1. (d)

2. (c)

3. (a)

4. (c)

5. (a)

6. (a)

7. (b)

8. (a)

9. (a)

10. (c)

11. (b)

12. (c)

13. (a)

14. (b)

15. (c)

16. (b)

17. (a)

18. (c)

19. (a)

20. (a)

21. (b)

22. (d)

ESE-2018 PRELIMS TEST SERIESDate: 01 October, 2017

23. (b)

24. (a)

25. (d)

26. (b)

27. (d)

28. (c)

29. (c)

30. (c)

31. (a)

32. (b)

33. (b)

34. (b)

35. (b)

36. (b)

37. (c)

38. (c)

39. (c)

40. (b)

41. (d)

42. (b)

43. (d)

44. (c)

45. (d)

46. (c)

47. (b)

48. (c)

49. (a)

50. (b)

51. (d)

52. (b)

53. (a)

54. (d)

55. (a)

56. (b)

57. (c)

58. (d)

59. (b)

60. (c)

61. (c)

62. (a)

63. (c)

64. (c)

65. (d)

66. (c)

ANSWERS

67. (a)

68. (a)

69. (a)

70. (b)

71. (b)

72. (a)

73. (c)

74. (d)

75. (a)

76. (c)

77. (b)

78. (c)

79. (b)

80. (a)

81. (a)

82. (a)

83. (a)

84. (c)

85. (a)

86. (d)

87. (a)

88. (c)

89. (c)

90. (c)

91. (b)

92. (c)

93. (d)

94. (c)

95. (d)

96. (d)

97. (c)

98. (b)

99. (d)

100. (b)

101. (b)

102. (b)

103. (c)

104. (d)

105. (c)

106. (c)

107. (b)

108. (a)

109. (a)

110. (d)

111. (b)

112. (a)

113. (c)

114. (d)

115. (a)

116. (b)

117. (b)

118. (a)

119. (c)

120. (d)

121. (a)

122. (d)

123. (d)

124. (a)

125. (b)

126. (a)

127. (a)

128. (c)

129. (a)

130. (a)

131. (c)

132. (c)

133. (d)

134. (a)

135. (a)

136. (a)

137. (a)

138. (c)

139. (a)

140. (b)

141. (b)

142. (a)

143. (a)

144. (a)

145. (a)

146. (a)

147. (b)

148. (a)

149. (b)

150. (b)

(2) (Test - 02)-01 October 2017

Sol–1: (d)Length of cut × depth of cut= length of chip × thickness of chip 76 × 0.2 = 61 × t

t = 0.25 mmSol–2: (c)

BUE always decreases tool life.Increasing the rake angle decreasesthe cutting forces, heat produced atthe tool tip, and therefore increasestool life. However, increasing the rakeangle to a large value, reduces thetool material available at the tool tipfor conducting heat generated, thusincreasing the tool-tip temperature.This would decrease the tool life, thusexplaining the existence of anoptimum value for the rake angle asshown in figure.

200

150

100

50

0 5 10 15 20Rake angle, deg

Tool

life

's

f = 0.127 mm/revV = 0.66 m/s

f = 0.508 mm/revV = 0.41 m/s

Tool-life as affected by rake angleand feed rate

Sol–3: (a)

ChipTool

FSFH

NS

FV WorkpieceF

N

R

Sol–4: (c)Increasing the side cutting edge angleincreases the chip contact length anddecreases chip thickness. As a result,the cutting force is dispersed on a

longer cutting edge and so tool life isprolonged.

Sol–5: (a)Sol–6: (a)

For max production rate:Total production time (T p) =machining time + tool changing time+ non-productive timeFor finding optimum cutting speedand optimum tool life for maximum

production rate, PdTdV = 0.

Optimum cutting speed

Vopt =

n

C

C1 1 tn

where tC = tool changing time inminutes

C = Machining constantn = Exponent depending on

cutting conditionswhere VTn = C

Sol–7: (b)Melting unit is an integral part in hotchamber Die casting.

Sol–8: (a)Sol–9: (a)Sol–10: (c)Sol–11: (b)

Elasticity is the property of a materialto regain its original shape afterdeformation when the external forcesare removed.Ductility is the property of a materialenabling it to be drawn into wire withthe application of a tensile force.The ductility is measured in terms ofpercentage elongation and percentagereduction in area.Malleability is a special case of ductilitywhich permits materials to be rolled orhammered into thin sheets.

Sol–12: (c)Sol–13: (a)

Chaplets are metallic supports keptinside the mould cavity to support the

(3) (Test - 02)-01 October 2017

cores. These are supposed to fuse withthe parent metal.

Sol–14: (b)Sol–15: (c)Sol–16: (b)

Sol–17: (a)

OB

CA

( )

Shear strain , that the materialundergoes

= AB AO OBOC OC OC

= cot tanSol–18: (c)Sol–19: (a)Sol–20: (a)Sol–21: (b)Sol–22: (d)

It is also called the hardness of thewheel. This designates the forceholding the grains. The grade of awheel depends on the kind of bond,structure of wheel, and amount ofabrasive grains. Greater bond contentand strong bond results in hardergrinding wheel. Harder wheels’ holdthe abrasive grains till the grindingforce increase to a great extent. Thegrade is denoted by letter grades asindicated in figure.

51 A 36 L 5 v 23

Manufacturer'soptionNature ofabrsive

Types ofabrasive

Grainsize

Grade Structure Type ofbond

Manufacturer'sreference

A– –Al O2 3

C–SiC

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

V VitrifiedS–SilicateR–RubberB–ResinoidE–Shellac

–Open

89

1011121314

01234567 Hard

CloseVeryfine220240280320400500600

80100120150180

3036465460

1012141620

Coarse Medium Fine

MediumSoft

Very soft Medium V e r yhardABCDEFGHKLMNOPQRSTUVWXYX

Sol–23: (b)Sol–24: (a)

The process of metal rolling is madepossible the friction that occursbetween the contact surfaces of the rollsand the part being rolled. At themoment of contact, two forces act onthe metal from the side of each roll,normal force P and the tangential forceP, where is the coefficient offriction between the metal and the rollsurfaces. The part would be dragged inif the resultant of horizontal componentof the normal force P and the tangentialforce P is directed in that direction. Inthe limiting case,

Psin = Pcos = tan

or = 1tanIf the angle of contact is greaterthan 1tan , the metal mold notenter the space between the rollsautomatically.

Sol–25: (d)Deep drawing : used for makingcylindrical shaped parts such as cups,shells etc from sheet metal.Drawing : used for producing wires,rods, tubesForging : plastic deformation ofmaterial between two dies to achievedesired configuration.Spinning : a disc or tube of metal isrotated at high speed and formed intoan axially symmetric part.

Sol–26: (b)Aluminium is an active metal and itreacts with oxygen in air to producea thin film of aluminium oxide on itssurface. The melting point ofaluminium oxide is almost three timesthat of pure aluminium. In addition,this aluminium oxide film will absorbmoisture from the air which causesporosity in the aluminium welds.

Sol–27: (d)Sol–28: (c)Sol–29: (c)

(4) (Test - 02)-01 October 2017

Sol–30: (c)Coarse grains are refined and finegrain structure obtained in case of hotworking. In hot working, the metalsare plastically deformed above theirrecrystallization temperature.In cold working, ultimate strengthand yield strength increase, hardnessincreases but ductility decreases.

Sol–31: (a)

O

CB

AXd

O´

B´ c

Feed per tooth

BC = 2 2

2 2 D DOB OC = d2 2

= d D dSol–32: (b)

Crater wear occurs at the tool-chipcontact area where the tool issubjected to friction force of themoving chip under heavy loads and hightemperatures. At these hightemperatures, the atoms in the toolcontinuously diffuse to the movingchip. The temperature is greatestnear the midpoint of the tool-chipcontact length, where the greatestamount of crater wear occurs due tointensive diffusion.

Sol–33: (b)Transfer molding, like compressionmolding, is a process where theamount of molding materials (usuallya thermosetplastic) is measured andinserted before the molding takesplace. The molding material ispreheated and loaded into a chamberknown as the pot. A plunger is thenused to force the material from thepot through channels known as asprue and runner system into themold cavities. The mold remainsclosed as the material is inserted andis opened to release the part from the

sprue and runner. The mold walls areheated to a temperature above themelting point of the mold material;this allows a faster flow of materialthorugh the cavities.This is an automated operation thatcombines compression-, molding, andtransfer-molding processes. Thiscombination has the good surfacefinish, dimensional stability, andmechanical properties obtained incompression molding and the high-automation capability and low cost ofinjection molding and transfermolding.

Sol–34: (b)Sintering is a heat treatment appliedto the compacted powder in order toimpart strength and integrity. Thetemperature used for sintering isbelow the melting point of metal.

Sol–35: (b)In green sand moulding process,uniform ramming leads to greaterdimensional stability in the castings.

Sol–36: (b)Orthogonal cutting:Cutting edge remains perpendicularto the cutting velocity. Direction ofchip flow velocity is normal to thecutting edge. Angle of inclination andchip flow velocity are zero. Cuttingedge is larger than the width of cut.Only two mutually perpendicularcomponents of cutting forces areacting on the tool cutting edge. Themaximum chip thickness occurs in themiddle. The tool prepares a surfacewhich is parallel to the work surface.

Sol–37: (c)Ceramic tools are produced by asintering process. The mainconstituent is Al2O3. Ceramic toolsretain their hardness up to 1400°C.They have better resistance toabrasion wear and crater formationthan cemented carbides. Also, theyexhibit a low coefficient of frictionwith most work materials. Due tothese reasons, ceramic tools can beused at substantially higher cuttingspeeds and therefore, help achievehigh production rates. Ceramic tool

(5) (Test - 02)-01 October 2017

tips are highly brittle. To givestrength to the tool tip, negative rakeangle is given.

Sol–38: (c)In the metals subjected to cold forming,with plastic deformation dislocationdensity increases because underapplied stress, dislocation sourcesbecome active. High density ofdislocations and thus increasedinteraction under applied loads duringcold working is the cause for increasein strength of the material.

Sol–39: (c)Crater wear occurs on the rake faceof the tool. The most significantfactors affecting the crater wear aretemperture at the tool-chip interfaceand the chemical affinity between thetool and the workpiece materials.The location of maximum depth ofcrater wear coincides with thelocation of maximum temperature atthe tool-chip interface. It is because,crater wear has been described interms of diffusion mechanism, that isthe movement of atoms across thetool-chip interface since diffusion rateincreases with increasing temperature,so crater wear increases astemperature increases.

Sol–40: (b)Sol–41: (d)

In capstan lathe, the capstan head ismounted on a ram or slide. It slides inthe ways provided on the saddle. Thesaddle can be fixed anywhere along thebed. The operating traverse of the toolis made by moving the slide and notthe saddle.

Sol–42: (b)Punching operation is used for makingholes in a sheet. In the punchingoperation, clearance is provided on thedie, while in the blanking operation,clearance is provided on the punch.

Sol–43: (d)Two high nonreversing millsIn this, the rolls ofequal size are rotatedonly in one direction.

+

+Three high mills(Reversing)In this, the direction ofmaterial movement isreversed after eachpass; the plate beingrolled is repeatedlyraised to the upper rollgap, rolled, and thenlowered to the lowerroll gap by elevatorsand various manipulators.

+

+

+

Four-high mills

+

+

+

+

It is based on theprinciple that small-diameter rolls lower rollforces and powerr e q u i r e m e n t s .Moreover, when broken,small rolls can bereplaced at less costthan can large ones.Cluster mills : On the sameprinciple, the diameter of working rollis very small.

Sol–44: (c)Roll forces can be reduced by any ofthe following 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–45: (d)

The thin sections of the casting solidifyfaster than the thicker sections. Due tothis, there will be uneven contraction,thereby giving rise to internal strainsin the casting, even develop cracks inthe casting. Hence, for rapidsolidification of heavy sections and toachieve directional solidification, chillsare commonly used.

(6) (Test - 02)-01 October 2017

Sol–46: (c)Sol–47: (b)Sol–48: (c)

All thermoplastic resins can beextruded. Thermosetting resins aregenerally not suitable for extrusion.In injection moulding, exact amountof material to fill the cylinder isdelivered. Further, injection diemoulding is generally limited toforming thermoplastic material.Extrusion moulding process is usedfor rods, tubes, pipes etc.

Sol–49: (a)• Most plastics have low density of

the order of 2.7g/cm3.• Plastics in general are soft.• Plastics have high electric and

thermal insulation properties.• Good resistance to shock and

vibrations.• High corrosion resistance and even

acid alcohol resistance.• High optical properties like

transparency.• Ease of fabrication• Good surface finish & less costlyUnfavourable Features• low tensile & compression strength• low in heat resistance• high hygroscopicity• considerably brittle at low

temperature• high deformation under load and

have low creep strengthSol–50: (b)

In die casting, the casting is producedby forcing molten metal under pressureinto a permanent metal mould. The diecasting is mostly used for castings ofnon-ferrous metals of comparatively lowfusion temperature. By this process,thin and complex shapes can be easilyproduced.Zinc die castings have high strength.Zinc in its pure form is a brittle, lowstrength metal, but when combined

with small amounts of aluminium andmagneisium, zinc die casting alloyshave high strength and hardness, goodductility and excellent castability.

Sol–51: (d)Lubrication is an important factor whenmanufactured by metal drawing, itsapplication can help control the drawingforces and metal flow. Lubrication willalso extend the life of the mold, reducetemperature and improve surface finish.

Sol–52: (d)In shell moulding process, the mould ismade up of mixture of dried silica andphenolic resin, formed into a thin half-mould shells which are clampedtogether for pouring metal. The sand isfirst mixed with either urea or phenolformaldehyde resin. Metal pattern isheated to 205 to 230°C in an oven andsprayed with silicon grease and kepton the top of the dump box. The dumpbox contains sand mixed withthermoplastic resin. The box is inverted,causing the sand mix to fall on the hotpattern. The resin melts and flows inbetween the grains of sand, acting as abond. After 30 seconds, a hard layer ofsand is formed over the pattern. Thenthe dumped box is inverted back to itsoriginal position. The pattern with athin shell is cured for two minutes at315°C. The shell is finally removed fromthe pattern by ejector pins. The twoshells are clamped together to form themould and placed in the flask withbacking sand.

Sol–53: (a)In ammonia vapour absorptionrefrigeration system, ammonia isabsorbed in water at low temperatureand after being heated by generatorthe ammonia vapour is released fromthe aqua ammonia solution. This isbecause the solubility of ammonia inwater decreases as the temperatureincreases. Ammonia is used as therefrigerant and water as absorbent.

Sol–54: (d)Sol–55: (a)

R – 1349: Tetrafluoroethane

(7) (Test - 02)-01 October 2017

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

Sol–56: (b)Sol–57: (c)

Slight superheating of vapour isdesirable at the suction of compressor,otherwise liquid refrigerant wouldwash away the lubricant oil ofcompressor thus increasing wear &tear.

Sol–58: (d)In split AC condenser is also locatedoutside the room.

Sol–59: (b)Desert air cooler cools air throughevaporation of water.

Sol–60: (c)Sol–61: (c)

COP = refrigerating effect

work done in compressor

=

1 4

2 1

300 150 150 5= =330 300 30

h hh h

p

3

41

W

2Q

h

1Q

2

Sol–62: (a)

1

Humidityratio

2

TemperatureIn winter air-condition ing, therequirement is heating andhumidification of air. Thus, wbt anddbt both will increase. Temperatureand humidity ratio both will increase.Humidity ratio will rise but relativehumidity can be lower.

Sol–63: (c)

Pb = a vP P

Pv = 1 03 1 0 03bar= =b aP P

Relative humidity

0 03 60%= =0 05

v

s

PP

Sol–64: (c)Psychrometric chart is drawn for fixedatmospheric pressure.

Sol–65: (d)Sol–66: (c)

Temperature and humidity ratio bothdecrease in process AB.

Sol–67: (a)Effective temperature is uniformtemperature of an imaginaryenclosure with the same heat transferby radiation and convection as inactual environment.The degree of warmth or cold felt bya human body depends mainly on thefollowing three factors:1. Dry bulb temperature2. Relative humidity, and3. Air velocityIn order to evaluate the combinedeffect of these factors, the termeffective temperature is employed. Itis defined as that index whichcorrelates the combined effects of airtemperature, relative humidity and airvelocity on the human body. Thenumerical value of effectivetemperature is made equal to thetemperature of still (i.e. 5 to 8 m/minair velocity) saturated air, whichproduces the same sensation ofwarmth or coolness as produced underthe given conditions.The practical application of theconcept of effective temperature ispresented by the comfort chart. Thischart is the result of research madeon different kinds of people subjectedto wide range of environmentaltemperature, relative humidity and airmovement by the American Society ofHeating, Refrigeration and Airconditioning Engineers (ASHRAE). Itis applicable to resonably still air (5to 8m/min air velocity) to situationswhere the occupants are seated at rest

(8) (Test - 02)-01 October 2017

or doing light work and to spaceswhose enclosing surfaces are at amean temperature equal to the air drybulb temperature.All points located on a given effectivetemperature line in the comfort chartdon’t indicate conditions of equalcomfort or discomfort. The extremelyhigh or low relative humidities mayproduce conditions of discomfortregardless of the existent effectivetemperature. The most desirablerelative humidity range is between 30and 70 percent. Further, it does nottake into account the variations incomfort conditions when there arewide variations in the mean radianttemperature. The effect of meanradiant temperature on comfort is lesspronounced at high temperature thanat low temperatures.Effective temperature takes intoconsideration the air velocity and isapplicable to reasonably still air withair velocity 5 to 8m/min.

Sol–68: (a)ADP = 12°C t2 = 15°C

BPF = 2

10 15=

t ADPt ADP

0.15 =1

15 1212t

ADP=12°C

1 2

t1 = 312 12 20 32= =0 15

C

Sol–69: (a)

T

s

T =T2 33 2

14

s =s4 3 s =s1 2

COP of the refrigeration systemworking on reversed Carnot cycle

(COP)R = 1

2 1

TT T

where T1 = lower temperatureand T2 = higher temperature

T1

T2

RW

COP = 5

COP = 1

2 1

TT T

5 =2

1

1T 1T

2

1

TT = 1 61 1.2

5 5

Sol–70: (b)Heating

coil t = 41°CH

t =

28°C

2t =

15°C

1

Coil Efficiency = 2 1

H 1

t tt t

= 28 15 13 0.5 or 50%= =41 15 26

Sol–71: (b)RSH = 100 kW,RLH = 25 kWOASH = 10 kWOALH = 10 kWBPF = 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 = 101.5 kWEffective Room Latent Heat (ERLH)

(9) (Test - 02)-01 October 2017

= RLH + BPF × OALH = 25 + 0.15 ×10 = 26.5 kW

Effective Room Sensible factor =ERSHF =

ERSH 101.5=ERSH ERLH 101.5 26.5

= 0.79Sol–72: (a)

DBT

W2 1

1 2 : Sensible cooling

2 1 : Sensible heating

During sensible cooling and sensibleheating, dew point temperatureremains constant.

Sol–73: (c)For dehumidification, the coiltemperature must be below the dewpoint temperature of incoming air.

Sol–74: (d)Sol–75: (a)

T

s

T =T2 33 2

14

s =s4 3 s =s1 2

COP of the refrigeration systemworking on reversed Carnot cycle

(COP)R = 1

2 1

TT T

where T1 = lower temperatureand T2 = higher temperatureThus, COP of the reversed Carnotcycle decreases on(i) increasing the higher temperature

and(ii) decreasing the lower temperature.

Sol–76: (c)Sol–77: (b)

The thermal efficiency of modifiedRankine cycle is higher because of highmean temperature of heat addition.

Sol–78: (c)Fusible Plug : sefety precautionagainst low water level in steam boilers.Safety valve : To prevent excessiveinternal pressure.Blow off cock : To remove sludge andother impurites from inside the boiler.

Sol–79: (b)Boiler mountings are fittings for thesafety of the boiler, and for completecontrol of the process of steamgeneration. The mountings form anintegral part of the boiler and aremounted on the body of the boiler itself.Ex- Safety valve, water level indicators,pressure gauge, fusible plug, steam stopvalve, feed check valve, blowoff cock.Boiler accessories are installed toincrease the efficiency of the steampower plants.Ex- Air preheater, Economiser,superheater, feed pump, Injector.

Sol–80: (a)The risk of radioactive hazard isgreatest in boiling water reactorbecause the coolant water boils in thecore of the reactor and no heatexchanger is used.

Sol–81: (a)Cadmium is generally used as materialfor control rods in nuclear reactor.

Sol–82: (a)Higher the value of mean temperatureof heat addition Tm, higher will be theefficiency.

Sol–83: (a)

T

s1

42

3

1–2: isentropic compression2–3: isobaric heat addition3–4: isentropic expansion4–1: isobaric heat rejection

Sol–84: (c)High air fuel ratio in gas turbinesreduces the exhaust gas temperature.

(10) (Test - 02)-01 October 2017

Sol–85: (a)Path of flue gasSuperheater Economiser Airpreheater Precipitator.

Sol–86: (d)A cooling tower is a heat rejection devicewhich rejects waste heat to theatmosphere through the cooling of awater stream to a lower temperature.Cooling towers generally use theevaporation of water to remove processheat and cool the working fluid to nearthe wet-bulb temperature of air.

Sol–87: (a)

Regeneration will increase efficiency,but has no effect on the power output.

Sol–88: (c)

DOR = m

m f

hh h

Sol–89: (c)

1

T

s5 pi

p0

2

3

4

6

Hence, we can see that the heat rejectedthrough the condenser increases thusreheating leads to increased condenserrequirements.

Sol–90: (c)

v = 1/n

2

1

p1 C C p

Hence, the volumetric efficiencydecreases as the pressure ratio isincreased.

W = n 1n

21 41

1

pn V Vp 1pn 1

Thus, if 2

1

pp

is more, then shaft power

increases.Sol–91: (b)

h1 = 3200 KJ/Kg

S

T1

2 2

h2 = 2400 KJ/Kg

2h = 2560 KJ/Kg

Actual work turbine efficiency=Isentropic work

= 1 2

1 2

h h 3200 2560 540= =h h 3200 2400 800

= 0.8 or 80%Sol–92: (c)

The Lenoir cycle consists of thefollowing processes: constant volumeheat addition (1–2); reversible adiabaticexpansion (2–3); and constant pressureheat rejection (3–1). The Lenoir cycle isapplicable to pulse jet engines.

p

v

Q1

Q23

2

1

pV =C

Sol–93: (d)A forced draft fan is used to force theair and products of combustion out ofthe stack. FD boilers operate with theair and combustion productsmaintained above atmospheric pressure.A fan at the inlet to the boiler systemprovides the necessary pressure to forcethe air and flue gas through the system.The static pressure is above theatmospheric pressure at the forced draftfan outlet.

Sol–94: (c)

32 2 T 100 32 2 T 48 8 = 800

2T – 104 = 6400 10064

T = 102°CSol–95: (d)

1 1 2AK=

2 1 2K A

(11) (Test - 02)-01 October 2017

K1 = K2

Sol–96: (d)General equation

2 2 2G

2 2 2qT T TKx y z =

1 T

t

just putting

T 0t gives poisson’s

equation,Sol–97: (c)

There can be no system possible which canviolate either of the two laws ofthermodynamics.

Sol–98: (b)

At r = rc, dQ 0dr

dQ 8r 32dr = 0

0 + 8rC = 32

rC = 4 = k [for cylinder]h

4 = 8h

h = 2W/m2KSol–99: (d)

Q =

f1 f 2

T T

T TTR R

T = TQ R

= 1 1400 40

100 40 20 40

T = 0.1 + 16000 + 0.5= 16000.6 °C

Sol–100: (b)

Tc = 2

wqrT4k

Tc =

2G

w 2Q rT

4kr

210 =

GQ45 22 14 4 2.47 100

22 14165 4 2.47 100 = QG

QG = 697 WSol–101: (b)

T1 = 110 – 60 x1T2 = 110 – 60 x2

T1 – T2 = –60x1 + 60(x2)

T = 60 (x2 – x1)= 60 × 0.75 = 45°C

Sol–102: (b)Sol–103: (c)Sol–104: (d)Sol–105: (c)

fin = sfin

c

AA

fin = finx

x =

s

c

A 2 7 7 30A 7 7

= 17.14Sol–106: (c)

J = bE G

= 0[black body]

J = 41 T

= 5.67 × 10–8 × (273)4

= 314.94 315W/m2

Sol–107: (b)

h h h1 h2m c (T T ] = c c c2 c1m c T T

For balanced counter flow heat exchanger.

h hm c = c cm c and variation of temperatureof fluids with respect to length isstraight line

45 – 15 = T – 10T = 40°C

(12) (Test - 02)-01 October 2017

T

45°

15°

40

x=0 x=L

L/2 L/2

15°

T1

10

40

1T 10L / 2 = 40 10

L2T1 – 20 = 302T1 = 50

T1 = 25°CSol–108: (a)

actual

max

QQ =

c c c2 c1m c T T53328

= 0.85

5232.5 [Tc2 –6] = 0.85 × 53328 5232.5 Tc2 – 31395.0 = 45328.8 5232.5 Tc2 = 76723.8

Tc2 = 14.66°CSol–109: (a)Sol–110: (d)Sol–111: (b)

Cmin = min[1.2 × 4.18, 2.5 × 1.005] =2.5125 kW/KMaximum rate = Cmin [Th1 – Tc1]

=2.5125 × [90 – 12]=195.98 kW

Sol–112: (a)

0.32 = 0.470.47 1

= NTU satisfiesNTU 1

and counter flow hence C = 1 i.e.balanced.

Sol–113: (c)1.6 × 1 × [50 – 25]

= 0.8 × 4 × [Tc2 – 18]Tc2 = 30.5°C

Sol–114: (d)No heat loss to the surrounding orentropy change of surrounding is zero.

Sol–115: (a)

Rf = 0 0

1 1U U

0U = 0.5U0

0.0004 =0 0

1 10.5U U

U0 = 21 2500 W/m K0.0004

Sol–116: (b)

t

h

LeLe =

d

d

0.05Re Pr D Pr0.05Re D

and

3 3

pc 7.785 10 1.78 10PrK 0.14

= 98.98Sol–117: (b)

Nu = convection heat transfer in

fluid layer 4.36 =conduction heat transfer

in fluid layer.

convection heat transfer 4.36 =

10

Convection heat transfer = 4.36 × 10= 43.6 kW

Sol–118: (a)

Qmax

Q

rO<rC rC rO>rCr

Sol–119: (c)Geometrical mean area of sphere

Ag = 0 iA A = 22 8 4m

(13) (Test - 02)-01 October 2017

Sol–120: (d)

Sol–121: (a)

2cT1

1cT2

2hT1hT

xThe counter flow HE with equal heatcapacity rates,

Ch = Cc

h1 h2T T = c2 c1T T

h1 c2T T = h2 c1T T

1 = 2 constant=

LMTD= 1 2

1 2ln /

=0 Underlined0

In equal heat capacity rate LMTDcan not be calculated due to equalityof temperature difference. HereAMTD is used in place of LMTD forheat transfer calculation purposes.

Sol–122: (d)Generally non-metals have very few freeelectrons as compared to metals. So non-metals has lower thermal conductivitythan metals.There are other factor for thermalconductivity e.g. porosity etc, but themain factor is number of free electrons.

Sol–123:(d)Sol–124:(a)

In a two high reversing mill the rollsrotate first in one direction and thenin the other so that the rolled metalmay pass back and forth through therolls several times. However these areexpensive compared to the non-reversing mills because of thereversible drive needed.

Sol–125:(b)Sol–126: (a)

The shaper is unsuitable for

generating flat surfaces on very largeparts because of limitations on thestroke and overhang of the arm. Thisproblem is solved in the planer byapplying the linear primary motionto the workpiece and feeding the toolat right angles to this motion. Theprimary motion is normallyaccomplished by a rack and piniondrive using a variable speed motor.Shapers are commonly used tomachine flat surface on smallcomponents and are only suitable forlow-batch quantities.

Sol–127: (a)Complex shapes are possible ininvestment casting because pattern iswithdrawn by melting it.

Sol–128: (c)In forging, rolling and extrusion, theforces applied are primarly compressive.

Sol–129: (a)Often the cutting speed and feeddetermined through optimization hasto be modified because only a finitenumbers of speed and feed steps areavailable on the machine tool. Factorswhich influence the choice of speedand feed are as follows:(i) Requirement of surface finish(ii) Requirement of dimensional

accuracy or tolerances(iii) Available forcepower on the

machine tool(iv) Available speed and feed steps on

the machine toolThe cutting speed calculated fromoptimizing equations may not beavailable on the machine tool becausespeeds are provided in a limitednumber of steps. In case of a steppeddrive, the nearest available machinespeed has to be used. Only if themachine is provided with a steplessdrive, the calculated optimum speedcan be used.

Sol–130:(a)Sol–131: (c)

Atomization involves mechanicaldisintegration of molten metal into fine

(14) (Test - 02)-01 October 2017

particles by means of a jet ofcompressed air, inert gases or water.

Sol–132: (c)Throttling is a constant enthalpyprocess, in which h1 = h2. Whether thetemperature and internal energychange in throttling process dependson whether the fluid behaves as anideal gas or not. Since the enthalpy ofan ideal gas is a function oftemperature alone, hence, T1 = T2 forthrottling process for ideal gas andU1 = U2.For an ideal gas, throttling processtakes place at(i) constant enthalpy(ii) constant temperature(iii) constant internal energyThus, for real gas, after throttlingchange in internal energy is not zero,

but is given by U2 – U1 = 1 1 2 2p p .A reversible cycle consist of reversibleprocesses only. Since VCS consists ofirreversible processes, viz. throttling,hence it can not be reversible. For acycle to be reversible all the processesshould be reversible.

Sol–133: (d)In the vapour absorption systemsusing water- lithium bromide, wateris used as refrigerant and a solutionof lithium bromide in water is used asabsorbent.Since water is used as refrigerantusing these systems, it is not possibleto provide refrigeration at sub-zerotemperatures. Hence, it is used onlyin applications requiring refrigerationat temperatures above 0°C. Hence,these systems are used for airconditioning applications. In ammoniabased vapour absorption system,ammonia is used as refrigerant andwater as absorbent. Further, the Li-Br-water systems operate under verylow (high vacuum) pressures, theammonia-water system is operated atpressures much higher thanatmosphere.

Sol–134: (a)The comfort depends on many factorssuch as type of clothing, age and sex,duration of stay, kind of activity, densityof occupants, climatic and seasonaldifferences. Hence the actual insidedesign temperatures selected in the airconditioner may not be suitable foroptimum comfort.

Sol–135:(a)Sol–136:(a)

Specific heat of water is higher andhence water has more cooling capacitythan air.

Sol–137: (a)

Heat pump used for heating is a definiteadvancement over the simple electricheater because a heat pump rejects moreheat (Q1 = Q2 + W) as compare to simpleelectric heater (W).Heat pump is more economical becauseit is giving more output (Q1) with sameinput (W) to both the devices.

Sol–138:(c)An economiser is a heat exchangerwhich raises the temperature of thefeedwater leaving the highest pressurefeedwater heater to about thesaturation temperature correspondingto the boiler pressure. This is done bythe hot flue gases exiting the lastsuperheater or reheater at atemperature varying from 370°C to540°C. The term economiser is usedbecause the throwing away of such hightemperature gases involved a great dealof energy loss. By utilizing these gasesin heating feedwater, higher efficiencyand better economy are achieved.

Sol–139:(a)The major shortcoming of a fire-tubeboiler is that definite size and pressurelimitations are inherent in its basicdesign i.e. the maximum size of the unitand the maximum operating pressureare limited. The tensile stress on thedrum wall is a function of the drumdiameter and the internal pressuregiven by

(15) (Test - 02)-01 October 2017

= pd2t

where = tensile stressp = gauge pressured = internal diameter of shellt = thickness of wall

Creep mechanism is elongation understress over a period of time, usually atelevated temperatures.

Sol–140:(b)Sol–141: (b)

When thermal conductivity is very highinternal conduction resistance is so low

LkA that entire fin becomes a lump mass

at base temperature.Sol–142: (a)

(rc) = kh

i.e. (rc) 1h

As h increases, rc decreases.Sol–143: (a)

The electrons, atoms and molecules ofall liquids, and gases and solids aboveabsolute zero temperature areconstantly in motion and thus radiationis constantly emitted as well as beingabsorbed or transmitted throughout theentire volume of matter, hence radiationis a volumetric phenomenon. For opaquesurface, radiation is considered to be asurface phenomenon because theradiation emitted by the interior regioncan never reach the surface and theradiation incident on such bodies isabsorbed within a few microns from thesurface.

Sol–144: (a)A baffle is a metal plate, usually in theform of the segment of circle, have holesto accomodate tubes, because a baffleplate support the tubes.

Sol–145: (a)More is the entropy generation, morewill be the irreverssibilities and more is

the area required for heat transfer.Sol–146: (a)

Heat exchnager are adiabatic devices inwhich heat exchange takes place hot andcold fluid without exchnaging heart withsurrounding. So applying SFEE for HeatExchanger,

Q + W = H PE KE

It results in H 0

hot fluid cold fluidH H 0

hot fluid cold fluidH H

Rate of enthalpy decrease of hot fluid =rate of enthalpy increase of cold fluid. Soreason correctly explains the assertion.

Sol–147: (b)We have LMTD calculation for eithercounter flow or parallel flow heatexchanger, it is not possible to haveLMTD for Multi Pass heat exchangersas they are those in which two fluidsflow alternately in parallel flow andcounterflow and hence LMTD expressionis not valid and for them we requirecorrection factor for LMTD calculation.

Sol–148:(a)As the material in hot working isabove the recrystallisationtemperature, any amount of workingcan be imparted since there is nostrain hardening taking place.Under the action of heat and force,when the atoms reach a certain higherenergy level , the new crystals startforming which is termed asrecrystallation. Recrystallisationdestroys the old grain struoturedeformed by the mechanical working,and entirely new strain free crystalsare formed.

Sol–149:(b)The rolls pull the material into rollgap through a net frictional force onthe material. Hence, the frictionalforce to the left side of the neutralpoint must be higher than the frictionforce to the right. Although, frictionis necessary for rolling material,energy is dissipated in overcomingfriction. Also, the increasing friction

(16) (Test - 02)-01 October 2017

increases the rolling force and powerrequirements. Furthermore, highfriction may damage the surface ofthe rolled products or cause sticking.In practice, a compromise is made byusing effective lubricants to lower thecoefficient of friction.

Sol–150: (b)In the indirect extrusion, a hollow ram

compresses metal through a die in adirection opposite to ram motion. Eitherthe ram is moved against a stationarybillet or the billet (hence container) ismade to move against stationary ram.The scrap is less (5 to 6% of billetweight) in indirect extrusion incomparison to the direct extrusion(18 to 20% of billet weight).