Conventional Question Practice Program - IES Masteriesmaster.org/public/archive/2016/IM-1462628192.pdf · Work done by atmosphere on body = F.ds or Work interaction = – F· ds =

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Conventional Question Practice ProgramDate: 30th April, 2016

1. (d)

2. (a)

3. (a)

4. (c)

5. (b)

6. (d)

7. (d)

8. (d)

9. (c)

10. (a)

11. (b)

12. (a)

13. (a)

14. (c)

15. (c)

16. (c)

17. (a)

18. (b)

19. (b)

20. (c)

21. (c)

22. (c)

23. (b)

24. (b)

25. (d)

26. (b)

27. (b)

28. (c)

29. (c)

30. (b)

31. (c)

32. (a)

33. (a)

34. (b)

35. (c)

36. (c)

37. (d)

38. (a)

39. (a)

40. (c)

41. (d)

42. (c)

43. (d)

44. (a)

45. (d)

46. (a)

47. (b)

48. (c)

49. (a)

50. (b)

51. (c)

52. (a)

53. (d)

54. (d)

55. (d)

56. (c)

57. (d)

58. (d)

59. (b)

60. (b)

61. (b)

62. (c)

63. (d)

64. (d)

65. (a)

66. (b)

67. (d)

68. (a)

69. (a)

70. (a)

71. (d)

72. (c)

73. (b)

74. (b)

75. (b)

76. (b)

77. (b)

78. (b)

79. (a)

80. (c)

81. (d)

82. (b)

83. (c)

84. (a)

85. (d)

86. (d)

87. (c)

88. (b)

89. (d)

90. (d)

91. (a)

92. (b)

93. (d)

94. (b)

95. (b)

96. (c)

97. (d)

98. (a)

99. (a)

100. (b)

101. (d)

102. (d)

103. (a)

104. (d)

105. (c)

106. (a)

107. (a)

108. (d)

109. (d)

110. (c)

111. (d)

112. (d)

113. (c)

114. (c)

115. (a)

116. (d)

117. (d)

118. (d)

119. (c)

120. (c)

ANSWERS

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(2) ME (Test-19), Objective Solutions, 30th April 2016

Sol–1: (d)Work done by atmosphere on body = F.dsorWork interaction = – F· ds

= – (0.5) (– 50)

W 25 J

Sol–2: (a)The final temperature in the tank wheninitiallly evacuated tank is connected topipe line.

T2 = × Temperature in pipe line= × T1= 1.4 × (350 + 273)= 872.2 K= 599.2°C

2T 600 C

Note :

It is advisable to remember the finaltemperature in the insulated & evacuatedtank which is charging thgough a pipe line.

Sol–3: (a)For reversible heat transfer, entropychange of universe is zero.

Suniverse = 1 2sys surr

đQ đQT T

= 0

Hence for this condition, T1 should beequal to T2

Hence, for reversible heat transfer, tem-perature difference causing heat transfertends to zero.

Sol–4: (c)There are two terms which require thereversibility condition as

p.dv and T.dsBut T.ds = dv + p.dV is true for reversibleas well as irreversible processes. It isrelationship among propeties so options (b)and (d) i.e.

T.ds = dV W and

Q = dV + p.dV are true for

reversible processes only.Sol–5: (b)

The energy and mass flow in centrifugalcompressor are show in figure below.

Systemboundary

mw

1

m

Hence both mass and energy can enterthe system boundary so it is open system.The other options detail are,1. Pressure cooker is closed system i.e.

heat enter the cooker but not mass.2. Bomb calorimeter- Isolated system.3. Manual Ice cream freezer- Isolated

system.Note: Pressure cooker, Bomb calorim-eter and manual ice cream freezer aredefined as various systems only duringprocess.

Sol–6: (d)A. Mercury in glass thermometer is

based upon volumetric expansion dif-ference of glass and mercury.

B. In thermocouple two junctions oftwo different specied material are putat different temperature and electricvoltage in induced.

C. Thermister- Here electric resistanceof conductor varies as temperaturevary.

D. In constant volume gas thermom-eter, the pressure varies as tempera-ture varies.

Sol–7: (d)Since triple point of water is reference intemperature measurement. So one kelvin

is defined as 1273.16 of temperature of

triple point of pure water.Sol–8: (d)

All gases except monoatomic gases showsan increases in specific heat at higher tem-perature. At the same time the difference

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(cp – cv = R) remains constant. But theirratio (cp/cv) decreases with increase in tem-perature.Dissociation of gases at higher tempera-ture (compression and heat addition) re-sults with heat absorption by moleculesi.e. reduction in temperature from normaltrend of compression with constant spe-cific heat. At the same time the dissociatedmolecules combines (during expansion andheat rejection) and release heat. The figurebelow shows the effect on otto cycle.

P

V

3

1

4

22

34

Cycle, 1–2–3–4 - constant specific heat.

Cycle, 1 2 3 4 - Variable specific heatand dissociation.

Sol–9: (c)Compression ratio of diesel engine,C.R. = 21Let Vs is the swept volume and Vc is theclearance volume.

Then, C.R. = s c

c

V VV = 21

s cV V = 21 Vc

Vs = 20Vc ...(1)Steady state air intake,

m = 330 10 kg sec.

Density of air ( ) = 1.0 kg/m3

Volume intake rate = mass intake rate

density

= 30 × 10–3m3/sec.In one second volume entering in theengine = 30 × 10–3 m3

Vs + Vc = 30 × 10–3

ss

VV20

= 330 10

Vs =330 20 10

21

=3200 10

7

Power generated = 15 kWIn one second work done by the engine =15 kJWork done = (Mean effective pressure) ×(Swept volume)15 × 103 joule = Pmep × Vs

mepP =3

315 10 7

200 10

= 3525 10 Pa

= 525 kPa

Sol–10: (a)By increasing temperature the density ofair decreases and mass flow rate reduces.This reduced air mass flow rate providesrich mixture and so efficiency reduces.

Sol–11: (b)The second law analysis can be applied toboth thermodynamic cycle as well asindividual processes.

dQT 0 for cycles

S2 – S1 2

1dQT for individual processes

The second law of thermodynamics tellsabout the irreversibilities of the processthrough the concept of entropy. Theirreversibility I is directly proportional tothe entropy generated.

I = 0 genT S

On the other hand, the first law analysismerely is an extension of the law of energyconservation and does not tell about theirreversibilities in the process.

Sol–12: (a)Work done in polytropic process pvn = Cwill be

W = 1 1 2 2p v p vn 1

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For an ideal gas,

W = 1 2mR T Tn 1

Change in internal energy,

U = mCV(T2 – T1) Heat transfer Q = U W

= 1 2v 2 1

mR T TmC T Tn 1

= 1 22 1

mR T TmR T T1 n 1

= 1 21 1mR T T

n 1 1

= 1 2

nmr T Tn 1 1

=

n work done1

Heat absorbed or rejected duringpolytropic process PVn = C is

Q = n work done1

Sol–13: (a)

During process 1–2,Q1–2 = 150 kJ, W1–2 = ?

During process 2 –1,Q2–1 = –50 kJ, W2–1 = 75 kJ

From first law of thermodynamics,U1–2 = Q1–2 – W1–2 = 150 – W1–2

U2–1 = Q2–1 – W2–1 = –50 – 75= –125

U1–2 + U2–1 = 0 150 – W1–2 + (–125) = 0 W1–2 = 150 – 125 = 25 kJ

Sol–14: (c)Maximum efficiency

max = 1000 300 0.7 or 70%=

1000

Heat input Q = 500 kJ Maximum work output

Wmax = Q max = 500 × 0.7 = 350 kJ

Minimum heat rejected= Q – Wmax = 500 – 350 = 150 kJ

Sol–15: (c)Work transfer in free expansion is zero,because in free expansion, the pressureagainst which the gas expands is zero.Hence, work = P V 0 .

Sol–16: (c)Work interaction = area enclosed in theP–V diagram or

(PST) = 1 PT ST2

= 1 2 PR 2 QR2

= 1 UR VR2 = 1 48 24 Nm=

2

because, or (WVUR) = UR×VR = 48 NmSol–17: (a)

Path function – their magnitudes dependon the path followed during a process.Example work, heat.Point function- they depend on the stateonly and not on how a system reachesthat state. All properties are pointfunctions.Extensive properties depend on the massof the substance, while intensiveproperties are independent of mass.

Sol–18: (b)

B C

A

F

D

E

1500

500

1 2 3 4 5Work done, W = area of ABCD

= 1 1000BC AD2

= 1 1000 3000kJ=2 42

Heat rejected, Q = area of ADEF

= AD AF 4 500 2000 kJ= =

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Efficiency = W

W Q

= 3000 0.60=

3000 2000

Sol–19: (b)

Maximum efficiency = 2

1

T1T

T 1

HE

T 2 = 293 K

= 873 KQ 1

W

Q 2

2931873

= 1 1

W 1Q Q

873 293

873

= 1Q

1Q = 873 1.505kW580

Heat rejection 2 1Q Q W=

= 1.505 – 1 = 0.505 kWSol–20: (c)

In large air conditioning plant, centrifugalcompressors are used to ensure largerefrigerant flow rate.

Sol–21: (c)Sol–22: (c)

Total heat to be extracted from the fishQ = 3600 × [(300–270) × 2 + 230 +(270 –250) × 0.5]= 3600×[60+230+10] = 3600×300 kJThe cooling takes place in 10 hours,so the rate of heat extracted

3600 300Q 30kW= =3600 10

Ideal COP = 250 5=300 250

Actual COP = 15 2.5=2

Power required

= Q 30 12kW= =actual COP 2.5

Sol–23: (b)COP of hermetically sealed compressor isless than that of the open compressorbecause heat is lost in the former.Open type compressor is driven by anexternal motor, either belt-driven or directlyconnected. Since this involves a shaftextending through crank case of compressor,it requires shaft seal. Opposite to this typeof compressor is the hermetically sealedcompressor, in which the motor isincorporated within the same housing asthe compressor. Thus, hermetically sealedcompressor has no shaft extending throughthe crankcase and therefore requires noseal.Further, hermetic type compressor issmaller, more compact, more free ofvibration, thus more silent and has itsmotor continuously and positivelylubricated. They have no belts which needadjusting and eventual replacement.

Sol–24: (b)

1

2

QQ = 1

2

TT

Q = Q2 + Winput

= 1000 KJ/min + 6 KW

=000 6 22.67kW=60

Q2 =1000 6 22.67 kW=

60

22.6716.67 = 1T

233

T1 = 316.86 K= 43.86°C

Sol–25: (d)All three methods mentioned in optionsare used for design of air duct system.

Sol–26: (b)Equivalent diameter is given by,

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D =4 Area

Wetted perimeter

=

4.ab 2ab2(a b) a b

Sol–27: (b)(i) Aspect ratio for rectangular duct should

not be high and should be close to one(aspect ratio: height/width).

(ii) Equal friction method gives equalpressure loss in various branches andno dampers are required to balance thesystem.

(iii)The static regain method is notsuitable for long ducts because the gainin pressure (or static pressure regain)is not possible completly in longbranches duct.

(iv)The velocity reduction method isusually adopted for simple designing ofduct.

Sol–28: (c)Ammonia absorption refrigeration systemmeans vapour absorption refrigerationsystem. The COP of VAR system is lessas compared to that of prectical VCRsystem. Hence vapour compression systemis superior to vapour absorptionrefrigeration systems. So, all statementsare correct about vapour absorptionrefrigeration system except statement 1.

Sol–29: (c)Azeotrope are mixture of two or morerefrigerant and behaves like pure substancein working range of refrigerator machinesi.e., no separation of its constitutents fromcondenser to evaporator temperature.

Sol–30: (b)High pressure on discharge side ofcompressor of the refrigerating machineis due to the improper heat removal ratein condensor. The reason could be(i) Lack of cooling water(ii) Temperature of cooling water being

high(iii)Dirty heat transfer surface of

condenserSol–31: (c)Sol–32: (a)

Knock in SI engine reduced by any factorwhich reduces peak temperature andpressure. So supercharging increases thesetwo parameters and high knock. By nature,straight chain hydrocarbon are moreknock prone than branched hydrocarbon.Retarded spark and high engine speed hassame effect on knock i.e. peak temperatureand pressure both are lowered becauseoccure during expansion stroke so reducedknock.

Sol–33: (a)Air-fuel ratio in various operation mode ofSI engine,Idling – 10:1 - rich mixtureCrusing – 15 :1 - lean mixtureMaximum power- richmixture -12.5Cold starting- Due to low temperature,high heat losses and leakage of mixturepast the valves, very rich mixture isrequired i.e. 4:1.

Sol–34: (b)The combustion diagram of SI engine-

TDC

12°12°

6°9°

Initialinjection

Newinjection

P

Angular speed of engine in degree persecond

=

2 N 18060

= 6 N= 6 × 1000= 6000 deg/sec

Rotation during first stage of

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combustion

1 = 1t

= 6000 × 10–3

= 6 degreeRotation during second stage of combustion

2 = 2t

= 6000 × 2 × 10–3

= 12 degreeSince spark advance initially is set at 15°btdc, so the peak pressure occure 3°(12 + 6 – 15) after tdc. The second stageof combustion starts 9° btdc.Now speed of engine is doubled andassuming second stage of combustionremains same, the rotation for first stagewill be

1 = = 1 12 t 12Nt

= 12 × 1000 × 10–3 =12° New spark advance

= 12 + 9= 21°

Sol–35: (c)Pre-ignition means start of combustionbefore regular spark from spark plug. Thisis equivalent to very high spark advance.In pre-ignition the combustion starts earlyand temperature and pressure are muchmore during compression. This results inhigher work of compression, decrease innet work of cycle and decrease in engineor fuel efficiency. But at the same time,due to incomplete combustion, fuel lossesincrease.

Sol–36: (c)

Inletvalve

Cyl

inde

r

Air

Masked inlet valve in CI engine pro-duces spiral flow called primary swirl

or simply swirl as air enters the cylin-der as shown in figure. This motion isnecessary for mixing of fuel and airproperly before combustion starts.

Sol–37: (d)The T-S diagram of Otto cycle, Dualcycle, and Diesel cycle,

2

1

7

T

a b s

35

4

6

The heat rejected is same in all cyclesi.e. area 1–7–b–a is common.So efficiency,

= rejected

added

Q1

Q

For maximum efficiency, Qadded shouldbe maximum. From T-s diagram it isclear that heat added is1. Otto cycle (1–2–4–7)

Qotto = Area a – 2 – 4 – 7 – b2. Dual cycle (1–2––3–5–7)

Qdual = Area a – 2 – 3 – 5 – b3. Diesel cycle (1–2–6–7)

Qdiesel = Area a – 2 – 6 – bHence Qotto > Qdual > Qdiesel.

So otto dual diesel

Sol–38: (a)The angular velocity of crank in degree/sec.

= 2 Nrad/sec60

=

2 N 180 deg/sec60

= 6 N deg/secThe rotation of crank in 1.5m/sec

= w.t. degree

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= 6 N × 1.5 × 10–3

= 6 × 2000 × 1.5 × 10–3

= 18° degree.Sol–39: (a)

The Diesel cycle p-V diagram:

2 3

4

1

V

P

Compression ratio

r = 1

2

VV

Fuel cut off ratio

= 3

2

VV

The efficiency is expressed as–

=

1

1111r

Sol–40: (c) The knock in CI engine is- If large

amount of fuel is available at timeof self ignition. The large amountcan be due to large ignition delay.

Long delay period can be due to lawvolatality and high self ignition tem-perature.

Low self ignition temperature allowsthe mixture to burn early and re-duces chances of diesel knock.

Because around 50% of air is forced inpre-chamber, so high compression pres-sure are required here.

Sol–41: (d)Diamond cutting tools are notrecommended for machining of ferrousmetals due to chemical affinity of diamondtool material with iron.

Sol–42: (c)

The internal gears can be best producedby using gear shaping with pinion cutter,next best produced by gear broaching.

Sol–43: (d)Grinding process is a finishing process andconsumes high specific energy as the metalremoval is less.

Sol–44: (a)During grinding of hard work pieces usinghard grinding wheels, the abrasivesbecomes blunt very fast and startedrubbing on the work. But because of highhardness of wheel the rubbing forces arenot sufficient to pull out the bluntabrasives and hence blunt abrasive willsimply rubbing without any machiningcalled as glazing of grinding wheel. It isrecovered by dressing operation.

Sol–45: (d)

V = DN 200 160=60 60

= 1675.52 mm/secMRR = fdV = 0.25 × 4 × 1675.516MRR = 1675.516 mm3/sec

Sol–46: (a)In deep drawing when the force is appliedby using punch on the metallic sheet, thetendency of the sheet is to lift upwardsdue to which wrinkles are produced aroundthe periphery. It can eleminated byincreasing blank holding force.

Sol–47: (b)

Q1

W

Q2

TH

HP/R

TL

The CoP of heat pump operating betweentemperature TH and TL,

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(COP)HP = H

H L

TT T

The COP of refrigerator,

(COP)R = L

H L

TT T

(COP)HP – (COP)R =

LH

H L

T TT T

(COP)HP – (COP)R = 1

Sol–48: (c)At tip of impeller eye, the possibility ofshock formation is high. To avoid shockformation M < 0.9.The velocity triangle at tip of impellereye.

1wV

u1

1a 1V V=

Mach number at tip of impellereye,

M = 1w

1

V

RT

Sol–49: (a)Consider a centrifugal compressor hav-ing no inlet guide vane and backwardcurved exit.The pressure rise in rotor–

rotPg

= 1 2

2 2 2 22 1 r r

1 u u V V2g

...(i)

Assuming constant flow velocity.

1fV = =

2f 1V V

At inlet to impeller,

u1

V1=Vf1

1A B

C

2r1V = 2 2 2 2

f1 1 1 1V u V u …(i)

At outlet of impeller,

V2

Vf2

u2

Vr2

D E

F

2V

2r2V =

22 2f2 2V (u V )

=2

2 21 2V (u V ) …(ii)

Subtracting equation (ii) from (i),

1 22 2r rV V = 2 2

1 1(V u ) –

2 22 2 21 2 w 2 w(V u V 2u V )

= 2 2

2 2 22 w w 1 22u V V u u

1 22 2 2 2r r 2 1V V u u = 2 2

22 w w2u V V

...(iii)Total pressure rise in compressure

stagePg = 2 1w 2 w 1V u V u

= 2 1w 2V u ( V 0)

Degree of reaction,

R =

rot

stage

PP

= 2 2

2

22 w w

2 w

2u V V2u V

R = w

2

V1

2u

Sol–50: (b)Chocking – It is maximum mass flowrate possible i.e. fixed rate for all pres-sure ratio at a given speed.Stall – It is a local flow separation fromsurface of blades causing pits.Slip – Due to inertia, the air deflectsbackward at high velocity causing re-duction in whirl velocity.

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Sol–51: (c)Let us redraw the velocity triangles withvarious angles

901 =V1

1

1rV

U1

(i)U2

V2

22

2rV

(ii)

Since 1 90= no inlet guide vanes. U1

and U2 , 1 2 and have very largevariation, the machine must be radial.the value of 2 is more than 90°, so themachine is forward curved.

Sol–52: (a)

A2kN-m

CB

1kN/m

2m 2mRA

RA

Taking moment about BMB= 0

RA × 4 – 2 – 21 4

2 = 0

RA = 2.5 kN

B.M. at C, MC =2

A2R 2 2 12

= 2.5 × 2 – 2 – 2= 1 kN-m Sagging

Alternative :

2w8l2

B.M. at C, MC =2w 2

8 2

l =21 4 1

8

= 1 kNm (Sagging)

Sol–53: (d) A BM

AMR3

BMR3

M3 SFD

Maximum SF = M3

Sol–54: (d)

L ML

ML

ML SFD

BMD

M/2

M/2

M

BM diagram is linear.Thus 1 & 3 are correct.

Sol–55: (d)W

L

L/3

In cantilever deflection under the point load

=3PL

3EI =

32W L3

3EI

=

38 WL81EI

Sol–56: (c)Area of circular cross-section = Area ofsquare cross-section

2d4 = a2

d = 1.128 aMoment resistance capacity = fy × Z

square

circular

ZZ =

3

3(a /6) 1.182

( d /32)

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Zsquare > Zcircular

Hence square section is stronger than circularsection. So more economical also.

Sol–57: (d)In rectangular beam

Max. shear stress = 32

Average shear stress

4 =3 V2 bd

4 =33 60 10

2 100 d

d = 225 mmSol–58: (d)

t = V(Ay)Ib

Ay = Moment of area about NA

varies only with ISol–59: (b)Sol–60: (b)Sol–61: (b)Sol–62: (c)Sol–63: (d)

Plastics require larger f inishingtolerances than metals. Heat generated af-fects dimensional stablity of theplastic. And generally many steps arerequired to ensure good heat disspitationduring machining.

Sol–64: (d)The vapour compression cycle on T-Sdiagram is given below.

S

T

14

3

2

In vapour compression cycle, expansion( 3 4 ) take place in expansion valve orthrottle valve. Throttling is an isenthalpicprocess. Hence it is adiabatic process with

constant enthalpy.Sol–65: (a)

From a series of experiments, stabler pro-posed a modified rule,

c = ki

where c = chip flow anglek = constant, varies from

0.9 to 1.0, based on work material andcutting conditions.

Sol–66: (b)The gear-shaper cutter is mounted on avertical ram and is rotated about its axisas it performs the reciprocating action. Theworkpiece is also mounted on a verticalspindle and rotates in mesh with theshaping cutter during the cutting operation.The relative rotary motions of the shapingcutter and the gear blank are calculated asper the requirement and incorporated withthe change gears.

Sol–67: (d)DN = 22

322 70 10 N7

= 22

Feed per minute = 100 × 120×0.05 =60mm/min.

Sol–68: (a)Shear angle is the angle between the shearplane and the direction of cutting velocity.

Sol–69: (a)Sol–70: (a)

n1 1V T = C

n2 2V T = C

n2 2V T = n

1 1V T

2

1

VV =

n1

2

T 1T 2

n2

1

TT

= 2

2

1

TT = 1/n 1/0.52 2 4= =

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Hence, T2 = 4T1

Sol–71: (d)Volumetric efficiency is the ratio of actualvolume of gas drawn into the cylinder tothe piston displacement. Volumetricefficiency is less than 100% mainly becauseof the following reasons:Firstly, the gas is heated during admissionto the cylinder. Secondly there is leakagepast valves and piston rings. Thirdly,there is re-expansion of the gas trappedin the clearance volume from the previousstroke.

Sol–72: (c)Criticality is a term that refers to thebalance of neutrons in the system.Subcritical refers to a system when theloss of neutrons is greater than theproduction rate of neutrons and thereforethe neutron population decreases as timegoes on. Supercritical refers to a systemwhere the production rate of neutrons isgreater than the loss rate of neutrons andtherefore the neutron population increases.When the neutron population remainsconstant, this means that there is aperfect balance between the productionrate and loss rate, and the nuclear systemis said to be critical.

Sol–73: (b)Under appropriate operating conditions,the neutrons given off by fission reactionscan breed more fuel from otherwise non-fissionable isotopes. The most commonbreeding reaction is that of Pu-239 fromnon-fissionable U-238. The term fastbreeder refers to the types ofconfigurations which can actually producemore fissionable fuel that they use. Thisscenario is possible because the non-fissionable U-238 is 140 times moreabundant that the fissionable U-235 andcan be efficiently converted into Pu-239by the neutrons from a fission chainreaction.Breeder reactor does not use moderator.

Sol–74: (b)The nuclear reactor consists of fissionprocess which occurs when a thermal

energy neutron is absorbed by the targetnucleus leading to division into two nucleiand emission of 2 or 3 neutrons apartfrom heat energy. These neutrons flyrandomly in all directions and are usuallyin the region of fast moving energyneutrons. The moderator is used to controlthe speed of these neutrons. Themoderator is used to control the speed ofthese neutrons so that they act usefullyin creating more fission, but many ofthese neutrons may simply get lost byflying off the reactor core and thus servingno useful purpose. This might hinder theprogression of a chain reaction which isvery necessary for the nuclear reactor. Inorder to reduce this process of neutronloss, the inner surface of the reactor coreis surrounded by a material which helpsto reflect these escaping neutrons backtowards the core of the reactor and thesematerials are known as reflectingmaterials. The reflecting materials shouldhave low absorption, high reflection,radiation stability and resistance tooxidation properties. Generally, lightwater, heavy and carbon are mostly usedas reflectors.

Sol–75: (b)

v = 1/n

2

1

p1 C C p

Hence, the volumetric efficiency decreasesas the pressure ratio is increased.

W = n 1n

21 41

1

pn V Vp 1pn 1

Thus, if 2

1

pp

is more, then shaft

power increases.Sol–76: (b)

The light water reactor uses ordinarywater which absorbs too many neutronsto be used with unenriched naturaluranium and therefore uraniumenrichment becomes necessary to operatesuch reactors.

Sol–77: (b)Coolant: High heat transfer coefficient

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(13) ME (Test-19), Objective Solutions, 30th April 2016Moderator: Low neutron absorptionFuel: Low radiation damageShield: High absorption of radiation

Sol–78: (b)

se

syt

(500)sut

(700)

a (MPa)

m (MPa)

Given :se = 350 MN/m2

m = max min2

m = 100 3002

m = 200 MN/m2

v = max min2

v = 300 1002

v = 100 MN/m2

Since nothing is mentioned above criterionso use first Soderberg equation :

1N =

vm

yt e

1N = 200 100

500 350

N = 1.46Now using Goodman equation

1N = vm

ut e

1N = 200 100

700 350

N = 1.75Taking larger of the two valuesFactor of safety 1.75

Sol–79: (a)

V

= 2 2 3 ˆˆ ˆx yi 2xy zj yz k Shear deformation rate,

yz =

1 w v2 y z

yz = 12 [–z3 + 2xy2]

yz = xy2 – 3z2

at (–2, –1, 2)

yz = (–2) (–1)2 – 32

2

yz 6

Sol–80: (c)

Tempering Both hardness andbrittleness arereduced

Austempering Austenite is convertedinto bainite

Martempering Austenite is convertedinto martensite

Sol–81: (d)Martensite is the hardest phase amongthe given structures. fine grains havehigher strength and hardness comparedto coarse grains. Spherodite structure haslower hardness compared to pearlite andmartensite.

Sol–82: (b)Vanadium improves endurance strengthof steel.

Sol–83: (c)From Bernoulli’s equation, piezometrichead + velocity head= constantSo for constant piezometric at two points,their velocity should be same.

Sol–84: (a)Euler's equation of motion are applicableto compressbile or incompressible, nonviscous fluids in steady or unsteady stateof flow.As we know Euler’s equation of motionconsiders only two forces namely thepressure force and fluid weight

g pF F

= ma

Euler’s equation of motion

g p uF F F

= ma

Navier-Stokes equationHence, Euler’s eqn can be derived fromNavier - Stokes equation.Therefore option (a) is correct.

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(14) ME (Test-19), Objective Solutions, 30th April 2016Sol–85: (d)

A stream line is an imaginary line withinthe flow for which tangent at any pointgives the direction of the velocity of flowat that point.

The Acceleration of fluid particle

a =

V V V Vu v wt y zx

Local or temporalacceleration

Convectiveacceleration

Both convective acceleration and temporalacceleration can co-exist.Statement 1 is wrong because streamlines canbe straight also and for that, there won’t beany normal acceleration.

Sol–86: (d) = 3xy ... (i)

= 2 23 (y – )2

x ... (ii)

(1)(1,3) = 23 (3) – 12 = 3 8

2 =12 units

(2)(3,3) = 2 23 (3) – (3)2 = 0

Q =(1 –2) = 12 – 0 = 12 units.

Sol–87: (c) m

Vd

= p

Vd

= w

w

1.6 d =

w

w

V 2d750

60

1.6 =60 2V

750

V = 10 m/sSol–88: (b)

Reynolds law is applicable in followingcases

(i) Incompressible fluid, flow in closedpipes (where viscosity has significanteffect)

(ii) Motion of fully submerged bodies like.submarine, aeroplanes, automobilesetc.

Sol–89: (d)Sol–90: (d)Sol–91: (a)Sol–92: (b)Sol–93: (d)Sol–94: (b)Sol–95: (b)Sol–96: (c)Sol–97: (d)Sol–98: (a)

Zeroth law of thermodynamics is basis oftemperature measurement. This lawmeasures the temperature based uponthermal equilibrium concept.

Sol–99: (a)The expression-

2

1p.dv

This expression is for displacement work.The integration of the expression is possibleonly if process 1-2 is quasi-static orreversible. The displacement work isassociated with closed system so theexpression gives work associated with nonflow reversible process.

Sol–100: (b)Silicon is strong graphitizer and promotesgraphization (i.e. decomposition of cement-ite to iron and graphite Graphitizer im-proves the fluidity of iron.

Sol–101: (d)Except martensite, all other are present inFe-C equilibrium diagrams. Margensite isproduced during heat treatment process inwhich austenite transform to martensite byquenching.

Sol–102: (d)Chromium, Tungsten and Molybdenum areferrite stabilizer.

Sol–103: (a)Annealing is softening process whichimproves ductility. Normalizing is coolingof material in air so it refines grain surfaces(due to faster cooling than annealing)Nitriding is employed with alloy steel inwhich different alloys forms their respective

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(15) ME (Test-19), Objective Solutions, 30th April 2016nitrides so surface hardness improves.Martempering is two stage cooling processavoiding the nost of TTT diagram to frommartensite. The two stage cooling minimizedistortion and quench crack.

Sol–104: (d)Tempering is done to transform the mar-tensite into tempered martensite (ductilityachieved).

Sol–105:(c)

The given combination of both reversibleengines is–

T1

HE1

Q1

W1

Q2

T3

HE2

T2

W2

Q3

Since efficiency of both engines, HE1 andHE2, are equal. So

1 = 11

2

1

T1T

= 3

2

T1T

2

1

TT = 3

2

TT

T2 = 1 2T T

Sol–106: (a)Sol–107: (a)

For an ideal or perfect gas,Pv = nRT

or, PvT = nR = constant

Hyperbolic expansion is Pv = constant,If T = constant for ideal gas, then idealgas equation becomes, Pv = nRT =constant. which is hyperbolic expansion.

Sol–108: (d)

Compressor

3

h

p Condenser

4 1

3

4

2

S =

c

Evaporator

P1

P2

The (p-h) diagram clearly indicates thaton subcooling, refrigeration effect increasesfrom (h1 – h4) to 1 4(h h ) but it has noeffect on work requirement. Hence CoPincreases.

Sol–109: (d)Higher compression ratio in SI engineincreases temperature of end mixture.This increased temperature reaches closeto auto ignition temperature of fuel andthe fuel burns quickly. This all results inhigher knocking tendency.

Sol–110: (c)

2 3

4

1V

P

The efficiency of diesel cycle-

= 11 11

1r

where cut off ratio-

= 3

2

VV

In diesel cycle, the load taking capacityis proportional to cut off ratio ' ' i.e.high cut off ratio means high load. Butat higher cut off ratio ' ' the efficiencymentioned in above expression reduces.The cut off ratio is also proportional tothe fuel added i.e. load taking capacity.

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Sol–111: (d)Higher load requires higher cylinderpressure and temperature. This higherpressure and temperature provide betterair fuel contanct and fuel burns easily.So knocky tendency reduces.

Sol–112: (d)Pre-chamber diesel engine have high airswirl and injector has one or two holesso the injection pressure is low. But onother hand, open chamber engine hasless air swirl and multihole injector ofsmall hole size. Since pre-chamber die-sel engines have generated compressionswirl i.e. very good air motion for mix-ing the fuel, so high injection pressureis required in open chmaber.

Sol–113: (c)There is no sticking friction in metalcutting.

Sol–114: (c)For bending without twisting, plane ofloading must contain one of the principalcentral axis of the section. In case thesection is having a plane of symmetry,the symmetrical plane contain theprincipal central axis.Thus assertion is correct the bending axiswill be perpendicular to the plane ofloading is the case. Hence reason is wrong.

Sol–115: (a)Often the cutting speed and feed determinedthrough optimization has to be modifiedbecause only a finite numbers of speed andfeed steps are available on the machine tool.Factors which influence the choice of speedand feed are as follows:

(i) Requirement of surface finish

(ii) Requirement of dimensional accuracyor tolerances

(iii) Available forcepower on the machinetool

(iv) Available speed and feed steps on themachine tool

The cutting speed calculated from optimizingequations may not be available on themachine tool because speeds are provided ina limited number of steps. In case of astepped drive, the nearest available machinespeed has to be used. Only if the machine isprovided with a stepless drive, the calculatedoptimum speed can be used.

Sol–116: (d)The shaper is unsuitable for generating flatsurfaces on very large parts because of limi-tations on the stroke and overhang of thearm. This problem is solved in the planerby applying the linear primary motion tothe workpiece and feeding the tool at rightangles to this motion. The primary motionis normally accomplished by a rack andpinion drive using a variable speed motor.Shapers are commonly used to machine flatsurface on small components and are onlysuitable for low-batch quantities.

Sol–117: (d)Machinability is a term indicating how thework material responds to the cuttingprocess. In the most general case, goodmachinability means that material is cutwith good surface finish, long tool life, lowforce and power requirements, and low cost.Machinability index km is defined by

kM = 60 60RV / V

where V60 is the cutting speed for the targetmaterial that ensures tool life of 60 min,V60 R is the same for the reference material.Reference materials are selected for eachgroup of work materials (ferrous andnonferrous) among the most popular andwidely used brands. Ifkm > 1, the machinability of the targetmaterial is better than this of the referencematerial, and vice-versa.

Sol–118: (d)Carburising is employed to ferrous mate-rial which has less percentage of carboncontent and then tempered.

Sol–119: (c)Heat can transfer from law temperature tohigh temperature in heat pump, refrigeratorsand air conditioners with the help of external

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(17) ME (Test-19), Objective Solutions, 30th April 2016power consumption. This power generationis other effect on surroundings because itis irreversible process.

Sol–120: (c)A solenoid valve is a electro-mechanicallyoperated valve. The valve is controlled byan electric current through a solenoid. Asolenoid valve is normally closed and thevalve will need constant flow of a current toremain open. At starting stage, motor drawshigh current and at this instant there is apossibility to drop the voltage across thesolenoid which leads to prevent the valvefrom opening. Hence never connect asolenoid valve directly to the motor leads.

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Conventional Question Practice Program - IES Masteriesmaster.org/public/archive/2016/IM-1462628192.pdf · Work done by atmosphere on body = F.ds or Work interaction = – F· ds =