answersiesmaster.org/public/archive/2016/im-1480342893.pdf · 2017. 5. 26. · c. helical gear: two...

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ESE-2017 PRELIMS TEST SERIESDate: 27th November, 2016

28. (b)

29. (b)

30. (b)

31. (b)

32. (d)

33. (d)

34. (b)

35. (a)

36. (c)

37. (c)

38. (c)

39. (d)

40. (c)

41. (b)

42. (c)

43. (a)

44. (d)

45. (d)

46. (b)

47. (c)

48. (a)

49. (c)

50. (d)

51. (b)

52. (a)

53. (d)

54. (c)

55. (a)

56. (b)

57. (a)

58. (b)

59. (b)

60. (d)

61. (c)

62. (a)

63. (b)

64. (b)

65. (a)

66. (d)

67. (b)

68. (b)

69. (d)

70. (a)

71. (c)

72. (c)

73. (a)

74. (b)

75. (b)

76. (c)

77. (d)

78. (d)

79. (d)

80. (c)

81. (a)

82. (c)

83. (c)

84. (d)

85. (d)

86. (d)

87. (d)

88. (a)

89. (a)

90. (b)

91. (c)

92. (a)

93. (d)

94. (c)

95. (d)

96. (b)

97. (a)

98. (d)

99. (a)

100. (d)

101. (c)

102. (b)

103. (b)

104. (a)

105. (d)

106. (a)

107. (a)

108. (b)

109. (d)

110. (c)

111. (b)

112. (b)

113. (b)

114. (b)

115. (b)

116. (c)

117. (c)

118. (a)

119. (b)

120. (c)

121. (d)

122. (b)

123. (d)

124. (d)

125. (c)

126. (b)

127. (d)

128. (b)

129. (a)

130. (b)

131. (b)

132. (b)

133. (c)

134. (c)

135. (a)

136. (a)

137. (d)

138. (d)

139. (a)

140. (a)

141. (a)

142. (a)

143. (a)

144. (b)

145. (a)

146. (d)

147. (a)

148. (d)

149. (c)

150. (d)

151. (c)

152. (c)

153. (a)

154. (a)

155. (b)

156. (b)

157. (d)

158. (d)

159. (a)

160. (c)

ANSWERS

1. (b)

2. (b)

3. (c)

4. (b)

5. (a)

6. (c)

7. (d)

8. (b)

9. (c)

10. (c)

11. (c)

12. (b)

13. (c)

14. (b)

15. (d)

16. (b)

17. (d)

18. (b)

19. (d)

20. (d)

21. (d)

22. (a)

23. (d)

24. (b)

25. (a)

26. (a)

27. (c)

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(2) ME (Test-12), Objective Solutions, 27th November 2016

Sol–1: (b)

Power source torque,

T = (800 200sin )N m

The coupled machine requires con-stant torque supply, so averagetorque,

Tavg =

2

0

1 T d =800 N.m2

Torque fluctuations,

T = avgT T

= 800 + 200sin 800

T = 200 sin

Maximum value of torque fluctuations,

maxT = 200, at = 90°

Angular acceleration (maximum),

max = = maxT 200Inertia 200

= 1 Rad/sec2

max = 1 rad/sec2 at = 90°.

Sol–2: (b)

0 90 180 270 360Crank angle in deg

Torq

ue in

N-m

1200

1000

800

Average energy of flywheel is workdonein cycle,

=

2

0T.d

= 1800×2 + ×400×22

= 2000 J

Energy fluctuation,

E =

3 /2

/2T

= 1 1200 10002

= 100

Coefficient of energy fluctuation,

Ke =

E 100%E

=

100 100% 5%2000

Sol–3: (c)

I = 1 Kg-m2, max 30rad / sec,

min = 20rad / sec.

During punching operation, maximum energyis supplied by flywheel, soCutting / punching energy

max1 T ×2

= 2 2max min

1 I2

max

1 T ×2 3

= 2 21 ×1× 30 – 202

= 250

Tmax =

2× 3 × 250 = 477.46 mm

Sol–4: (b)Disk type Flywheel,

2dmk = 21mr

2

kd =r2

For rim type,mkr = mr2

kr = r

d

r

kk

=12

kd = rk2

Sol–5: (a)To control the fluctuation of energy of anengine as prime mover, only flywheel isrequired. Because the fluctuation of speedneeds to be controlled are in a cycle only.The variation/fluctuation of speed over largenumber of cycle is not matter of concern.Governor controls the speed variation overlarge number of cycles in constant speedmachinary.

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Sol–6: (c)A. Law of correct steering

cot cot = cb

B. Displacement relation of Hooke’sjoint tan tan cos =C. Relationship between kinematic pairs

and links i.e. Grubler relationf = 3(n – 1) – 2j

D. Displacement equation for reciprocat-ing engine piston

x = 2 2n n sinR 1 cosn

Sol–7: (d)We know that the maximum peripheralspeed of a flywheel.

v =

max. tensile stressdensity of material

1. From this equation it is clear thatcoefficient of fluctuation does not comein picture. So it does not decide the rimdiameter. It is a performance parameternot design.

2. The stress involved in above expressionis due to centrifugal force which is massdependent.

Sol–8: (b)Maximum fluctuation of energy of flywheel,

E = 2 2max min

12

I = Emax – Emin

Sol–9: (c)Since the given gear train is epicyclic geartrain, So

3/3/

Arm C Gear A Gear B0 +1 40 x 4x

3xy x + y y4

1. Fix the arm and CCW one revolutionto A.

2. Rotate whole mechanism by x-revolution.

3. Rotate arm y-revolutions

ArmC rotates at, y = 200 rpm

Gear A is fixed, x + y = 0

x = –y = – 200 rpm

Rotation of Gear B

34 43xy – = 200– × –200

= 200 + 150 = 350 rpm.

Sol–10: (c)

1

2

3

36

Sum

96

Annulus

Planet

1. Radius of Annulus,

r3 = r1 + 2r2

Module of all gears must be same.

3mt2

= 12

mt mt2

t3 = t1 + 2t296 = 36 + 2t2t2 = 30

2. 3 to 4 planet gears are provided toepicyclic gear train for balancingpurpose.

3. Input and output are on either sun gearor arm or annulus.

4. Module of all mating gear must be same.Sol–11: (c)

T =602

T =103

M

x=35mm

m=1mmN =2400 rpm1

T1 T4

N =100 rpm4

The speed ratio in riverted gear train

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1

2

NN =

Product of teeth ofdriven gear

Product of teeth ofdriver gears

1

2

NN =

2 4

1 3

T TT T

2100100 =

4

1

70 TT 10

4

1

TT = 3 ...(i)

Distance between axes,

x = 1 2mT mT2 2

T1 + T2 = = =2x 2 40 80m 1

T1 + 70 = 80T1 = 10

T4 = 30

3 4MT MT2 2

= x

M(T3 + T4) = 2x Module of gear 3 and 4,

M = 3 4

2xT T

= =

2 40 8010 30 40

= 2 mmSol–12: (b)

1. Worm gears are used for very high speedreduction in single stage i.e. of the orderof 50 and more.

2. The velocity ratio in worm drive

=Angle turned by gearAngle turned by worm

=lead( )

worm wheel dia(d)

=2d

Independent of helix angle of worm given as

=1

leadd

d1-diameter of worm.

Sol–13: (c)

A. Spur Gear: Connecting two parallel,coplanar shafts and teeth are parallel toshaft axis. So A–2.

B. Bevel Gear: It connects two nonparallel coplanar and intersecting shafts.But these two intersecting shafts needsnot be perpendicular. So B–1.

C. Helical Gear: Two parallel, coplanarshafts. Teeth are inclined with shaft axiscalled as helix angle. So C–3.

D. Mitre Gear: Connecting mutuallyperpendicular shafts. The size of twogears i.e. bevel are equal. So D–4.

Sol–14: (b)

C D

A BTA

NA NB

TB

This the case of reverted gear train which isalso a compound gear train. So

Speed ratio, A

B

NN

= Product of teeth of driven gear (BandC)Product of teeth of driving gear (A &D)

= B C

A D

T TT T

Distance between axes,RA + RC = RD + RB

DA + DC = DD + DB

Module

m = DT

A CmT mT = D BmT mT

TA + TC = D BT T

So option ‘b’ is not correct.Sol–15: (d)

The only criterion for reverted gear train isthe co-axiality of input and output shaft. Soreverted gear train is:

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2

3

ccw

1

4

cwcw

So statement 1 and 4 are correct. But option‘d’ has statement 4 only. So it will becorrect.

Sol–16: (b)If number of gears in simple train is even,then the direction of output rotation is reverseof input rotation. In case of odd numbers ofgears, the output rotation is same.For example-1. 4-gear simple train-

It has two gears at input and output and2-idler.So the sequence of rotation is-

Input Output

Idler

c cc c cc

2. 5-Gear simple trainIdlers

Inputc

output

c cc c cc

Sol–17: (d)Speed ratio in compound train

= Speed of driverSpeed of driven

= Product of teeth of driven gearProduct of teeth of driving gear

or

= Product of teeth of followerProduct of teeth of drivers

Sol–18: (b)

Pressure angle = 14.5

Nos. of teeth t = 46Pitch circle diameter = 27.6 cm

1. Module, m = dt

= 27646 = 6

mm

2. Circular pitch = dt

= 19.67

3. Addendum = one module = 6 mm

4. Diametral pitch = =d 46t 276 = 3/18

Sol–19: (d)Stub 20° involute system has followingadvantage over 20° full depth involute system-1. Because of small addendum and more

dedendum of mating gear, interferenceis avoided.

2. Since the height of teeth is small sothey are stronger.

3. Due to small height of teeth, lessamount of metal is cut so the cost islow.

4. The minimum number of teeth-

tmin =

w2

2a

1 3sin 1

For full depth gear, aw = 1For stub teeth, aw = 0.8 Stub teeth gear has less minimum

teeth to avoid interference.Sol–20: (d)

B 20

C

A100

Action C A BFix arm C 0

revolutions of A 0 5xy revolutions of y y x y 5xwhole mechanism

x x

Arm C makes 4-revolution, y = 4Gear A is fixed,

x + y = 0x = –y = –4

Revolutions made by ‘B’,y–5x = 4 – 5x–4

= 4 + 20 = 24

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Sol–21: (d)

A

P

S t = 200

t = 40

Action A S PFix arm A 0

revolutions of sun S 0 5xy revolutions of y y x y 5xwhole mechanism

x x

Sun gear ‘S’ is fixed, x+y = 0Arm ‘A’ turn one revolution,

y = 1x = –1

So revolutions of planet gear’P’,

y–5x = 1 – 5(–1)

= 1 + 5 = 6Sol–22: (a)

300 M.m

100 cmR2

R1

A B

Since the shafts are in position R1 + R2 = 0Taking moment about A.

2100R100 = 300

R2 = 300N R1 = – R2 = – 300 N

Sol–23: (d)Secondary Force,

Fs =

2mr cos2

nwhere ‘n’ is ratio of length of connecting rodto length of crank as,

n = /rSo as ‘n’ decreases, secondary forceincreases

Sol–24: (b)

The schematic of reciprocating mechanismis,

m = 1.2 kg

l = 200 mm r = 50 mmN = 900 rpm

Angular speed of crank

= 2 N 2 × 900=60 60

= 30 rad/sec.

n = l 200 = 4r 50

Shaking force means unbalance force.

=

2 cos2mr cos +

n

=

2 11.2×0.05× 30 1+4

= 21.2× 0.05 ×900 ×1.25= 666.2 N

Sol–25: (a)m = 0.5 kg, k = 5cmk1 = 6 cm, = 4 22 10 rad/sec

The correction couple

ΔT = 1I IwhereI1 - mass moment of inertia of two massequivalent system = 2

1mk

I - mass moment of inertia of rod = mk2

T = 2 2 –4 40.5× 6 – 5 ×10 × 2×10

= 0.5 (36 – 25) × 2 = 11 N-m.Sol–26: (a)

9 kg

Shaft

B1 B2

100 cm50 cm50 cm 50 cm

50 cm

For static balance,

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B1 × 50 + B2 × 50 = 9 × 50

B1 + B2 = 9 ...(i)

Taking moment about B1 end,B2 × 50 × (100 + 50) = 9 × 50 × 50

B2 = 3kg B1 = 9 – B2 = 9 – 3 = 6 kg

Sol–27: (c)

For dynamic balance of masses rotating indifferent planes,

No radial or centrifugal force i.e. centreof mass on the rotation axis.

No unbalance couple

Sol–28: (b)

The direct and reverse crank method is usedto balance the radial engine.

In this method, the reverse crank which isimaginary rotates with same magnitude ofangular velocity as direct actual crank butin reverse direction.

Since secondary crank speed is double ofprimary crank. So the secondary reversecrank will be at 2 × 45 = 90° anticlockwiseposition.

Sol–29: (b)

Let the system is rotating at angular velocity' ' and weight will cause centrifugal force-

= 2W eg

So the schematic of situation is,

WR

A

e L

B

bx

W

xR

2W eg

2W eg

The reactions ‘R’ at bearings will beequal because of symmetry,Taking moment about any support i.e. Aor B

2 2W We L x e x RLg g

= 0

R = 2W e L - 2x

g L

2x + b = L

L – 2x = b

R = 2W beg L

Sol–30: (b)

r =600 26

2

= 313 mm

bp =2 N (W S)r

60 1000

=

2 600 (340) 313

60 1000= 6.6866 kW

Sol–31: (b)P

2

3

4

1v2 v1

r = 8v1 = 0.9m3/kg

r =1

2

vv

v2 = 30.9 0.1125 m /kg8

=

Work output,

W = mean eff. pressure × 1 2(v v )

The net heat interaction per cycle iswork output because change in internalenergy is zero.

1575 = mp (0.9 0.1125)

pm =31575×10

(0.9 0.1125)

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= 2000 kN/m2

= 20 bar

Sol–32: (d)

m = 80%, BP = 60 kW

Mechanical efficiency-

m =BP

PI

0.8 = 60PI

PI = 60 75kW0.8

=

Friction Power,

FP = P BPI= 75 – 60 = 15 kW

Sol–33: (d)Brake power = 40 kWIndicated power IP = ?

Mechanical efficiency,

m = BPIP

IP = 40 50kW0.8

=

Friction power = IP – BP = 50 – 40 =10 kWSince friction power is constant and at75% rated load, the brake power,

BP = 0.75 × 40 = 30 kW

IP = BP + friction power= 30 + 10 = 40 kW

Mechanical efficiency,

m = 75=30 %40

Sol–34: (b)The brake power of engine,

BP = 10 kW

bth = 30%

CV = 40,000 kJ/kg

Let fuel consumption rate = fm

Brake thermal efficiency

bth =f v

BPm C

Fuel consumption rate.

fm = 10 kg/sec0.3 40000

= 8.33 × 10–4 kg/sec= 3 kg/hr

Sol–35: (a)Morse test on 2-cylinder, 2-stroke en-gine-The complete equation for morse test-

P FP I = BP

(IP1 + IP2) – (FP1 + FP2) = BP1 +BP2

TotalBP= BP1 + BP2 = 9 kWCase- I : Spark cut off in cylinder-1 IP1 = 0 and

BP = 4.25From equation (i)

(0 + IP2) – (FP1 + FP2) = 4.25Putting this value in equation...(i)

IP1 + 4.25 = 9 IP1 = 4.75 kWCase-II : Now cylinder-2 spark is cutoff-

IP2 = 0 and

BP = 3.75From equation (i)

(IP1 + 0) – (FP1 + FP2) = 3.75Putting it in equation (i)

IP2 + 3.75 = 9IP2 = 9 – 3.75 = 5.25 kW

Total IP of engine when both sparkare on- IP = IP1 + IP2

= 4.75 + 5.25 = 10 kW

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Mechanical efficiency-

m = BP 9 90%IP 10

= =

Sol–36: (c)

In performance analysis, various param-eters are measured by various instru-ments,

A. BHP - Dynamometer i.e. A–3. Thedynamometer may be hydraulic, fric-tion type etc.

B. Speed - Tachometer i.e. B–2. Tachom-eter may be mechanical, electricalor optical.

C. Calorific Value - Bomb calorimeter.i.e. C–1

D. Exhaust emission - Flame ionizationdetector is one of several instrumentsor methods to analyse exhaust emis-sion.

Sol–37: (c)1. Supercharging reduces the delay period

of the CI engine which tends to reducethe chance of knocking in CI engine. Italso reduces self ignition temperatureof fluid. So smooth operation.

2. Due to supercharging heat generationand heat transfer increases. Due towhich there is greater tendency tooverheat the piston crown and the seatand edges of exhaust valves.Toovercome this problem the valve overlapis greater.

3. Supercharging for CI engine is reachedby thermal and mechanical loading dueto above reason.

Sol–38: (c)1. Due to supercharging power output for

given engine increases due to increasein density of air.

2. Supercharging is done in aircraft tocompensate the loss of power due toaltitude as density of air decreases athigh altitude.

3. Supercharging increases the knockingin SI engine due to increase intemperature of mixture.

4. Mechanical efficiency of superchargedengine is better not quite high comparedto naturally aspirated engines.

Sol–39: (d)The oxides of nitrogen (NOx) in petrol engineexhaust are due to availability of oxygenand high temperature. The recirculation offraction of exhaust gases reduces hightemperature so NOx formation reduces. Allother statements provide favourablecondition for NOx formation.

Sol–40: (c)1. Supercharging of diesel engine in-

creases mechanical efficiency despiteincrease in friction power. The rea-son being large increase in brakemean effective pressure i.e. brakepower.

2. The reduction of smoke in super-charged engine at overload is due toavailability of oxygen in sufficientquantity. In naturally aspirated en-gine, this level of smoke is high.

Sol–41: (b)The control of black smoke in dieselengine is-1. At lower load, the amount of fuel

injected is less so less black smoke.2. Maintaining injection system perfect

means right time injection becausedelayed injection results in lowerpower and black smoke due to in-complete combustion during expan-sion stroke.

3. Higher cetane number diesel meansearly combustion i.e. lower self igni-tion temperature. So fuel has enoughtime to burn if air is available andlower black smoke emission. So,third statement is wrong.

Sol–42: (c) Rich mixture to produce maximum

power produces maximum tempera-ture also. This maximum tempera-ture promotes NOx emission in ex-haust i.e. A–2.

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Wall quenching and wall depositesresults in extinguishing the flamewhich gives hydrocarbon emission inexhaust i.e. B–1.

Tetraethyl lead which is used to re-duce knock produces more and dan-gerous emission in exhaust i.e. C–4.

Reduction in compression ratio in-creases wall quenching area volumeratio which increases CO and hydro-carbon in exhaust i.e. D–3.

Sol–43: (a)In three way catalytic converter or over-all emission control, platinum and pal-ladium are used to oxidise HC and COand Rhodium reduces NOx into nitrogenand oxygen.

SecondaryAir

HC/COElement(Pt/Pd)NOx element

(Rhodium)

Sol–44: (d)

Three way catalytic converts controlsonly HC, CO and NOx and not SPMemission. SPM are big compounds con-taining carbon and cannot be control ifthey are out of engine cylinder. They areminimized by ensuring steady engineoperation.

Sol–45: (d)

Formation of NOx depends only uponhigh temperature and oxygen avail-ability. Coolant has no effect.

Type of coolant has no effect on NOxformation but rate of circulation af-fects NOx formation by controllingthe cylinder temperature.

It is Rhodium which reduces NOxinto nitrogen and oxygen not plati-num.

Sol–46: (b) Evaporative losses are completely

unburnt hydrocarbon from evapora-tion of fuel because it happens be-fore combustion.

Blow by emission is leakage of com-bustion products from piston ringswhich is not only CO and suspendedparticulates only but also others.

Since source of CO and NOx is com-bustion, so 100% CO and NOx arefrom exhaust. HC emission haveseveral sources i.e. evaporation, ex-haust etc.

Suspended particulate is in exhaustof CI engine due to nature of fuel.

Sol–47: (c)Unburnt hydrocarbon (HC) and oxide ofnitrogen (NOx) in presence of sun lightcauses photo chemical smog during win-ter. This is a complicated chemical re-action.

Sol–48: (a)1. Stirling cycle

P

Isoc

horic

2

3

Isothermal

Isothermal 4

V1

Isochoric

Stirling cycle consist of two isothermaland two isochoric processes.

2. Vapour compression cycle

T

s

Compressorentry

5

4

2

1

3

The vapour at entry of compressive atstate ‘1’ is saturated. This is idealsituation but in practice. The state ‘1’ is

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in dry staturated state to avoid anychance of liquid entry.

3. Diesel cycle

P

V

23

4

1

Isobaric

Isoc

horic

AdiabaticAdiabatic

Sol–49: (c)

The Ericsson cycle consist of two isobarsand two isothermals as shown below:

P

2 3

41

V

Isothermals

Sol–50: (d)

The Otto cycle has separate work and heattransfer as shown below:

P3

We

42

Wc

Q2

V

Q1

1

Process 1-2 - adiabatic no heat transferand only require compression work.

Process 2-3 since no volume change sowork transfer is zero because

1 2w =2

1p dV

dV = 0

1 2w = 0

This process is constant volume heataddition.

Process 3-4 - Adiabatic expansion, noheat transfer and only work output.

Process 4-1 - constant volume heatrejection and work output is zero becausedV = 0.

Sol–51: (b)

1. First figure on T-s diagram shows somephase changes so it is Rankine cycle.

2. Second figure has two isometric and twoadiabatic process so it is Otto cycle.

3. Third figure has two adiabatics, oneisobar and one isochore so it is dieselcycle.

4. Fourth figure has two isotherms and twoadibatics so it is Carnot.

Hence the correct sequence is-

Rankine - Otto - Diesel - CarnotSol–52: (a)

A. Otto cycle efficiency

3P

2

1

4

V

0 = 1k

11r

=1

2 1 4

1 2 3

V T T1 1 1V T T

= =

So A – 3B. Diesel cycle

P

V

32

4

1

The expression for efficiency

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d = k

1 1)1r 1

Where kr - compression ratio = V1/V2

- cut-off ratio = V3/V2

So B - 4C. Carnot cycle - The expression for

efficiency-

c = min

max

T1T

D. Brayton cycle - The efficiency expression

b = 1

3 21 P /P

=1

p

1i

r

P3 4

2 1

VSo D – 2

Sol–53: (d)A. The cyle has two constant volume and

two adiabatic isentropic process, so itwill be Otto cycle i.e. A–2.

B. The cycle has two constant volume andtwo isothermals. So it is Stirling cyclei.e. B-3.

C. The cycle has two isobars and twoisothermals so it is Ericsson cycle i.e.C-4.

D. The cycle has two isobars and twoadiabatic so it is Brayton or Joule cyclei.e. D-1.

Sol–54: (c)

2P

3

V

1

PV C=

The ideal cycle thermal efficiency

= net

1

WQ =

2 3 3 1

1 2

W WQ

=

2 2 3 31 3 1

v 2 1

P V P VP V V

1C T T

P1 = P3 and V1 = V2

2 2 3 3 1 3 1 1 3 1

2 1

P V P V P V V P V VR T T

=

2 2 3 3 1 3 1 1 1 3 1

2 2 1 1

P V P V P V P V P V VP V P V

=

1 3 1

1 2 1

P V V1

V P P

Note: It is pulse jet engine cycle or Lenoircycle.

Sol–55: (a)

A. Two isothermals and two adiabaticconstitute Carnot cycle as shown below:

2

3T2

P

14T1

VPressure 2-3 and 4-1 are isothermalsand 1-2 and 3-4 are adiabatics. Becauseadiabatic process has more slope thanisothermals on p-V diagram. So A-3.

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B. Two isothermals and two isochores orisometerics constitute Stirling cycle asshown below:P

3

2

1

4

T1

T3

VProcess 1-2 and 3-4 are isothermals and2-3 and 4-1 are constant volume i.e.isochores or isometric. So B-4

C. Two adiabatics and two constant volumeprocess constitute Otto cycles shownbelow:

Process 1-2 and 3-4 are adiabatics. SoC-1.

D. Two adiabatic and two constant pressureprocess constitute Brayton or Joule cycleshown below.

V

P

4

32

1

pVx = C

Process 1-2 and 3 -4 are adiabatic. SoD-2

Sol–56: (b)

Otto = 1

11kr

Thus, as increases, Otto will alsoincrease.

Sol–57: (a)Efficiency of diesel cycle

=

k

11 s 11

kr s 1

Efficiency of otto cycle = 1 – 111

r

Since, efficiency of both these cycles are

equal, so

1–

k

11 s 1

kr s 1

= 1

11r

ks 1k s 1

= 1

ks k 1s 1 = 0Sol–58: (b)

For the same compression ratio and sameheat input minimum heat is rejected inotto cycle and maximum heat is rejectedin diesel cycle.T

3

3

3 2

44

6 6 6 S5

41

2

Sol–59: (b)Ig = 3000 kJ/m2-hIb = 2000 kJ/m2-hId = Ig – Ib = 1000 kJ/m2-h

Total radiation flux incident on tiltedsurface (IT)

IT = Ibrb + Idrd= (2000 × 1.1) + (1000 × 0.9)

IT = 3100 kJ/m2-hSol–60: (d)

rr =1 cos

2

= 1 cos0.2 and 0.022

rr = 0.2 × 0.02 = 0.004Sol–61: (c)

a = 2kL/cose

= 16 0.002/cos60e

a = 0.064e

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Sol–62: (a)Bottom loss coefficient (Ub)

Ub = K 0.02L 0.1

Ub = 0.2 W/m2COverall heat transfer coefficient (UL)

UL = t b sU U U

= 6.6 + 0.2 + 0.2UL = 7W/m2C

Sol–63: (b)Both factors will decrease collector efficiencyfactor.

Sol–64: (b)Volume change should be small during phasechange.

Sol–65: (a)Sol–66: (d)

rd = 1 cos2

rd = 1 cos 602

rd = 1 1 / 2 32 4

Sol–67: (b)Solar time = (standard time) 4 (standardtime longitude – longitude of location) +Equation of time.Since Mumbai lies in Eastern hemisphereso negative sign will be taken in firstcorrection.Solar time = 1430 hr – 4 (82.5 –72.5)minutes + (–5) minutes= 1430 hr – 40 minutes – 5 minutes= 1345 hr

Sol–68: (b)

C =a

1sin

where a is half the value of acceptance

angle.

C =1

sin60

C =23

Sol–69: (d)Compound parabolic collector is non-imaging device, has large acceptance angleand requires only intermittent tracking.

Sol–70: (a)For Re > 2000Nu = 0.023 Re0.8 Pr0.4 in above situation

Sol–71: (c)Value of FR increases on increasing massflow rate through collector.

Sol–72: (c)

AF = L DF cos F sin

= 7 + 8 = 15°

= D Dx F cos F sin

= DF sin15 x cos15

Sol–73: (a)Base silicon has high reflection coefficientof order of 0.33 to 0.54

Sol–74: (b)

max = sc oc

T c

FF I VI A

High value of Isc are obtained with lowband gap material whereas high value of(FF) and Voc are obtained with high andgap material. Therefore maximumefficiency is function of band gap energy.Highest efficiency values are obtained forsilicon and gallium arsenide.

Sol–75: (b)Battery for solar PV system should havelow self discharge rate.

Sol–76: (c)

Poutput = 50 W, conversion 10% 0.1

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IT = 1000 W/m2

We know that

max = output

T

PI A

A = output

T max

PI

A = 250 0.5 m1000 0.1

Sol–77: (d)Performance of amorphous silicondegrades by 10% to 20% by exposure tolight.

Sol–78: (d)It consist of two or three blades withaerofoil cross section.

Sol–79: (d)Darrieus rotor is vertical axis windmachine

Sol–80: (c)Relation between power coefficient andaxial induction factor is given asCp = 4a(1–a)2

Maximizing the value of Cp

pdcda = 0

8a(1 – a) + (1 – a2) × 4 = 0 (1 – a) (4 – 12a) = 0

a = 1 or a = 13

Value of ‘a’ cannot be equal to 1

a = 13

Sol–81: (a)According to Betz theory, no wind turbinecan extract more than 59.3% of energyavailable in wind.

Sol–82: (c)Typical composition of producer gas onvolumetric basis is:

N2 : 50 – 54%, CO – 20 – 22%, H2 – 15– 18%, CO2 : 9 – 11%, CH4 : 2 – 4%

Sol–83: (c)Both gravitational pull and rotationalmovement is responsible for tidalphenomena.

Sol–84: (d)Reduction is gain of electrons andoxidation is loss of electrons.

Sol–85 (d)All these components are part of a controller.

Sol–86 (b)Industrial robots known for positioning flexibilityhave an undesired characteristic of elasticflexibility. The elastic flexibility impairs thepositional accuracy in assembly tasks.The beam-like structure used in most of therobots leads to considerable bending anddeflection. finite element method (FEM)technique permits static and dynamic analysisof structure and can be used for estimatingdeflection and stresses at various locationsoptimisation of structure, for estimating naturalfrequencies etc.

Sol–87 (d)Many commercially available industrial robotare widely used in manufacturing and assemblytasks such as:

Spot/arc welding.

Material handling

Parts assembly.

Paints spraying

Loading and unloading NC (numericallycontrolled) machines.

Space and undersea exploration.

Handling hazardous materials

Prosthetic arm research.

Sol–88 (a)Pneumatic sensors may be contact lessdevices or contact sensors. They can be usedas proximity sensors, touch sensors, orpressure or force sensors.

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Sol–89 (a)Since the arm movements in different robotconfigurations are different, the work volumesor work envelopes of different coordinatesystems are also different.

Sol–90 (b)

Sol–91 (c)

Repeatability of the robot link = ± 3 = ± (3 × 0.06)= ± 0.18 mm

Sol–92 (a)Manipulator is a park of the robot.

Sol–93 (d)

Electric – Preferred drive system in today’srobots

Hydraulic – Known for their higher powerand lift capacity.

Sol–94 (c)Six-axis robots move forward and back, upand down and can yaw, pitch and roll to offermore directional movement than SCARAs. Thisis suitable for complex movements thatsimulate a human arm– for e.g. reaching undersomething to grab a part and place it in aconveyor. The additional range of movementalso lets six axis robots service a larger volumethan SCARAs can.

Sol–95 (d)Sol–96 (b)

Roll motion also known as wrist swivel,enables rotation of wrist

Yaw, also called wrist yaw, facilitatesrightward or leftward swivelling movementof the wrist.

Sol–97 (a)Pneumatic drives–These systems usecompressed air to move the robot arm. Thepneumatic systems may employ a linearactuator or rotary actuators.

Sol–98: (d)Electrical system

R

IV1 V2

Thermal system,

T1 T2

qtR

kA

l

So voltage is analogus to temperature.

Current is heat flow,

Electrical resistance is analogous tothermal resistance.

The capacitance or charge carryingcapacity of conductor is equivalent tothermal capacity (i.e. specific heat)

Sol–99: (a)The schematic showing layers

k1 k2 k2

1 2 3

t1 t2 t3h1

kn

n

tn hn

iki

Conduction resistance of ith layer

Ri = i

i

tk A

The thermal circuit of composite wall,

1

1h A

1

1

tk A

2

2

tk A

n

n

tk A 2

1h A

Equivalent resistance for unit area, A = 1

1H

= 1 2 n

1 1 2 n A

t t t1 1.....h k k k h

= n

1 nn 1

t1 1h k h

=

i

i

H = n

1 nn 1

1t1 1

h k h

=

i

i

Sol–100:(d)

0.3m 0.3m

outercasing

120°C

1000°C

T= 48°C

IES M

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The furnance outside casing is made metalicor other way the resistance offered by it isnegligible, The heat transfer,

q = 1 2

1 2

T TR R

=

1 2

1 2

1000 120L L

k A k A

= 880

0.3 0.33 1 0.3 1

= 800

0.1 1= 727.3 W

But, q = h (T2 – T1)

727.30 = h (120 – 48)

h =727.3

72= 10.1 W/m2K

Sol–101:(c)

T1

T2

The peak in temperature profile is possiblein these two cases only i.e. the giventemperature distribution is possible inunsteady state and heat generation.

But the temperature profile variation is alsoin variable conductivity but peak is not threreas shown below.

T1

T2

3

1

2

Thermal conductivity for-Case-I,

k = 0k (1 T)

Case-2, k = k0 = constant

Case-3, k = 0k 1 T

Sol–102:(b)

T1

T2

One dimensional heat transfer equation insteady state from fouriers law,

q = kdTd

x

From Fourier’s law, the heat flux,

q = dTkd

x

Thermal conductivity varies linearlywith temperature as,

k = 0k (1 T)

dT1 Td

x

= 0

qk

Integrating it-2TT

2

=0

q ck

x

2 2TT

=

0

2q 2ck

x

Case-I Taking (+)ve sign-

22

2T 1T

= 20

2q 2c 1k

x

21T

= 20

2q 2c 1k

x ...(i)

Case-II Taking (–)ve sign-2 2TT

=

0

2q 2ck

x

22

2T 1T

= 20

2q 2c 1k

x

21T

= 20

2q 2c 1k

x...(ii)

The curve of these two equations on T-x diagram

IES M

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T

Case-II

III

Case-I

x

Hence in case-I upper segment of pa-rabola is same as given in question, sothe actual variation of thermal conduc-tivity-

k = 0k (1 T)

Note that the lower segment of case-IIparabola is for k = 0k (1 T)

Sol–103:(b)

cr r=

q

r

Rth

rcr r=

The concept of critical insulation says– Atcritical radius the thermal resistance isminimum and heat loss is maximum. So if

Radius of conductor is less than criticalradius and upon increasing radius byincreasing insulation thickness, the heatloss is increased as shown in figure.This situation is beneficial in electricalconductor and have more current.

Sol–104:(a)

n321

0r1

r2k1

k2k3kn

rn

r2

Heat transfer through cylinder,

Q = 1 2

2 1

1

T Tnr / r2 k L

l

= 1 2

2 1

1

2 L(T T )nr / r

k

l

The same heat will pass through all layers.So for n layers

=

=

=

1 n 1k n

n 1

n nn 1

2 LT Tr1 log

k r

Sol–105:(d)A. Radial heat conduction in sphere

221 d dTr

dr drr = 0

2

2d T 2 dT

r drdr = 0

B. One dimensional heat conduction inplane slab

2

2d Tdx

=1 T

x

C. For cylinder, radial conduction only

d dTrdr dr = 0

2

2d T 1 dT

r drdr = 0

D. The governing equation of constantcross-section fin-

22

2d md

x = 0

where ' ' is temperature differencebetween surface and surroundings

Sol–106:(a)Since heat dissipated from free end from isnegligible so,

q = x p

dTkdx

=

0 = x

dTkdx

= l

k cannot be zero, so

x

dTdx

= l

= 0

Hence temperature gradient at tip of fin iszero.

IES M

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Sol–107:(a)

Base

A 1ms=2

A 2mf = 2

Fin

50°C

h 20W/m = 2K

30°C

Assuming infinitely long fin

The heat loss from fin,

Qt = 0 cQ hpkA

Effectiveness,

=c

kphA

0.75 =c

kphA

ckp 0.75 h·A

= 0.75 20 1

= 3.354

Total heat transfer from fin

Qt = 0 ck p hA

= 20 3.354 20 1 Qt = 400 W

Sol–108:(b)Since enclosure consists of four surfacesnamely 1, 2, 3, 4.So, sum of all view factors,

F11 + F12 + F13 + F14 = 10.1 + 0.4 + 0.25 + F14 = 1

F14 = 1 – 0.75 = 0.25Reciprocity Theorem,

A1F14 = A4F41

F41 = 1 14

4

A F 4 0.25A 2

=

F41 = 0.5

Sol–109:(d)Emissive power of a surface,

E = 4T

T1 = 27°C = 300 K

E1 = 4300 ...(i)Now temperature of body increased to–

T2 = 627°C = 900 KThe emissive power

2E = 4.900 ...(ii)

Dividing these two equations,

2

1

EE

=44

4900 900

300300

=

= 34 = 81Sol–110: (c)

D1

D21

2

We know that,F22 + F21 = 1

F12 = 1 (Convex surface-1)Reciprocating theorem,

A1F12 = A2F21

F21 =1 1

2 2

A D 11A D 3

= =

F22 = 1 – F21

=113

= 23

Sol–111: (b)Fluid flowing over flat plate-

x

Nu = 100= 0.5 mx

P

Thermal conductivity of flowing fluid -

k = 0.025 W/mK Nusselt Number-

Nu = hkx x

IES M

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100 = h 0.5*0.025x

Coefficient of convection,

hx = 22.5 5 W/m K0.5

=

Sol–112: (b)Assuming no loss of heat from plate byradiation, so

500 W

/m2

30°C

2m25°C

2m

Convection heat loss = solar heat gain500 × 4 = h × 2 × 4 × (30 – 25)

h =500 4

8 5

= 50 W/m2K

Sol–113: (b)

T

u

Ts

The laminar flow over vertical plate is shownin figure. The Nusselt number for naturalconvection-

Nu = 0.25C(Gr Pr)

The Grashof’s number-

Gr =

2 3 3

2 2g TL g TL

Since all other parameter except

sT( T T ) are constant

The Nusselt number-

Nu =

0.253

2g TLC Pr

Nu 0.25T

Nu = 0.25C T ...(i)Initial temperature difference-

1T = 180 – 20 = 160°C

Nu1= 0.251C 160 ...(ii)Final temperature difference -

2T = 30 – 20 = 10°C

Nu2= 0.251C 10 ...(iii)From equation (ii) and (iii)

1

2

NuNu

=0.25160 2

10

Nu2= 1Nu 32 162 2

Sol–114: (b)Flow over flat plate

th

U

Hydrodynamic boundary layer

h = 0.5 mm

Dynamic viscosity- = 25 × 10–6 Pa-seccp = 2.0 × 103

Thermal conductivityk = 0.05 W/mK

Prandtl number-

Pr = 6 3

pc 25 10 2 10k 0.05

= 3 250 10 1

5

Thermal boundary layer,

t = h0.33Pr

= 0.330.5

1 = 0.5 mm

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Sol–115: (b)Shell and tube heat exchanger-

T

Length

Th1

Th2

Tc1

Tc2

Hot fluid exhaust gases.

hm = 5kg/secCh = 1000 J/kg-K

Heat capacity rateCh = h hm C = 5 × 1000 = 5000 kg/KCold fluid -water

cm = 4180 J/kg-KCc = 2 kg/sec

Heat capacity rate.Cc = c cm C

= 4180 × 2= 8360 W/K

Cmin= Ch = 5000 W/KHeat transfer area A = 24 m2

Overall heat transfer coefficientU = 500 W/m2K

Number of transfer units-

NTU =min

U.A 500 24C 5000

= 2.4Sol–116: (c)

Double pipe HE,

Parallel Flow Arrangement

ColdHot

Hot water-

hm = 10 kg/secThi = 80°C

Cold water

cm = 20 kg/secTc i = 20°C

The temperature profile in parallel flowarrangement.80°C

Th

20°C

Hot

x

ColdTc

When length is very highTh = cT 40 C=

Counter flow arrangement

Cold

80°C

20°C

ThHot

x

Tc

Assuming no loss of heat

h h hm c 80 T = c c cm c T 20

10 × 4.2 (80 – Th) = 20 × 4.2 (Tc –20) (80 – Th) = 2(Tc – 20)Since mass flow rate of hot water isless so its temperature reduction willbe large as compared to temperatureincrease in cold fluid. Hence, first hotfluid will reach 20°C.

Th = 2080 – 20 = 2(Tc – 20)Tc – 20 = 30Tc = 50°C

Sol–117: (c)

Double pipe counter flow HE

Hot fluid - oil

IES M

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hm = 1000 kg/hch = 20 J/kg-K

Cold fluid

cm = 1250 kg/hcc = 16 J/kg-K

1hT1

2cT2hT

2

1cT

Hot125°C

Cold 125°C150°C Hot

150°C

75°C Cold

Assuming no loss of heat,

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

1000 × 20 (150–125) = 1250 × 16(75 – c1T )

20000 × 25 = 20000 (75 – Tc1) Tc1 = 75 – 25 = 50° LMTD of HE,

LMTD = 1 2

1 2ln /

1 = Th1 – Tc2 = 150 – 75 = 75°C

2 = Th2 – Tc1 = 125 – 50 = 75

1 = 2

LMTD = AMTD

= 1 22

= 75°CSol–118: (a)

Hot180°C

30°CCold

110°CCold

Hot160°C

HE

Since in heat exchanger, no loss of heat

1 2h h h hm c T T = 2 1c c c cm c T T

Ch (180 – 160) = Cc (110 – 30)

Ch × 20 = Cc × 80

Ch = 4 Cc

From this equation i t is clear thatCmax = Ch and Cmin = Cc

Heat capacity ratio

C = min

max

CC =

c

h

CC =

14

= 0.25

Sol–119: (b)Since heating and cooling both sides aregases, so both heat capacity rates are equal

Cmin = Cmax

In counter flow design, effectiveness,expression in this situation is,

=NTU

1 NTU

0.8 =NTU

1 NTU1 + NTU = 1.25 NTU

NTU =1 1 4

1.25 1 0.25= =

Sol–120:(c)We know that heat capcity ratio,

C = min

max

CC

In phase change (boiling and condensationboth) Cmax is very very high or infinite socapacity ratio is zero.Then the expression for effectiveness of heatexchanger in both parallel and counter flowchanges to

= 1 – e– NTU

Sol–121:(d)

In regenerative heat exchanger, the heatis transfered from hot to cold fluidintermitently. Here a heat transfermedium is used which first exposed tohot fluid then to cold fluid and this waytransfer heat. So Ljungstrom air heateris regenerative heat exchanger used inthermal power plant as air preheater.

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We know that the temperature distributionin cyclindrical wall is logarithmic i.e.

T(r) = C1ln r + C2

The temperature distribution inspherical conduction is hyperbolic i.e.

T(r) = 12

C Cr

Water cooling lower are direct con-tact type HE i.e. air and water comein contact for heat exchange.

Sol–122:(b)A. Fourier law governs conduction, heat

transfer,

Q = –dTk Adx

B. Stefan Boltzman law governs radiationheat transfer,

E = 4T

C. Newton's law of cooling governsconvection heat transfer

Q = h A T

D. Ficks law governs diffusion mass transferas Fourier law conduction,

mA

=dCDd

x

Sol–123:(d)The correct expressions of variousdimensionless number and their matchesare,A. Schmidt Number

Sc =Kinematic Diffusivity ( )

Mass Diffusivity (D)

= D

A – 3

B. Thermal diffusivity ( )

=p

kC

B – 4

C. Lewis number, Le

= Thermal Diffusivity( )Mass Diffusivity (D)

=

p

kC D

C – 1

D. Sherwood Number – It is equivalent toNusselt number in convection,

= mh LD

D – 2Sol–124:(d)

25 mm dia.

1.75 m/sec

500 mm

100 mm

Since the nozzle lies in horizontal plane.So from continuity equation

A1V1 = A2V2

2 2

2(0.100) 1.75 = (0.025) V4 4

V2 =

20.1001.75 ×0.025

= 1.75 × 16= 28 m/sec

Sol–125:(c)

x

y

A = (0.5 – 0.2x)

Discharge,Q = A.V

V =QA

…(i)

Discharge is varying with time as, Q = (0.5 + 0.2t)

Velocity, V = (0.5 + 0.2t)(0.5 0.2x)

Local acceleration

IES M

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x 0

V 0 + 0.2×1 0.2= =t 0.5 0.2×0 0.5

= 0.4 m/sec2

Sol–126:(b)

S1x

h

S2

V

Applying Bernoulli’s equation at centre lineof tube

P + S2hg + xS1g = 21 1

1P S V S g(x h)2

P + S2hg = 21 1

1P + S V + S hg2

(S2 – S1)hg = 21

1S V2

V2 = 2 1

1

(S S )2hgS

V =

2

1

S2hg 1S

Sol–127:(d)The equivalent pipe has same friction lossas given combination of pipes having samedischarge.Friction loss in pipe

h f = 24fL V

D 2g

= AV

= 2D V4

hf =2 2

2 2 52

4fL Q 64fL Q× = ×2gD 2g D

D4

Now we have to find equivalent length interms of D

H = 1 2 3f f fh + h + h

2 2e

2 5 2 5 5 5L64fQ 64fQ L L 4L= + +

2g D 2g D (D / 2) (2D)

e5 5

L 1 4= 1+ 32 +32D D

Le =

1L 33 +8

Le =133 L8

Sol–128:(b)In steady state

Q1 + Q2= Q4 + Q3

A1V1 + A2V2 = A4V4 + A3V3

50 × 10 + 50V2 = 70 × 5 + 80 × 550V2 = 350 + 400 – 500

V2 =25050

= 5 cm/sec

Sol–129:(a)U

(U u)

ActualBoundary

Effective Boundary

u

y

It is the distance move by boundary surfaceinto the flow such that actual momentumflux would be same if the ideal fluid flowpast the boundary. The expression formomentum thickness,

=

0u u1 dy

U U

Sol–130:(b)We know that the thickness of boundarylayer at a distance x from leading edge,

U

x

x

5 5= =x Re Ux

IES M

ASTER


= 5xUx

x

At x = 2m

= 2mm

2

4

2=4

4 = 2 2

= 2 2 mm

Sol–131:(b)The laminar flow between two plates, onestationary other moving is called Couetteflow after the name of M.F.A couette.

Velocity Profile

Sol–132:(b)

Co-axial cylinder Viscometer: It isbased upon Newton law of viscosity. Thegap between two cylinders is very small.

Capillary tube viscometer: HagenPoiseuille viscosity.

Say-bolt viscometer: In this viscometerviscous fluid falls through capillary. Thisfalling of fluid is called efflux typeviscometer.

Falling sphere viscometer: Here asmall sphere is gently droped in liquidat rest and distance is measured in timei.e. velocity.

Sol–133:(c)Turbulent viscosity or eddy viscosity inturbulent flow is given by

t =

2 uLy

Kinemetic eddy viscosity,

t =

2t u= L

y

This turbulent viscosity is not property offluid but characteristic of flow. It's value can

be many times of viscosity of fluid ( ) orzero depending upon type of flow i.e.turbulent or laminar.

Sol–134:(c)The dimension of various parameters,(A) Dynamic viscosity, ()

= N·s/m2

=2

2MLT T

L= ML–1 T–1

(B) Chezy’s Roughness coefficient ‘C’.

Velocity, V = C mi

LT–1 =12CL

C 1

12L T

(C) Bulk modulus of elasticity

B =3 2

3 2V P L MLT

V L L

=

=2

2MLT

L

= ML–1T–2

(D) Surface tension, = N/m

=2MLT

L = MT–2

Sol–135:(a)Since the ship is Heating on surface so itis Froude model.

Frp = Frm

p m

p m

V V=gL gL

Vm =

mp

p

LVL

= 115

25

=155 = 3 m/sec

IES M

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Sol–136:(a)We know that force on propeller in fluid(neglecting compressibility)

F = L2V2(Re)

For both model and prototypes, Reynoldnumber is same.

m P2 2 2 2

m m m P P P

F F=L V L V

Since VP = 10 m/s,Lm : LP 1 : 10Vm = 5 m/sm = P (air in both cases)

P2 2 2 2m m

F50 =L 5 (10L ) 10

FP =50 100 100

25

= 20000 N

Sol–137:(d)In skew shaft or spiral gears, the contactbetween gears in point contact. Due to thispoint contact, these gears are used in lightduty applications only.

Sol–138:(d)

The epicyclic gearboxes are compactbecause of gears on moving axes andmore than one pair of gears remain incontact. While in spur gearbox, (simpleand compound) one pair of hears hassingle reduction only.

Sol–139:(a)The basic function of differential gears inmotor vehicle is to run the back wheels atdifferent speed at turns. Generally it isprovided at back axle as it plays its rolewhile taking a turn.

Sol–140:(a)Danger of lifting the wheel from track is dueto hammer blow cause by unbalance forcewhich is perpendicular to stroke line ofpiston or perpendicular to track.

Sol–141:(a)Since the critical speed is characteristic ofa system and so the natural frequency. But

instatical and dynamical balanced system(ideal condition), the natural frequency isassumed to be zero and period of oscilationas infinite. So any unbalanced anddisturbing force to cause vibration is zero.

Sol–142:(a)B

r

A

R 2mr cos 2MR cos

2MR sin

O

M

It whole reciprocating primary force isbalance i.e.

2mr cos = 2MR cos

mr = MRThe balancing mass M has one morecomponent of force perpendicular tostroke as.

2MR sinThis force component will cause ham-mer below on rail and after certainvalue of angular velocity ' ' the wheelwill start lifting from rail.

Sol–143: (a)

Overcooling in engine is always avoided.because it may result in higher frictionlosses due to increase viscosity of lubri-cants. In addition to higher viscosity,cooling losses are also increase. Theseall factors results in lower efficiency.

Sol–144: (b)Generally supercharging is preferred inCI engine to have more output and re-duced knock tendency. But at the sametime, knocking is accelerated in super-charging of SI engine due to increasedpeak temperature and pressure.

Sol–145: (a)Lead in the fuel will be deposited on thereduction catalyst or it may corrode thecatalyst. So the catalyst will be shortlived and will not be effective, in en-gines using lead fuels in long period.

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Sol–146: (d)In total emission control or three waycatalytic conver ter , first NOx is reducedand HC and CO are oxidised as shownin figure below:

SecondaryAir

HC/COElement(Pt/Pd)NOx element

(Rhodium)

Sol–147: (a)Both statements are correct and reasoncorrectly explains the assertion.

Sol–148: (d)For a given compression ratio, thermalefficiency of otto cycle is higher thanthat of diesel cycle.For the same compression ratio andsame heat input minimum heat isrejected in otto cycle and maximum heatis rejected in diesel cycle.

T 3

6 S5

412

33

4

66

4

2

Sol–149: (c)

upperarm

Forearm

Elbow joint

Shoulder joint

Trunk

The work envelope of the robot with jointedspherical coordinate system approximates aportion of a sphere. This configuration canoperate in a large working volume for its sizegiv ing greater reach and thus greaterapplication flexibility.

Sol–150: (d)A joint used in robots is a lower pair formedbetween two links.

Sol–151: (c)Heat cannot be added or removedsimultaneously in packed bed storage.

Sol–152 (c)

Generally biomass has calorific value of18–21 MJ/kg, which is less than 29.7 MJ/kg of CV of ethanol.

Biomass gasification means incompletecombustion of biomass resulting inproduction of combustible gases consistingof carbon monoxide (CO), hydrogen (H2)and traces of methand (CH4). this mixturecalled producer gas, can be used to runinternal combustion engines, (both SI andCI), can be used as a substitute for furnaceoil in direct heat applications, and otherapplications. However, ethanol or biogasproduction sequences selected biomassmaterials, hence biomass gasification ismore attractive.

Sol–153 (a)

Solar water heating systems are most likelyto be cost effective for facilities with waterheating systems that are expensive tooperate, or with operations such aslaundries or kitchens that require largequantities of hot water. Hence, thesedevices are preferred for water heatingsystems.

Sol–154: (a)

Both statements are correct and reasoncorrectly explains the assertion.

Sol–155:(b)

The reflective coating of heat insulatingmaterial is applied inside surface (con-ditioned space) of room. This will pre-vent heat leakage from window. Such

IES M

ASTER


heat reflected glases are also called asheat mirror.Window pane glass transparent to solarradiation depends upon type of climateoutside and purpose of building. In In-dia, window pane glasses are made aslow transparent as possible.

Sol–156:(b)The ‘n’ surface enclosure,

45 6

3

2

1n

Assume no leakage of radiationF11 + F12 + F13 + F14 + ......+ Fin = 1

n

1ii 1

F=

= 1

This is also called summation ruleBoth statements are individuallycorrect.

Sol–157:(d)The schematic of boiling.

satT

critical flux

Hea

t flu

x q

When heat flux ‘q’ increases beyond criticalvalue, the temperature between solid surfaceand liquid increases sharply not decreases.After critical value the boiling regimechanges from nucleate boiling to filmboiling. In film boiling a film of vapour isformed between surface and liquid and heattransfer is mainly through radiation.

Sol–158:(d)The Nusselt number Nu is defined as-

Nu = h ConductanceResis tancek ConvectionResistance

x

For unity of Nusselt number, the con-duction and convection resistance shouldbe of same order. This condition boundto happen in special fluid e.g. Binghamfluid. So in sluggish flow of highly vis-cous fluid. But in normal working fluide.g. air, water etc, the Nusselt numberis always greater than unity.

Sol–159:(a)With gradual closure of valve the velocityjust below valve i.e. at B increases to veryhigh as shown in figure below.

B

V

Due to very high velocity pressure at B.decreases to very low value and pressureat A is constant or increase a little. Due toreduction of pressure at B, the mercury levelrise to B.

Sol–160:(c)Power transmitted through a pipe,

P = gQh

=

2fg D V(H h )

4

P =

32 4fL Vg D HV

4 D 2g

Differentiating it with respect to 'V',

22dP 4fL V= gD H 3 × ×

dV 4 D 2g = 0

H – 3hf = 0

h f =H3

As head loss deteriorates, the velocityreduces. But at maximum powertransmission head loss is one-third. Sovelocity will not be maximum.

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answersiesmaster.org/public/archive/2016/im-1480342893.pdf · 2017. 5. 26. · c. helical gear: two...

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