200 - {T_2}}}{{1200 - 621.537}}\)

T₂ = 679.38 K

Concept:

Humidity ratio (w) is the mass of water associated with each kilogram of dry air. It is known as specific humidity or absolute humidity.

\(\omega = \frac{{{m_v}}}{{{m_a}}} = 0.622\left( {\frac{{{P_v}}}{{{P_t} - {P_v}}}} \right) \)

Relative humidity (ϕ):

It is defined as the ratio of partial pressure of water vapour (pv)in a mixture to the saturation pressure (ps) of pure water at the same temperature of the mixture.

\(\phi=\frac{p_v}{p_{vs}}\)

Calculation:

\(\phi = \frac{{{P_v}}}{{{P_v}_s}} \Rightarrow 0.55 = \frac{{{P_v}}}{{4.246}} \Rightarrow {P_v} = 2.3353;kPa\)

\(w = 0.622\left( {\frac{{{P_v}}}{{{P_t} - {P_v}}}} \right) =0.622\left( {\frac{{2.3353}}{{100 - 2.3353}}} \right) = 0.0149;kg/Kg;dry;air\)

= 14.9 gm/ Kg dry air

Concept:

One litre of air contains 21% O₂ and 79% N₂, so 1 litre of air = [O₂ + 3.76 N₂]

Now stoichiometric reaction

CH₄ + 2[O₂ + 3.76 N₂] → 2H₂O + 2 × 3.76 N₂ + CO₂

50% excess air is supplied

CH₄ + 1.5 × 2 [O₂ + 3.76 N₂] → 2H₂₀ + CO₂ + 3 × 3.76 N₂ + O₂

\(% ;of;{N_2} = \frac{{3 \times 3.76}}{{2 + 1 + 1 + 3 \times 3.76}} = 73.82% \)

Concept:

Summation rule of view factor

F₁₁ + F₁₂ + F₁₃ .........=1

For flat surfaces F₁₁ = 0

A1 F12 = A2 F21

Calculation:

Taking surroundings as surface 3

From Summation rule,

F₂₁ + F₂₂ + F₂₃ = 1; F₂₂ = 0

F₂₁ + F₂₃ = 1

A₁ F₁₂ = A₂ F₂₁

6 × 0.5 = 5 x F₂₁

F₂₁ = 0.6

F₂₃ = 1 – F₂₁ = 0.4

In this Question, the rate of heat transfer from surface 2 to surroundings (Surface 3) is asked (Net heat transfer between 2 and 3 is not asked in this problem) therefore

\({{\rm{Q}}_{23}} = {\rm{\sigma ;}}{{\rm{A}}_2}{{\rm{F}}_{23}}T_2^4\)

= 5.67 × 10^-8 × 5 × 1 × 600⁴ × 0.4 = 14696.64 W

= 14.69 kW

Cylindrical rod dia = 10 mm

Rate of heat generation q_g = 4 × 10⁷ W/m³

Temperature distribution in a cylindrical rod with uniform heat generation is given by

\({T_0} - {T_\infty } = \frac{{{q_g}{R^2}}}{{4K}}{\left( {1 - \left( {\frac{r}{R}} \right)} \right)^2}\)

Where T₀ = center temp

Now T₀ = C_entre, so r = 0

\(\therefore {T_0} - {T_\infty } = \frac{{{q_g}{R^2}}}{{4K}} = \frac{{4 \times {{10}^7} \times {{\left( {0.005} \right)}^2}}}{{4 \times 25}} = 10\)

Concept:

Crippling load:

\(P = \frac{{{\pi ^2}EI}}{{L_e^2}}\)

If Rod is free to expand

Free elongation ΔL  = LαΔT

∴ ΔL/L = αΔT

Here Rod is fixed at both ends so strain will be zero but due to restriction given by fixed ends, there will be reaction force at the end.

That reaction force per unit area is known as thermal stress. It is given by- 

\(σ = E\epsilon\)

or, σ = E(ΔL/L) = Eα(ΔT)

Calculation:

Given:

At the time when the rod just starts to buckle, the load required to buckle will be equal to the reaction force due to thermal stress.

i.e. P_thermal = P_buckling

\(\therefore P_{thermal}=\sigma A\Rightarrow E\alpha(\Delta T)\frac{\pi d^2}{4}\)

∴ P_thermal = σA ⇒ EAα(ΔT)

\(P_{buckling} = \frac{{{\pi ^2}EI}}{{L_e^2}}\)

∵ both ends are fixed, \(L_{e}=\frac{L}{2}\)

\(\therefore P_{buckling} = \frac{{{4\pi ^2}EI}}{{L^2}}\Rightarrow\frac{4\pi^2E}{L^2}\times\frac{\pi d^4}{64}=\left(\frac{\pi^3E}{16L^2}\right)d^4\)

∵ Pthermal = Pbuckling

\(E\alpha(\Delta T)\frac{\pi d^2}{4}=\left(\frac{\pi^3E}{16L^2}\right)d^4\)

\((\Delta T)=\left(\frac{\pi^2}{4\alpha L^2}\right)d^2\)

\(\therefore;\Delta T \propto d^2\)

Concept:

δImpact = δstatic × IF

\(IF = 1 + \sqrt {1 + \frac{{2H}}{{{\delta _{static}}}}} \)

Calculation:

Given:

Cross-sectional area (A) = 100 mm2, length (L) = 1 m, Young’s modulus (E) = 100 GPa, H = 10 mm

This is a case of impact loading

\({\delta _{static}} = \frac{{wL}}{{AE}} = \frac{{mgL}}{{AE}} = \frac{{100 \times 10 \times 1}}{{100 \times {{10}^{ - 6}} \times 100 \times {{10}^9}}} = 1 \times {10^{ - 4}}m = 0.1;mm\)

δ_Impact = δ_static × IF

\(IF = 1 + \sqrt {1 + \frac{{2H}}{{{\delta _{static}}}}} \)

\(IF = 1 + \sqrt {1 + \frac{{2 \times 10}}{{0.1}}} = 15.177\)

δImpact = δstatic × IF

∴ δ_Impact = 0.1 × 15.177 = 1.517 mm

Concept:

Total KE of rolling body = \(\frac 12\) mv²  + \(\frac 12\) I w²

where m = mass, v = velocity of centre of mass, I = mass moment of inertia and w = angular velocity

For circular disc : \(I = \frac {mR^2}{2}\)

For circular Rim: I = mR2

Calculation:

Given:

\({m_A} = {m_B} = m\)

\({I_A} = m{R^2};{I_B} = \frac{{m{R^2}}}{2}\)

mgh = K.E of disk A = K.E of disc B

\(mgh = \frac{1}{2}m{V^2} + \frac{1}{2}I{\omega ^2}; = \frac{1}{2}mV_A^2 + \frac{1}{2}m{R^2}\frac{{V_A^2}}{{{R^2}}} = ;\frac{1}{2}mV_B^2 + \frac{1}{2} \times \frac{1}{2}m{R^2}.\frac{{V_B^2}}{{{R^2}}}\)

\(mV_A^2 = \frac{3}{4}mV_B^2\)

\({V_A} = \sqrt {gh} \)

\({V_B} = \sqrt {\frac{4}{3}gh} \)

\(\frac{{{V_A}}}{{{V_B}}} = \frac{{\sqrt {gh} }}{{\sqrt {\frac{{4gh}}{3}} }} = \sqrt {\frac{3}{4}} \)

Concept:

\(Strain=\frac{{{\rm{\Delta }}l}}{l} = \frac{{{{\epsilon_{xx}}} + {{\epsilon_{yy}}}}}{2} + \frac{{{{\epsilon_{xx}}} - {{\epsilon_{yy}}}}}{2}\cos 2\theta + \frac{{{\gamma _{xy}}}}{2}\sin 2\theta \)

Calculation:

Given:

ϵxx = 0.001, ϵyy = 0.002, γxy = 0.003, l = 5 units

Since, \(cos \theta=\frac{x}{l}\)

\(cos 36.86=\frac{4}{l}\)

l = 5 unit

tan θ = ¾ ⇒ θ = 36.86°

\(Strain=\frac{{{\rm{\Delta }}l}}{l} = \frac{{{{\epsilon_{xx}}} + {{\epsilon_{yy}}}}}{2} + \frac{{{{\epsilon_{xx}}} - {{\epsilon_{yy}}}}}{2}\cos 2\theta + \frac{{{\gamma _{xy}}}}{2}\sin 2\theta \)

\( = \frac{{0.001 + 0.002}}{2} + \frac{{0.001 - 0.002}}{2}\cos 2\theta + \frac{{0.003}}{2}\sin 2\theta \)

\(\frac{{{\rm{\Delta }}l}}{l} = 2.8 \times {10^{ - 3}}\)

Δl = 5 × 2.8 × 10^-3 = 0.014

Deformed length = l + Δl = 5 + 0.014 = 5.014 units

Concept:

The S-N curve on logS vs logN graph is given below,

Apply linear interpolation for S-N curve,

\(\frac{{{{\log }_{10}}σ - \log {σ _e}}}{{{{\log }_{10}}N - {{\log }_{10}}{{10}^6}}} = \frac{{\log 0.9;{σ _{ut}} - \log {σ _e}}}{{\log {{10}^3} - \log {{10}^6}}}{{;;;;;}} \ldots \left( 1 \right)\)

Where, 

σe = Endurance strength, N = lif of the component, σut = ultimate tensile strength of the component

Calculation:

Given:  

σ_u = 600 N/mm²

σ_en = 250 N/mm²

Co-ordinates of points are:

A → (log (0.8σ_u), 3)

B → (log S, 4)

C → (log σ_en, 6)

Now Slope AB = Slope AC

\( \Rightarrow \frac{{\log \left( {0.8{\sigma _u}} \right) - \log S}}{{3 - 4}} = \frac{{\log \left( {0.8{\sigma _u}} \right) - \log \left( {{\sigma _{en}}} \right)}}{{3 - 6}}\)

\( \Rightarrow \log S = \log \left( {0.8{\sigma _u}} \right) - \frac{{\log \left( {0.8{\sigma _u}} \right) - \log \left( {{\sigma _{en}}} \right)}}{3}\)

S = 386.19

Concept:

The change in length or axial displacement of the free is,

\(δ L=\frac{PL}{AE} \)

where, P = applied load, L = length of bar, A = cross-section area, E = Modulus of elasticity 

Here, the modulus of elasticity varies along the length.

Therefore, \(δ L = \mathop \smallint \limits_0^L \frac{P}{{AE}}dx\)

Calculation:

Given:

P = 10 kN = 10⁴ N, L = 1 m, A = 100 mm² = 10^-4 m², E = 100e^-x GPa = 10¹¹ e^-x Pa

 \(δ L = \mathop \smallint \limits_0^1 \frac{{{{10}^4}}}{{{{10}^{ - 4}} \times \left( {{{10}^{11}}{e^{ - x}}} \right)}}dx\)

⇒ \(δ L = \frac{{{{10}^4}}}{{{{10}^{ - 4}} \times {{10}^{11}}}}\mathop \smallint \limits_0^1 {e^x}dx = {10^{ - 3}} \times \left( {{e^x}} \right)_0^1\)

\(δ L = {10^{ - 3}}\left( {{e^1} - {e^0}} \right) = 1.71 \times {10^{ - 3}}m\)

∴ δL = 1.71 mm

Concept:

Let S is suffix for sun, P for planet and D for outer ring.

N_D = 0

N_S = -100 (Taking CW as positive)

T_S = 50, T_P = 25, T_D = 100

Now y + x = -100        ----(i)

\(y - x\left( {\frac{{50}}{{25}}} \right)\left( {\frac{{25}}{{100}}} \right) = 0 \Rightarrow y - \frac{x}{2} = 0\)        ----(ii)

From (i) & (ii)

\(x = \frac{{ - 200}}{3},;y = \frac{{ - 100}}{3}\)

\({N_P} = y - 2x = \frac{{ - 100}}{3} + \frac{{400}}{3} = 100\)

\({N_{arm}} = y = \frac{{ - 100}}{3}\)

\(\frac{{{N_P}}}{{{N_{arm}}}} = \frac{{100}}{{\frac{{ - 100}}{3}}} = - 3 = 3\left( {magnitude} \right)\)

Explanation:

Taking mass moment of inertia about 0

\(I = \frac{{m{L^2}}}{{12}} + M{\left( {\frac{L}{2} - \frac{L}{3}} \right)^2} + m \times {\left( {\frac{{2L}}{3}} \right)^2}\)

\( = \frac{{M{L^2}}}{{12}} + \frac{{M{L^2}}}{{36}} + \frac{{4M{L^2}}}{9} = \frac{{2M{L^2}}}{9}\)

Balancing Torque about 0

\(I\alpha = K \times \frac{{2L}}{3} \times \left( {\frac{{2L}}{3}\theta } \right) + K \times \frac{L}{3} \times \left( {\frac{L}{3}\theta } \right)\)

\(\frac{{2M{L^2}}}{9}\frac{{{d^2}\theta }}{{d{t^2}}} = \frac{{5k}}{{2M}} = \omega _n^2\theta \)

\({\omega _n} = \sqrt {\frac{{5k}}{{2M}}} \)

Concept:

\(\tan \alpha = \frac{4}{3} \Rightarrow \alpha = 53.13^\circ \)

\({V_{connecting;rod}} = 1 \times \cos \theta = \frac{4}{5}\frac{m}{s}\)

\({V_{connecting;rod}} = {w_{crank}} \times r\)

\(\frac{4}{5} = {w_{crank}} \times 3\)

\( \Rightarrow {w_{crank}} = \frac{4}{{15}} = 0.266\frac{{rad}}{s}\)

Alternate solution:

Applying Kennedy’s theorem at I₂₄

w₂ (I₂₄I₁₂) = V_A = V_B = 1

w₂ (I₂₄I₁₂) = 1

\( \Rightarrow {w_2} = \frac{1}{{{I_{24}}{I_{12}}}} = \frac{1}{{OB\tan \theta }} = \frac{1}{{5 \times \frac{3}{4}}} = 0.266\frac{{rad}}{s}\)

Concept:

For the deep drawing operation, diameter (D) of the blank needed to make a cup diameter (d) and height (h) in single pass,

\( ⇒ D = \sqrt {{d^2} + 4dh} ;;;;; \ldots \left( 1 \right)\)

The drawing load,

⇒ F = π × D× t × σy      … (2)

Work done by punch to draw = W = F × h      … (3)

Where,

t = Sheet metal thickness

σy = Yield strength of the sheet

h = Height of the cup

Calculation:

Given:

h = 10; d = 15mm

\( D = \sqrt {{d^2} + 4dh} \)

D² = d² + 4dh

D² = 15² + 4(15) (10)

D = 28.72 mm

Explanation:

Programming key letters in CNC

N – Sequence Number

G – Preparatory function

X – X axis designation

Y – Y axis designation

Z – Z axis designation

R – Radius designation

F – Feed rate designation

Important G codes

G00 – Rapid Transverse

G001 – Linear Interpolation

G02 – Circular Interpolation (CW)

G03 – Circular Interpolation (ACW)

G04 – Dwell

Concept:

Surface roughness

The machine element or parts retain the surface irregularities left after manufacturing.

The surface irregularities are usually understood in terms of surface finish, surface roughness, surface texture, or surface quality.

Centre-line average value:

\(CLA\left( {R_a} \right) = \frac{{{H_{max}}}}{4} \)

Calculation:

Given:

H_max = 20 μm

⇒ \(CLA\left( {R_a} \right) = \frac{{{H_{max}}}}{4} = \frac{{20}}{4} = 5;\mu m\)

Concept:

Let x are the units of P and y are the unit of Q.

Objective function Max. Z = 100 x + 80 y

Subjected to constraints

5x + 3y ≤ 150

x + y ≤ 40

Shaded region is the feasible region. Finding the corner points coordinates of B.

As B is intersection of

x + y = 40

5x + 3y = 150

On solving these coordinates of B is (15, 25)

Now, finding value of objective function at all corner points

Z(A) = Z(0, 40) = 3200

Z(B) = Z(15, 25) = 3500

Z(C) = Z(30, 0) = 3000

Z(O) = Z(0, 0) = 0

So Maximum value is at B (15, 25). The no. of units of P is 15

Concept:

In the SPT rule, jobs are sequenced in increasing order of their processing time i.e. minimum time first and maximum time in last.

Note: If two jobs have the same process time then job having less due date will be processed first.

Calculation:

According to shortest processing time, the flow sequence is

Average flow time \( = \frac{{124}}{4} = 31\)

Additional Information

Shortest Processing Time (SPT) Rule

Sequencing the jobs in a way that the job with least processing time is picked up first followed by the one with the next smallest processing time and so on is referred to as SPT sequencing and achieves the following objectives simultaneously :

• minimizing mean lateness
• minimizing mean lateness time
• minimizing mean waiting time
• minimizing the mean number of tasks waiting as in-process inventory

Earliest due date (EDD) rule

• Jobs are sequenced in order of increasing due dates.
• It minimizes maximum Tardiness.
• Tardiness and lateness depend on Due date.
• The number of tardy jobs is related to tardiness which is related to the due date.​

First Come First Serve (FCFS): Jobs are arranged as per they arrived

Critical Ratio (CR):

\(C.R = \frac{{Due;date}}{{Procesisng;Time}}\)

Jobs are arranged as per the ascending order of Critical Ratio.

Concept:

Compulsory approach = \(\sqrt{d(D~-~d)}\)

Length of travel = Length of block + Approach + over travel

(Length of travel)min = Length of block + compulsory approach

where,

d = depth of cut, D = Diameter of cutter, X = Compulsory approach

Calculation:

Given:

D = 34 mm, depth of cut (d) = 2 mm, length = 200 mm

Approach = Over travel (given)

\( = \sqrt {d\left( {D - d} \right)} = \sqrt {2\left( {34 - 2} \right)} = 8mm\)

Length of tool travel =Block Length + Over travel + Approach = 200 + 8 + 8 = 216 mm

\(Estimated;time = \frac{{length;of;tool;travel}}{{feed}} = \frac{{216}}{{18}} = 12;mm;\)

Concept:

Bernoulli's Equation

\(\frac{{{P_1}}}{{\rho g}} + \frac{{v_1^2}}{{2g}} + {h_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{v_2^2}}{{2g}} + {h_2}\)

Calculation:

Given:

Applying Bernoulli’s equation between (1) and (2).

v₁ = 0.5 m/s

h₁ = 200 mm

h₂ = 0

P₁ = P₂ = P_atm

\(\frac{{{P_1}}}{{\rho g}} + \frac{{v_1^2}}{{2g}} + {h_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{v_2^2}}{{2g}} + {h_2}\)

\(\frac{{{{0.5}^2}}}{{2 \times 9.8}} + \frac{{200}}{{1000}} = \frac{{v_2^2}}{{2 \times 9.8}} + 0\)

⇒ v₂ = 2.042 m/s

Explanation:

Let speed is x m/min

\(xT_1^{0.1} = 150\)

\( \Rightarrow {T_1}={\left( {\frac{{150}}{x}} \right)^{\frac{1}{{0.1}}}};and;{T_2} = {\left( {\frac{{300}}{x}} \right)^{\frac{1}{{0.3}}}}\)

Now \({T_2} > {T_1} \Rightarrow {\left( {\frac{{300}}{x}} \right)^{\frac{1}{{0.3}}}} > {\left( {\frac{{150}}{x}} \right)^{\frac{1}{{0.1}}}}\)

For solving, \({\left( {\frac{{300}}{x}} \right)^{\frac{1}{{0.3}}}} = {\left( {\frac{{150}}{x}} \right)^{\frac{1}{{0.1}}}}\)

\({\left( {\frac{{300}}{x}} \right)} = {\left( {\frac{{150}}{x}} \right)^3}\)

\(\frac{{{x^{3}}}}{{{x}}} = \frac{{{{150}^{3}}}}{{{{300}}}} \Rightarrow x = 106.066\frac{m}{{min}}\)