ANSWERS - IES Master · 2. (b) Beam engine mechanism is an inversion of four-bar mechanism, whereas whitworth quick return mechanism is an inversion of single slider crank mechanism.

1. (d)

2. (b)

3. (c)

4. (d)

5. (c)

6. (d)

7. (a)

8. (d)

9. (b)

10. (b)

11. (b)

12. (c)

13. (c)

14. (a)

15. (b)

16. (d)

17. (b)

18. (a)

19. (d)

20. (d)

21. (d)

22. (a)

23. (d)

24. (b)

25. (a)

ESE-2019 PRELIMS TEST SERIESDate: 11th November, 2018

26. (b)

27. (c)

28. (d)

29. (b)

30. (d)

31. (a)

32. (c)

33. (d)

34. (c)

35. (b)

36. (a)

37. (c)

38. (c)

39. (d)

40. (a)

41. (c)

42. (d)

43. (b)

44. (b)

45. (c)

46. (c)

47. (c)

48. (c)

49. (b)

50. (a)

51. (c)

52. (d)

53. (b)

54. (b)

55. (c)

56. (b)

57. (b)

58. (d)

59. (c)

60. (c)

61. (a)

62. (b)

63. (c)

64. (c)

65. (c)

66. (d)

67. (d)

68. (b)

69. (a)

70. (b)

71. (b)

72. (c)

73. (c)

74. (d)

75. (c)

ANSWERS

76. (a)

77. (a)

78. (d)

79. (d)

80. (d)

81. (b)

82. (d)

83. (a)

84. (a)

85. (b)

86. (c)

87. (d)

88. (d)

89. (d)

90. (b)

91. (b)

92. (a)

93. (a)

94. (b)

95. (b)

96. (c)

97. (d)

98. (c)

99. (a)

100. (a)

101. (a)

102. (c)

103. (b)

104. (d)

105. (c)

106. (b)

107. (c)

108. (a)

109. (b)

110. (d)

111. (c)

112. (a)

113. (c)

114. (b)

115. (a)

116. (d)

117. (a)

118. (a)

119. (d)

120. (c)

121. (a)

122. (b)

123. (c)

124. (a)

125. (c)

126. (a)

127. (a)

128. (a)

129. (b)

130. (b)

131. (d)

132. (d)

133. (c)

134. (b)

135. (a)

136. (b)

137. (a)

138. (d)

139. (d)

140. (a)

141. (d)

142. (a)

143. (a)

144. (b)

145. (a)

146. (a)

147. (a)

148. (c)

149. (a)

150. (d)

IES M

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(2) MS+TOM+MD+PPE

1. (d) A body in motion will be subjected to Corioli’sacceleration when that body is restrained torotate while sliding over another body.

2. (b) Beam engine mechanism is an inversion offour-bar mechanism, whereas whitworth quickreturn mechanism is an inversion of singleslider crank mechanism. Oldham coupling andelliptical trammel mechanisms are inversionsof double slider crank mechanism.

3. (c) Klein’s construction is a graphical method forthe kinematic analysis of an IC enginemechanism. The velocity diagram representsvelocity of crank, velocity of piston and velocityof piston relative to crank. From this, the velocityof any point on the connecting rod can also befound out. The acceleration diagram is aquadrilateral and the four sides representcentrifugal acceleration of the crank, radialcomponent of acceleration of connecting rod,tangential component of acceleration of theconnecting rod and the acceleration of thepiston.From this, the angular acceleration of theconnecting rod can be found out.

4. (d)

5. (c) The horizontal component which exerts lateralpressure on the bearing is F sin .

6. (d)

7. (a) Given VR = 1,4

m = 6 mm, C = 225mm

VR = 2 1

1 2

N T1N 4 T T2 = 4T1

and C =

1 21 2 m T Td d2 2

225 = 1 26 T T

2

225 2

6= T1 + 4T1

T1 =

225 2 156 5

T2 = 60

number of teeth on gear wheel = 60

d2 = mT2 = 6 × 60 = 360 mm

So, Base circle radius 2br = 2d cos2

= 360 cos 22

2 = 180 × 0.927

= 167 mm

8. (d)9. (b) For wilson Hartnell governor

2a

bs y b4s2 x a

=

C2 C1

2 1

F Fr r

2as b4 2002 a

=

400 2000.2 0.1

2

absa

= 2400 N/m

10. (b) Different mechanism obtained by fixing differentlinks of slider crank chain are as follows.

1. Crank is fixed Rotary engine, whitworthquick return mechanism.

2. Connecting rod is fixed Oscillatory engine,slotted crank mechanism

3. Slider is fixed Pendulum pump, Handpump.

11. (b) The instantaneous centres I12 and I14 remainin the same place for all configurations of themechanism and are called the f ixedinstantaneous centres as the joints are ofpermanent nature. The instantaneous centresI23 and I34 move when the mechanism moves;they are called permanent instantaneouscentres. The instantaneous centre I13 and I24are neither fixed nor permanent as they varywith the configuration of the mechanism.

12. (c) The resultant unbalanced force is

2 2 2 2 2W r (1 c) cos c sing

=

2 22 2 2W 2 2r 1 cos sin

g 3 313. (c) Velocity of any point on a link with respect to

another point (relative velocity) on the samelink is always perpendicular to the line joiningthese points on the configuration (or space)diagram.

VQP = Relative velocity between P & Q

= VP – VQ always perpendicular to PQ.

14. (a) According to Grashof’s law “For a four barmechanism, the sum of the shortest andlongest link lengths should not be greaterthan the sum of remaining two link lengths ifthere is to be continuous relative motionbetween the two links.

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[ME], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-08) (3)

15. (b)16. (d) Corrected pairs are, P-2, Q-6, R-5, S-1

17. (b) Since grassof’s law is satisfied

s p q

Drag - crank mechanism (also known asdouble-crank mechanism) will be obtained byfixing shortest link.

18. (a) Centripetal acceleration, ac = 2r

Tangential acceleration, at = r

Net acceleration, aN = 2 2c la a

a = 2 22r r

19. (d)epicycloid

Hypocycloid

D

C*

20. (d)21. (d)22. (a)

1

2

3

Degree of freedom,

F = 3 1 2j h

= 3(3–1) – 2 × 2 – 1 = 1

where – no. of links

j – no. of lower pairs

h – no. of higher pairs

23. (d)24. (b) Follower velocity in various cases,

1. Cycloidal motion, V =

2h

2. SHM, V =

h h1.570

23. Uniform velocity motion.

V =

h

So the cycloidal motion has maximumvelocity, then SHM and the least velocity is

in uniform velocity motion.25. (a)

26. (b)

Given, TA = 30, TB = 40, NA = 240 (CW),Na = 80 (CW)

B40

30A

a

We know

A

B

TT =

B a

A a

N NN N

3040

=

BN 80240 80

NB – 80 =

30 160 120

40

NB = –120 + 80 = –40

NB = 40 (CCW)

27. (c)

= 0.5 rad/s2, T = 1000 N-m, 1 = 0,t = 15 s

2 = 1 t

= 0 + 15 × 0.5

= 7.5 rad/s

T = I

I =

2T 1000 2000 kg m0.5

Kinetic energy of the flywheel after 15 secondswill be

K.E = 22

1I2

= 21 2000 7.52

= 56250 J or 56.25 kJ

28. (d)Given, P = 400 kW, N = 100 rpm,

Cs = ± 0.5% = 0.01

CE = 0.1

Work done per cycle is given by

IES M

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(4) MS+TOM+MD+PPE

Wcycle = 360 60P 400 10N 100

= 240000 N-m

Coefficient of fluctionation of energy is given by

CE =

cycle

EW …

[ E = Maximum fluction of energy]

0.1 = E

240000 E = 24000 Nm = 24 kN-m

29. (b)Given,

2 = 5 rad/s

3 = 2 rad/s

I12 1 I14

I34I23

I13

42

3

1 2

34

I13

I I12 13

I I14 34

Both the instantaneous centre

(I12 and I13) lies on opposite side of I23 so, thereangular velocities are unlike in nature.

23 =

2 3 5 ( 2) 7 rad / s

30. (d)31. (a)

Total Energy, E = K.E + P.E

= 2 2 21 1 1mv I 3K x2 2 2

= 2 2 21 1 3mx I Kx2 2 2

= 2 2 2 2 21 1 3mr I Kr2 2 2

Now, for conservative system,

E = constant

dEdt

=

dE d 0d dt

2 2mr I 3Kr = 0

2 2mr I 3Kr = 0

2Im 3K

r = 0

Comparing with eq eq eqm c k = 0, we get

eqm = 2

Imr and Keq = 3K

32. (c)Number of natural frequencies = Number of eigenvalues of stiffness matrix.As the order of matrix = 2 × 2

No. of eigen values = 2Hence ‘c’ is correct.

33. (d)

For resonance, Natural f requency n( ) =

Excitation frequency ( ) .

Here,

F = 0F sin t = 200sin50t = 50 rad/s

Keq = 1 2K K[parallel connection]

= 5 52 10 1 10

= 53 10 N m

n = eqKm

m =5

eq2

K 3 10 120Kg50 50

Hence ‘d’ is correct.

34. (c)

The intercept of the direction vector onx, y & z axis are 1/2, 1 and 0 respectively.Taking reciprocal of these intercepts we have(210)

35. (b)Chemicals attack atoms within grain bound-aries more because grain boundaries are inhigher energy state than those in the grainsbecause atoms are not as closely packedon grain boundaries.

36. (a)

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A. Nickel: Corrosion resistance

B. Chromium: Hardenability

C. Tungsten: Heat resistance

D. Silicon: Magnetic permeability

37. (c)Manganese steel or mangalloy or Hadfield steelhaving manganese content from 11% to 14%is extremely wear resistant steel used in earthmoving equipments and mining industry.

38. (c)Air has very low cooling rate followed by, fusedsalt, oil and water in increasing order.

39. (d)

Time, temperature, Transformation (TTT)diagrams indicates transformation of austeniticphase.

40. (a)

41. (c)Being a semiconductor with very good electricalproperties, silicon and germanium are widely usedin electrical industry.Hence ‘c’ is correct.

42. (d)German silver variers in composition, thepercentage of three elements ranging approximatelyas follows : copper, from 50% to 61.6%, zinc,from 19% to 17.2%, Nickel from 30% to 21.1%.The proportions are always specified in commercialalloys. Hence there is no silver.Hence ‘d’ is correct.

43. (b)The effectiveness of adding carbon to harden thesteel depends on lattice spacing, crystal structure,shape and distribution of carbide in iron.Also adding more carbon after a limit will makethe steel brittle rather hard.Hence (b) is correct.

44. (b)Work hardening also known as strain hardening orcold working, is the strengthening of a metal byplastic deformation. This strengthening occursbecause of dislocation movement and dislocationgeneration withing the crystal structure.Hence ‘b’ is correct.

45. (c)Toughness: Energy absorbed before fracturein a tension test.Endurance strength: Fatigue loading.Resistance to abrasion: Hardness.Deflection in beam: Moment area method.

46. (c)47. (c)48. (c)

In vertical hyperbolic profile there is an advantageof superior strength and greatest resistance tooutside wind loading compared to other shape.

49. (b)Cooling efficiency,

= (Incoming water temp) (Exiting water temp)(Incoming water temp) (WBT of ambient air)

= 35 27 8 0.5714 57.14%35 21 14

50. (a)51. (c)

52. (d)

Since boiler operates at saturated state, heataddition and rejection will be isothermal as wellas isobaric.

T = C

T = C

S

T

53. (b)

e

1nd

vols

P1 C C

P

11.50

min vol640 1 C C1

0 = 1 + C – 16C = 1 – 15C = 0

C = 1/15

IES M

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(6) MS+TOM+MD+PPE

54. (b) Degree of reaction

= Enthalpy drop in moving blade

enthalpy drop in stage

55. (c)Volume of air when converted to 1 bar pressure& 15°C is referred as free air delivery.

56. (b)

No. of stages = In (overall pressure ratio)

ln (Pr essure ratio per stage)

pressure ratio per stage, 2

1

PP =

12

1

TT

= 1.4

1.4 1327 27327 273

= (2)3.5 = 3(2) 2 = 8 × 1.414 = 11.31

No. of stages = ln (113.13 / 1)

ln (11.31) = 2

57. (b)2

max cos

energy input = 21

1 C2

max maxW energy input

= 2 2 2 21 1

1 1cos C C cos2 2

Qsupplied = Wmax + Qrejected

2 2 21 1 rejected

1 1C C cos Q2 2

Qrej = 2 2 21 1

1 1C C cos2 2

= 2 21

1 C sin2

58. (d)For minimum work of compression

3 2

2 1

P PP P

23 21

1P PP

= 2 1 16(4)

(0.5) 0.5 = 32 bar

59. (c)

IC engine power plant has higher weight topower ratio than gas turbine power plants.

(Tmax)IC engine = 2500°C

(Tmax)gas engine = 1200°C

IC engine plants are more polluting than gasturbine plants

60. (c)

Air being a diatomic gas is used as workingfluid in Brayton cycle.

cp & cv are constant and not a function oftemperature during the cycle.

61. (a) Since degree of reaction = 0.50 = 50%

This turbine is parson’s reaction turbine

1r 2V V & 21 rV V

1r 2V 100 m / s V

2r 1V 300 m / s V

Energy input to the blade is given by

2 12 22 2 2 2r r1 V VV 300 300 100

2 2 2 2

= 45000 + 40000 = 85000 J/kg = 85 KJ/kg

62. (b)

brayton regenerative = 1

1P

3

T1 (r )T

Pr

1T

3T

only statement (4) is correct.

63. (c)Pulse Jet is an intermittent combustion engineand it operates on a cycle similar to that usedin IC engine. In pulse Jet engine cycle heataddition occurs at constant volume & expansionoccurs at constant entropy and therefore it canbe compared with an ideal otto cycle than Braytoncycle.

64. (c)

1

2

34

5

6P = C

P = C

P = C

P = CTurbine

NozzleP = C

Combustion chamber

Ram compressionin diffuserCompressor

T

s

T-s diagram for a Turbo Jet engine.

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65. (c)A regenerative rankine cycle is possible withinfinite number of reheater’s which leads toefficiency equal to carnot efficiency.

66. (d) Heat added = h2 – h5 = Tm (s2 – s5)

where Tm is the mean temperature of heataddition.

(2800 – 200) = Tm (4.240 – 1.240)

Tm = 866.66 K = 593.66°C

67. (d)

Circulation ratio of riser circuit should not bemore than 25 otherwise riser design willbecome uneconomical.

Circulation ratio of downcomer circuit shouldnot be less than 6 otherwise riser tubes willfail by overheating.

68. (b)Degree of reaction of axial flow compressor =

2w

2

V1 0.60

2u

2

1001 0.602u

2 2

50 501 0.60 0.40u u

u2 = 125 m/s

u2 = DN60

= D 1193 125

60

D = 2m

69. (a) topping bottomingoverall 1 11

T0.60 1 1 1 0.40

T0.60 1 0.601

T 0.60 0.401

T 0.333 33.3%

70. (b)

Inverted type have proper drainage of condensedsteam among all types of superheater & reheater.

gas flow

71. (b)h3 = mh1 + (1 – m)h2

= 0.20 × 2800 + (1 – 0.20) × 1100

= 1440 KJ/kg

72. (c)

enthalpy drop in turbine, plantenthalpy drop in boiler

800 0.402000

Ienthalpy drop in all turbine, plant

Total enthalpy drop in heat addition

1200 0.482500

II

I

plant 0.48 1.2plant 0.40

73. (c)

sm = 5400 kg/hour

= 5400 1.5 kg / sec3600

Turbine work = 1.5 × 2200 = 3300 KW

Pump work = 1.5 × 200 = 300 KW

Net work = 3300 – 300 = 3000 KW

Specific steam consumption

= 3600 3600 1.2 kg / kw hr

Net work 3000

74. (d)

Back work ratio = Compression work

Turbine work

BWR =

p 2 1 2 1

p 3 4 3 4

mc T T T Tmc T T T T

In an ideal brayton cycle

3 11 44

2 3 2

T .TT T TT T T

BWR = 2 1

3 13

2

T TT TTT

=

2 1

3 2 1

2

T TT T T

T

BWR = 2

3

TT

IES M

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(8) MS+TOM+MD+PPE

75. (c)

p opt

brayton r ) = min

max

T1T

= (127 273)11000

=4001 1 0.63 0.37

1000

carnot = min

max

T 127 2731T 1000

= 1 – 0.4 = 0.60

brayton w max

carnot

=

0.37 0.610.60

76. (a)Terminal temperature difference = saturationtemperature of bled steam – feed water exittemperature.

77. (a)

mobESP

AV1 exponentialQ

where

A is the collecting plate area, Q is the volumetricflow rate of air for each plate and Vmob is theeffective migration velocity of particles

As mob ESPA , V , Q

78. (d)

T

s

1

24

3

Now, power = a T Cm (w w )

2000 = p 3 4 p 2 1a c T T c T Tm

= am 1 11200 600 600 300

2000 = am 600 300

am 6.66 kg / s

79. (d)

effectiveness = a 2

4 2

T T 0.80T T

aT 500900 500

= 0.80 TTa = 820 K

Turbine work compressor workHeat supplied

=

p 3 4 2 1

p 3 a

mc T T (T T )mc T T

1200 900 500 300 26.3%1200 820

80. (d)

Pre

ssur

e ra

tio

Chockinglimit

operatingrange

Surge limit

mminm maxm

81. (b)At low speeds jet propulsion has poor fueleconomy than IC engine.

82. (d)The losses drift & blowdown occur in coolingtowers of a steam power plant.

83. (a)

Due to irreversibility ideal actual( h) ( h) duringexpansion process in turbine.

84. (a)

Efficiency of regenerative cycle, 1

1p

3

T1 rT

For an ideal regenerative brayton cycle aspressure ratio increases, efficiency of the plantdrops.

85. (b)

For speed less than 200 m/s reciprocatingengine has higher propulsive efficiency. From200 m/s to 1000 m/s turbojet engine aremore efficient. For speed greater than 1000m/s, rocket engine are most efficient.

86. (c) Positive contact clutches. Once coupled cantransmit large torque with no slip.

In general, positive clutches are rarely usedas compared with friction clutches. Howeverthey have some important application wheresynchrnous operation is required like powerpresses and rolling mills.

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87. (d)88. (d) Although needle bearings are considered as

a variety of cylinderical roller bearings. Theyare essentially not the same.

The needles, which are considerably longerthan their diameter, cannot be manufacturedwith the same degree of accuracy, this resultsin high friction in needle bearings.

89. (d)90. (b) Insufficient tightness of the outer race in the

housing seat may cause ‘creep’. In bearingterminology, creep is slow rotation of the outerrace relative to its seating.

91. (b) Axles do not used for power transmission likeshaft. So it is designed using simple bendingmoment only.

92. (a) It is known as pitting action which is a kind ofsurface fatigue failure. It occurs when loadingexceeds the surface endurance strength.Surface fatigue in case of ball bearing is knownpitting.

93. (a) Maximum principle stress theory is used fordesigning a brittle material as these materialfails under tensile stress / Normal loadingcondition.

94. (b)95. (b)96. (c) In case of bolt of uniform strength we do

(i) Drill a bolt i.e. shank area is reduced to rootarea.

(ii) Reduce the diameter of shank of boltcorresponding to that of minor diameter.

97. (d)Tapered roller can bear axial as well as radialload.

But, r

a

F 1F

. So option (d) is correct.

98. (c) Power transmission = (T ) = C

Since as vary, T also vary, so shaft willexperience constant bending and varyingtorsional stresses.

99. (a) Spiral gears are used for connecting non-parallel and non-intersecting shafts for powertransmission.

100. (a) It is due to wedge shape film of oil whichcase bearing function in a hydrodynamic sliderbearings.

101. (a) It is advantageous to use hydro-dynamicjournal bearing as they have high load carryingcapacity at high speed due to hydrodynamicpressure developed by the film.

102. (c) A little variation in centre distance of gearsdoes not affect the velocity ratio if the teethprofile of gears are involute where as incycloidal profile of teeth, exact centre distanceshould be maintained.

103. (b)

ZNP

Thin film lubrication

Mixed lubrication

Thick filmlubrication

FigureIn thickfilm Hydrdynamic journal bearing, the

coefficient of friction decreases with increasein load i.e. decrease in bearing characteristic

number ZNP

.

104. (d)1. At high temperature heat treatment, all the

chromium carbide particles are redissolved.

2. so that chromium carbide formation is minimal

3. which has greater tendency to form carbidesthan chromium, so chromium reamins in solidsolution.

105. (c)The dislocation density in a metal increaseswith deformation or cold work, due to dislocationmultipl ication or the formation of newdislocations.

106.(b) The case obtained by cyniding is of high wearresistant and endurance limit as compared tocarburising.

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(10) MS+TOM+MD+PPE

107.(c) Since the martensitic transformation does notinvolve dif fusion, i t occurs almostinstantaneously, the martensite grains nucleateand grow at a very rapid rate. Thus themartensitic transformation rate, for all practicalpurposes, is time independent.

108.(a) Increasing the carbon content in the steelincreases the strength.

the strength of fine pearlite is greater thancoarse pearlite of same carbon steel.

109.(b)

Knoop test is a microhardness test because ofthe light load used; hence it is suitable for verysmall or thin specimens and for brittle materialssuch as gemstones, carbide and glass.Because the impressions are very small, thistest is also used to measure the hardness ofindividual grains in a metal.

110.(d)

Because of presence of cementite, fracturedsurface appear white, hence the name as whitecast iron.

Malleable cast iron is formed after heat treatingwhite cast iron. Heat treatments involve heatingthe material up to 800–900°C and keep it forlong hours, before cooling it to roomtemperature.

111.(c)

112. (a)Idle gears have no effect on the train value orvelocity ratio of gear trains. It only used forchanging the direction of rotation of the drivenshaft.

113. (c)

Quick return ratio,

= 3

Q

P

O2

O1

360 3

90

In right angle 1 2O O P ,

length of crank, O1P = 1 2(O O )cos / 2

= 220 × cos(90/2) = 155.56 mm

114. (b)x

m

Applying newton’s law, mx A xg

Agx x 0

m

Natural frequency : fn =

1 Ag Hz

2 m115. (a)

1.0

0.8

0.6

0.4

0.2

0

0

xx

0 1 2 3 4 5

10

5

2

1Criticallydamped

nt

Fig. Displacement-time curves of over damped and critically damped system

Door closer is a overdamped system because inthis system it need some time to come back toequilibrium.

Energy method can be used in conservativesystems only. Damped system is not aconservative system.

116. (d)

Natural frequency, n = k 10000m 1

= 100 rad/s

Damping factor, n

c 402m 2 1 100

= 0.2

Damped vibration frequency,

d = 2n1

= 21 0.2 100 = 97.98 rad/s

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117. (a)118. (a)

I = m.k2 = 450 × (0.5)2 = 112.5 kg/m2

pV 60r 30

= 2 rad/s

= 200 rad/s

Gyroscopic couple, C = p

= 112.5 × 200 × 2 = 45 KN-m

ViewRear(tail) Nose

a’

b’b

ao

b'a' is the reaction couple and is perpendicularto oa in the limit. This couple acts in the verticalplane and tends to raise the nose and depressthe tail of the aeroplane.

119. (d)For static balancing

2 2c cm r mr cos mr sin ...(i)

mrcos 4 50 cos30 2 100 cos150 0

mr sin 4 50 sin30 2 100 sin150 200

2 2cm 100 0 200

mc = 2 kg

cmr sintanmr cos

c200tan0

ctan

c = 90° or 270°

Numerator is negative, so angle will be in 3rd or4th quadrant.

c = 270°

120. (c) Secondary or imaginary crank makes an anglewith the line of storke equal to twice that ofthe engine crank.

121. (a) When the acceleration or deceleration is verysmall or change in velocity is very slow, theadditional inertia force is practically zero andinertia governor in effect, become a centrifugalgovernor.

122. (b)123. (c) The damping force is proportional to instant

velocity as,

Fd =dxcdt

This instant velocity is not related to ampli-tude any way. So the damping force is notproportional to amplitude.

124. (a)125. (c) A pantograph is a bar linkage used to produce

paths exactly similar to the ones traced outby point on the linkage. The paths soproduced are usually, on an enlarged orreduced scale.

126. (a) As gears for skew shafts have point contact,so they are not used for high powertransmission.

127. (a)128. (a) In ideal cycle, internal f riction and

irreversibilities are assumed to be zero,therefore any heat rejection will definitelydecrease the entropy.

129. (b)

130. (b)

131. (d) Multistage compression is very useful methodfor compressors where the overall pressurerise required is large.

Multistaging with intercooler reduces workinput and temperature. So, reduces lubricationand cooling required.

132. (d) Approach is the difference in temperaturebetween the exit temperature of cooling waterand entering air wet bulb temperature. Naturaldraught cooling towers are more dependenton temperature gradients between air & water.

When temperature gradient is high i.e.approach is high, natural draught coolingtowers are more efficient.

133. (c) Design considerations for reheaters &superheaters are same as they both are heatexchangers. Design temperature forsuperheater & reheater is same but designpressure in reheater is only 20–25% ofsuperheater design pressure.

134. (b)

IES M

ASTER

(12) MS+TOM+MD+PPE

135. (a) Gas turbine power plant’s efficiency dependsupon compressor & turbine efficiency.Whereas efficiency of thermal plant dependsonly upon turbine efficiency since pump workis negligible in comparison to turbine work.

136. (b) Proper turbulence (so that the fuel moleculesmeet the oxygen molecules) will help incomplete burning of coal.

Since the complete mixing of the fuel and airis virtually impossible, excess air must besupplied to ensure complete combustion.

137. (a) When the lining is put into service, wearoccurs. Therefore a major portion of life ofclutch comes under the uniform wear criterion.

138. (d)139. (d) In hydrodynamic journal bearing, lifting support

pressure to journal generated usinghydrodynamic action of shaft / journal onlubricant.

140. (a)141. (d)

For cone clutch, torque transmitting capacity

Tcone = p

4sin

(D+d) ...(1)

For single plate dutch, torque transmitting capacity

Tplate = p D d4

for same dimensions,

t cone

t plate

TT =

1 1 4.62sin sin 12.5

so, (Tt)cone > (Tt)plate

142.(a)143.(a) The galvanic series represents, the relative

reactiv ities of a number of metals andcommercial alloys in seawater. The alloys nearthe top are cathodic and unreactive andwhereas those at the bottom are most anodic(active).

144.(b) During recrystallization, grain-boundary motionoccurs as the new grain nuclei form and thengrow. It is believed that impurity atomspreferentially segregate at and interact with

those recrystallized grain boundaries so as todiminish their (i.e., grain boundary) mobilities,this results in a decrease of the recrystallizationrate and raises the recrystal l izationtemperature.

145.(a) The higher the carbon content, the lower thecorrosion resistance of stainless steel. Thereason is that the carbon combines withchromium in the steels and forms chromiumcarbide, which lowers the passivity of the steel.

146.(a) Recrystallization depends on the degree of priorcold work; the higher the amount of cold work,the lower the temperature required forrecrystallization to occur. The reason for thisinverse relationship is that as the amount ofcold work increases, the number of dislocationsand the amount of energy stored in thedislocations also increase. The stored energysupplies some of the energy required forrecrystallzation.

147. (a)It is necessary for a link to be resistant body totransmit power and motion.

A belt act as a rigid body, when it is subjectedto tensile force and deformations are inpermissible limits, so the belt-drive acts as aresistant body.

148. (c)Ackermann steering gear has only turning pairswhere as Davis steering gear has slilding pairswhich means more friction and easy wearing.So, Ackermann steering gear is preferred.

It fulfills the fundamental equation of gearing atthe middle and the two extreme positions andnot at all positions.

149. (a)Involute profile depends on base circle. Increasingcentre to centre distance between gears doesnot effect the base circle, so law of gearing holds.

150. (d)The speed variation due to variation in turningmoment of engine is controlled by flywheel.Governor is used to control the variation due toload-speed characteristic of engine.

ANSWERS - IES Master · 2. (b) Beam engine mechanism is an inversion of four-bar mechanism, whereas whitworth quick return mechanism is an inversion of single slider crank mechanism.

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