Conventional Question Practice Program - IES Masteriesmaster.org/public/archive/2016/IM-1462627731.pdf · R (2) ME (Test-3), Objective Solutions, 12th March 2016 Sol–1: (c) Using

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Conventional Question Practice ProgramDate: 12th March, 2016

1. (c)

2. (c)

3. (a)

4. (c)

5. (d)

6. (d)

7. (b)

8. (d)

9. (a)

10. (b)

11. (b)

12. (b)

13. (c)

14. (c)

15. (b)

16. (d)

17. (c)

18. (c)

19. (a)

20. (b)

21. (d)

22. (b)

23. (b)

24. (c)

25. (c)

26. (b)

27. (d)

28. (b)

29. (c)

30. (b)

31. (b)

32. (a)

33. (b)

34. (c)

35. (a)

36. (c)

37. (a)

38. (d)

39. (a)

40. (c)

41. (b)

42. (c)

43. (c)

44. (a)

45. (b)

46. (b)

47. (d)

48. (b)

49. (a)

50. (c)

51. (b)

52. (c)

53. (c)

54. (b)

55. (d)

56. (c)

57. (b)

58. (b)

59. (d)

60. (a)

61. (c)

62. (d)

63. (c)

64. (c)

65. (b)

66. (c)

67. (d)

68. (d)

69. (d)

70. (a)

71. (c)

72. (d)

73. (d)

74. (a)

75. (b)

76. (b)

77. (c)

78. (a)

79. (a)

80. (a)

81. (d)

82. (d)

83. (d)

84. (a)

85. (b)

86. (a)

87. (d)

88. (b)

89. (a)

90. (c)

91. (c)

92. (b)

93. (d)

94. (a)

95. (c)

96. (a)

97. (b)

98. (a)

99. (d)

100. (b)

101. (d)

102. (c)

103. (b)

104. (c)

105. (a)

106. (c)

107. (a)

108. (b)

109. (c)

110. (a)

111. (a)

112. (a)

113. (d)

114. (d)

115. (c)

116. (b)

117. (d)

118. (a)

119. (a)

120. (a)

ANSWERS

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(2) ME (Test-3), Objective Solutions, 12th March 2016

Sol–1: (c)Using steady flow energy equation

21 1

1 1 2V gzm h Q

2000 1000

=

22 2

2 1 2V g zm h W

2000 1000For adiabatic, system, Q1-2 = 0

W1–2 = m

2 21 21 2

1 2g Z ZV Vh h

2000 1000

=

2 210 20 2010 0.1 1 102000 1000

= 10 (–0.1 – 0.15 + (–0.2)= –4.5 kJWork done on system = 4.5 kJ

Sol–2: (c)Given,

P = 300 kPaP0 = 100 kPa

Maximum useful work,Wmax = P V – Work lost overcoming

atm resistance= 0P V P V

= 0P P V

= (300 – 100) × 0.01

P0

P= 2 kJ

maxW 2 kJ

Sol–3: (a)

T = 350 K1

Q1

Q2

W = 2.5 kW

Solar radiation

T = 315 K2

HE

Efficiency of heat Engine,

HE = 1 2

1

T TT

= 3151350

HE = 0.1

0.1 = 1

WQ

Q1 = W0.1

= 2.50.1

Q1 = 25 kWLet ‘A’ be the area of collector.

0.6 A × 0.5 = 25

A =25

0.6 0.5

2A 83.33 m

Sol–4: (c)

T

s

900 K

300 KLoss in Availability

Entropy change of hot reservoir is :

hS = QT = 9000

900

= –10 kJ/KEntropy change of cold reservoir :

cS = 9000300 = 30 kJ/K

Loss in availability = T0 c hS S

= 300 [30 – 10]

LAE 6000 kJ

Sol–5: (d)The total four cycles are possible underthe given constraints. Two at constantvolume and two at constant pressure asshown in figure.

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1

2

3

P

V

Pmax

Pmin

VcV

= c

T = cPV =c

1

2

3

P

V

Pmax

Pmin

Vc

P = c

T = cPV=c

1

23

P

V

Pmax

Pmin

Vc

V = c

T = cPV=c

1

2

3

P

V

Pmax

Pmin

Vc

P = c

T = cPV = c

Sol–6: (d)Pressure and velocity diagram along thelength of impulse turbine.

P

V

Movingblades

x

Nozzle

The figure clearly indicates that pressureis constant and velocity reduces in mov-ing Baldes of impulse turbine.

Sol–7: (b)A correctly designed convergent-divergentnozzle working at designed load is always

choked. Isentropic condition never existany practical system because of friction.

Sol–8: (d)In pulverised-fuel-fired boiler, heat trans-fer from the burning fuel to the walls ofthe furnace is predominantly by radiation.

Sol–9: (a)Equivalent evaporation,

= 1 feed

fgatm

h h mh

where h1 = enthalpy of steam producedhfeed = enthalpy of water at feed

hfgatm = enthalpy of evaporation atatmospheric pressure.

= 2000 2426 1682258

Equivalent evaporation 2000 kg/hr

Sol–10: (b)Sol–11: (b)

Hints :-1. In impluse turbine 1 2W W . So it is not

impulse turbine.2. In centrifugal compressor 1 2u u .3. In axial compressor generally entry and

exit are axial.4. But reaction turbine, if reaction is 50%,

the triangle symmetrical and C1 = W2and C2 = W1, so the turbine appears to bereaction.

Sol–12: (b)Work done in steady flow process =

vdP

So work done in turbine is much greaterthan that of in pump because in turbine,working fluid is steam and in pumpworking fluid is water because volume (v)is very less in pump.

Sol–13: (c)

Since the expression involved in gasthermometer is ideal gas equation as

PV = RT

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This equation is valid even for real gasesat low pressure and high temperature.So in constant volume gas thermometer,

P = R TV

P T or T PSo at low pressure, the temperature ofgas is directly proportional to its pres-sure at constant volume.

Sol–14: (c)The first law of thermodynamics,

Q = W UU = Q W

Change in internal energy, withtemperature,

dUdT =

dQ dWdT dT

= 2 – (2 – 0.1T)

dUdT = 0.1 T

So change in internal energy when tem-perature varies from 100°C to 150°C.

U = 0.1 150

100T dT

= 0.12 × (1502 – 1002)

= 625 kJSol–15: (b)

A mixture of ice, water and steam is aheterogeneous system because more thanone phases exist.

Sol–16: (d)H = U + PV

dH = dU + PdV + vdP dH = TdS + vdP[ for a closed and reversible process,TdS = Q = dU + PdV]

Sol–17: (c)Zeroth law of thermodynamics: If twobodies are in thermal equilibrium with a

third body, then they are also in thermalequilibrium with each otherFirst law of thermodynamics: The lawof conservation of energySecond law of thermodynamics:Spontaneous processes occur in a certaindirectionThird law of thermodynamics: Theentropy of a pure crystalline substance iszero at absolute zero temperature

Sol–18: (c)Ideal gas thermometer is independent ofsubstance used in its construction.

Sol–19: (a)

According to clausius inequality

S > dQT

For adiabatic process, dQ 0T

and

irreversible process, S > 0.Sol–20: (b)

In a cyclic process, the net change ininternal energy is zero.

Sol–21: (d)The mixture of air and liquid air is nota pure substance since the relativeproportions of oxygen and nitrogen differin the gas and liquid phases inequilibrium.

Sol–22: (b)

100 K 150 K 200 K

Q1 Q2 Q3 = .300 kcal

ER

W = 50 kcalQ1 + Q2 + Q3–W = 0

Q1 + Q2 = W – Q3 = 50 –300 = –250Sol–23: (b)

The expansion of fluid in convergent-divergent nozzle on T-s diagram.

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1

T

22s

P1

P2

s

• Process 1-2s is isentropic reversibleprocess. So the heat drop-

= h1 – h2s

• Process 1–2 is adiabatic with friction.So the heat drop

= (h1 – h2)

(h1–h2)< (h1–h2s). So heat drop reduces.The entropy increases. Since the processis adiabatic the stagnation temperaturewill remain same but stagnationpressure reduces due to irreversibilities.

Sol–24: (c)

In convergent nozzle for compressibleflow, the shock can not occur becausefor shock formation super-sonic flow ismust. But in convergent nozzle, super-sonic flow is not possible.

The sonic condition is possible atminimum cross-section i.e. throat at endof nozzle. If the flow of throat is sonicand pressure at exit is reduced further,there will be expansion wave as shownbelow

P

P

P

x

sonic condition

expansion wave

Sol–25: (c)

s

h P0

inlet

Nozzleexit

Nozzle inlet

pressure

TotalPexit

Since the difference between total (stag-nation pressure) and static pressure isdue to kinetic head. At exit and inlet ofnozzle, the kinetic head are different buttotal head should be same if the flow isisentropic. But in given problem totalhead is decreasing which representlosses. So losses in nozzle,

= (Ptotal)inlet – (Ptotal)outlet

= 186 – 180

= 6 kPaSol–26: (b)

Four row velocity compounded impulseturbine, the ratio of work output fromeach stage from first to forth stage.7 : 5 : 3 : 1So the work output from last stage,

Stage 4 =

1 6400 kW

7 5 3 1

= 640016 = 400 kW

Similarly for third stage

Stage 3 = 3 640016

= 1200 kWSol–27: (d)Sol–28: (b)

Anticipatory gear is used in steamturbine speed regulation i.e. turbinegoverging.

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Labyrinth – The sealings of thismaterial are used between turbinecasing and shaft to make leakageproof.

Inverted T arrangement – Bladeof turbine are attached with rotorwith this arrangement.

Deaerator – This removes the non-condensible gases from steam circuitof steam turbine.

Sol–29: (c)Propulsive efficiency

2 2=Jet velocity v1 1Flight velocity u

1 2 2 / 32800 1 211400

= 66.6%Sol–30: (b)

1. As compared to turboprop engine theturbojet engine can operate at higheraltitude and higher speed. Becauseat high altitude, density air reducesso mass flow rate through propeller.But in turbojet, the ramming actiontakes care of density reduction.

2. It lower speed and lower altitude, theturboprop engine are more efficient.

Sol–31: (b)The thermal efficiency of modifiedRankine cycle is higher because of highmean temperature of heat addition.

Sol–32: (a)Pressurized heavy water reactor:Natural uranium.Gas cooled reactor: GraphiteFast breed reactor: No moderatorSodium cooled reactor: Magnetic pump

Sol–33: (b) = 1/2( )n

where n = nozzle efficiency and

= velocity coefficientSol–34: (c)

The due to supersaturation the specificvolume at exit is considerably less thenthe specific volume at exit with isentropicflow.Hence, the mass flow rate increases dueto supersaturation.The exit velocity per supersaturated flowwill be less than the exit velocity forisentropic flow because the difference inenthalpy reduces.

Sol–35: (a)

*

1

pp =

121

Sol–36: (c)

DOR = m

m f

hh h

Sol–37: (a)

Reheat factor:

Cumulative heat drop of the stagesTotal adiabatic heat drop

Internal efficiency:

Total work done on the rotorTotal adiabatic heat drop

Degree of reaction:

Enthalpy drop in moving bladesEnthalpy drop in the stage

Rankine efficiency:Adiabatic heat drop

Heat suppliedSol–38: (d)

1aV2aV

zwV1wV

1rV

2rV1V

Tangential force = 1 2w wm V V

= 1 × 910 = 910 N

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Axial thrust = 1 2a am V V = 1 × 4 = 4 N

Sol–39: (a)

W =

2 1

2 22 2 2 2r r1 2 1 1V VV V u u

2 2 22Since u1 = u2, v1 = 236 m/s,

1rV 126 m/s, V2 = 232 m/s,

2rV = 132m/s.

W = 2 2 2 21 1236 126 232 1322 2

= 38.1 kWSol–40: (c)

Isentropic enthalpy drop

= 280 338kW=0.92 0.9

Sol–41: (b)In the case of flow of a fluid through axialflow turbine, if the flow has axial, radialand tangential velocity components, the pathtraced by a fluid particle during the flowis a helix of changing radius. Theassumption that one section of the turbineblade could be taken as representative ofthe fluid flow lines all over the blade, cannot be valid since a centripetal force isrequired to maintain the fluid in thehelical path of constant radius. The largerthe tangential velocity component, thehigher will be the centripetal force needed.The fluid path therefore tends to moveoutward radially and the helix radiustends to increase considerably as the fluidflows from stage to stage.

Sol–42: (c)De Laval Turbine: Simple impulseturbine.Curtis Turbine: Velocity compoundedimpulse turbineParsons Turbine: 50% Reaction turbine

Sol–43: (c)The Lenoir cycle consists of the followingprocesses: constant volume heat addition(1–2); reversible adiabatic expansion(2–3); and constant pressure heat rejection

(3–1). The Lenoir cycle is applicable topulse jet engines.

p

v

Q1

Q23

2

1

pV =C

Sol–44: (a)When in the two-phase condition, thevolume varies as

v = f fgxv v

Thus, volume increases as the drynessfraction increases. As, the steam getssuperheated, then the pressure remainsconstant, but the temperature increases

f g

g

T

sSince v T. for constant pressureSo, volume of superheated steam will morethan the volume of the saturated steam.

Sol–45: (b)Free body diagram.

1000 N800 mm

480 mm

100 mmN

NTaking moment about hinge.

1000 × 800 = N 100 N 480

= 0.2 × 100 × N + N × 480 N = 1600 Newton

Braking torque,

TB = N 200

= 0.2 × 1600 × 200= 64 × 103 N-mm

BT 64N m= -

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Sol–46: (b)

1

2

LL = 2

1

3e

e

PP

L2 = 1

2

3e

e

PP

× L1

where, L1 and L2 are life in millionrevolution

L2 = 39800

4900

× 61000 60 3000

10

L2 = 1440 million revolution

=61440 10

2000 60

hrs

2L 12,000 hours=

Sol–47: (d)Statement I

dAA = 2 dV

M 1 V

For nozzle, velocity increases, so dV 0V

Further, fluid velocity is greater than localsound velocity, so M > 1.

dA 0A

, hence duct must diverge

Statement IIFor diffuser, velocity decreases, hencedV 0V

and fluid velocity is greater thanlocal sound velocity, so, M > 1

dV 0,V

hence duct must converge.

Sol–48: (b)Shear strength of shaft,

= 316T

d In this question strength means

maximum possible torque transmitted.

T =3d

16

...(1)

W = 2d Lg

4 ...(2)

T dW 4 Lg

Sol–49: (a)Total slip will occur in a belt drive if theangle of rest is zero.

Sol–50: (c)

MPa25015050

TimeFrom the given diagram,

Max. value of stress max = 250 MPa

Min. value of stress min = 50 MPa

Amplitude = max min2

= 250 502 = 100 MPa

Stress ratio =

min

max = 50

250 = 0.2

Option (c) is correct.Sol–51: (b)

Stress amplitude,

a = max min2

= 400 2002 = 100 MPa

a 100MPa

Sol–52: (c)T 0

k

ri

r0

Ti

L

The thermal resistance in the cylindricalwall of the tube,

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0

ith

rnrR = 2 kL

l

Sol–53: (c)The schematic of a helix along with vari-ous parameters,

md

lead,N

W

P

N

The expression for efficiency

=

sin 2 sinsin 2 sin

where is helix angle and is frictionangle which is constant.

For maximum value of efficiency ' '

sin 2 must be maximum.

So,

2 = 2

Helix angle-

=

12 2

=

4 2Maximum efficiency,

max =

1 sin1 sin

Sol–54: (b)

The number of contact surfaces inthe multiplate clutch is given by,

= n1 + n2 – 1

where n1 is number of disc on drivingshaft and n2 on driven shaft.

The above formula is explained byfollowing illustration

1 2 3 4Driving

n =21

Drivenn =32

So driven shaft has 3-plates and drivingshaft 2 plates. Total number of contactplates-

= n1 + n2 – 1 = 2 + 3 – 1 = 4Sol–55: (d)

Given, refrigerant R-134.

Comparing with R-(n – 1)(m + 1)p

n – 1 = 1

n = 2

m + 1 = 3

m = 2

p = 4

Chemical formula is

Cn Hm Fp Clq

Where,2zn + 2 = m + p + q

4 + 2 = 2 + 4 + q q = 0

Hence, chemical formula is

2 2 4C H F

Sol–56: (c)

R744 CO2

R290 C3H8

R502 CHClF2 + CClF2CF3

R718 H2O

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The R-502 is an azeotropic mixture of R-22 and R-115.

Sol–57: (b)Mixing of moist air with superheatedwater vapour without condensation

m3

m1w1

w3

Superheated

water vapourNo condensation

Since second stream is superheated watervapour and no dry air Dry air mass flow rate in first stream

a1m = 1

1

m 10.1 10kg/sec= =1 1 0.01

The specific humidity of moist air

1 = 0.01 kg wv/kg.d.a. The mass flow rate of water vapour infirst stream

= 0.01×10 kg wv/sec= 0.1 kg wv/sec

After mixing total vapour flow rate= 0.1 + 0.1 = 0.2 kg/sec

Total dry air flow rate= 10 kg/sec

Final specific humidity

= 0.2 0.02kg wv/ kg.d.a.=10

Sol–58: (b)

This problem is based upon thepsychrometric chart.

WBT

DPT

Humidityratio ( )

500DBT

Ta Tb

100%

RH

80% R

H70

% RH

ab

It is clear from the chart that the dew

point temperature (Ta) is less than wetbulb temperature (Tb).

Sol–59: (d)Length of cut × depth of cut

= length of chip × thickness of chip 76 × 0.2 = 61×t t = 0.25 mm

Sol–60: (a)

ChipTool

FSFH

NS

FV WorkpieceF

N

R

Sol–61: (c)Increasing the side cutting edge angleincreases the chip contact length anddecreases chip thickness. As a result, thecutting force is dispersed on a longer cuttingedge and so tool life is prolonged.

Sol–62: (d)It is also called the hardness of the wheel.This designates the force holding the grains.The grade of a wheel depends on the kindof bond, structure of wheel, and amount ofabrasive grains. Greater bond content andstrong bond results in harder grindingwheel. Harder wheels’ hold the abrasivegrains till the grinding force increase to agreat extent. The grade is denoted by lettergrades as indicated in Fig.

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

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Very soft Medium Very hardABCDEFGHKLMNOPQRSTUVWXYX

Sol–63: (c)Ceramic tools are produced by a sinteringprocess. The main constituent is Al2O3.Ceramic tools retain their hardness up to1400°C. They have better resistance toabrasion wear and crater formation thancemented carbides. Also, they exhibit a lowcoefficient of friction with most workmaterials. Due to these reasons, ceramictools can be used at substantially highercutting speeds and therefore, help achievehigh production rates. Ceramic tool tips arehighly brittle. To give strength to the tooltip, negative rake angle is given.

Sol–64: (c)Crater wear occurs on the rake face of thetool. The most significant factors affectingthe crater wear area temperture at the tool-chip interface and the chemical affinitybetween the tool and the workpiecematerials.The location of maximum depth of craterwear coincides with the location ofmaximum temperature at the tool-chipinterface. It is because, crater wear hasbeen described in terms of diffusionmechanism, that is the movement of atomsacross the tool-chip interface since diffusionrate increases with increasing temperature,so crater wear increases as temperatureincreases.

Sol–65: (b)Taylor equation for tool life is

VTn = C log v + n log T = log cThus, on log-log scale, it is a straight linevariation between log V and log T.

Sol–66: (c)If production costs are being plotted againstcutting speed, the cost decreases as cuttingspeed increases, because more workpiecesare being produced. However, a point isreached where production cost begins toincrease with additional increase in speedbecause of shorter tool life. This point isthe minimum cost speed. If production rateis plotted against cutting speed on the same

graph, it can be seen that as cutting speedincreases, the production rate also increasesto a point of maximum production and thendecreases rapidly. Note that the speed formaximum production is faster than thespeed that produces minimum cost perpiece. If maximum production is desired,one cutting speed is indicated, but if thelowest unit machining cost is desired, alower speed must be selected. In any case,the theoretical best cutting speed will liebetween points of minimum cost andmaximum production.

Production maximumproduction

costminimum

cost

Cutting speed

Sol–67: (d)

L

h

Time for drilling the hole = L minutesfN

where f = feed rate in number, N = rotationalspeed of drill in rpm, L = total length of tooltravel, = depth of holeL = hThe traverse distance beyond the hole istermed as break through distance and isrequired because of the conical shape of thedrill. This value ‘h’ or conical height of the

drill = D

2tan

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where D = dia of the twist drill = lip angle

Sol–68: (d)Stroke length = 250 mm = 250 × 10–3 m= 0.25 mThe shaping machine makes 30 doublestrokes per min.Hence, the length travelled per minute is0.25 × (30 × 2) = 15m/min

Sol–69: (d)Hobbing cannot be used for machininginternal gears or gears with shoulders andflanges because of the clearance needed forthe hob.

Sol–70: (a)Most of the heat goes to chip. (about 60%).

Sol–71: (c)Carbides tool material uses negative rakeangles due to strong in compression.

Sol–72: (d)Amount of energy consumption for unitvolume of metal removal are –Grinding > Reaming > Milling > Turning

Sol–73. (d)

d = 18 mmd = 24 mm

L = 200 mm1 L =100 mm2

E = 200 GN/m2 = 200 × 103 N/mm2

Total contraction of bar due to temperature= Total expansion of bar due to force P

(L1 + L2) t = 1 2

1 2

PL PLA E A E

300 × 12.5 × 10–6 × 30 =

2 2P 200 100E (24) (18)

4 4P = 26943.77 N

Max.stress in the bar =min

PA =

226943.77

(18)4

= 105.88 N/mm2

= 105.88 MN/m2

Sol–74: (a)

Stress

Endurancelimit

No. of cycles of loadingwhich causes fatiguefailure.

for ferrousmetals

Sol–75: (b)

We know, E = 2G(1 + ) … (i)

E = 3K(1 – 2 ) …(ii)Equations (i) (ii)

1 =

2 G (1 )3 K (1 2 )

Given K = 2G

1 =

2 G (1 )3 2G (1 2 )

= 27

Sol–76: (b)

All the wires are connected with rigid barso deformation in all the wires all be same.force in outer bars is F1 and in centralbar is F2

1F LAE = 2F (L / 2)

AE

F1 = 2F2

Force in central wire is more so it will yieldfirst, At yielding

F2 = fy . A

F1 = yf A2

Wy = 1 2 1 yF F F 2f AAt the time of collapse all the three wires willbe yielded so force in each bar = fy.A Wc = 1 2 1F F F = 3 fy . A

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c

y

Collapse load (W )Yielding load (W ) = y

y

3f .A 32f .A 2

Sol–77. (c)

Radius of Mohr’s circle in plane straincondition

R =

2 2x y xy

2 2

=

262 12 10(0)

2

= 6 × 10–6

Diameter = 2R = 12 × 10–6

Sol–78: (a)

Stresses are given on two mutuallyperpendicular planes and associated shearstress on these planes is zero so givenstresses are principal stresses.

Sol–79: (a)NTU is given by ratio of UA and minimum heat capacity ratio (Ch or Cc)

= min

UAC

NTU is measure of size (because of area‘A’ term) of heat exchanger.

Sol–80: (a)The heat dissipation ‘q’ and total thermalresistance (conduction and convection)Rth variation with radius-

cr r=

q

r

Rth

rcr r=

In resistance variation, upto critical radius,the thermal resistance decreases and heatdissipation increases. At critical radius,thermal resistance is minimum and heatdissipation is maximum.

Sol–81: (d)The composite sphere is shown in figure.

2k

kO

r1r2

r3

T1 T2

T3

Equal thickness means(r2 – r1) = (r3 – r2)

1

3

rr = 0.8

Since in steady state heat flow is same inboth layers, so,

1 2

2 1

2 1

4 k T Tr rr .r

=

2 3

3 2

2 3

4 2k T Tr rr .r

2 2

1 22 1

T Tr rr r =

2 3

2 33 2

T T2r rr r

r1(T1 – T2) = 2r3 (T2 – T3)

1 2

2 3

T TT T =

3

1

r2r =

120.8 = 2.5

Sol–82: (d)Slip between driver and belt,

S1 = 1%Slip between belt and follower,

S2 = 3% The velocity ratio in belt drive,

2

1

NN = 1

1 22

d 1 s 1 sd

where d1 and d2 are pitch diameter at cen-tre of belt (d + t) d1 = d2

2

1

NN = 1 21 s 1 s

= 1 0.01 1 0.07= 0.99 × 0.97 = 0.9603

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Sol–83: (d)In 20° stub teeth, the gear is smaller thanthe full depth. The addendum is 0.8 mod-ule. So1. Interference is reduced due to small

addendum.2. Smaller teeth are stronger (low bending

moment due to smaller height)3. Low production cost due to reduced

material removal for small height.4. Since (aw = 0.8) addendum coefficient

is reduced, so minimum number ofteeth to avoid interference is reduced.

Sol–84: (a)Since axle generally supports the rotatingelements so it is subjected to bendingmoment due to transverse load. But sometime it is subjected to twisting moment e.g.auto mobile rear axle.So by definition, the axle undergoes bend-ing moment due to transverse load only.

Sol–85: (b)Balance delay is an assembly line is= 100 – line efficiency (in percentage)

Sol–86: (a)Inputs to MRP are MPS, Bill of material,supplier lead time.

Sol–87: (d)The dispatching function in productionplanning is concerned with starting theprocess. It gives necessary authority so asto start a particular work, which hasalready been planned under routing andscheduling.

Sol–88: (b)Gantt chart is a very useful and in loadingand scheduling, dispatching and progress-ing. The chart consists of a simple gridformed by a series of horizontal and verti-cal lines. The vertical lines divide horizon-tal lines into small squares representingunits of measurement which may be days,weaks or months. The horizontal lines di-vide the vertical lines into sections whichare used to represent either operations (jobschedule) or to represent work centres(load chart). Thus, Gantt chart may be usedto show at a glance the scheduling of

various operations involved in a job or itmay be drawn to show work ahead of eachmachine or work centre wrt time i.e. indays or weaks.

Sol–89: (a)R-717 (Ammonia): Ammonia greatestapplication is found in large and commercialreciprocating compression systems where hightoxicity is secondary.R-744 (CO2): The principal refrigeration useof carbon dioxide is same as that of dry ice.This is not used for compression purpose.R-11 (CCl3 F): R-11 is exclusively used inlarge centrifugal compressor systems of 200TR and above. But, can also be used inreciprocating compressors.

Sol–90: (c)A psychrometer or sling psychrometer canmeasure directly wet-bulb temperature anddry-bulb temperature because it consists ofa DBT and WBT thermometer only. Dewpoint temperatrue is obtained bypsychrometric chart.

Sol–91: (c)The process in case A and case B are shownon psychrometric chart.

DBT

HumidityRatio( )

Case A

WBT LineCase B

DPT Line

When air is saturated adiabatically thenwe get wet bulb temperature. If air is satu-rated isobarically then we get dew pointtemperature.

Sol–92: (b)Given,P0 = 101.325 kPaPv = 1.344 kPaPvs = 3.36 kPaHumidity Ratio,

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= v

o v

P0.622P P

=

1.3440.622101.325 1.344

= 0.00836 kg W.V./kg dry airRelative humidity,

= v

vs

P 1.344 0.4P 3.36

= 40%

Sol–93: (d)There are three air changes per hour,so mass flow rate of air.m =V × No. of changes × Density

= 10 3 1.23600

m = 0.01 kg/secSo sensible heat load,

SH = m cP T= 0.01 × 1 × 20 = 0.2 kW

Sol–94: (a)We know that the effective by pass factorof more than one similar coils.

Xc = xn

where,X By-pass factor of single coiln No. of coils Xc = (0.6)3 = 0.216

Sol–95: (c)When an electric current is applied acrossthe junction of two dissimilar metals, heatis removed from one of the metals andtransfered to the other. This is the basis ofthermoelectric refrigeration. It uses a prin-ciple called the Peltier effect. The Peltiereffect is the reverse of the Seebeck effect.Hence COP in thermoelectric refrigerationis a function of.(i) Electrical conductivity of materials.(ii) Peltier coefficient(iii) Temperature at node and(iv) Thermal conductivity of materials cold

junctions.

Sol–96: (a)Since the desired condition requires coolingfrom 45ºC dbt to 25° dbt and reducing rela-tive humidity from 70% to 50%, the prac-tical arrangement would be cooling anddehumidification.

Sol–97: (b)Latent heat of ammonia is significantlyhigher compared to other refrigerants henceit reduces mass of refrigerant required pertonne of refrigeration.

Sol–98: (a)heat removed from refrigerator Q1 = 1 kJ

so, COP = heat removedwork done

5 = 1Qw

W = 0.2 kJ

efficiency of heat engine = work doneheat input

0.A = Wheat input

heat input = W 0.2 0.5kJ= =0.4 0.4

Sol–99: (d)

Saturation

curve

Outdoorcondition

Mixtureconditionenteringthe coolingcoil

ADP5

42

6

31

GSHF line

Supply airfrom coolingcoil

RSHFline

Dry bulbtemperature

Fig. Grand sensible heat factor

The intersection of RSHF and GSHF linesgives room supply air condition.In the diagram, point 3 represents themixture condition of air entering the coolingcoil. When the mixture condition enters thecooling coil, it is cooled and dehumidified.The point 4 shows the supply air from the

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cooling coil. When the point 3 is joined withpoint 4, it gives grand sensible heat factorline (GSHF line). This line when producedupto the saturation curve, it gives theapparatus dew point (ADP).

Sol–100: (b)1 ton = 3.5 kW

COP =refrigeration

power required

power required

= refrigeration 3.5COP COP

N = 3.5

Sol–101: (d)

Sol–102: (c)Ideal refrigerant mixture obeys Raoult’s lawin liquid phase and Dalton’s law in vapourphaseRaoult’s Law

Pv1 = x1 P1 sat

Pv2 = x2 P2 sat

where x1 and x2 are mole fraction of compo-nents 1 and 2 in solution and P1 sat andP2 sat and are saturation pressures.Dalton’s Law

Pv1 = y1 Ptotal

Pv2 = y2 Ptotal

where y1 and y2 are vapour phase molefractions of components 1 and 2 respectively.

Sol–103: (b)From the energy balance,

1 21 w 2 w1 2m m m mh h

= 33 w 3m m h

1 21 w 2 wp 1 p 2m m m mc t c t

= 33 w p 3m m c t

t3 =

1 2

3

1 w 2 w1 2

3 w

m m m mt tm m

Sol–104: (c)Psychrometric chart is drawn for fixedatmospheric pressure.

Sol–105: (a)Sol–106: (c)

Definitely condenser is an importantequipment in steam power plant otherwiseoutput will be very low.For the same mass flow rate and pressurerise, pump requires very less power ascompared to compressor because volumeof liquid is less as compared to that ofvapour or steam of same mass.

Sol–107: (a)In regenerative the heat added in lowtemperature range is reduced whichincreases overall heat additiontemperature.

Sol–108: (b)Since both fuel and oxidiser in rocketare in rocket body and does not take airfrom atmosphere for combustion. So itcan operate in vacuum.In rocket highly compressed and heatedair comes out from rocket body. So it isa pure reaction machine. If air comesfrom outside and leaves system at highvelocity. This process will impart someimpulse due to change in momentum.This action is absent in rocket.

Sol–109: (c)The turboprop engines fills the gapbetween aircraft driven by reciprocatingengine (used for low speed and lowaltitude) and turbojet engine used forhigh speed and high altitude). So theyare used for medium speed and mediumaltitude with good performance.The reasons for reduction in efficiencyof turboprop engine of higher speed are

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- profile drag of blades, separation offlow from blade surface, swirl and slipof stream over blades of propeller i.e.shock formation.

Sol–110: (a)In closed cycle gas turbines, the workingfluid is generally other than air havinghigher value of specific heat ratio. In opencycle gas turbine, the working fluid is airand the products of combustion areexpanded upto the atmospheric conditionsince they cannot be used any more.Thus, the working fluid is replacedcontinuously.

Sol–111: (a)For the isentropic flow of an ideal gasthrough a convergent-divergent nozzle,mass flow rate per unit area is maximumwhen M = 1. Further, M = 1 occurs onlyat the throat and nowhere else and thishappens only when the discharge ismaximum. The nozzle is said to be chokedbecause it is incapable of allowing moredischarge even with further decrease inexhaust pressure.If the convergent-divergent duct acts asnozzle, in the divergent part also, thepressure will fall continuously to yield acontinuous rise in the velocity. Thus, fora certian mass flow rate, with the decreasein pressure, density decreases at a ratefaster than the rate at which areaincreases, as a result of which velocitycontinues to increase. This is true onlyfor a compressible fluid.

Sol–112: (a)

FF

Assertion is correct because if both endsare free and an axial force in applied atboth the ends, the system will be inequilibrium but if a displacement is giventhere would be rigid body movement ofthe member. This will not change thestress in the member because stress isdeveloped due to resistance which doesn’texist in this case.Reasoning is correct because rigid body

displacement without acceleration has noeffect on elastic deformation.

Sol–113: (d)In Counter flow HE, the heat transfer fromhot fluid to cold is at more uniform tem-perature difference than parallel. In paral-lel flow configuration the temperature dif-ference at inlet is very high and there ishuge entropy generation i.e. loss of avail-able energy which can never be recovered.So due to less entropy generation, thecounter flow HE are more efficient thermo-dynamically.At the same time, the LMTD of counterflow configuration is more than parallelflow.

Sol–114: (d)

Generally the non-metals have very fewfree electrons as compared to metals.So non-metals have lower thermalconductivity than metals.

There are other factor for thermalconductivity e.g. porosity etc, but themain factor is number of free electrons.

Sol–115: (c)In short open belt drive, when difference inpulley diameter is high, the lap angle onshort pulley is small. Due to this small lapangle at small pulley, there are chances ofslip. So to avoid slip at small pulley, anidler is provided on slack side as shownbelow.

IdlerIt does not change anything except lap angleand friction.

Sol–116: (b)Both Statement I and II are correct butneed of explanation does not arise becuaseboth statement are different and

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independent.

Sol–117: (d)During starting and stopping of journalbearing the speed of bearing is low andhydrodynamic pressure generated by rota-tion will not be sufficient to support theload. So there is metal to metal contactand bearing may demage so to avoid thisdemage the hydrostatic bearing is providedfor these operating conditions of Journalbearing.

Sol–118: (a)Production planning and control incontinuous production is usually for simplerthan that is job or batch production.Extensive effort is required for detailedplanning before production starts, but bothscheduling and control need not be elaborateusually. The output is either limited byavailable capacity or regulated within evenlimits to conform to production targetsbased on periodic sales forecast. Detailedplanning is required for setting the rate ofproduction from line and arranging thevarious production centres on the lineaccordingly.

Sol–119: (a)A heat pump is superior to an electric re-sistance heater because a heat pump re-jects more heat (Q1 = Q2 + W) as compareto electric resistance heater (W).

Sol–120: (a)The energy loss due to fr iction isproportional to the square of the velocity ofa fluid (steam). Since the fluid velocity isthe highest for a 2-row Curtis stage andlowest in the 50% reaction stage, so theenergy loss due to friction in the reactionstage is the least while that in the Curtisstage is the highest. The energy loss in theimpulse stage will be in between these twovalues. Therefore the efficiency of thereaction stage will be the highest and thatof the Curtis stage will be the lowest, whilethat of impulse stage will lie in between.

reaction(50)%stage simple impulse stage

2 row Curtis .

Conventional Question Practice Program - IES Masteriesmaster.org/public/archive/2016/IM-1462627731.pdf · R (2) ME (Test-3), Objective Solutions, 12th March 2016 Sol–1: (c) Using

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