GATE ME 2020 Official Paper: Shift 1 (Previous Year Paper)

The correct answer is Option 3 i.e. Crocodile

"Crocodile tears" is a phrase that means - to put on an insincere show of sorrow. When someone feigns sadness they 'cry crocodile tears' a phrase that comes from an old myth that the animals cry while eating. 

Complete sentence: He is known for his unscrupulous ways. He always sheds Crocodile tears to deceive people.

The correct answer is Option 1 i.e. more fast

When two qualities of the same noun are compared, more + positive degree adjective is used. In this sentence, we are comparing the two qualities of the England fast bowler i.e. 'fast' and 'accurate' for such comparison we will use comparative degree ( more + positive degree adjective ).

∴ Use ‘more fast’ rather than ‘faster’.

Let's see an example:

• Incorrect- She is prettier than wise.
• Correct- She is more pretty than wise.

Complete Sentence: Jofra Archer, the England fast bowler, is more fast than accurate.

The correct answer is option 3 i.e. Growth.

From the first part of the analogy, it becomes clear that a noun is needed after its corresponding Verb. 'Build' is a verb and it is related to Building (noun) that means a roofed and walled structure built for permanent use (as for a dwelling) or the art or business of assembling materials into a structure. Similarly, Grow is also a verb and it is related to 'Growth' i.e. the process of growing.

The rest of the options are verbs and are not an appropriate fit for our analogy.

The correct answer is option 2 i.e. despite what I have said

“Having said that” is a transitional phrase that has become more and more common in spoken language. When people say, “Having said that” it is a signal that they are going to say something which will contrast or disagree with what they said a moment ago.

Let's see an example to develop more clarity:

• Their work has been fairly good. Having said that, I still think there's room for improvement.

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

• In general, If Then [x] = n (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Example:

Calculation:

∀ x ϵ (-∞, ∞)

Also, y = [x] for x ϵ [1, 4]

For x ϵ [1, 2), [x] = 1 i.e. y = 1

For x ϵ [2, 3), [x] = 2 i.e. y = 2

For x ϵ [3, 4), [x] = 3 i.e. y = 3

We can plot it as:

∴ Area under y for xϵ [1, 4] = (1 × 1) + (1 × 2) + (1 × 3) = 6

The correct answer is Option 4 i.e. ​Funds raised through voluntary contributions on web-based platforms.

After reading the above paragraph one can easily comprehend that a large number of people voluntarily (proceeding from the will or from one's own choice or consent) invested in small amounts for the crowd-funding, now let's go through each option one by one.

• Option 1 states that- "Funds raised through unwilling contributions on web-based platforms.", this option is incorrect because the contribution was made willingly in small amounts by a large number of peoples.
• Option 2 states that- "Funds raised through large contributions on web-based platforms.", this option is incorrect, it has been clearly mentioned in the passage that a large number of people invest small amounts. (Large contributions are mentioned nowhere in the paragraph)
• Option 3 states that- "Funds raised through coerced contributions on web-based platforms.", Coerced means to compel to an act or choice or to achieve by force or threat, people were not forced to make a contribution in this crowd-funding they willingly invested small amounts. Hence, this option is incorrect as well.

Explanation:

All different possible codes using α and β are:

αβ, βα, αα, ββ

Given:

P is coded as αα

Q is coded as αβ

∴ R and S can be coded as βα and ββ respectively.

Concepts:

Let us consider sequence a1, a2, a3 …. an is an G.P.

• Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
• nth term of the G.P. is an = arn−1
• Sum of n terms = s = \(\frac{{a;\left( {{r^n} - 1} \right)}}{{r - 1}}\); where r >1
• Sum of n terms = s = \(\frac{{a;\left( {1 - {r^n}} \right)}}{{1 - r}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{}}\frac{{\rm{a}}}{{1{\rm{}} - {\rm{r}}}}{\rm{}}\); |r| < 1

Where a is 1^st term and r is common ratio.

Explanation:

8 + 88 + 888 + 8888 + ….

Taking 8 common from each term

⇒ 8 (1 + 11 + 111 + 1111 + ….)

Multiplying and dividing the equation by 9

⇒ \(\frac{8}{9}{\rm{;}}\left[ {9 + 99 + 999 + 9999{\rm{;}} + {\rm{;}} \ldots .} \right]\)

⇒ \(\frac{8}{9}{\rm{;}}\left[ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + \left( {10000 - 1} \right) + {\rm{;}} \ldots .} \right]\)

⇒ \(\frac{8}{9}{\rm{;}}\left[ {\left( {10 + {{10}^2} + {{10}^3} + {{10}^4} + {\rm{;}} \ldots + {{10}^{\rm{n}}}} \right) - \left( {1 + 1 + 1 + 1 + {\rm{;}} \ldots .{\rm{;}} + {\rm{;n}}} \right)} \right]\)

The above equation is in geometric progression (GP) with a ratio of 10.

∵ the sum of G.P of with first term a, geometric ratio r and number of terms n is given by \(\frac{{{\bf{a}}\left( {{{\bf{r}}^{\bf{n}}} - 1} \right)}}{{{\bf{r}} - 1}}\)

⇒ \({\rm{;}}\frac{8}{9}{\rm{;}}\left[ {\left{ {\frac{{10\left( {{{10}^{\rm{n}}} - 1} \right)}}{9}} \right} - \left( {\rm{n}} \right)} \right]\)

⇒ \(;\frac{{80\left( {{{10}^{\bf{n}}} - 1} \right)}}{{81}} - \frac{{8{\bf{n}}}}{9}\) 

Trick:

You can also use the formula to find the sum of such problems.

\(\frac{{\rm{N}}}{9};\left[ {\left{ {\frac{{10\left( {{{10}^n} - 1} \right)}}{9}} \right} - \left( n \right)} \right]\)

Here N is the single-digit no which is repeating

For example: In this question, N = 8

Given y = x^m where m > 1 & y = x^1/m and 0 ≤ x ≤ 1

Let us consider m = 2

In that case:

y = x² graph can be plotted as

[Slope: \(\frac{dy}{dx}=2x ⇒ x \uparrow \rightarrow \frac{dy}{dx}\uparrow\)]

Also y = x^0.5 is represented in graph as:

[Slope: \(\frac{dy}{dx}=0.5x^{-0.5} ⇒ x \uparrow \rightarrow \frac{dy}{dx}\downarrow\)]

∴ y = x^m & y = x^1/m is correctly represented by option 1) i.e.

• y = xm ⇒ increasing slope
• y = x1/m ⇒ decreasing slope

Concept:

Average success rate of one school = Total no of passed students/Total no. of students who appeared

The average success rates of these four schools = Sum of Average success rate of all four schools/Total no. of schools

Therefore, The average success rates of these four schools will be 

\( \frac{{\sum \frac{{Total;no.;of;passed;students}}{{Total;no.;of;students}}}}{{4;}}; \times 100;\)

Calculation:

The average success rates of these four schools will be 

\(\Rightarrow \frac{{\left{ {\left( {\frac{{280}}{{500}}} \right) + \left( {\frac{{330}}{{600}}} \right) + \left( {\frac{{455}}{{700}}} \right) + \left( {\frac{{240}}{{400}}} \right)} \right}}}{4};{\rm{;}} \times 100=\frac{0.56 + 0.55 + 0.65+0.6}{4} \times 100=59 %\)

∴ The average success rates of these four schools = 59.0%

EXPLANATION:

Let A, B and C be the matrices a, b, and c be scalars and the sizes of matrices are such that the operations can be performed then

The multiplication of real-valued square matrices of the same dimension is:

Associative, which means that for any three matrices A, B, and C of the same dimension, the following holds true: (AB)C = A(BC)

Not Commutative, which means that the order of the matrices matters. In other words, AB is not necessarily equal to BA, unless one or both of the matrices are diagonal or scalar matrices.

Not always positive definite, since the determinant of the product of two matrices is equal to the product of the determinants of the matrices, and the determinant of a matrix is positive if and only if the matrix is invertible. Therefore, the product of two invertible matrices is invertible, and hence positive definite. However, the product of two non-invertible matrices may not be invertible, and hence not positive definite.

The order of the multiplication matters, so it is not always possible to compute the matrices. That is, AB is not necessarily equal to BA, unless one or both of the matrices are diagonal or scalar matrices.

Additional Information 

Properties of Matrix addition and scalar multiplication:

Commutative Property of addition

• A + B = B + A

Associative Property of addition

• A + (B + C) = (A + B) + C
• A + O = O + A Where O is the appropriate zero matrix

Distributive Property of addition

• c(A + B) = cA + cB
• (a + b)C = aC + bC
• (ab)C = a(bc)

Concept:

\(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 - {e^{ - c\left( {1 - x} \right)}}}}{{1 - {xe^{ - c\left( {1 - x} \right)}}}}} \right)\) = 0/0 form

Hence apply L – Hospital rule

Calculation: 

Now,

By applying L – Hospital rule, i.e.

By differentiating both the numerator and denominator

\(\frac{{\frac{d}{{dx}}\left( {1 - {e^{ - c(1 - x)}}} \right)}}{{\frac{d}{{dx}}\left( {1 - x{e^{ - c(1 - x)}}} \right)}} = \frac{{\left( { - c{e^{ - c(1 - x)}}} \right)}}{{\left( { - cx{e^{ - c(1 - x)}} - {e^{ - c(1 - x)}}} \right)}} = \frac{{\left( {c{e^{ - c(1 - x)}}} \right)}}{{\left( {cx{e^{ - c(1 - x)}} + {e^{ - c(1 - x)}}} \right)}} = \frac{c}{{cx + 1}}\)

\(\therefore \mathop {\lim }\limits_{x \to 1} \frac{c}{{ 1 + cx}} = \frac{c}{{1 + c}}\)

Concept:

We know that

L{f(t)} = F(s) then f(t) = L^-1F(s)

Calculation:

\({\rm{L}}\left( {\sin {\rm{at}}} \right) = \frac{{\rm{a}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}}}\)

Similarly,

\({\rm{L}}\left( {\sin {\rm{\omega t}}} \right) = \frac{{\rm{\omega }}}{{{{\rm{s}}^2} + {{\rm{\omega }}^2}}}\)

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

Concept:

The complex function f(z) = u (x, y) + iv (x, y) is to be analytic if it satisfy the following two conditions of Cauchy-Reiman theorem.

• \(\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}};;and;\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)
• \(\frac{{\partial u}}{{\partial x}},\frac{{\partial u}}{{\partial y}},\frac{{\partial v}}{{\partial x}},\frac{{\partial v}}{{\partial y}};{\rm{are;continuous;function;of;x;and;y}}.\)

f(z) = log z

\(\frac{{\partial }}{{\partial z}}f(z) = \frac{1}{z}\)

Here, the function f(z) is analytic except z = 0. Since the function is not defined for these two values.

But in question it is asked at all points hence f(z) = log z is not analytic at all points.

If at a joint 3 members meet out of which 2 are collinear then force in 3^rd member will be zero provided there is no load/reaction at that joint.

∴ F_GC = 0, F_FB = 0 and F_HD = 0

i.e. forces in the member GC, FB, HD is equal to zero

Now, considering joint E:

∵ F_EH = R_E

There is no horizontal force.

∴ F_DE = 0

Also considering joint D, we can say that force in member CD = 0

Concept:

In vibration system, transmissibility is given by

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

Where F_T = maximum force transmitted to the ground

\(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi \omega }{{{\omega }_{n}}} \right)}^{2}}}}\)

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

K = spring stiffness, m = mass

Calculation:

Given, Harmonic excitation force: F(t) = F₀cos(ωt)

F_T = F_O

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

∴ Transmissibility, ϵ = 1

Also, \(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{w}{{{w}_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{w}{{{w}_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi w}{{{w}_{n}}} \right)}^{2}}}}=1\)

Let \(\frac{\omega }{{{\omega }_{n}}}=x\)

So, 1 + (2ξ x)² = (1 – x²)² + (2ξ x)²

(1 – x²)² = 1

1 – x² = ± 1

Taking +ve sign x² = 0

∴ Taking –ve sign

x² = 2

Putting value of x:

(ω/ω_n)² = 2

\(\frac{\omega }{{{\omega }_{n}}}=\sqrt{2}\)

∴ ω = √2ω_n

∵ Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

\(\omega =\sqrt{\frac{2K}{m}}\)

Concept:

General Mohr Circle for biaxial stress condition:

Principal stresses are given by

\({{\sigma }_{1,~2}}=\frac{{{\sigma }_{x}}+{{\sigma }_{y}}}{2}~\pm ~\sqrt{{{\left( \frac{{{\sigma }_{x}}-{{\sigma }_{y}}}{2} \right)}^{2}}+\tau _{xy}^{2}}\)

Radius of circle (Mohr’s circle) \(=\sqrt{{{\left( \frac{{{\sigma }_{x}}-{{\sigma }_{y}}}{2} \right)}^{2}}+\tau _{xy}^{2}}\)

Calculation:

Given state of stress is represented by

∴ σ_x = 10 MPa. σ_y = 10 MPa τ_xy = 0

For this the Mohr Circle will be a point:

 

∴ σ₁ = σ₂ = 10 MPa

Radius of circle (Mohr’s circle) will be zero.

The co-ordinate of the centre of Mohr’s circle on σ-axis at a distance of 10 units from the origin.

Concept:

According to Grashof’s Law:

(1) S + L ≤ P + Q (Class I mechanism)

There will be at least one complete revolution between the two links.

• If the shortest link is fixed: Double Crank Mechanism
• If any of the adjacent links of the shortest link if fixed: Crank rocker mechanism
• If the link opposite to shortest link is fixed: Double Rocker Mechanism

(2) S + L > P + Q (Class II mechanism)

• Only a double rocker mechanism is possible.

Calculation:

Given:

For getting a crank rocker mechanism, any of the adjacent links of the shortest link should be fixed and S + L ≤ P + Q.

Here fixed link is PS = 400 mm which is not the shortest link.

Let's start with incorrect options first.

1. Now consider the length of PQ = 350 mm

• S + L = 300 + 600 = 900
• P + Q = 400 + 350 = 750
• S + L > P + Q.
• Grashof’s Law not satisfied

2. Now consider the length of PQ = 300 mm

• S + L = 300 + 600 = 900
• P + Q = 400 + 300 = 700
• S + L > P + Q
• Grashof’s Law not satisfied

3. Now consider the length of PQ = 200 mm

• S + L = 200 + 600 = 800
• P + Q = 400 + 300 = 700
• S + L > P + Q
• Grashof’s Law not satisfied

4. Now consider the length of PQ = 80 mm

• S + L = 80 + 600 = 680
• P + Q = 400 + 300 = 700
• S + L < P + Q
• Grashof’s Law is satisfied
• Shortest link is PQ (80 mm). Link adjacent to shortest link is fixed i.e. PS. So it will give crank rocker mechanism.

The resultant force on helix gear i.e. F_N can be resolved as:

ϕ = pressure angle

ψ = helix angle

i) F_T i.e. Tangential force

ii) F_R i.e. Radial force

iii) F_a i.e. Axial force

• F_a induces axial stress in the shaft
• F_T and F_R are responsible for bending of shaft in two plane.
• Due to F_T tangential shear stress will be induced in the shaft.

Explanation:

• At 1539°C, iron cools from liquid to solid, it is in BCC structure and it is in δ-Ferrite stage.
• After cooling in temperature range from 1400°C to 910°C it contains FCC structure which is also known as γ-Austenite.
• After again cooling below 910°C, α-ferrite phase will start.
• 768°C temperature line in diagram is known as Curie point temp line, because it distinguishes the metal property as Magnetic and non-magnetic.
• Above 768°C, the metal is non-magnetic and below 768°C the metal is magnetic.

Concept:

Heat treatment:

• Heat treatment operation can be defined as heating a metal or alloy to various definite temperatures, holding these for the various time durations and cooling at various rates to get the desired grain structure and property in the metal or alloy.

 

Concept:

A grinding wheel consists of the abrasive that does the cutting, and the bond that holds the abrasive particles together.

A standard marking system is used to specify and identify grinding wheels.

The following is the sequence of arrangement:

Abrasive type – Grain size – Grade of bond – Structure – Bond type

 

• Abrasive type: ‘A’ for aluminium oxide, ‘C’ for silicon carbide
• Silicon Carbide: It is less hard than diamond and less tough than aluminium oxide. It is used for grinding of material of low tensile strength like cemented carbide, stone and ceramic, grey cast iron, copper, brass, bronze, aluminium, vulcanized rubber, etc.
• Aluminium Oxide: Aluminium oxide is tough and fracture resistant. It is preferred for grinding of materials of higher tensile strengths like steel; high carbon and high-speed steel and tough bronze.
• Grain size: They are indicated by a number ranging from 10 (coarse) up to 600 (very fine)
• Grade of bond: The grades range from ‘A’ indicating light or ‘soft’ bond to ‘Z’ indicating a firm or ‘hard’ bond
• A soft bond permits the grains to break away more readily than a hard one and should be used where the abrasive becomes rapidly dulled or blunt, i.e., when grinding hard materials.
• A hard bond retains the abrasive grains longer and should be used when grinding soft materials.
• Structure: This structure is indicated by a number from 1 to 12. The higher numbers indicate a progressively more open structure. It is the optional specification.
• Bond type: V – Vitrified, S – Silicate, B – Resinoid, R – Rubber, E – Shellac, O – Oxychloride

Solution:

Rough cutting of brass, silicon carbide should be used as abrasive, therefore options with Al₂O₃ as abrasive are eliminated.

Also, brass is a soft material so, the hard wheel is required for its grinding ‘J’ lies in (A-H) range i.e. for the soft wheel so, it can’t be used whereas Q represents hard wheel.

So, we will go for grinding wheel with specification C30Q12V

Concept:

Crash time is the minimum activity duration to which an activity can be compressed by increasing the resources and by increasing the direct cost.

The slope of the line gives the amount of increase in the direct cost per unit time for crashing an activity.

Concept:

Cost Slope: The direct cost curve is a curve that can be approximated by a straight line, depending upon the flatness of the curve. The slope of this straight is the cost slope. It is very helpful in the project cost analysis.

\({{\rm{C}}_{\rm{s}}} = \frac{{{\rm{Crash;Cost}} - {\rm{Normal;Cost}}}}{{{\rm{Normal;Time}} - {\rm{Crash;Time}}}}\)

The activity with minimum cost slope should be crashed first, as it results in optimum crashing of the project.

Concept:

Froude number is the ratio of inertial force to the gravitation force.

\(\text{Froude }!!~!!\text{ number}=\text{ }!!~!!\text{ }\sqrt{\frac{\text{Inertia force }!!~!!\text{ }}{\text{Gravitational force }!!~!!\text{ }}}\)

Froude number has the following applications:

• Used in cases of river flows, open-channel flows, spillways, surface wave motion created by boats
• It can be used for flow classification

Important Points

\(\text{Reynold's number}=\frac{\text{Inertia force}}{\text{Viscous force}}\)

\(\text{Euler }!!~!!\text{ number}=\sqrt{\frac{\text{Inertia force}}{\text{Pressure force}}}\)

\(\text{Weber }!!~!!\text{ number}=\sqrt{\frac{\text{Inertia force}}{\text{Surface Tension force}}}\)

\(\text{Mach }!!~!!\text{ number}=\sqrt{\frac{\text{Inertia force}}{\text{Compressibility force}}}\)

Different dimensionless numbers:

• \(Weber;Number = \frac{{Inertia;force}}{{Surface;tension;force}}\)
• \(Reynold's;number = \frac{{Inertia;force}}{{Viscous;force}}\)
• \(Mach;number = \frac{{Inertia;force}}{{Elastic;force}}\)
• \(Euler's;number = \frac{{Inertia;force}}{{Pressure;force}}\)
• \(Froude;number = \frac{{Inertia;force}}{{Gravity;force}}\)
• Biot number  → Ratio of internal thermal resistance to boundary layer thermal resistance
• Grashof number  → Ratio of buoyancy to viscous force
• Prandtl number  → Ratio of momentum diffusivity (ν) and thermal diffusivity (α). Pr =ν/α
• Reynolds number  → Ratio of inertia force to viscous force
• Nusselt number → Ratio of Convective heat transfer to Conduction heat transfer
• \(Nu = \frac{{h{L_c}}}{k}\)

Concept:

If velocity is given as \({\rm{\vec V}} = {\rm{u\hat i}} + {\rm{v\hat j}} + {\rm{w\hat k}}\)

Then for incompressible flow continuity equation has to be satisfied,

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

Calculation:

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

u = 2(x² – y²)

v = v

w = 3

Putting the values in continuity equation

\(\frac{{2\partial \left( {{{\rm{x}}^2} - {{\rm{y}}^2}} \right)}}{{\partial {\rm{x}}}} + \frac{{\partial {\rm{v}}}}{{\partial {\rm{y}}}} + \frac{{\partial \left( 3 \right)}}{{\partial {\rm{z}}}} = 0\)

⇒ \(4{\rm{x}} + {\rm{;}}\frac{{\partial {\rm{v}}}}{{\partial {\rm{y}}}} + 0 = 0\)

⇒ \(\frac{{\partial {\rm{v}}}}{{\partial {\rm{y}}}} = {\rm{;}} - 4{\rm{x}}\)

⇒ v = -4xy + f(x,z)

∴ Out of the given options, option 2 matches our answer.

Explanation:

Joule – Thomson coefficient:- When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

\(\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}\)

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Concept:

Combined equations of first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = C_vdT

dh = C_pdT

C_v = specific heat at constant volume, C_p = specific heat at constant pressure

Calculation:

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = C_vdT

∴ 1^st equation becomes: Tds = C_vdT

\({\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}\)

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = C_pdT

Tds = C_pdT

\({\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}\)

Therefore, ratio of slope of constant pressure and constant volume lines at point of intersection:

\(\frac{{{{\left( {\frac{{dT}}{{ds}}} \right)}_{p = c}}}}{{{{\left( {\frac{{dT}}{{ds}}} \right)}_{v = c}}}} = \frac{{\frac{T}{{cp}}}}{{\frac{T}{{{c_v}}}}} = \frac{{{C_v}}}{{{C_p}}}\)

Concept:

\(\overrightarrow {{\rm{A}}.} \left( {{\rm{\vec B}} \times {\rm{\vec C}}} \right)\) is the scalar triple product which is defined as \(\det \left( {\vec A;\vec B;\vec C} \right);\)

Calculation:

Given:

\(\vec A = 2\hat j - 3\hat k,\vec B = - 2\hat i + \hat k;and;\hat C = 3\hat i - \hat j\)

⇒ \(\overrightarrow {{\rm{A}}.} \left( {{\rm{\vec B}} \times {\rm{\vec C}}} \right) = {\rm}\left| {\begin{array}{*{20}{c}} {{\rm{;;;}}0}&{{\rm{;;}}2}&{ - 3}\ { - 2}&{{\rm{;;}}0}&{{\rm{;;;}}1}\ {{\rm{;;;}}3}&{ - 1}&{{\rm{;;;;}}0} \end{array}} \right|\)

⇒ \(\overrightarrow {A} \left( {\vec B \times \vec C} \right) = \left( { - 2} \right)\left( { - 3} \right){\rm}-{\rm}3\left( 2 \right)\)           

⇒ \(\overrightarrow {{\rm{A}}.} \left( {{\rm{\vec B}} \times {\rm{\vec C}}} \right) = 0\)

∴ \(\overrightarrow {{\rm{A}}.} \left( {{\rm{\vec B}} \times {\rm{\vec C}}} \right) + 6 = 6\)

Concept:

ω₁ = maximum angular speed, ω₂ = minimum angular speed,

ΔKE = Change in Kinetic energy, I_f = Moment of Inertia of Flywheel

Therefore, \({\rm{\Delta KE;}} = \frac{1}{2}; \times {I_f} \times \left( {\omega _1^2 - \omega _2^2} \right)\)

Calculation:

ω_{1 =} 110 rad/sec, ω₂ = 100 rad/sec, ΔKE = 1.05 kJ = 1050 J

1050 = \(\frac{1}{2}; \times {I_f}; \times \left( {{{110}^2} - {{100}^2};} \right)\)

\({I_f} = 1;kg.{m^2}\)

Concept:

A balanced rigid disc which has four identical masses, A, B, C and D

These masses balance each other in the system, A balance C and vice versa in X-direction while B balances D and vice versa in Y-direction.

If any of the mass get detached, there will be an unbalance force

∴ Unbalance force = mrω²

Where, m is the mass, r is the radius where mass is lying, and ω is the angular velocity of the mass.

Calculation:

Given, m = (mass in kg) = 10 g = 10 × 10^-3 kg

r(radius) = 100 mm = 0.1 m, ω (Angular speed) = 10 rad/s

∴ Unbalance force = 10 × 10^-3 × 0.1 × 10² = 0.1 N

Concept:

Clearance while using a punch & die in metal sheet formation is given by,

\(C = 0.0032 + \sqrt \tau\)

Where τ is the shear strength of sheet metal  

\(\therefore C \propto \sqrt \tau\)

τ ∝ HRC

Where HRC (Hardness Rockwell C)

Thus, \(C \propto \sqrt {HRC}\)

Calculation:

Given, Hardness of sheet in first case = 250 HRC, Clearance provided for sheet = 1 mm, Hardness of sheet in second case = 400 HRC

\(\frac{{{C_2}}}{{{C_1}}} = \frac{{\sqrt {HR{C_2}} }}{{\sqrt {HR{C_1}} }}\)

\({C_2} = {C_1} \times \sqrt {\frac{{HR{C_2}}}{{HR{C_1}}}} = \left[ {1 \times \sqrt {\frac{{400}}{{250}}} } \right]\)

C₂ = 1.265 mm

Hence, clearance provided in second case = 1.265 mm

There are 5 men and 3 women. Company is hiring to fill four managerial vacancies.

Probability that at least one woman will be selected = 1 - Probability that no woman is selected

Probability that no woman is selected = \(\frac{^5C_4 \times ^3C_0}{^8C_4}=\frac{5\times 1}{70}=\frac{5}{70}\)

Probability that at least one woman will be selected = 1 - Probability that no woman is selected = 1 - (5/70) = 65/70 = 0.93

For perfect intercooling in compressor the intermediate pressure for two stages is given by

For perfect intercooling T₃ = T₁

\({P_i} = \sqrt {{P_2}{P_1}}\)

Where P₁ = Intake pressure and P₂ is the delivery pressure

r_p = pressure ratio

\({r_p} = \frac{{{P_2}}}{{{P_1}}}\)

⇒ P₂ = r_p × P₁

Intermediate pressure is given by:

\({p_i} = \sqrt {{p_2}{p_1}} ;\left( {\frac{{{T_3}}}{{{T_1}}}} \right)\frac{n}{{2\left( {n - 1} \right)}}\)

Calculation:

Given,  P₁ = 100 kPa,  r_p = 6

P₂ = 6 × 100 = 600 kPa

Intermediate pressure \(= \sqrt {100 \times 600} = 244.95;kPa\)

Concept:

For counter flow heat exchanger:

Log mean temperature difference is given by:

\(LMTD = \frac{{{\rm{\Delta }}{T_i}; - ;{\rm{\Delta }}{T_e}}}{{\ln \left( {\frac{{{\rm{\Delta }}Ti}}{{{\rm{\Delta }}Te}}} \right)}}\)

Where ΔT_i = T_hi – T_ce

ΔT_e = T_he - T_ci

T_hi & T_ci = Temperature at which hot or and cold water enters respectively

T_he & T_ce = Temperature at which hot air & cold water exits respectively

Calculation:

Given, T_hi = 102°C, T_he = 65°C

T_ci = 25°C, T_ce = 42°C

ΔT_i = T_hi – T_ce = 102 – 42 = 60°C

ΔT_e = T_he - T_ci = 65 – 25 = 40°C

Log mean temperature difference is given by

\({L_{MTD}} = \frac{{{\rm{\Delta }}{T_i} - {\rm{\Delta }}{T_e}}}{{\ln \frac{{{\rm{\Delta }}Ti}}{{{\rm{\Delta }}Te}}}}\)

\(\therefore LMTD = \frac{{60 - 40}}{{\ln \left( {\frac{{60}}{{40}}} \right)}} = 49.3^\circ C\)

Concept:

\({\rm{Number;of;interval}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{;}}\)

Where,

b is the upper limit, a is the lower limit, h is the step size

According to Simpson's \(\frac{1}{3}\) rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{3}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_2} + {{\rm{y}}_4} + {\rm{;}} \ldots } \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3} + {\rm{;}} \ldots } \right)} \right]\)

Calculation:

Given:

f(x) = x|x|

a = -1, b = 1.4, h = 0.6

\({\rm{Number;of;interval}} = \frac{{1.4 - \left( { - 1} \right)}}{{0.6}} = 4\)

 

According to Simpson's 1/3^rd rule

\(\mathop \smallint \limits_{ - 1}^{1.4} x\left| x \right|dx = \frac{{\rm{h}}}{3}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_4} + 2\left( {{{\rm{y}}_2}} \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3}} \right)} \right]\)

\(\mathop \smallint \limits_{ - 1}^{1.4} x\left| x \right|dx = \frac{{0.6}}{3}\left[ { - 1 + 1.96 + 2\left( {0.04} \right) + 4\left( { - 0.16 + 0.64} \right)} \right]{\rm{;}}\)

\(\therefore \mathop \smallint \limits_{ - 1}^{1.4} x\left| x \right|dx = 0.592{\rm{;}}\)

Concept:

According to the Gauss Divergence theorem:

\(\iint {\overrightarrow {f.} \overrightarrow {ds} } = \iiint {div;\overrightarrow {f;} dv} \)

where, \(div~ \vec f = \frac{{\partial {f_1}}}{{\partial x}} + \frac{{\partial {f_2}}}{{\partial y}} + \frac{{\partial {f_3}}}{{\partial z}}\)

The given vector field is 

\(\vec f\left( {x,y,z} \right) = \frac{x}{{{{\left[ {{x^2} + {y^2} + {z^2}} \right]}^{\frac{3}{2}}}}}\hat i + \frac{y}{{{{\left[ {{x^2} + {y^2} + {z^2}} \right]}^{\frac{3}{2}}}}}\hat j + \frac{z}{{{{\left[ {{x^2} + {y^2} + {z^2}} \right]}^{\frac{3}{2}}}}}\hat k\)

Now, \({\rm{Vector;field}} = {\rm{\vec f}}\left( {x,y,z} \right) = \frac{{x\hat i + y\hat j + zk}}{{{{\left[ {{x^2} + {y^2} + {z^2}} \right]}^{\frac{3}{2}}};}}\)

Since position vector \(\vec r\)

\(\therefore \vec r = x\widehat i + y\widehat j + z\widehat k\)

\(\begin{array}{l} \Rightarrow {\rm{;r}} = \sqrt {{x^2} + {y^2} + {z^2}} \ \end{array}\)

\(\Rightarrow {r^2} = {x^2} + {y^2} + {z^2}\)

Calculation:

\({\rm{Vector;field}} = {\rm{\vec f}}\left( {x,y,z} \right) = \frac{{x\hat i + y\hat j + zk}}{{{{\left[ {{x^2} + {y^2} + {z^2}} \right]}^{\frac{3}{2}}};}}= \frac{{x\hat i + y\hat j + zk}}{{{{(r^2)}^{\frac{3}{2}}};}}=\frac{{x\hat i + y\hat j + zk}}{{{{(r^3)}}}}\)

\(\therefore div,\vec f = \frac{\partial }{{\partial x}}\left[ {\frac{x}{{{r^3}}}} \right] + \frac{\partial }{{\partial y}}\left[ {\frac{y}{{{r^3}}}} \right] + \frac{\partial }{{\partial z}}\left[ {\frac{z}{{{r^3}}}} \right]\)

\(\therefore div,\vec f = x\left( { - 3} \right){r^{ - 4}}\left( {\frac{{\partial r}}{{\partial x}}} \right) + \frac{1}{{{r^3}}} + y\left( { - 3} \right){r^{ - 4}}\left( {\frac{\partial r }{{\partial y}}} \right) + z\left( { - 3} \right){r^{ - 4}}\left( {\frac{{\partial r}}{{\partial z}}} \right) + \frac{1}{{{r^3}}}\)

\(\therefore div;\vec f = \left( { - 3} \right)x{r^{ - 4}}\left( {\frac{x}{r}} \right) + \left( { - 3} \right)y{r^{ - 4}}\left( {\frac{y}{r}} \right) + \left( { - 3} \right)z{r^{ - 4}}\left( {\frac{z}{r}} \right) + \frac{3}{{{r^3}}}\)

\(\therefore div,\vec f = \frac{{ - 3}}{{{r^5}}}\left[ {{x^2} + {y^2} + {z^2}} \right] + \frac{3}{{{r^3}}} = \frac{{ - 3;}}{{{r^5}}}\left[ {{r^2}} \right] + \frac{3}{{{r^3}}}\)

\(\therefore div,\vec f = \frac{{ - 3}}{{{r^3}}} + \frac{3}{{{r^3}}} = 0\)

Now, according to the theorem

\(\iint {\overrightarrow {f.} \overrightarrow {ds} } = \iiint {div;\overrightarrow {f;} dv} =\iiint {div;\overrightarrow {0} dv} =0\)

\(\iint {\overrightarrow {f.} \overrightarrow {ds} } = 0\)

Concept:

According to Maximum Shear Stress theory, failure occurs when the maximum shear stress from a combination of principal stresses equals or exceeds the value of its shear strength.

Calculation:

Given:

Shear strength of the material = 20 MPa

Side of Square cross-section = 4 cm = 40 mm

Diameter fo circular cross-section = 4 cm = 40 mm

Case 1: Tensile load

Given:

Load = 80 kN = 80 × 10³ N

\({\rm{Normal;stress}}\left( {\rm{\sigma }} \right) = \frac{{{\rm{load}}}}{{{\rm{area}}}} = \frac{{80 \times {{10}^3}}}{{{{40}^2}}} = 50;MPa\)

\({\rm{Maximum;Shear;Stress}}\left( {{\tau _{max}}} \right) = \frac{\sigma }{2} = \frac{{50}}{2}=25~MPa\)

∴ Maximum Shear stress (τ_max ) = 25 MPa > 20 MPa

∴ The specimen will fail.

Case 2: Compressive load

Given:

Load = 80 kN = 80 × 10³ N

\({\rm{Normal;stress}}\left( {\rm{\sigma }} \right) = \frac{{{\rm{load}}}}{{{\rm{area}}}} = \frac{{80 \times {{10}^3}}}{{{{40}^2}}} =50;MPa\)

\({\rm{Maximum;Shear;Stress}}\left( {{\tau _{max}}} \right) = \frac{\sigma }{2} = \frac{{50}}{2}=25~MPa\)

∴ Maximum Shear stress (τ_max ) = 25 MPa > 20 MPa

∴ The specimen will fail.

Case 3: Torsional load

Given:

Torque applied = 64π N-m = 64π × 10³ N-mm

\({\rm{Maximum;Shear;Stress;}}\left( {{\tau _{max}}} \right) = \frac{{16T}}{{\pi {d^3}}} = \frac{{16 \times 64\pi \times {{10}^3}}}{{\pi ; \times {{40}^3}}}\)

∴ Maximum Shear stress (τ_max ) = 16 MPa < 20 MPa

∴ The specimen will not fail.

Case 4: Bending load

Given:

Bending moment applied = 320N-m = 320 × 10³ N-mm

\({\rm{Section;modulus;of;square;cross}} - {\rm{section;}}\left( Z \right) = \frac{{sid{e^3}}}{6} = \frac{{{{40}^3}}}{6};m{m^3}\)

\({\rm{Maximum;Normal;stress;}}\left( {{\sigma _{max}}} \right) = \frac{{\rm{M}}}{{\rm{Z}}} = \frac{{6{\rm{;}} \times {\rm{;}}320{\rm{;}} \times {\rm{;}}{{10}^3}}}{{{{40}^3}}} = 30MPa\)

\({\rm{Maximum;Shear;stress}}\left( {{\tau _{max}}} \right) = \frac{{{{\rm{\sigma }}_{{\rm{max}}}}}}{2} = \frac{{30}}{2}{\rm{;}}=15~MPa\)

∴ Maximum Shear stress (τ_max ) = 15 MPa < 20 MPa

∴ The specimen will not fail.

Concept:

The total energy of the chord and block will remain constant. When the chord has stretched the part of the kinetic energy of the block gets converted into the potential energy of the chord.

The kinetic energy of the block is given by \(KE = \frac{1}{2}m{v^2}\)

The potential energy of the chord is given by \(PE = \frac{1}{2}k{x^2}\)

where m is the mass of the block v is the velocity of the block and K is the stiffness of the chord and x is the elongation of the chord

Calculation:

Given,  Mass of block (m) = 2 kg, Rod stiffness (k) = 5 N/m, Change in length of rod (x) = 0.4 m

v₁ = initial velocity = 1.5 m/s and let v₂ be the final velocity

Using conservation of energy

\(\frac{1}{2}mv_1^2 = \frac{1}{2};mv_2^2 + \frac{1}{2}k{x^2}\)

\(\frac{1}{2} \times 2 \times {\left( {1.5} \right)^2} = \frac{1}{2} \times 2 \times V_2^2 + \frac{1}{2} \times 5 \times {\left( {0.4} \right)^2}\)

v₂ = 1.36 m/s

Considering the forces in whole truss:

So, we can observe that member maximum load is on member PR

So, considering member PR

Resultant compressive load at point P and R

\(= \sqrt {{{\left( {\frac{P}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{P}{{\sqrt 2 }}} \right)}^2}} = P\)

Now, resultant diagram

So minimum value of P for which any member will buckle

(Euler’s crippling load)

\(P = \frac{{n{\pi ^2}E{I_{PR}}}}{{\ell _{PR}^2}}\)    

Where (EI)_PR = 4EI, n = 1 (for hinge joint)

\({\ell _{PR}} = \sqrt 2 ;\ell\)

\(\therefore P = \frac{{{\pi ^2}\left( {4EI} \right)}}{{{{\left( {\sqrt 2 \ell } \right)}^2}}}\)

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

When we will rotate disc by θ, rod will rotate by angle ϕ as disc is pure rolling on the string

\(\therefore L\phi = r\theta \)

\(\phi = \frac{{r\theta }}{L}\) (∵ \(r = \frac{L}{4}\) as given in question)

\(\therefore \phi = \frac{\theta }{4}\)

Moment of Inertia of disc about point D:

\({I_D} = m{r^2} + \frac{{m{r^2}}}{2} = \frac{3}{2};m{r^2}\) (Using parallel axis theorem)

Now, by energy method:

\(\frac{{d{E_{total}}}}{{dt}} = 0\)

E_total = Energy of the disc due to rotation + Energy of the springs

\({E_{total}} = \frac{1}{2}{I_{disc}} \times ;{\dot \theta ^2} + \frac{1}{2}k;{\left( {2r\theta } \right)^2} + \frac{1}{2}2k{\left( {\frac{L}{2} \cdot \phi } \right)^2}\)

\({E_{total}} = \frac{1}{2}\left( {\frac{3}{2}m{r^2}} \right){\dot \theta ^2} + \frac{1}{2}k;{\left( {2r\theta } \right)^2} + k{\left( {\frac{L}{2} \cdot \frac{\theta }{4}} \right)^2}\)

\(\frac{{d{E_{total}}}}{{dt}} = 0\)

\(\frac{3}{4}~m{{r}^{2}}\times 2\dot{\theta }~\overset{\ddot{\ }}{\mathop{\theta }},+\frac{1}{2}k~\left( 4r \right)\times 2\theta ~\dot{\theta }+\frac{k{{L}^{2}}}{64}\times \left( 2\theta ~\dot{\theta } \right)=0\)

\(\frac{3}{2}m~\left( \frac{{{L}^{2}}}{16} \right)\overset{\ddot{\ }}{\mathop{\theta }},+\frac{4{{L}^{2}}}{16}k\theta +\frac{2}{64}k{{L}^{2}}\theta =0\)

\(\left( \frac{3m}{32} \right)\overset{\ddot{\ }}{\mathop{\theta }},+\left( \frac{9k}{32} \right)\theta =0\)

\(3m~\overset{\ddot{\ }}{\mathop{\theta }},+9k~\theta =0\)

\(\overset{\ddot{\ }}{\mathop{\theta }},+\frac{9k}{3m}~\theta =0\Rightarrow ~\overset{\ddot{\ }}{\mathop{\theta }},+\frac{3k}{m}~\theta =0\)

Comparing with standard equation

\(\because ~\overset{\ddot{\ }}{\mathop{\theta }},+\omega _{n}^{2}~\theta =0\)

\(\omega _{n}^{2}=\frac{3k}{m}\)

\({{\omega }_{n}}=\sqrt{\frac{3k}{m}}\)

Concept:

Projected length

\(L =\sqrt{R\text{ }!!\Delta!!\text{ }h}\)

Where, Δh = Reduction or Draft, R = Radius of roller

Also, Δh = μ²R

Where μ = coefficient of friction

Calculation:

Given, h_i = 40 mm, h_f = 20 mm, D = 200 mm

\(R=\frac{\text{D}}{2}=100~mm\)

Δh = h_i - h_f

Δh = 20 mm

Projected length, \(L=\sqrt{R\text{ }!!\Delta!!\text{ }h}\) 

\(\Rightarrow L=~\sqrt{100\times 20}\)

⇒ L = 44.7213 mm

∵ Δh = μ²R

\(\mu =\sqrt{\frac{\text{ }!!\Delta!!\text{ }h}{R}} =\sqrt{\frac{20}{100}}\)

∴ μ = 0.4472

Given,

Processing time, P.T₁ = 60 min = 1 hour

Production rate = 4/hr = 4 units per hour

Second case:

Production rate = 4/hour

But processing time reduced by 30%

∴ New processing time, P.T₂ = (0.7 × 1) hour

New Production per hour = 4 × 0.7 = 2.8 units.

Thus, Work in progress has been reduced and decrease in

\(WIP=\frac{4-2.8}{4}\times 100=30%\)

Lumped parameter analysis:

Internal/conductive resistance is very little as compared to surface convective resistance and the temperature distribution is given by

\(\frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~}=~{{e}^{\left( \frac{hA}{\rho V{{C}_{p}}} \right)t}}\)

\(\frac{hA}{\rho V{{C}_{p}}}t=\ln\left( \frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~} \right)\)

where

• T_i = Initial temperature of beat at t = 0,
• T = Temperature of body at any instant ‘t’ sec
• T_∞ = Ambient fluid temperature
• h = heat transfer coefficient, ρ is the density of the metal bead

For sphere: \(\frac{V}{A} = \frac{{Volume;of;body}}{{surface;area}} = \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = \frac{r}{3}\) 

V/A = r/3

Calculation:

Given, C_p = 400 J/kgK, ρ = 8500 kg/m³, t = 4.35 sec, r = 0.5 mm

T_i = 100°C, T_∞ = 20°C, T = 28°C

Using equation (1) and putting the given values:

\(\ln\left( {\frac{{100 - 20}}{{28 - 20}}} \right) = \left( {\frac{{h \times 3 \times 1000}}{{0.5 \times 8500 \times 400}}} \right) \times 4.35\)

\(\Rightarrow h = \frac{{2.302 \times ;0.5 \times 8500 \times 400}}{{3000 \times 4.35}} = 299.87;W/{m^2}K;\)

∴ Heat transfer coefficient = 299.8 W/m²K

Let λ be the parameter of exponential distribution

So, for exponentially distributed random variables X & Y

Expected Value or Mean \(E\left( X \right) = \frac{1}{\lambda }\)

Variance: \(Var;\left( X \right) = \frac{1}{{{\lambda ^2}}}\)

Var (X + Y) = Var X + Var Y + 2 Cov (X, Y)

Where Cov (X, Y) = Covariance of X and Y

Correction coefficient is given by:

\(r = \frac{{Cov;\left( {X,;Y} \right)}}{{\sqrt {Var\left( X \right)Var\left( Y \right)} }}\)

Calculation:

Given: E(X) = E(Y) = 0.5

\(\frac{1}{{{\lambda _1}}} = \frac{1}{{{\lambda _2}}} = 0.5\)

λ₁ = λ₂ = 2

\(Var;\left( X \right) = \frac{1}{{\lambda _1^2}} = \frac{1}{{{2^2}}} = 0.25\)

\(Var;\left( Y \right) = \frac{1}{{\lambda _2^2}} = \frac{1}{{{2^2}}} = 0.25\)

Also it is given, Z = X + Y

And Var (Z) = 0

∴ Var (X + Y) = 0

Var X + Var Y + 2 Cov (X, Y) = 0

On putting values:

0.25 + 0.25 + 2 Cov (X, Y) = 0

\(Cov;\left( {X,;Y} \right) = - \frac{{0.5}}{2} = - 0.25\)

Correction coefficient is given by:

\(r = \frac{{Cov;\left( {X,;Y} \right)}}{{\sqrt {Var\left( X \right)Var\left( Y \right)} }} = \frac{{ - 0.25}}{{\sqrt {0.25 \times 0.25} }} = ; - 1\)

Concept:

f(z) = ϕ + iψ

ϕ = Real part, ψ = Imaginary part

If f(z) is analytic function

\(\left. {\begin{array}{*{20}{c}} {\frac{{\partial \phi }}{{\partial x}} = \frac{{\partial \psi }}{{\partial y}}}\ {\frac{{\partial \psi }}{{\partial x}} = - \frac{{\partial \phi }}{{\partial y}}} \end{array}} \right} \to C.R;equation\)

\(\therefore d\psi = \frac{{\partial \psi }}{{\partial x}}dx + \frac{{\partial \psi }}{{\partial y}}dy\)

\(d\psi = \frac{{ - \partial \phi }}{{\partial y}}dx + \frac{{\partial \phi }}{{\partial x}} \ dy\)

Calculation:

Given, f(z) = x² – y² + i ψ (x, y)

\(\phi = {x^2} - {y^2}\)

∵ f(z) is analytic function

\(d\psi = \frac{{ - \partial \phi }}{{\partial y}}dx + \frac{{\partial \phi }}{{\partial x}} \cdot dy\)

\(\frac{{\partial \phi }}{{\partial y}} = ; - 2y,;\frac{{\partial \phi }}{{\partial x}} = 2x\)

\(d\psi = 2y;dx + 2x;dy\)

\(\smallint d\psi = 2\smallint d\left( {xy} \right)\)

ψ = 2 xy

Given, z = 1 + i

Comparing it with z = x + iy, we get:

∴ x = 1, y = 1

\(\therefore {\left( \psi \right)_{\left( {1,;1} \right)}} = 2 \times 1 \times 1=2\)

ψ = 2 when z = 1 + i

Concept:

For disc type clutch:

Torque transmitted by clutch is given by:

T = μ × W × R_m

Where μ = coefficient of friction, W = force/load to transmit torque, R_m = mean radius

For New clutch, uniform pressure theory is applied

\({R_m} = \frac{2}{3}\left[ {\frac{{R_o^3 - R_i^3}}{{R_o^2 - R_i^2}}} \right]\)

For Old clutch, uniform wear theory is applied

\({R_m} = \frac{{{R_o} + {R_i}}}{2}\)

Calculation:

Given, Inner diameter, D_i = 50 mm, R_i = 25 mm

Outer Diameter, D₀ = 250 mm, Ro = 125 mm

For the new clutch, uniform pressure theory is applied

∴ T₁ = μ F_i × R_m1

\({R_{{m_1}}} = \frac{2}{3}\left[ {\frac{{{{125}^3} - {{25}^3}}}{{{{125}^2} - {{25}^2}}}} \right]=86.11~mm\)

For Old clutch, uniform wear theory is applied

T₂ = μ₂ F₂ × R_m2

\({R_{{m_2}}} = \frac{{125 + 25}}{2} = 75;mm\)

Also ∵ T₁ = T₂

∴ From (1) & (2)

\(\frac{{{F_1}}}{{{F_2}}} = \frac{{{R_{{m_2}}}}}{{{R_{{m_1}}}}} = \frac{{75}}{{86.11}} = 0.871\)

\(\frac{{{F_1}}}{{{F_2}}} = 0.871\)

Concept:

In flat-faced follower radius of curvature at the point of contact between cam and follower is:

Minimum radius of curvature (R) =  displacement(y) + base radius(rb) + acceleration \((\frac{{{d^2}y}}{{d\theta }}) \)

\(∴ R = y + {r_b} + \frac{{{d^2}y}}{{d{\theta ^2}}}\)

Calculation:

Given,

Displacement equation for flat face follower

y = 4(2πθ – θ²)       -----(1)

In flat-faced follower radius of the armature at the point of contact between cam and follower is:

\(R = y + {r_b} + \frac{{{d^2}y}}{{d{\theta ^2}}}\)      ----(2)

From (1)

\(\frac{{dy}}{{d\theta }} = 4;\left( {2\pi - 2\theta } \right)\)

\(\frac{{{d^2}y}}{{d\theta }} = - 8\)

We can rewrite equation (2) as:

\(R = {r_b} + \left( {y + \frac{{{d^2}y}}{{d{\theta ^2}}}} \right)\)

Value of \(\left( {y + \frac{{{d^2}y}}{{d{\theta ^2}}}} \right)\) will be minimum when y is minimum

y will be minimum when θ = 0° or 2π i.e.

(when you will substitute this values in the given displacement equation you will get y = 0)

∴ y_min = 0

Now,

\({R_{min}} = {\left( {{r_b}} \right)_{min}} + {\left( {y + \frac{{{d^2}y}}{{d{\theta ^2}}}} \right)_{min}}\)

R_min = 40 mm (given)

∴ 40 = (r_b)_min + (0 - 8)

(r_b)_min = 48 min

∴ The minimum radius of base circle (r_b) is 48 mm

Concept:

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

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

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

The primary shear force on any bolt is given by:

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

The secondary shear force is given by:

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

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

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

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

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

Calculation:

\({P_{SA}} = {P_{SB}} = {P_{SC}} = {P_{SD}} = \frac{{Load}}{{No.;of;Bolts}} = \frac{{10}}{4} = 2.5;kN\)

Let \(l = {l_1} = {l_2} = {l_3} = {l_4} = \sqrt {{{50}^2} + {{50}^2}} = 50\sqrt 2 ;mm\)

e = 300 + 100 = 400 mm

Secondary shear force –

\({{\rm{F}}_B} = \frac{{Pe{l_1}}}{{4l_1^2}} = \frac{{Pe}}{{4{l_1}}} = \frac{{10 \times 400}}{{4 \times 50 \sqrt 2}} = 14.14 ~kN\)

Resultant force at B:

\({R_{B}} = \sqrt {P_B^2 + F{{_B}^2} + 2;{P_B}F_B\cos \theta } \)

\(\Rightarrow R_B = \sqrt {{{\left( {2.5} \right)}^2} + {{\left( {14.14} \right)}^2} + 2 \times 2.5 \times 14.14\cos 45^\circ } = 16;kN\)

Concept:

Net hydrostatic force exerted by fluid on flat surface, F = ρ × g × h̅ × A

Where, ρ = density of fluid, g = acceleration due to gravity, A = Area of the flat surface, h̅ = depth of centroid from free surface

This force is assumed to be acting at a point.

Centre of the pressure of rectangular lamina from free surface,

\(CP = \frac{2}{3}h\) (h = height of rectangular lamina)

Calculation:

For left side of the barrier

Area = 4 m × 1 m = 4 m², \(\bar h = \frac{4}{2} = 2;m,\) 

Center of pressure from top surface of the water, \(C{P_1} = 2 \times \frac{4}{3} = \frac{8}{3}m\) 

F₁= ρ × g × h̅ × A = 1000 × 10 × 2 × 4  = 80 kN

For right side of the barrier

Area = 1 m × 1 m, \(\bar h = \frac{1}{2} = 0.5;m\),

Center of pressure from top surface of the water \(C{P_2} = 2 \times \frac{1}{3} = \frac{2}{3}m\)

F₂ = ρ × g × h̅ × A = 1000 × 10 × (1 × 1) × 0.5 = 5 kN             

Distance of centre of pressure from bottom end:

\(C{P_1} = \frac{h}{3} = \frac{4}{3}\)

\(C{P_2} = \frac{h}{3} = \frac{1}{3}\)

These forces will act as point load on the barrier. If the barrier is treated as a cantilever and the fixed end of this cantilever beam is on the bottom side of the tank, then the loading conditions will look like the figure drawn below

F₁ = 80 kN, F₂ = 5 kN

From the above-simplified diagram:

A moment about the fixed end of barrier,

\(M = 80 \times \frac{4}{3} - 5 \times \frac{1}{3} = 105;kN - m\)

Since**, point** B of the beam is hinged, so it becomes point of contraflexture and bending moment at point B becomes zero. So, we can calculate bending moment on either side of point B and equate it to zero.

The point of application of uniformly distributed load will be at its centre, i.e., 2 m from point B and its value = 10 × 4 = 40 kN

So, considering

R.H.S. and calculating moment about point B

M_B = 40 × 2 - R_c × 4 = 0

R_c = 80/4 = 20 kN

∴ Reaction at point C = 20 kN

Concept:

Cutting speed of the milling cutter is given by, V = πDN m/min

Where, D = Diameter of cutter, N = speed in rpm

In Face milling:

Length of cut (L) = Workpiece length + Approach + Over travel + Radius of tool

Time required for milling (single pass):

\(t = \frac{{Length;of;cut;\left( L \right)}}{{Table;feed}}\)

Table feed is given by, F = f × Z × N

Where f is the feed per tooth, Z is the number of tooth and N is the r.p.m. of the cutter

Calculation:

Given, cutter speed = 35 m/min, length of the workpiece = 300 mm, Approach and over travel = 5 mm, Diameter of the cutter = 100 mm, ∴ radius of the tool = 50 mm

Therefore,                         

35 = π × 0.100 × N

N= 111.4084 rpm             

Length of cut (L) = Workpiece length + Approach + Over travel + Radius of tool

L = 300 + 5 + 5 + 50 = 360 mm

Time required for milling (single pass)

t = \(\frac{{Length;of;cut;\left( L \right)}}{{f; \times ;Z; \times ;N}} = \frac{{360}}{{0.1 \times 20 \times 111.4084}}\)

∴ t = 1.615 min per pass

Since depth of cut is 5 mm, therefore for 25 mm of cut, 5 passes are required.

Total machining time = time per pass × number of pass

t = 1.6157 × 5 = 8.078 min

Concept:

In basic shaft system, the zero line lies on the maximum size of the shaft.

The allowance is minimum clearance between the shaft and hole, i.e. the distance between the Lower limit of the hole and higher limit of the shaft.

If we draw the given dimensions then,

From the above diagram:

Maximum size of hole or upper limit (UL) of hole = 50 + 0.003 + 0.002 = 50.005 mm

Concept:

I = 12 + 0.2t

After time, 't', Next infinitely small time 'dt' let heat deposited 'dQ'

∴ dQ = 2.50 × 10^-2 (mm³/A.s) × 12 + 0.2t × dt

As we have to convert this 's' to 'min'

∴ dQ = 2.50 × 10^-2 (mm³/A × min) × 12 + 0.2t × dt

Considering cathode efficiency of 15%

dQ = 2.50 × 10^-2 × 60 × (12 + 0.2t)dt × 0.15 mm³

∴ In 20 min, \(Q = \int dQ= \displaystyle\int^{20}_0\)  2.50 × 10^-2 × 60 × (12 + 0.2t)dt × 0.15

\(= 0.225[12 +0.1t^2]^{20}_0\) = 0.225[12 × 20 + 0.1 × 20²] mm³

= 63 mm³

As area os 125 cm²

Plating thickness,  \(t = \frac{63}{125×(100)}=0 .00504 \ mm=5.04\ \mu m\)

Concept:

Cutting shear plane angle (ϕ) is given by

\(\tan \phi = \frac{{r\cos α }}{{1; - ;r\sin α }}\)

Where r is chip thickness ratio,

\(r = \frac{{Length;of;chip}}{{Uncut;chip;length}} = \frac{{Thickness;of;uncut}}{{Thickness;of;chip}}\)

Calculation:

Given: Length of chip = 100 mm, Uncut chip length = 250 mm, Rake angle, α = 20°

\(\therefore r = \frac{{100}}{{250}}=\frac{2}{5}=0.4\)

Now,

\(\tan \phi = \frac{{0.4\cos 20^\circ }}{{1 - 0.4\sin 20^\circ }}=\frac{{0.3758}}{{0.86319}}= 0.4354\)

∴ ϕ = tan^-1 (0.4354)

∴ ϕ = 23.53°

Concept:

Rolling force = Average flow stress × Projected Length × Width

\({\rm{True;Strain}} = \ln \left( {\frac{{Final;length}}{{Initial;length}}} \right)\)

\({\rm{Projected;Length}} = \sqrt {{{\left( {{\rm{\Delta H}}} \right)}_{max}} \times Radius;of;Roller} \)

Important Point:

In rolling initial volume is equal to final volume and width remain constant till it is not mentioned in questioned that width is stretching.

Calculation:

Now,

Final Volume = Initial Volume

L_f × W × t_f = L_i × W × t_i

L_f × 200 × 15 = L_i × 200 × 20

\(\frac{{{L_f}}}{{{L_i}}} = \frac{{20}}{{15}} = \frac{4}{3}\)

\({\rm{True;Strain;}} = \ln \left( {\frac{4}{3}} \right)\)

∴ True strain = 0.2876

Now,

According to the given relationship:

\({\rm{Average;flow;stress;}} = {\rm{;Material;strength;coefficient;}} \times {\rm{;}}\frac{{{{\left( {True;strain} \right)}^n}}}{{\left( {1 + n} \right)}}\)

\(\therefore {\rm{Average;flow;stress;}} = 210 \times \frac{{{{\left( {0.2876} \right)}^{0.25}}}}{{1 + 0.25}}\)

∴ Average flow stress = 123.03 MPa

Now,

(ΔH)_max = Initial thickness – Final thickness

(ΔH)_max = 20 – 15

∴ (ΔH)_max = 5 mm

Now,

\(L = \sqrt {{\bf{\Delta }}{H_{max}} \times R} \)

\(\begin{array}{l} \therefore L = \sqrt {5 \times 450} = 47.43;mm \end{array}\)

Now,

Rolling force = Average flow stress × Projected Length × Width

∴ Rolling force = (123.03 × 47.43 × 200) N = 1167.165 × 10³ N

∴ Rolling force = 1167.16 kN

Concept:

Linear programing is a graphical representation of the simple mathematical problem.

• Let x be daily production of A
• Let y be daily production of B

Objective function:

Z_max = 2000x + 3000y

Case 1: For shweta:

Constraint:

• "daily production of A has to be at least as much as B" ⇒ x ≥ y

• "upper limit for production of B is 10 kg per day" ⇒ y ≤ 10

• "Shveta’s production of A to less than 20 kg per day" ⇒ x ≤ 20

• At O (0,0) ⇒ Z = 0

• At A (20, 0) ⇒ Z = 40,000

• At B (20, 10) ⇒ Z = 2000 × 20 + 3000 × 10 = 70,000

• At D (10, 10) ⇒ Z = 2000 × 10 + 3000 × 10 = 50,000

Zmax = 70,000

 

Case 2: For Ashok:

Constraint:

• "daily production of A has to be at least as much as B" ⇒ x ≥ y
• "upper limit for production of B is 10 kg per day" ⇒ y ≤ 10
• "Ashok’s production of A to less than 15 kg per day" ⇒ x ≤ 15

At O (0,0) ⇒ Z = 0

At A (15, 0) ⇒ Z = 2000 × 15 + 3000 × 0 = 30,000

At B (15, 10) ⇒ Z = 2000 × 15 + 3000 × 10 = 60,000

At C (10, 10) ⇒ Z = 2000 × 10 + 3000 × 10 = 50000

Zmax = 60,000

Now,

Difference in daily revenue of Shweta & Ashoka = 70,000 – 60,000 = 10,000 Rupees

∴ The answer will be 10 as it is asked in a thousand Rupees.

Concept:

The Annual cost of the inventory is given as

TC = Material Cost + Holding Cost + Downtime cost 

The down cost means the cost associated with the changeover, this change over may be due to the time taken in loading the machine, loading-unloading the raw material bins, or simply the time which is taken to make the system start working upon the new inventory.

Material cost = Cost per piece (C) × Annual usage value

Holding cost = \(\frac{Q}{2} \times C \times \frac{r}{{100}}\)

Where r is the holding cost rate per year

Q is the inventory ordered at a time

Downtime cost for one order = Cost of downtime/hr × Change over time

The downtime cost and the holding cost are the two variable costs here, therefore for the optimum cost they should be equal to each other 

\(\frac{Q^*}{2} \times C_h = \frac{D}{Q^*}\times C_d\)

\(EOQ\left( {{Q^*}} \right) = \sqrt {\frac{{2D{C_d}}}{{{C_h}}}} \)

Where D is the demand/year, C_h is Holding cost/unit/year, C_d is the downtime cost/order

Calculation:

Given, for case (i)

Q = 1000, C = Rs. 10, r = 10% and Cost of downtime = 200/hr, Change over time = 2 hrs

Then,

\(C_h=10\times\frac{10}{100}=Rs;1/unit/year\)

\(C_d=200\times2=400/order\)

\(\left( {{Q^*}} \right) = \sqrt {\frac{{2D{C_d}}}{{{C_h}}}} = \sqrt {\frac{{2 \times 1000 \times 400}}{1}} = 894.427;Units\)

Total cost for Case 1

\({T_{c1}} = \frac{{{Q^*}}}{2} \times {C_h} + \frac{D}{{{Q^*}}} \times {C_d} + D \times C\)

\({T_{c1}} = ;\frac{{894.427}}{2} \times 1 + \frac{{1000}}{{894.427}} \times 400 + 1000 \times 10 = Rs.;10894.427\)

Case (ii)

800 units are ordered from vendor X then total cost from vendor X, only the units will change other conditions (holding and downtime cost) will remain the same as the vendor is the same 

For quantities ordered from the vendor x

\(Q_x^* = \sqrt {\frac{{2 \times 800 \times 400}}{1}} = 800;units\)

Total cost for vendor x 

\({T_{cx}} = \frac{{Q_x^*}}{2} \times {C_h} + \frac{D}{{Q_x^*}} \times {C_d} + D \times C;\)

\({T_{cx}} = \frac{{800}}{2} \times 1 + \frac{{800}}{{800}} \times 400 + 800 \times 10 = Rs.;8800;\)

For vendor Y

Q = 200, C = Rs. 5, r = 100% and Cost of downtime = 200/hr, Change over time = 6 minutes

Then,

\(C_h=5\times\frac{100}{100}=Rs;5/unit/year\)

\(C_d=200\times\frac{6}{60}=20/order\)

\(Q_y^* = \sqrt {\frac{{2 \times 200 \times 20}}{{50}}} = 40;units\)

\({T_{cy}} = \frac{{Q_y^*}}{2} \times {C_h} + \frac{D}{{Q_y^*}} \times {C_d} + D \times C\)

\({T_{cx}} = \frac{{40}}{2} \times 5 + \frac{{200}}{{40}} \times 20 + 200 \times 5 = Rs.;1200\)

Total cost for the second case

T_c2 = T_cx + T_cy = 8800 + 1200 = Rs. 10000

\(% reduction = \frac{{{T_{c1}} - {T_{c2}}}}{{{T_{c1}}}} \times 100 = \frac{{10894.427 - 10000}}{{10894.427}} \times 100 = ;8.209% \)

Note:

If all the quantities are ordered through a single order then, 

Calculation:

Given, for case (i)

Q = 1000, C = Rs. 10, r = 10% and Cost of downtime = 200/hr, Change over time = 2 hrs

Then,

\({T_{c1}} = 1000 \times 10 + \frac{{1000}}{2} \times 10 \times \frac{{10}}{{100}} + 200 \times 2 = Rs.;10900\)

Case (ii)

800 units are ordered from vendor X then total cost from vendor X, only the units will change other conditions will remain the same as the vendor is the same 

\({T_{cx}} = 800 \times 10 + \frac{{800}}{2} \times 10 \times \frac{{10}}{{100}} + 200 \times 2 = Rs.;;8800\)

For vendor Y

Q = 200, C = Rs. 5, r = 100% and Cost of downtime = 200/hr, Change over time = 6 minutes

\({T_{cy}} = 200 \times 5 + \frac{{200}}{2} \times 5 \times \frac{{100}}{{100}} + 200 \times \frac{1}{{10}} = Rs.;;1520\)

Therefore \({T_{c2}} = 8800 + 1520 = Rs.;10320\)

Now the percentage reduction in cost in case (ii)

\(% reduction = \frac{{{T_{c1}} - {T_{c2}}}}{{{T_{c1}}}} \times 100 = \frac{{10900 - 10320}}{{10900}} \times 100 = 5.321;% \)

Since %reduction in EOQ model is more compared to the single order therefore the correct answer will be 8.209%.

Concept:

Kinetic energy of a fluid is due to virtue of its motion.

\({\rm{K}}.{\rm{E;}} = \frac{1}{2}m{v^2}\)

Mass flow rate through a small section = ρ vdA = ρ v2π r.dr

Calculation:

\(K.E = \mathop \smallint \nolimits_o^R \frac{1}{2}\left( {\rho v2\pi r.dr} \right){v^2} = \mathop \smallint \nolimits_o^R \frac{1}{2}\rho {v^3}2\pi r.dr\)

\(K.E = \mathop \smallint \nolimits_o^R \frac{1}{2}\rho 2\pi r.{\left[ { - C\left( {{r^2} - \frac{{{D^2}}}{4}} \right)} \right]^3}dr\)

\(\because v = - C\left( {{r^2} - \frac{{{D^2}}}{4}} \right) = - C\left( {{r^2} - {R^2}} \right) = C\left( {{R^2} - {r^2}} \right)\)

\(\therefore K.E = \mathop \smallint \nolimits_o^R \rho \pi r{\left[ {C\left( {{R^2} - {r^2}} \right)} \right]^3}dr\)

\(\therefore K.E = \mathop \smallint \nolimits_o^R \rho \pi {C^3}{R^6}{\left[ {\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)} \right]^3}rdr\)

Now,

∵ (a - b)³ = a³ – b³ – 3ab (a - b)

\(\therefore K.E = \rho \pi {C^3}{R^6}\mathop \smallint \nolimits_o^R {\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)^3}rdr\)

\(K.E = \rho {C^3}\pi {R^6};\mathop \smallint \limits_0^R \left[ {1 - \frac{{{r^6}}}{{{R^6}}} - \frac{{3{r^2}}}{{{R^2}}}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)} \right]rdr\)

\(K.E = \rho {C^3}\pi {R^6}\mathop \smallint \limits_0^R \left[ {r - \frac{{{r^7}}}{{{R^6}}} - \frac{{3{r^3}}}{{{R^2}}}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)} \right]dr\)

\(K.E = \rho {C^3}\pi {R^6}\left[ {\frac{{{r ^2}}}{2} - \frac{{{r^8}}}{{8{R^6}}} - \frac{3}{{{R^2}}} \cdot \frac{{{r^4}}}{4} + \frac{3}{{{R^4}}} \cdot \frac{{{r^6}}}{6}} \right]_0^R\)

\(K.E = \rho {C^3}\pi {R^6}\left[ {\frac{{{R^2}}}{2} - \frac{{{R^2}}}{8} - \frac{3}{4}{R^2} + \frac{1}{2}{R^2} - 0} \right]\)

\(K.E = \rho {C^3}\pi {R^6}\left[ {\frac{{4{R^2} - {R^2} - 6{R^2} + 4{R^2}}}{8}} \right]= \rho {C^3}\pi {R^6} \cdot \frac{{{R^2}}}{8}= \frac{{\rho {C^3}\pi }}{8};{R^8}\)

\(K.E = \frac{{\rho {C^3} \pi }}{8}\frac{{{D^8}}}{{256}}\)

Concept:

For an incompressible, non-viscous fluid.

Bernoulli’s statement or equation is valid

\(\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + z = constant\)

 ∵ It is mentioned in the question that air density is constant. Therefore, Bernoulli’s equation is valid

Calculation:

Mass flow rate through nozzle remain constant

\(\begin{array}{l} \therefore {\left( {\rho ;A;V} \right)_{inlet}} = {\left( {\rho ;A;V} \right)_{outlet}}\ 0.1 \times {V_i} = 0.02;{V_O} \end{array}\)

V_O = 5 V_i

∴ V₂ = 5 V_i

Bernoulli’s equation for nozzle

\(\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}};;;;;\because \left[ {{z_1} = {z_2}} \right]\)

\(\frac{{0.36 \times {{10}^3}}}{{1.2 \times 9.81}} + \frac{{V_1^2}}{{2g}} - \frac{{{{\left( {5{V_1}} \right)}^2}}}{{2g}} = \frac{{{P_2}}}{{\rho g}}\)

\(\frac{{0.36 \times {{10}^3}}}{{1.2;g}} - \frac{{24;V_1^2}}{{2g}} = \frac{{{P_2}}}{{\rho g}}\)

P₂ = P_atm = 0 (Gauge)

∴ V₁ = 5 m/s  V₂ = 25 m/s

∵ P₂ = P_O = P_atm = 0 (Gauge)

But at point O there will be a pressure due to motion of fluid which will exert a pressure on the plate.

Now,

Bernoulli’s equation between exit point & point O

\(\frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} = \frac{{{P_3}}}{{\rho g}} + \frac{{V_3^2}}{{2g}}\)

\({P_3} = \frac{1}{2}\rho V_2^2\)

\({P_3} = \frac{1}{2} \times 1.2 \times {\left( {25} \right)^2}=375 ~Pa\)

∴ P₃ = 0.375 kPa

Concept:

Compressor is a mechanical device which is used to increase the pressure of gas (compressible fluid)

Calculation:

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}}\)

\({T_2} = 310;{\left( 6 \right)^{0.2857}}\)

∴ T₂ = 517.22 K

\({\eta _{isentropic}} = \frac{{{h_2} - {h_1}}}{{{h_2}' - {h_1}}} = \frac{{{W_{ideal}}}}{{{W_{actual}}}}\)

\({W_{actual}} = \frac{{{C_p}\left( {{T_2} - {T_1}} \right)}}{{{\eta _{min}}}}\)

\({W_{actual}} = \frac{{1.005\left( {517.21 - 310} \right)}}{{0.85}}\)

∴ W_act = 245 kJ/kg

∴ The difference between enthalpies is nothing but the W_actual and it is 245 kJ/kg.

Concept:

For a constant pressure process

Work done \(= \mathop \smallint \limits_{{v_1}}^{{v_2}} p dV = p \left( {{V_2} - {V_1}} \right)\)

\(W = mR\left( {{T_2} - {T_1}} \right)\)

Calculation:

Given:

\({V_2} = 2{V_1}\)

For process 1 to 2

\(\frac{{{T_2}}}{{{T_1}}} = \frac{{{V_2}}}{{{V_1}}}\)

T₂ = 2 × 400

T₂ = 800 K

W = 1 × 287 (800 - 400) = 114800 J

∴ W = 114.80 kJ

Concept:

\(Specific;Steam;Consumption = \frac{{3600}}{{{W_{net}}}}\)

\(Specific;Steam;Consumption = \frac{{3600}}{{({W_T} - {W_{P}}})}\)

Calculation:

Given:

W_T = 903 kJ/kg, W_comp = 3 kJ/kg

Now,

\(Specific;Steam;Consumption = \frac{{3600}}{{({W_T} - {W_{P}}})}\)

\(SSC = \frac{{3600}}{{903 - 3}} = \frac{{3600}}{{900}}=4\)

∴ SSC = 4 kg/kW.hr

Concept:

Kaplan is an axial flow reaction turbine. It is a low head turbine with specific speed above 300 unit.

Calculation:

\(\tan \beta = \frac{{{C_f}\left( {flow;velocity} \right)}}{{{C_b}\left( {blade;velocity} \right)}}\)

\({\rm{Blade;Velocity;}} = \frac{{\pi dN}}{{60}}\)

\({\rm{Blade;Velocity;}} = \frac{{\pi \times 3 \times 300}}{{60}}\)

∴ Blade Velocity (C_b)= 47.12 m/s

Now,

\({\rm{Volume;flow;rate;}} = \frac{\pi }{4}\left( {D_1^2 - D_n^2} \right) \times {C_f}\)

\(100 = \frac{\pi }{4}\left( {{4^2} - {2^2}} \right) \times {C_f}\)

∴ C_f =10.61 m/s

∴ Flow velocity (C_f) =10.61 m/s

Now,

\(\tan \beta = \frac{{10.61}}{{47.12}}\)

\(\begin{array}{l} tan\beta = 0.225\ \beta = {\tan ^{ - 1}}\left( {0.225} \right) \end{array}\)

∴ β = 12.69°

Concept:

Indicated power is the power actually developed by the engine cylinder i.e. net work done during the cycle.

Brake power is the power available at the crank shaft.

\({η _{mechanical}} = \frac{{B.P}}{{I.P}}\)

Calculation:

Given: r = 8, Qsupplied = 10 kW, γ = 1.4, η_m = 0.8, B.P = ?

Efficiency of Otto cycle

\({η _0} = 1 - \frac{1}{{{r^{γ - 1}}}}\)

\({η _0} = 1 - \frac{1}{{{8^{1.4 - 1}}}}\)

\({η _0} = 0.5647\)

Now,

\({η _0} = \frac{{{W_{net}}}}{{{Q_{supplied}}}}=0.5647\)

W_net = 0.5647×10 kN

∴ W_net = 5.647 kN

Now,

\({η _{mechanical}} = \frac{{B.P}}{{I.P}}\)

∴ B.P. = 0.8 × 5.647 kW

∴ B.P = 4.517 kW