Conventional Question Practice Program · 2017-05-26 · R – 600a: Isobutane Sol–13: (c) If f g...

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Conventional Question Practice ProgramDate: 2nd April, 2016

1. (b)

2. (c)

3. (d)

4. (a)

5. (a)

6. (c)

7. (d)

8. (b)

9. (c)

10. (b)

11. (d)

12. (a)

13. (c)

14. (b)

15. (b)

16. (a)

17. (b)

18. (c)

19. (a)

20. (a)

21. (d)

22. (c)

23. (a)

24. (a)

25. (c)

26. (b)

27. (b)

28. (a)

29. (a)

30. (b)

31. (c)

32. (d)

33. (d)

34. (a)

35. (a)

36. (c)

37. (a)

38. (b)

39. (c)

40. (b)

41. (c)

42. (b)

43. (a)

44. (b)

45. (a)

46. (d)

47. (a)

48. (b)

49. (b)

50. (c)

51. (c)

52. (a)

53. (c)

54. (b)

55. (d)

56. (c)

57. (d)

58. (b)

59. (a)

60. (b)

61. (d)

62. (c)

63. (a)

64. (a)

65. (a)

66. (b)

67. (d)

68. (b)

69. (b)

70. (c)

71. (c)

72. (c)

73. (d)

74. (d)

75. (d)

76. (b)

77. (b)

78. (c)

79. (a)

80. (b)

81. (c)

82. (d)

83. (a)

84. (b)

85. (a)

86. (c)

87. (c)

88. (a)

89. (b)

90. (c)

91. (b)

92. (a)

93. (a)

94. (d)

95. (d)

96. (c)

97. (c)

98. (b)

99. (c)

100. (b)

101. (a)

102. (b)

103. (c)

104. (b)

105. (c)

106. (a)

107. (c)

108. (c)

109. (a)

110. (c)

111. (a)

112. (b)

113. (a)

114. (c)

115. (d)

116. (c)

117. (c)

118. (b)

119. (c)

120. (a)

ANSWERS

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(2) ME (Test-10), Objective Solutions, 2nd April 2016

Sol–1: (b)Since R-22 have limited miscibility inlubricating oil and it again reduces astemperature goes down so it is notrecommended at very low temperature.

Sol–2: (c)Roll forces can be reduced by any of thefollowing means: Reducing friction; Using smaller-diameter rolls,to reduce

the contact area; Taking smaller reductions per pass, to

reduce the contact area; rolling at elevated temperatures, to

lower the strength of the material.Sol–3: (d)

It is the process of heating of air withoutadding or removing any moisture content.

t1

12

t2

(t2 > t )1

1 = 2

It is clear from psychrometric chart thatrelative humidity of the air decreases.

Sol–4: (a)Thermal conductivity of gases mainlydepends on the intra-molecular collisions.In case of gases as the temperature increasesthe intra-molecular collisions increaseswhich are sole responsible for heat conductionin gases. So conductivity of gases increaseswith increase in temperature.

Sol–5: (a)In shell and tube type heat exchangernormally, liquid flows through tubes andgas/steam with low heat transfer coefficienton the shell side. So that the overall heattransfer coefficient can be increased byincreasing velocity of liquid.

Sol–6: (c)

1. Statement 1 is fundamental law ofgearing.

2. As centre distance changes, the pressureangle also changes i.e. increased centredistance has higher pressure angle orvice-versa.

3. Statement 3 is wrong because in com-pound gear train, the velocity ratio de-pends on product of teeth of driving gearsdivided by product driven gear. Hencenumber of teeth of intermediate gearsalso play role.

4. Statement 4 is definition of epicyclic geartrain.

Sol–7: (d)Speed of gear A relative to armC = NA – NC; Speed of gear B relative toarm C = NB – NC since the gears A and Bare meshing directly, they will revolve inopposite directions

B C

A C

N NN N

= A

B

TT

Since gear A is fixed, hence NA = 0

B C

C

N NN

= A

B

T 40 2= =T 20

NB – NC = 2NC NB = 3NCSo, for three revolutions of arm C, NB = 3× 3 = 9 revolutions.

Sol–8: (b)For meshing gears, both pressure angle andthe module must be same.

Sol–9: (c)base circle radius = pitch circle radius ×cos

Sol–10: (b)Air: Aircraft always use air refrigerationadvantages.Ammonia: Ammonia greatest applicationis found in large industrial scale refrig-erating.Carbon Dioxides: Carbon dioxide is alsocalled as dry ice and is used as direct con-tact freezing of food.

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R-11: It is exclusively used in large cen-trifugal compressor systems of 200 TRand above due to the low density and highspecific volume

Sol–11: (d)From human body heat equation.

M = W + Q + SHeat change between man and hisenvironment isM – W = Q + S

Note:The rate at which human body producesheat is known as metabolic rate (M). Thisheat is utilized in three ways.

WRate of Work done by the man

Q Rate of heat loss by man in terms of con-vection, radiation and evaporation.

S Rate of heat stored in the body tomaintain the temperature of the body.Hence,

M = W + Q + SSol–12: (a)

R – 1349: TetrafluoroethaneR – 152a: DifluoroethaneR – 290: PropaneR – 600a: Isobutane

Sol–13: (c)If fg is the film coefficient of heat transferthrough the air film, we have for sensibleheat transfer between the air and the wettedsurface

sdQ = fg(t – ts) dA

• Also, if Kw is the diffusion coefficient ofwater vapour through the air film, andif sfgh is the latent heat at ts, then forlatent heat transfer between theunsaturated air and wetted surface,

LdQ =

sfg vh dm

where vdm = Kw(W – Ws) dA

so that LdQ = Kw(W – Ws) sfgh dA

Thus, total heat transfer,

Kw =g g

ee P p

f ffor L 1

L C C

dQ = gg s

e p

ff dAt t

L C

sfg sh dAW W

= sfggp s

p e

hfdAC t t W WlsC L

= s sfg fggp p s s

p e e

h hfdAC t W C t WC L L

For Le = 1, dQ = gs

p

fdAh h

C

The quantity (h–h s) is the enthalpypotential. Thus, the driving force for thetotal heat transfer is the difference in theenthalpy h of unsaturated air and theenthalpy h s of saturated air at thetemperature of the wetted surface.

Sol–14: (b)RSH = 100 kW,RLH = 25 kW

OASH = 10 kWOALH = 10 kW

BPF = 0.15Hence, Room sensible heat factor

RSHF = RSHRSH RLH

= 100 0.8=100 25

Effective Room sensible Heat (ERSH) = RSH+ BPF × OASH = 100 + 0.15 × 10 = 101.5kWEffective Room Latent Heat (ERLH) = RLH+ BPF × OALH = 25 + 0.15 × 10 = 26.5kW Effective Room Sensible factor =

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ERSHF = ERSH 101.5=ERSH ERLH 101.5 26.5

= 0.79Sol–15: (b)

When the activity level is high, therequired indoor temperature will belower. Similarly, when air velocityincreases, effective temperature, alsoincreases.

Sol–16: (a)The conditioned air supplied to the roommust have the capacity to take upsimultaneously both the room sensible heatand room latent heat loads. The point S onpsychrometric chart represents the supplyair condition and point R represents theroom design condition. The line SR is calledroom sensible heat factor line (RSHF line).The slope of the line gives the ratio of theroom sensible heat (RSH) to the room latentheat (RLH). Thus, the supply air havingits conditions given by any point on thisline will satisfy the requirements of theroom with adequate supply of such air.

Dry bulb temperature

Roomdesigncondition

Sp. h

umid

ity

ADPS

Supply aircondition

R

RSHFline

RTH

RSH

RLH

td1 td2

Sol–17: (b)Wet bulb temperature is the lowesttemperature that can be reached byevaporating water into the air. On a hot day,when the wbt is low, rapid evaporation andhence cooling takes place at the skin’s surface.As the wbt approaches the air temperature,less cooling occurs and the skin temperaturemay begin to rise. When the wet bulbtemperature exceeds the skin’s temperature,no net evaporation occurs and the body

temperature can rise quite rapidly.Fortunately, the wet bulb temperature isalmost always considerably below thetemperature of the skin.

Sol–18: (c)

BPF

(1-BPF)

Saturationcurve

Outdoorcondition

Mixturecondition

entering the cooling coilRoom supply

air condition

ADP5 4

6BPFBPF

(1-BPF)

(1-BPF)

Roomdesign

condition

RSHF line

3

b

4 Alignmentcircle

Dry bulb temperature

4'

Sp. h

umid

ity

SHF

scal

e

1

Sol–19: (a)The colour of halide torch in case of leakageof Freon will change to green.

Sol–20: (a)The critical temperature of water is 374°C.Refrigerant Critical

temperature(°C)Water 374CO2 31Freon 12 112Ammonia 133

Sol–21: (d)During chemical dehumidification, as theair comes in contact with thesechemicals, the moisture gets condensed outof air and gives up latent heat. Due tocondensation, the specific humiditydecreases and the heat of condensationsupplies sensible heat for heating the airand thus increasing its dry bulbtemperature. Thus, wet bulb temperatureremains constant. Since wbt line andenthalpy line are parallel to each other inpsychrometric chart hence enthalpy alsoremains constant.

Sol–22: (c)Steam spray into air: Heating andhumidificationAir passing over a coil carrying steam:Sensible heating

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Air passing over a coil havingtemperature less than dew point:Cooling and dehumidificationAir passing over a coil havingtemperature above the dew point butbelow the wbt: Sensible coolingDehumidification will take place only whenthe cooling coil has temperature lower thanthe dew point temperature of air.

Sol–23: (a)

1tdp1

dry bulb temp.

spec

ific h

umid

ity

The point at which, a line parallel to thedbt line or at constant humidity ratio, cutsthe saturation curve is called dew pointtemperature. For saturated air, dbt, dpt &wbt are same. Dehumidification can beachieved along with both heating andcooling. At constant enthalpy, wbt remainsconstant even with addition of moisturebecause constant enthalpy line and constantwbt lines are parallel in psychrometricchart.

Sol–24: (a)Sol–25: (c)Sol–26: (b)Sol–27: (b)

In exponential smoothing forcast.Ft = Ft–1 + (Dt–1 – Ft–1)

where,0 < < 1

High value of ‘ ’ means more weightagefor immediate forcast and ‘ ’ more re-sponsive (i.e. unstable). Less value of ‘ ’means relatively less weightage for im-mediate forcast and ‘ ’ is less responsive(i.e. stable).

Sol–28: (a)Month Demandforecast

for cupsJanuary 500400

February 600– (475)March – –

Smoothing coefficient = 0.75

Ft = t 1 t 1D 1 f

Ffeb = 0.75 × 500 + (0.25) × 400= 475 cups

Fmarch = 0.75 × 600 + 0.25 × 475= 568.75 or

marchF 569 cups

Sol–29: (a)F = 6.9 + 2.9 t

F1 = 6.9 + 2.9 × 1F1 = 9.8F2 = 12.7F3 = 15.6F4 = 18.5F5 = 21.4

Absolute Deviation = i iD F

= |10 – 9.8| + |13 – 12.7|+ |15 – 15.6| + |18 – 18.5|+ |22 – 21.4|

Absolute Deviation 2.2

Sol–30: (b)Given,Production rate,

p = 1000 pins / monthDemand or consumption rate,

d = 500 pins / monthLot size,

Q = 1000 pins

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Qm

Qp

p-d

tp

d

Max. inventorylevel

Inventorylevel

timeT

Maximum inventory level,

Qm = p dQp

=1000 5001000

1000

mQ 500 pins

Sol–31: (c)Sol–32: (d)

Setup cost is associated with cost of bring-ing a shut-down production system intostarting position. It includes, maintenancecost, cost associated with arrangement ofworkers, material, Tools, etc.

Sol–33: (d)The demand in the time series analysis isa function of T (underlying trend), C(Cyclic variatings within the trend), S(Seasonal variation within the trend) andR (Random variations). Time seriesdecomposition is a method where the abovefour variations are isolated individuallyand their effect on forecast is studied.

Sol–34: (a)In ABC inventory control, inventory itemsare classified into A, B and C categorydepending on their usage value. Usagevalue is given by number of units requiredmultiplied by cost per unit.For A category, inventory is kept almostnil and frequent review is done as usagevalue is high and have high stock outcost, on the other hand, for C categoryitems high inventory is kept as usuagevalue is low, so inventory is reviewed after

long period because of low stock out cost.Sol–35: (a)

When re-order level is greater than orderquantity, the number of order outstandingat any time is never more than one.

Sol–36: (c)Average Outgoing Quality (AOQ)represents the maximum percentagedefective in the outgoing product. AOQdepends on incoming quality, sample andlot size and probability of acceptance.

AOQ =

aP N mP

N

where Pa is probability of acceptanceP is incoming quality.N = lot size and m = sample size.

P N mN gives the fraction defective.

Therefore,AOQ = 0.95× 0.01=0.0095

Sol–37: (a)

2 cm

k = 40 W/mKm

insulation

ki = 0.1 W/mK

metal rod

Critical radius of insulation in cylindricalshape

cr = Thermal conductivity of insulation

Convective heat coefficient

= k 0.1h 5 = 2 cm

So thicknessof critical insulation= 2 – 1 = 1 cm

Sol–38: (b)The Fourier’s law,

Q = dTkAd

x

The assumptions in this law are,

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(1) Steady-state conditions exist.(2) Thermal conductivity of medium of

heat transfer is constant.(3) Heat flow is one dimensional.(4) Uniform wall temperature has no

meaning in this expression but thecross-section of medium perpendicu-lar to heat flow has uniform tem-perature to ensure one dimensionalflow.

Sol–39: (c)The number of transfer unit i.e. NTU isdefined as–

NTU =min

UAC

Cmin – Product of mass flow rate multi-plied specific heatU – Overall heat transfer coefficient.Both Cmin and U have nothing to do withgeometric parameter but related to flowparameters so NTU depends upon area‘A’ i.e. size of heat exchanger along withCmin and U.

Sol–40: (b)The heat capacity ratio–

C = min

max

CC

Since heat lost by hot fluid is equal toheat gained by cold fluid. From thisconcept it is concluded that a fluid havingsmall specific heat has maximumtemperature change so

hT = 150 – 75 = 75°C

cT = 125 – 25 = 100°C Heat transferred from hot fluid to cold

fluid,

h hC T = c cC T

h

c

CC = c

h

TT

cT > hT

hC < cC

Cc is minimum and Ch maximumheat capacity ratio

C = c h

h c

C T 75= =C T 100

= 0.75

Sol–41: (c)W

CL

MA MB

Fixed End Moments

WL8

WL8

MFAB = FBAWL M8

W

Free body BMD = WL4

WL/8EI

WL/8EI

W

+

WL/8 WL/8

Using moment area method between A &B about C.

c = Moment of area of M/EI diagram

c = 1 L WL 2 L WL L L2 2 4EI 3 2 8EI 2 4

c = 3 3 3WL WL WL

48EI 64EI 192EI

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Sol–42: (b)As the given shafts are arranged in se-ries, so the torque will be the same oneach part MN, NO and OP. Deflection oneach part is given by

M N O P

TT = 10 N-m

MN =MN

Tq

= 1020 = 0.5 rad

NO =NO

Tq

= 1030

= 13 rad

OP =OP

Tq

= 1060

= 16 rad

Total deflection,

MP = 1 1 12 3 6 = 6

6

MP 1rad

Sol–43: (a)State of stress at a point on shaft

xy

xy

45°Axis of shaft

= x y x y

2 2

cos 2 + xy sin 2

at, = 45°

= 0 – 0 + xy sin 90

xy

= x y

2

sin2 + xy cos 2

at, = 45°;

= 0 + xy cos 90

0

Maximum shear stress will act perpen-dicular to the axis

Sol–44: (b)Slenderness ratio,

10 mm20 mm

L

S = lengthof columnleastradiusof gyration

S =m in

LI

A

S =min

ALI

= 3

0.02 0.011.0.02 0.01

12

S = 346.41 S 346

Sol–45: (a)

CA B

ba

3 m1 m

Torque T is applied at point C from end AT = TA + TB

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600 = TA + TB … (i)

and CA = CB

A ACT LGJ = B CBT L

GJTA × 1 = TB × 3 … (ii)

From eq. (i) and (ii)TA = 450 NmTB = 150 NmAlternative :

TA = bTa b

= 36004

= 450 Nm

TB =aT

a b

= 16004

= 150 Nm

Sol–46: (d)Diameter of solid circular shat is ds anddiameters of hollow circular shaft are d1and d2 (where d1 > d2)From torsion formula

r

=T GJ L

=T r

J

For the same material and same torsionalstrength max and T and same in boththe shaft.

max =

s 1

4 4 4s 1 2

d dT T2 2

d d d3232

s4s

dd = s

4 41 2

dd d … (i)

3sd =

43 21

1

ddd

d1 > dsexternal dia of hollow shaft is greater than

solid shaft.From eq. (i)

4s

s

dd =

4 4 2 2 2 21 2 1 2 1 2

1 1

d d (d d )(d d )d d

ds·As =2 21 2

H1

d dAd

As =

22

H 11

Hs

dA dd

K Ad

Q d1 > ds

22

11

ddd

> sd K 1

AH < AsQ Weight of shaft = A × r × Lor Weight A

Hence weight of hollow shaft is less thansolid shaft.

Stiffness of shaft, T GJL

For the same length of shaft and same G

T

J

hollow

solid

(Stiffness)(Stiffness) =

4 41 2

4s

d d 32d 32

=

4 41 2

4s

(d d )d

= 1

s

d 1d

[From eq. (i)]

Hence stiffness of hollow shaft is greaterthan the stiffness of solid shaft.

Sol–47: (a)

l leff =2l l leff = l

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l leff = 2l l leff =2l

A — 2 ; B — 4C — 1 ; D — 3

Sol–48: (b)For compressive stress

P MyA I

< 0

2 4

D( Pe)P 2D D4 64

< 0

e <D8

Within limit of D Dto8 8

Sol–49: (b)The region in which load acts but therewould be no tension. The region is calledKern.Kern of rectangular section– RhombusKern of I section – RhombusKern of Hollow circular – CircleKern of Square – Square

Sol–50: (c)

=3

464PR n

Gd

n[Given that all other factors are same forboth springs]

A

B

=

A

B

nn

A

B

=

12

Sol–51: (c)Spring of 10 kN/m stiffness and 40 kN/mstiffness are in series.

Equivalent stiffness = 1 110 40

= 10 4010 40

=

8 kN/mNow all the springs having stiffness 20

kN/m 30 kN/m and 8 kN/m are inparallel, so equivalent stiffness will be 20+ 30 + 8 = 58 kN/m.

Sol–52: (a)

Hoop stress h = pD2t

Longitudinal stress l = pD4t

Hoop strain h = h l = pD 2

4tE

Hoop strain is along the circumference

h = change in perimeter

perimeter = D

D

= D

D

D = 2

hpDD 24tE

Sol–53: (c)• Shear stress is zero on an element on the

inner wall so stress generated are principalstresses.

• The constants of Lame’s equation arepositive for internal pressure.

• Stresses on an element on the outer wallare bidirectional (hoop stress andlongitudinal stress).

Sol–54: (b)For a thin cylinder, longitudinal strain,

= hσ /mE

=

pd 12 –4tE m

Now, h

=

21m12m

= m 2

2m 1

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Sol–55: (d)

p

h max( )

h min( )

r

p

Stress distribution inH.C.P.V. subjected to I.P.

Variation of stress in thick cylinderdue to internal pressure

Sol–56: (c)

Stiffness of the spring (k) = W

= 4

3Gd

8D Nwhere, G = Shear Modulusd = Diameter of spring wireD = Diameter of springN = Number of turns

so, 1KN

If N is halved by cutting the spring, stiff-ness doubles.Option (c) is correct.

Sol–57: (d)

max = w 38 WDk

d

where kw is wahl’s correction factor

kw is given by 4C 1 0.6154C 4 C

where C is the spring index.Kw takes into account the curvature effectand the direct shear stress factor.

Sol–58: (b)BCC:Li, Na, K, Rb, Cs, Fr, Cr, Fe( iron and iron), Mo, Nob, Ta, W, V, Zr,TiFCC: Ac, Al, Ca, Ce, Cu, Au, lr, Pb, Ni,

Pd, Pt, Fe ( iron), Rn, Rh, Ag, Sr, Th, YbHCP: Be, Cd, Co, Mg, Os, Re, Ti, Zn, Zrand HeAmorphous : Glass charcoal

Sol–59: (a & c)Both (a) & (c) are correctCrystal structure : APFSimple cubic : 0.52BCC : 0.68FCC : 0.74HCP : 0.74

Sol–60: (b)Three equal axes, equally inclined, a = b =

c, 90

Sol–61: (d)Elasticity: is the property by virtue ofwhich material is able to return to itsoriginal shape on unloading.Malleability: is the property by virtueof which material is able to undergoesplastic deformation under compressivestress.Ductility: is the property by virtue ofwhich material is able to undergoes plasticdeformation under tensile load.Plasticity: is the property by virtue ofwhich material is able to deformnonelastically without fracture.

Sol–62: (c)(i) Neoprenes are used as wire and cable,

chemical tank linings, belts hoses, sealsand gaskets. It has excellent ozone, heatand weathering resistance and also goodoil resistance.

(ii) Bakelite has excellent thermal stability toover 150°C and maybe compounded witha large

(iii)Foamed Polyurethane are commonly usedas cushions in auto mobiles and furnitureas well as in packaging and thermalinsulation.

(iv)Araldite has excellent combination ofmechanical properties and corrosionresistance, dimensionally stable, good

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adhesion, relatively inexpensive and goodelectrical properties.

Sol–63: (a)Sol–64: (a)

Polyesters are thermosetting plastic formedby condensatin polymerization ofDicarboxylic acid and dihydroxy alcohols.

Sol–65: (a)Connecting rods – forgingPressure vessels – weldingMachine tool beds – castingCollapsible tubes – extrusion

Sol–66: (b)The temperature of heat source

T2 = T2 1

2

1

VV

= 295 (32)0.4

= 295 × 4= 1180 K

T = 1180 K1

T =295 K2

HE

Q1

W

Q2

The Carnot engine,

1

1

QT = 2

2

QT

Q2 = Q1 · 2

1

TT

= 52 × 2951180

= 13 kJSol–67: (d)

For 100% propulsive efficiency

1.0 =j

2uv u

vj = u

Thrust, developed by turbojet–

F = a jm v u = 0

Sol–68: (b)Propulsive efficiency of turbojet,

P = a

j a

2 Flight velocity,VJet velocity, V Flight velocity,V

= a

j a

2VV V =

j

a

2V

1V

Sol–69: (b)Assumption of air-standard Otto cycle :

1. The working substance is air and it be-have as an ideal gas.

2. The working substance is of fixed mass(closed system analysis).

3. The working fluid does not undergoes anychemical change.

4. The specific heat of working fluids remainsconstant ( constant).

5. All the process are both internally andexternally reversible.

6. Intake process is constant volume heataddition and exhaust process is constantvolume heat rejection process.

Sol–70: (c)Stirling cycle consist of two reversibleisotherms and two reversible isochores. Incase of stirling cycle heat is suppliedduring isothermal expansion and isochoricprocess in which temperature increaseswhile in case of Carnot cycle heat issupplied only during isothermal expansion.Hence, Due to heat transfer at constantvolume process, the efficiency of stirlingcycle is less than that of Carnot cycle.However, a regenerative arrangement instirling cycle may have efficiency equal toCarnot cycle.

Sol–71: (c)

=output

Heat supplied

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Q = 3500.34

Q = 1029.4 kJ/kg

W2 = 2Q

W2 = 1029.4 × 0.51

2W 525 kJ/kg

Sol–72: (c)A stream-lined body is defined as that bodywhose surface coincides with the stream-lines, when the body is placed in a flow.In that case the separation of flow willtake place only at the trailing edge, i.e.,the flow separation is suppressed.Though the boundary layer will start atthe leading edge, will become turbulentfrom laminar, yet it does not separate uptothe rear most part of the body in case ofstream-lined body.Thus behind a stream-lined body, wakeformation zone will be very small andconsequently pressure drag will be verysmall.

Sol–73: (d)A source flow is the flow coming from apoint and moving out radially in all direc-tions of a plane at uniform rate.The radial velocity ur at any radius r isgiven by

ur = q2 r

where, q is volume flow rate per unit depthEquation of Potential FunctionThe velocity components in cylindricalpolar co-ordinates in terms of velocitypotential function are :

ur = r

& u = 1r r

ur = velocity in radial direction

u = velocity in tangential direction

r

= q2 r

d = q2 r dr

q lnr2

..... (q is constant)

Sol–74: (d)Navier-stokes equation represents the con-servation of momentum.

Sol–75: (d)Given, a solid cylinder.

D1 = 100 mmh1 = 50 mmh2 = 25 mm

Since, volume of cylinder remains constantbefore and after forging.

21 1D .h

4 = 2

2 2D .h4

D2 = 11

2

hDh

= 100 2 mmPercentage change in diameter

2 1

1

D D 100D

= 100 2 100 100

100

= 41.42

2 1

1

D D 100 41.42 %D

Sol–76: (b)In thread rolling blank diameter shouldbe equal to pitch diameter of the thread.

Sol–77: (b)Sequence of forging operation –Fullering Edgeing Bending Blocking cum finishing Cutting

Sol–78: (c)Shrinkage allowance on pattern is providedto compensate for solid state shrinkageonly i.e. when the temperature of solidphase drops from freezing to roomtemperature. Riser compensates thematerial/metal for liquid and solidificationshrinkage.

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Sol–79: (a)The area of base metal adjacent to theweld pool is called heat affected zone. Since,this area is subjected to a hightemperature for considerable period of timeas compare to the rest portion of basemetal, so it’s microstructure changes anddifferent from base metal or prior towelding.

Sol–80: (b)For neutral flame :O2 : C2H2 = 1 : 1C2H2 = 10 ltrs.

2O 10 ltrs.

Sol–81: (c)Velocity potential function does not existin rotational flow.Vorticity exists in both rotational andirrotational flow but for irrotational flowits value is 0.

Sol–82: (d)

Stream function = 2 33x y

v = 6xx

v =2,2x

= 12

u = 23yy

u =2,2

12y

Resultant velocity,

V = 2 2u v

= 2 212 12

= 16.97 17Sol–83: (a)

Darcy Weisbach friction factor in the case

of pipe flow for laminar flow, is given by

f =e

64R

... (i)

So, friction factor will remain unchanged.Sol–84: (b)

Shear velocity is a fictitious quantityhaving dimension of velocity.

V* = 0

; t0 = boundary

shear stress. It is a mathematical aid for fluid flow

(turbulent and laminar) analysis.Sol–85: (a)

In case of turbulent flow on a flat plate

x

= 1/5ex

0.376R

x

1/51

x

Hence, relative thickness x

of turbulent

boundary decreases with distance x.Sol–86: (c)

m

*

U Uu

= 5.75 log10yR + 3.75

This equation is valid for both smooth andrough pipe. Now, at

y = R, u = Umax = Um

m

*

U Uu

= 3.75

Sol–87: (c)Froude No : Related to Inertia Force andgravity forceEuler number : Related to Inertia force

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and pressure force.Reynolds number : Related to Inertiaforce and viscous forceMach number : Related to Inertia forceand Elastic force.

Sol–88: (a)Specific gravity is unit less quantity, it isthe ratio of specific weight of a fluid to thespecific weight of a standard fluid

Dimensions of specific gravity = M0 L0 T0

Coefficient of viscosity

= 1 1kg M ML Tm S LT

Kinematic viscosity

=2

2 1 0 2 1m L T M L Ts

Stress = 1 2F ML TA

Sol–89: (b)Given

Lr = 1100

Fm = 0.10 NHere froude law will be applicable

Fr = armr

Vr = rL

tr =r r

rr r

L L KV L

ar = r2r

L 1t

fr = mrar = 3r rL (1)

m

p

FF = Lr

3

Fp = 0.1 × 31

(1 /100) Fp = 100 kN

Sol–90: (c)Since the pipes are connected in parallel,head loss will be same in both the pipes

2

5Q

12.1Dfl

= C

Q2D5

1

2

QQ

=5/2

1

2

DD

=5/22D

D

1

2

Q5.66

Q

Sol–91: (b)As per stoke’s law, Drag coefficient (CD)for sphere is given as

CD = 24Re {when Re < 0.2}

240 =24VD

240 = 3 324

(20 10 ) (5 10 )

= 10–3 m2/s= 10 cm2/s

= 10 stokesSol–92: (a)

A rising curve for the trajectory of ball

FL

More pressure

Less pressure

Left Direction of motion Right

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Sol–93: (a)The thermal stresses in the boiler areminimised because the temperature rangein boiler is reduced. The efficiencyimproves because the average temperatureof heat addition increases. For a givenpower, a large capacity boiler is required.A deaerater is a special case of open feedwater heater which is specifically designedto remove non-condensable gases from thefeedwater.

Sol–94: (d)Let ‘5’ be the point of combustion chamberinlet due to the regenerator.Temperature at exit of compressor,T2 = 550 K.Temperature at exit of turbine,T4 = 800 K,Effectiveness of regenerator, 0.8

Then, =

5 2

4 2

T TT T

T5 = 2 4 2T T T= 550 + 0.8 × (800 –550)= 550 + 0.8 × 250 = 750K

1

2

3

4

P = c

P = c

T

s

5

Sol–95: (d)1

T

s5 pi

p0

2

3

46

Thus, due to reheating, the specific outputand hence cycle efficiency increases.However the quality of steam increasesdue to which blade erosion decreases. Forreducing the turbine speed, compoundingis done.

Sol–96: (c)High air fuel ratio in gas turbines reducesthe exhaust gas temperature.

Sol–97: (c)The factors on which efficiency depends are1. Material of the nozzle2. Workmanship of the manufacture of nozzle3. Size and shape of the nozzle4. Reynolds number of flow5. Angle of divergence of divergent portion6. Nature of fluid flowing and its state7. Turbulence in fluid and its state.

Sol–98: (b)dAA = 2 dVM 1

V

and dAA = 2

2dp 1 Mv

When M < 1,A decreases P decreasesA increases P increases

When M > 1,A decreases P increasesA increases P decreases

Since M > 1, so for velocity to decreasefrom supersonic to subsonic, the areashould also decrease.

Sol–99: (c)Gas turbine with infinitely large numberof stages during compression andexpansion becomes ericsson cycle.

V

P1

2

34 T = C

T = C

Stirling cycle

1

2

3

4V = c

P = cT

sAtkinson cycle

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P

Ericsson cycleV

1

23

4

T = C

T = C

2 3

1 4V

P

pV =cpV =c

Brayton cycleSol–100: (b)

Superheat regionA

B Wilson line

Decrease in heat drop

Increase in entropy

h

h11

23

44́

h2

Saturation line(x=1)

s

Sol–101: (a)

The nozzle efficiency, n is defined as theratio of the actual enthalpy drop to theisentropic enthalpy drop.

n = 0 1

0 1s

h hh h

to

ho

s

1s 1p1

po

Velocity coefficient is defined as theratio of actual exit velocity to the exitvelocity when the flow is isentropicbetween the same pressures,

= 1

1s

VV

The convergent part of a nozzle is usuallysharp, while the divergent part is gradual.Most of the friction loss occurs in thedivergent portion of the nozzle. If thesemi-divergence angle is large, therewill be flow separation from the wall withthe formation of eddies, which entailsenergy loss. If the angle is small, the

length of nozzle becomes large to providethe desired exit flow area, causing moreenergy loss due to friction. Usually, varies from 5° to 8°.

Sol–102: (b)Sol–103: (c)

*1A * * V = 1 1 1A V

Density ratio = 1

* 2.45=

Velocity ratio = 1

V * 0.8=V

1AA * =

1 1

* V *V

21

2

d4d *

4

= 2.45 × 0.8 = 1.96

1d = 1.96 d * = 70 mm

Sol–104: (b)Single stage impulse: De LavalPressure compounding: RateauVelocity compounding: CurtisReaction turbine: Parsons

Sol–105: (c)Sol–106: (a)

In order to have a constant angular velocityratio for all positions of gears, the commonnormal at the point of contact between apair of teeth must always pass through thepitch point. This is the fundamentalcondition which must be sastisfied whiledesigning the profiles for the teeth of gearwheels. It is also known as law of gearing.In case of involute teeth, common normalintersects the line of centres at the fixedpoint called the pitch point. Therefore, theinvolute teeth satisfy the fundamentalcondition of constant velocity ratio.

Sol–107: (c)R11 (CCl3F): It has ozone depletion potential(ODP) of 1 and greenhouse potential (GP)of 3300. It is a CFC (Chlorofluorocarbons).R12 (CCl2F2): Its is a CFC having ODP of

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1 and GP of 10000.R22 (CHClF2): It is an HCFC, with muchlower ODP of 0.05, hence much less of athroat to ozone layer. It has GP of 1100.The hologenated chlorofluorocarbons(HCFC) have less ODP than CFC

Sol–108: (c)Sol–109: (a)Sol–110: (c)

Heat pipe is device to carry away heatfrom highly complicated areas whereconventional heat removal methods arenot applicable, e.g. computer chip etc.In this system, principles of latent heattransfer or phase change is used to takeaway the heat from heat generatingcomplicated small areas. So theconductance of heat pipe is very high ascompared to normal conductor due to phasechange of heat carrying medium.

Sol–111: (a)In a two high reversing mill the rolls rotatefirst in one direction and then in the otherso that the rolled metal may pass back andforth through the rolls several times.However these are expensive compared tothe non-reversing mills because of thereversible drive needed.

Sol–112: (b) From torsion formula

r

=T GJ L

=T rJ

d2ds

d1

They have same weight Area of x-section will be same for the two

shaft

2Sd

4

=2 21 2(d d )4

2Sd = 2 2

1 2d d ... (i)From (i) we can say that d1 > dS

Assuming same max. shear stress in bothcase

h

s

TT =

hmax

1

Smax

S

J(d / 2)

J(d / 2)

h

s

TT = sh

s 1

dJJ d

h

s

TT =

4 4s1 2

s 1

dd d .d 4 d

h

s

TT =

3 41 2 1

3s

d d / dd

... (ii)

From (i) & (ii)We can say that

Th > Tsalso power transmitted by shaft

P = TP T

So Ph > PsSol–113: (a)

Rankine theory takes into account, thefailure by both crushing and buckling soit is generally used for intermediatecolumns. However it can be used for bothshort column and long column and it givessatisfactory results).According to Euler’s theory,

Buckling stress, = 2

2P EA

where is slenderness ratio.From the above it can be realized that the

stress at failure PA according to the

formula will be high.When the slenderness ratio is small. Butthe stress at failure can not be greater

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than the crushing stress for the columnmaterial. Hence when the slendernessratio is less than a certain limit. Euler’sformula gives a value of the buckling loadeven greater than the crushing load.Hence Euler’s theory gives higher valuesfor buckling loads in intermediate columns.

Sol–114: (c)Refining of grain size means having smallgrain which don’t permit movement ofdislocation easily so strength increases

Sol–115: (d)Notch sensitivity of gray cast iron is zerobecause of presence of graphite flakes withhave itself many notches so practicallyinsensitive for stress concentration.

Sol–116: (c)Sol–117: (c)Sol–118: (b)

Because of high enthalpy drop in veloc-ity compounding turbine (i.e. one 2-rowCurtis stage is equivalent to eight 50%

reaction, two single stage impulse tur-bine) and higher density of steam, ini-tial stages of modern turbine are veloc-ity compounding and subsequent stagesare pressure compounding.Statement II says excessive leakage inhigh pressure zone i.e. Curtis stage. Thisis also true but not the sole criterion.The governing criterion is – having allpressure stages or reaction stages, thelength and size of turbine become verylarge i.e. cost increases.

Sol–119: (c)Rocket engine are used in space researchas well as in millitary applications (Mis-siles). These engine can travel in spacebecause they do not require oxygen forcombustion from atmosphere. Theseengines are moved by reaction force andhave very low specific impulse.

Sol–120: (a)Sprues of converging passage is used toavoid aspiration effect. Due to entrapementof air in casting creates blow holes.

Conventional Question Practice Program · 2017-05-26 · R – 600a: Isobutane Sol–13: (c) If f g...

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Transcript of Conventional Question Practice Program · 2017-05-26 · R – 600a: Isobutane Sol–13: (c) If f g...