1. (a)

2. (b)

3. (d)4. (b)

5. (a)

6. (c)

7. (a)

8. (c)

9. (b)

10. (d)

11. (b)

12. (a)

13. (c)

14. (d)

15. (c)

16. (b)

17. (c)

18. (d)

19. (d)

20. (c)

21. (d)

22. (c)

23. (d)

24. (b)

25. (a)

26. (d)

27. (c)

28. (c)

29. (d)

30. (b)

31. (c)

32. (c)

33. (c)

34. (c)

35. (b)

36. (b)

37. (b)

38. (b)

39. (b)

40. (c)

41. (c)

42. (b)

43. (d)

44. (b)45. (d)

46. (b)

47. (b)

48. (b)

49. (d)

50. (b)

51. (d)

52. (a)

53. (c)

54. (d)

55. (b)

56. (c)

57. (d)

58. (a)

59. (a)

60. (d)

ESE-2018 PRELIMS TEST SERIESDate: 19th November, 2017

ANSWERS

61. (d)

62. (d)

63. (c)

64: (a)

65. (b)

66. (d)

67. (a)

68. (d)

69. (b)

70. (d)

71. (c)

72. (a)

73. (b)

74. (c)

75. (d)76. (c)

77. (c)78. (d)

79. (a)

80. (a)

81. (b)

82. (a)

83. (c)

84. (b)

85. (a)

86. (a)

87. (a)

88. (a)

89. (c)

90. (b)

91. (b)

92. (a)

93. (d)

94. (a)

95. (b)

96. (a)

97. (b)

98. (c)

99. (b)

100. (d)

101. (c)

102. (a)

103. (d)

104. (d)

105. (a)

106. (c)

107. (d)

108. (d)

109. (b)

110. (a)

111. (d)

112. (b)

113. (b)

114. (b)

115. (c)

116. (d)

117. (b)

118. (b)

119. (b)

120. (b)

121. (b)

122. (b)

123. (a)

124. (d)

125. (b)

126. (c)

127. (b)

128. (b)

129. (b)

130. (a)

131. (c)

132. (b)

133. (d)

134. (c)

135. (b)

136. (c)

137. (b)

138. (b)

139. (b)

140. (d)

141. (b)

142. (d)

143. (b)

144. (b)

145. (d)

146. (d)

147. (b)

148. (c)

149. (c)

150. (a)

(2) (Test - 09)-19th Nov. 2017

Sol–1: (a)Methods for reducing rolling force• Reduce friction• Use small diameter rolls• Use small reduction per pass• Apply front and back tension

Sol–2: (b)• Hydraulic presses apply maximum force

at each point of stroke• Toggle type mechanical presses apply

maximum force 20° below top deadcentre and this maximum force acts forconsiderable amount of time (dwell time)

Sol–3: (d)• In swaging solid rods bars or tubes is

subjected to radial impact forces.• It is used to produce tapers or points on

round bars or tubes• Rifling in gun barrel is made by swaging

a tube over a mandrel having spiralgrooves.

Sol–4: (b)

• Voids or crack formation is suppresseddue to compressive hydrostaticstresses.

• It is suitable for high strengthsuperalloys.

• Measures must be taken to preventcomplete ejection of the product whichis often referred to as blowout.

• For elevated temperature extrusion,wax, glass and polymers can be usedas fluid medium such as vegetable oil(castor oil) is used.

Sol–5: (a)

• It can be forward or backward or both.Sol–6: (c)Sol–7: (a)Sol–8: (c)

Wrinkling is a defect which isprimarily caused by insufficient blankholding force.

Sol–9: (b)

Sol–10: (d)Weldability of stainless steel can beimpoved by heating the metal above1100°C and quench immediately afterwelding and using low carbon contentand using Ti,Ta,Nb in filler rod.

Sol–11: (b)

• Filler material is not used.

• It is a low voltage and high currentprocess.

• Strength of joint also depends uponpressure.

Sol–12: (a)

Stre

ngth

Optimumclearance

Sol–13: (c)

2RI 0.6 = 1 × I2

1002 × 0.6 = I2

I = 100 0.6

= 77.5 ASol–14: (d)

When iron oxidizes at hightemperature then it releases heatwhich melts the metal and it isexpelled from the cutting zone by thekinetic energy of oxygen gas stream.

Sol–15: (c)

Fluidity is dependent upon manyfactors such as

Freezing temperature Composition (Mushy zone) But the most important controlling

(3) (Test - 09)-19th Nov 2017

factor is usually the pouringtemperature or the amount ofsuperheat.

Sol–16: (b)Fluidity affects the followingparameters.

1. Minimum section thickness that canbe cast.

2. Maximum length of a thin section.

3. The fineness of details of casting.

4. The ability to fill mold cavities.Sol–17: (c)

Metals Solidificationshrinkage inpercent

Gray CI –1.9

Low carbon steel 2.5 - 3.0

Zinc 3.7

Magnesium, High carbon steel 4

Copper 4.9

White cast iron 4 – 5.5

Aluminium 6.6Sol–18: (d)

Solid metal contraction is often calledpattern maker’s contraction becausecompensation for these dimensionalchanges should be made when the moldcavity or pattern is designed.

Sol–19: (d)Electric arc furnace can be used toproduce high quality metal of almost anydesired composition.

Sol–20: (c)Vertical milling machine :Spindle head cannot be swivelled but itcan move in vertical direction.Bed type milling machine :Bed is fixed spindle head containing aspeed gearbox travels vertically along theguideways of the machine column.Rotary milling machine :Worktable rotates but workpiece does not.

Planetary milling machine :Motion imparted to cutter is planetary type.

Sol–21: (d)

tan = tF 1000 1000 1

Fc 1.732 1000 3 3

= 11 1, tan 30

3 3

Sol–22: (c)Resultant of Fc and FT is the diameterof Merchant’s circle.

FC

FTR

1 mm = 50 N 20 mm = 50 × 20 N = 1000 N

2 2C TF F = 1000

(2FT)2 + (FT)2 = 10002

5FT2 = 1000 × 1000

FT = 200 1000 5 4 10 1000

= 5 2 100 200 5

FC = 400 5

Sol–23: (d)

Gear ratio = lead of feed screw40

lead of helix

= L40 L2

= 80 : 1Sol–24: (b)

Characteristics of creep feed grinding Depth of cut is high Dressing tool must be used

continuously Feed of table is very slow

Infeed of grinding wheel must beadjusted in order to compensate wheelwear.

(4) (Test - 09)-19th Nov. 2017

Sol–25: (a)

As increases, also increases and chipthickness decreases and cutting force alsodecreases.

r = chip thickness ratio = 1

2

tt

F = cutting forc

Fcr

Sol–26: (d)The grinding wheel is specified asmanufacture prefix; abrasive – Grain / gritsize – grade – structure – Bonding materialmanufacturer suffix8 – 24 coarse30 – 60 medium70 – 180 fine220 – 600 very fineSince 65 is not a standard marking schemefor grit size hence answer is none.

Sol–27: (c)In drills rake angle at centre on cuttinglips is highly negative hence cutting forceis highest and it gradually increases tohigher positive values at the periphery ofdrill on cutting lips, hence cutting force islower.

Sol–28: (c)If up milling is applied then the surfacewhich is machined first is oxide free andhence it is easy to cut.If down milling is applied then thesurface which is machined first containsoxide which are very hard and abrasivehence it is difficult to machine.

Sol–29: (d)

= cot tan( ), 0

= cot tan

= cos sinsin cos

= 2 2cos sin

sin cos

= 1 1

1sin cos 2 sin cos2

= 2 2cosec2 2 1.06 2.12

sin2

= 2.12

Sol–30: (b)Since it is a shaft basis system theminimum deviation i.e. fundamentaldeviation is zero.

Sol–31: (c)Since the maximum size of shaft is lessthan the minimum size of hole so it isa clearance fit.

Sol–32: (c)

Sol–33: (c)

Generative type process planning.

Sol–34: (c)

Sol–35: (b)

Sol–36: (b)

Machining centre is a group ofautomatic machine tools.

It includes an automatic tool changerand a table that clamps the workpiecein place, this table rotates to setworkpiece against the tool.

Sol–37: (b) Diffraction grating : Measurement

of gear pitch Optical flat : Measurement of very

small displacement. Auto collimaters : Small angular

deviations on long flat surfaces. Laser micrometer : Online

measurement of moving parts.

Sol–38: (b)GivenS = Rs. 5,00,000, V = Rs. 4,00,000,

(5) (Test - 09)-19th Nov 2017

F = Rs. 60,000

'P' ratioV =

S V 5,00,000 4,00,000 0.2S 5,00,000

(BEP)Sale = Fixed cost 6,00,00 Rs. 300000'P' 0.2ratio

V

Sol–39: (b)F = Rs 60,000, v = 15/unit,s = Rs. 25/unit x = 9000 units.At break evenpoint

BepF x v = Bepx s

xBep = F 60,000 6000 units

s v 25 15

Margin of safety “(MOS)”sale

= s(x – xBep) = 25 (9000 – 6000)= Rs. 75000When production is 9000 unitsProfit ‘P’ = sx – (F+vx) = x (s – v) – F = 9000 (25 – 15) – 60,000 = Rs. 30,000

Sol–40: (c)

Target sales = Fixed overhead Target profit

contribution marg in per unit

=

20,000 5,000 500 unit

100 50Sol–41: (c)

‘P/V’ ratio =Increase in profitIncrease in sales

=

20,000 10,000 0.1200000 100,000

or ‘P/V’ ratio =10%Margin of safety for period ‘I’ is given by

(MOS)I = I(Pr ofit) 10,000'P / V ' ratio 0.1

= Rs, 100000Break–even sales = Sales – MOS

= 100000 – 100000= Rs. 0.00

Sol–42: (b)Given,

D = 2000 units per year,C = Rs. 10, C0 = Rs. 125

Ch = 20% of C = 0.2 × 10 =Rs. 2 per year

EOQ ‘Q*’ =

0

h

2DC 2 2000 125C 2

= 500 unitsTime between two consecutive order

t = Q * 500 yearD 2000

or t = 500 12 3 months2000

Sol–43: (d)P = 4000 parts/day,

d = 4000 2000 parts / day

2C0 = Rs. 1000Ch = Rs. 4/unit, D = 4,00,000

Optimal Lot size,

‘Q*’ = 0

h

2D C PC (P d)

=

2 400,000 1000 4000

4 (4000 2000)= 20,000 units

Length of each production run,

tp = Q * 20000 5 daysP 4000

Sol–44: (b)FP = Rs. 50,000,vp = Rs. 10 per unit

Total cost = FP + vpx = FQ + vQx50000 + 10x = 20000 + 25x

x =

50000 20000 2000 units

25 10

Sol–45: (d)

A

B

C D

E

F GL= 0E = 0

L= 10E = 10

L= 26E = 26 L= 31

E =31

L= 5E =9

E= 10L =13

E=20L=20

1210 6

8115

7

5

13

10

(6) (Test - 09)-19th Nov. 2017

Earliest expected completion time of theproject is 31 days.

Sol–46: (b)

A

BF G H

EC

D

L=0E=0

L=8E=5 L=29

E=29L=34E=34 L=39

E=39

E=7L=7

E=6L=8 E=16

L=16

6

5 813

7

510

2098 12

5

Free float = iji EEj (E t )

or (FF)C–E = 16 – (6 + 8) = 2

Sol–47: (b)Given,

t0 = 6 days, tm = 8 days,tp = 10 days

Expected time, ‘tE’ = 0 Pt 4tm t

6

=

6 4 8 10 8 days6

Sol–48: (b)Given, Fn–1 = 80, Dn–1 = 76, 0.25

Fn = n 1 n 1 n 1F (D F )

= 80 + 0.25 (76 – 80)= 79 units

Sol–49: (d)Using four months moving average, theforecast of sales in the next month will be

F =

70 85 65 100 804

Sol–50: (b)

i ii i

Demand Forecast ErrorTime period

(D F )(D ) (F )1 12 11 12 16 14 23 15 17 24 24 20 45 20 23 3

Runing sum forecast error = i i(D F )

= 1+2–2 + 4 – 3 = 2

Sol–51: (d)

Inter arvial rate = l' ' 10 min/ customer

Inter service rate 1' ' = 4 min/customer

Utilization factor ‘ ’ =

(1 / ) 4 0.4(1 / ) 10

Probability of zero customer is the system orprobability that a newly arrived customerdoesn’t have to wait = 1 1 0.4 0.6

Sol–52: (a)

Given, = 8 custmer/hr,

= 12 custmer/hr

Utilization factor, =

8 2

12 3Expected system length

‘LS’ =

2 / 3 221 13

Expected queue length

‘Lq’ =

2 2(2 / 3) 1.331 1 2 / 3

Sol–53: (c)

5

2

3

1 2

4

6

22

78

4

32

3

9

Path1 – 3 – 5 – 6 = 191 – 3 – 4 – 5 – 6 = 221 – 3 – 4 – 6 = 151 – 3 – 2 – 4 – 5 – 6 = 231 – 3 – 2 – 4 – 6 = 161 – 3 – 2 – 5 – 6 = 151 – 2 – 4 – 5 – 6 = 151 – 2 – 4 – 6 = 81 – 2 – 5 – 6 = 7Correct option is 1–3–4–6 = 15

(7) (Test - 09)-19th Nov 2017

Sol–54: (d)Annual revenue = 38500Total expenses = 30000Tax Rate =40%

Depreciation amount = 3200Total earning after tax

= (38500 – 30000) ×

100 40 5100100

The net cash flow that year = 5100 +5200 = 8300

Sol–55: (b)In value engineering

Value of the product=

Function or performance or qualitycost

Sol–56: (c)As the number of sampling planincreases chances of moving error andaverage number of unitsinspected decreases.

Sol–57: (d)• For the same compression ratio Otto

cycle is more efficient than the Dieselcycle

• As a result of detonation in an I.C.engine rate of rise of pressure attainsvery high value and heat transfer lossto cylinder walls increases henceefficiency and power output decreases

• Most of high speed compressionengines operate on Dual combustioncycle

Sol–58: (a)• The operation of forcing additional air

under pressure in the engine cylinderis known as supercharging

• Scavenging is usually not done todecrease knocking in petrol engine asit increase detonation in the petrolengine.

Sol–59: (a)

The inlet valve opens at 20°C beforetop dead center and closes at 35°Cafter the bottom dead center.

Sol–60: (d)

Sol–61: (d)Compression loss in the I.C. enginesoccurs due to following reasons-

• Leaking piston rings• Use of thick head gasket• Clogged air-inlet slots• Increase in clearance volume caused

by bearing-bushing wearSol–62: (d)

Excess quantities of sulphur in dieselfuel do not cause detonationphenomena, so option (d) is false

Sol–63: (c)The knock in diesel engine occurs dueto instantaneous and rapid burning ofthe first part of the charge.

Sol–64: (a)At high elevations, engines operatingwith a supercharger develop more powerthan engines without a supercharger.

Sol–65: (b)From given datas:

Heat input = 46000 × 0.2= 9200 kJ/min

Brake power (B.P.) = 40 kW= 2400 kJ/min

Heat loss in cooling water= 1600 kJ/min

Other combined heat losses (exhaustgases mixture and others)

= 9200 – 2400 – 1600= 5200 kJ/min

Sol–66: (d)

bsfc = m 0.150BP

fuelm = (0.150) (BP)

= 0.150 × 150= 22.5 kg/hr

Fuel consumption per cylinder =

22.5 3.75 kg/hr

6Sol–67: (a)Sol–68: (d)

(8) (Test - 09)-19th Nov. 2017

Sol–69: (b)Sol–70: (d)Sol–71: (c)

Willan’s line method is used to findfriction power at a particular speed. Atlight loads as many readings as possibleshould be taken as there may be slightcurvature at light loads.

Sol–72: (a)

BP =

2 NT 2 1800 10 1.885 kW60 60

FP = 2 – 1.885 = 0.115 kW

Percentage loss = 0.115 100 6.1%1.885

Sol–73 (b)

bsec =

3 3

fCV m 43 10 5 10BP 80

= 2.687

Sol–74: (c)

2 212 kg 32 kg 44 kgC O CO

2 2 22H O 2H O4 32 361 kg 8 kg 36

Stoichiometric Air fuel ratio

= 32 80.82 0.18

12 10.23

= 15.77Sol–75: (d)

For lumped parameter analysis problems

Bi = chLInternal conduction resistance 0.1external convective resistance K

So from above equation it is clear that allconditions are true.

Sol–76: (c)

=s

VCh A

1 =3

2

48000 400R3

350 4 R

R = 4350 3 3.28125 10 m8000 400

= 0.328125 mm

diameter (D) = 2R

=2 × 0.328125 = 0.65625 mmSol–77: (c)

x = 1/25x

Re

= 1/25x

u x

=12 1/2 1/25x u

Sol–78: (d)

uN = 0.5 0.330.664 Re Pr

h =0.330.5

pCU L0.664 KL K

From above relation, we can see thatonly Option d is wrong.

Sol–79: (a)

E =0

E d

=[200 × (11 –3)] + [400 × (25 – 11)]

=7200 W/m2

Sol–80: (a)

max T = 2898

T = 2898 5796K0.5

Sol–81: (b)The heat capacity ratio–

C = min

max

CC

Since heat lost by hot fluid is equal toheat gained by cold fluid. From this

(9) (Test - 09)-19th Nov 2017

concept it is concluded that a fluidhaving small specific heat has maximumtemperature change so

hT = 160 – 85 = 75°C

cT = 135 – 35 = 100°C Heat transferred from hot fluid to cold

fluid,

h hC T = c cC T

h

c

CC =

c

h

TT

cT > hT

Ch > Cc

Cc is minimum and Ch maximum, soHeat capacity ratio

C = c h

h c

C T 75 = 0.75= =C T 100

Sol–82: (a)

T(x) = C0 + C1x + C2 x2

dTdx = C1 + 2C2 x

x 0

dTdx

= C1 = – 250°C/m and

x 1m

dTdx

= C1 + 2C2 = – 250 + 2x – 40

= – 330 °C/m

From energy balanced in a wall,

in gen outQ Q Q = storedd Qdt

genx 0 x 1

dT dTkA q A kAdx dx

= stored

d Qdt

– 50 × 5 × – 250 + 500 × 5 × 1 + 50 × 5

× – 330 = storedd Qdt

Rate of change of energy storage in wall= – 17.5 kW

Sol–83: (c)

=9.98, 0.15

USL =10 + 0.1 = 10.1, LSL = 10 – 0.1= 9.9

CPL =LSL 9.98 9.9 0.1778

3 3 0.15

CPU =USL 10.1 9.98 0.2667

3 3 0.15

Process capability index

CPK = min(CPL , CPU)

= 0.1778 0.18Sol–84: (b)

1 2t : t 1 : 2 , dT1 = dT2 (given)

1

2

qq = 1 1 2

2 2 1

k dT dx 6 :1k dT dx

Sol–85: (a)

t1 = 10 cm, k1 = 0.7 W/mK

t2 = 5 cm, k2 = 0.5 W/mK

k3 = 0.05 W/mK

Q1 =1 2

1 2

Tt t

k A k A

Q2 =31 2

1 2 3

Ttt t

k A k A k A

1

2

QQ

= 4

= 30.14 0.1 t / 0.050.14 0.1

t3 = 0.036 m = 3.6 cm

(10) (Test - 09)-19th Nov. 2017

Sol–86: (a)

Heat conducted= Heat convected

dTkAdx

= shA(T T )

dTdx = s

h T Tk

= –63157°C/cmSol–87: (a)

Eb = 4T

= 8 45.67 10 (800)

= 23.2 kW/m²

A = 2 2D 0.2

= 20.1257m

time = 5 × 60 = 300 sec.

Q = 23.2 × 0.1257 × 300= 874.87 kJ

Sol–88: (a)

r cQ d Q d qd = VcdT

dTd = r cQ Q q

Vc

Vc = r cQ Q qdT / d

=4000 5000 2000

0.5

= 2000 J/°CSol–89: (c)

Total radiation absorbed = 500 – 150 –225

= 125 W/m²

= 125 0.25500

Sol–90: (b)

11 12 13F F F = 1

by symmetry F12 = F13 = 0.5

by reciprocity

F21 = 112

2

A 2LF 0.5A L

= 0.71

Sol–91: (b)Thermal resistance without shield

=1 0.8 1 0.81

0.8 0.8

= 1.5Thermal resistance with shield shouldbe

= 10 × 1.5 = 15

15 = s

s

11 0.8 1 0.81 10.8 0.8

s = 0.142Sol–92: (a)

Radiation occur in gases too.Sol–93: (d)

Direct view factor = N(N 1)2

(6 1)6 152

Sol–94: (a)Thermal conductivity of electricalconductors in generally high due topresence of free electrons. Thermalconductivity of gases increase withtemperature k T whereasconductivity of liquid is function oftemperature density and otherthermophysical properties.

Sol–95: (b)The strong heat conducting property ofdiamond is due to the covalent bondingbetween its atoms.

Sol–96: (a)

q = dTkAdx

dT1 = dT2

q1 = q2

1dT0.5W/mK×A×

1 m = 22

dTk A0.5

(11) (Test - 09)-19th Nov 2017

k2 = 0.5 × 0.5

= 0.25 W/mKSol–97. (b)

Only 1 dimension heat conduction existsand no heat generation is considered.

Sol–98. (c)Equivalent emissivity for radiationexchange between completely enclosedbody small compared to enclosing body(subscript 1 for enclosed body) = 1

1 = 0.5Sol–99: (b)

factor =

3 31 2

1 1 3 3 3 3 2 2

11 11 11 1

F F

F1–3 = F3–2 = 1

Factor =

31 2

1 3 2

111 12 2

= 1 2 3

2 3 1 2 1 3 1 2 32 2

Sol–100: (d)

Area of air heater

A = 60 m2

Overall heat transfer coefficient

U = 100 W/m2K

Heat capacity rate of both hot and coldfluid,

Ch = Cc = 2000 W/K

Number of Transfer Units i.e. NTU

NTU = min

UAC

= 100 60

2000

= 3

Sol–101: (c)Objectives of method study are: establishing standard procedures

development of better layout byestablishing method involvingsmooth and continuous flow ofmaterial

devising safe work method andthus maintaining safe workingconditions

effective utilisation of men,material, machinery

efficient material handling efficient planning

Sol–102: (a)During different development stagesof a layout the following methods maybe used.1. Process Flow Charts: They

show, how different componentparts assemble, in sequence ofoperations to form sub-assemblieswhich in turn lead to assemblies(finished products).

2. Material Movement Patterns:The flow pattern of materials-in-process is traced and layout isbuilt around it.

3. Layout Analogues: They covertwo-dimensional cutouts ortemplates and three-dimensionalmodels.

Templates: They are used to developplant layout. Templates or cutoutsshow the plan of the various facilitiesand the building. They show theactual floor space utilization. Thetemplates can be placed and attachedand the building. They show theactual f loor space util ization.Templates save a lot of time andlabour which otherwise would bespent in making drawings for eachalternative plant layout arrangement.They visually present variouscharacteristics, advantages andlimitations of a layout. A two-dimensional template gives machineoutline and its details whereas a blocktemplate shows the boundary of themaximum projected area of themachine. Templates, though simpleand inexpensive, do not give realsituation effect which is obtainedthrough the use of three-dimensionalmodels or block models.

(12) (Test - 09)-19th Nov. 2017

Two dimensionaltemplate

Blocktemplate

Travel Chart: A travel chart as thename suggests is a chart or record ofthe amount of travel by the materialin-process while going from machineto machine or from one departmentto another. The amount of traveldepends upon the frequency ofmovements between sections ordepartments. A travel chart helpsimproving the existing plant layout.

Gantt Chart: It is used in projectmanagement. It shows activitiesdisplayed against time. It is used totrack project schedules.

Sol–103:(d)It speficifies1. The sizing and timing of

production orders for specificitems.

2. The sequencing of individuals jobsand

3. the short term allocation ofresoucesto individual activitiesand operation.

Sol–104:(d)Product layout is generally used insystems where a product has to bemanufactured or assembled in largequantities. Further, the machineryand auxilliary services are locatedaccording to the processing sequenceof the product. Thus, if the productdesign or process change occurs, thelayout may become obsolete thus,there is less flexibility and hence thevariety of products is low. It involvesless in-process inspection, incomparison to process layout.

Sol–105:(a)Product layout: Where a largequantity of products is to be produced.Process layout: Where a large varietyof products is manufactured.

Combined layout:Where item is beingmade in different types of sizes.Fixed position: Where too heavy orhuge layout item is used as rawmeterial.

Sol–106:(c)Dispatching function in productionplanning and control refers toauthorising a production work orderto be launched.

Sol–107: (d)Capital for purchase of raw materials,payment of wages to workers,transport cost etc. are called workingcapital. The floating or workingcapital is required to meet the currentexpenses on raw materials and stores,incidental to marketing products, andalso to meet the outstanding dues inrespect of goods supplied and lastlyfor paying wages and other expensesfor day-to-day requirements.Thus, working capital is themanagement of inventories andpayment of wages etc.

Sol–108: (d)

2

14

3T

S

h4 = 65 kJ/kg, h1 = 165 kJ/kg

h2 = 185 kJ/kg

Refrigeration capacity ‘Rc’ = 5 kW

Energy performance ratio (Co-efficient ofperformance)

‘EPR’ = Desired effect

work input

= 1 4

2 1

h h 165 65 5h h 185 165

(13) (Test - 09)-19th Nov 2017

Power consumed by compressoris given by

P = LQ 5 1 kWCOP 5

Sol–109: (b)

HP

TH

TL

W

Given,

H

L

TT =1.5

(COP)HP = H

H LH L

H

T 1T TT T

T

= L

H

1 1 1.5 3T 1 0.511 1.5T

Sol–110: (a)

When the refrigerant is saturatedhydrocarbon,

Refrigerant, R – (m – 1) (n + 1)P

Whose n + p + q = 2m + 2

CmHnFpClq

R – 11 or R – 011

m – 1 = 0 m = 1

n + 1 = 1 n = 0

p = 1

n + p + q = 2m + 2

0 + 1 + q = 2 × 1 + 2

q = 3

Cm Hn FP Clq CFCl3

For R – 134

m – 1 = 1 m = 2

n + 1 = 3 n = 2

p = 4

n + p + q = 2m + 2

2 + 4 + q = 2 × 2 + 2

q = 0

CmHnFpClq C2H2F4

Simillarly R – 12 is CF2Cl2 and R – 22 isCHF2Cl.

Sol–111: (d)

Sol–112: (b)

Since the specific heat of liquid (Cp)lis always larger than that of vapour(CP)v ,(T4 – T5) < (T2 – T1), that meansdegree of subcooling is always lessthan degree of super heating.

3

16

4T

S

25

Sol–113: (b)

Given,

COP = 5, W = 20 kW

COP = ERW

RE = COP × W = 5 × 20 = 100 kW

or RE = 100 28.57 TR3.5

Sol–114: (b)

Given,

TG = 177 + 273 = 450 K

To = 27 + 273 = 300 K, COP = 3

(14) (Test - 09)-19th Nov. 2017

COP of vapour absorption system is givenby

COP = G OE

G O E

T TTT T T

3 = E E E

E E E

T T 150 T450 300450 300 T 450(300 T ) 3(300 T )

9 × 300 – 9TE = TE

TE = 9 300 270 K

10

or 270 – 273 = –3°C

Sol–115. (c)

Given, Pvs = 4.25 kPa, P = 1 bar = 100 kPa

mv = 2.5 kg

As air is saturated, relative humidity is100% i.e. Pv = Pvs = 4.25 kPa

Specific humidity

' ' = v

v

0.622 P 0.622 4.25P P 100 4.25

= 0.0276 kg/kg of dry air

But, = v

a

m 0.0276m

mass of dry air,

‘ma’= vm 2.5 90.58 kg0.0276 0.0276

Sol–116: (d)

Given, RSHF = 0.75, RLH = 40 KW

RSHF = RSH

RLH RSH

0.75 = RSH

RSH 40

or, RSH = 40 0.75 120 kW

0.25

Sol–117. (b)

Sol–118: (b)

Sensible heat load = 0.0204 × cmm ×T

= 10 15 50.0204 2 (45 25)

60

= 10.2 kW

Sol–119: (b)

t = 40°C3

t = 25°C2 t = 15°C1

By pass factor = 3 2

3 1

t t 40 25 0.6t t 40 15

Coil efficiency = 1 – BPF = 1 – 0.6 = 0.4

Sol–120: (b)

RSHF = RSHRTH

RSH = RSHF × RTH = 0.6 × 120 = 72 kW

RLH = RTH – RSH = 120 – 72 = 48 kW

RLH = 50 × cmm × 1 2( )

48 = 50 × 90 × (0.012 – 2 )

2 = 480.012 0.00133 kg / kg of dry air

50 90

Sol–121: (b)

Binding constraint means theconstraint should form boundary onthe feasible region in the direction ofoptimal movement of objectivefunction, (i.e. away from origin in caseof maximisation problem and towardsorigin in case of minimisationproblem)

Sol–122: (b)Sol–123: (a)Sol–124: (d)

Gantt charts are used to solvesequencing problems graphically.

(15) (Test - 09)-19th Nov 2017

Sol–125: (b)Sol–126: (c)

On an average as much as 15% to 22%money saving are achieved byselecting a proper preventivemaintenance program. But preventivemaintenance will not completelyeliminate catastrophic failures(sudden breakdown) but will reduceits frequency of occurance.

Sol–127: (b)The best parameter to compare dieselengine of different size is specificbrake fuel consumption i.e. fuel con-sumption per unit brake power. Thevolumetric efficiency depends upondesign of components also not only onsize parameters. Air fuel ratio is notparameter to compares same categoryof engines. Swept volume of enginehas no meaning in comparison of per-formance of different engine.

Sol–128: (b)Sol–129: (b)

Exact estimation of cooling load in airconditioning plant gives optimum designwhich runs at minimum total cost andgives maximum comfort condition.The external load is due to outside heatsource or gain e.g. heat leakage throughwall, ventilation air load etc. Theinternal load is due to internal heatsource or gain e.g. occupant, applienceslights etc.

Sol–130: (a)Sprues of converging passage is used toavoid aspiration effect. Due toentrapement of air in casting createsblow holes.

Sol–131: (c)Sol–132: (b)Sol–133: (d)

USM must uses soft tool due to highvibration.

Sol–134: (c)Sol–135: (b)

Single start threads are used forfastening purposes due to easilytightning and losening with smallamount of force.

Sol–136: (c)Abrasives used in grinding are harderthan milling cutters.

Sol–137: (b)

Sol–138: (b)

Sol–139: (b)Sol–140: (d)

• Method of governing used in dieselengine is quality governing

Sol–141: (b)

Sol–142: (d) does not play any role in steady stateheat conduction.

Sol–143: (b)

mx

se m1

m2m3

m m m3 2 1> >

x

s

Sol–144: (b)Sol–145: (d)

Window pane is transparent to onlyshort wavelength solar radiation,wavelength of radiation is inverselyproportional to temperature of emittingobject.

Sol–146: (d)Rate of heat conduction depends ongeometry, thickness, material of mediumand temperature difference across themedium.

Sol–147: (b)Insulation addition upto critical radiusincreases heat transfer above which heattransfer decreases.

Sol–148: (c)

Ozone layer is situated in stratosphere.

Sol–149: (c)

Sol–150. (a)