Conventional Question Practice Program ANSWERSiesmaster.org/public/archive/2016/IM-1462632660.pdfquick return mechanism is an inversion of single slider crank mechanism. Oldham coupling

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Conventional Question Practice ProgramDate: 7th May, 2016

1. (c)

2. (b)

3. (c)

4. (c)

5. (d)

6. (d)

7. (b)

8. (c)

9. (d)

10. (a)

11. (b)

12. (b)

13. (c)

14. (c)

15. (c)

16. (a)

17. (b)

18. (c)

19. (c)

20. (a)

21. (b)

22. (b)

23. (a)

24. (d)

25. (a)

26. (d)

27. (d)

28. (a)

29. (b)

30. (c)

31. (b)

32. (a)

33. (c)

34. (a)

35. (c)

36. (c)

37. (d)

38. (d)

39. (a)

40. (c)

41. (b)

42. (b)

43. (c)

44. (d)

45. (a)

46. (c)

47. (a)

48. (c)

49. (a)

50. (d)

51. (b)

52. (b)

53. (c)

54. (c)

55. (c)

56. (c)

57. (b)

58. (d)

59. (d)

60. (b)

61. (d)

62. (b)

63. (c)

64. (b)

65. (b)

66. (a)

67. (a)

68. (c)

69. (a)

70. (b)

71. (b)

72. (c)

73. (a)

74. (b)

75. (c)

76. (d)

77. (c)

78. (a)

79. (c)

80. (d)

81. (c)

82. (d)

83. (b)

84. (c)

85. (a)

86. (d)

87. (c)

88. (d)

89. (c)

90. (c)

91. (d)

92. (c)

93. (a)

94. (d)

95. (c)

96. (d)

97. (b)

98. (c)

99. (c)

100. (c)

ANSWERS

101. (b)

102. (c)

103. (a)

104. (d)

105. (c)

106. (c)

107. (c)

108. (a)

109. (c)

110. (c)

111. (a)

112. (c)

113. (d)

114. (c)

115. (b)

116. (a)

117. (d)

118. (d)

119. (d)

120. (d)

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(2) ME (Test-21), Objective Solutions, 7th May 2016

Sol–1: (c)This is Roberts mechanism and point Pwill trace an approximate straight line.

Sol–2: (b)Beam engine mechanism is an inversion offour-bar mechanism, whereas whitworthquick return mechanism is an inversion ofsingle slider crank mechanism. Oldhamcoupling and elliptical trammel mechanismsare inversions of double slider crankmechanism.

Sol–3: (c)The given arrangement has no. of links

5= and no. of joints j = 5, no. of pairsp = 5Applying the equation, 2p 4,= LHS = 5, RHS = 2 × 5 – 4 = 6Thus, LHS < RHS

Further, applying the equation, j = 3 22

LHS = 5, RHS = 3 5 2 5.5=2

Hence, LHS < RHSSince, LHS < RHS, hence, it isunconstrained kinematic chain, i.e. therelative motion is not completelyconstrained. This type of chain is of littlepractical importance.

Sol–4: (c)

A prismatic pair can be considered as thelimiting case of a revolute pair. When one ofthe links of a kinematic chain is fixed, thechain is known as mechanism. Other thannumber and types of links, kind of motionsthat the mechanism transmits is veryimportant.

Sol–5: (d)Sol–6: (d)Sol–7: (b)Sol–8: (c)Sol–9: (d)Sol–10: (a)

(i) Increasing engine rpm will increase theturbulence inside the S.I. engine and

the flame front will spread faster inall directions. So detonation willdecrease.

(ii) Detonation will be increased byadvancing the spark timing. Whenspark will take place earlier, it willexist for more time. Thus more energywill be taken by the end charge andtemperature will be higher and higherdetonation.

(iii)Make fuel air ratio very rich willreduce tendency of detonation becauseevoporation of fuel will reduce peaktemperature.

Sol–11: (b)Since spark ignition (SI) engine works onOtto cycle shown below,

P 3

24

1

VThe compression ratio,

r = 1 4

2 3

V V 9=V V

The efficiency of SI engine-

= 111

r

= 1.5 1 0.51 11 1=

9 9

= 1 21 66.7%= =3 3

Sol–12: (b)Since stage I of combustion is independentof engine speed, so it is time dependent.The crank angle rotation during first stageof combustion i.e. 1 ms.

= 31000 360 1 1060

= 6°Now speed is doubled, the crank rotationnow,

= 6 × 2 = 12°

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Because for same performance secondstage cannot be changed on pressure -rotation diagram. So the spark should beadvanced by 6° more.

Sol–13: (c)To avoid knocking, the exhaust valveshould not be in the end gas (mixture)region. The end gas region or exhaustvalve are the hottest part of combustionchamber. So to avoid or minimizeknocking, the spark should be near theexhaust valve. This location has anotheradvantage of avoiding pre-ignition.

Sol–14: (c)Since 4-stroke engines have one powerstroke in two revolutions and 2-strokeengines have power stroke in eachrevolution. So theoretically for same powerfrom both engines of equal size cylinder,the rpm of 4-stroke should be double of2-stroke engine i.e. 4-stroke rpm as 1000and 2-stroke rpm as 500.

Sol–15: (c)The p-v diagram of cycle,

P 3

24

1

VVc Vs

Clearance VolumeVc = 25 cc

Swept volume,Vs = 300 cc

Volumetric efficiency,

v =volume of air inducted

swept volume

=s

VV

Volume of air inducted

= v sV

= 0.78 × 300= 234 cc.

Sol–16: (a)The factors which decrease the knock-ing tendency in SI engine, increase theknocking tendency in CI engine. So1. Decreased compression ratio results

in longer delay period due to poorevaporation and large amount of fuelaccumulates before combustion startsi.e. higher knocking tendency.

2. At lower engine speed, enough timeis available for mixing and evapora-tion. So the delay period reduces interms of crank rotation and knocktendency reduces.

3. Retarded injection timing allows fuelinjection at higher temperature andburns instantly. So reduced knock-ing.

4. Supercharging increases inlet tem-perature and reduced delay periodand reduced knocking.

Sol–17: (b)

43

2

1

0

P=const

AB

C

V=con

st

56

7

a b cs

T

Since heat input is same, so area underT-S diagram will be same. Area 0-2-C-c-0

= Area, 0-3-M-B-b-0= Area, 0 -4-A-a-0

So thermal efficiency

= rejected heat1given heat

The rejected heat invarious cycles

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Otto – area 0-1-5-c-0Diesel – area 0-1-7-a-0Dual – area 0-1-6-b-0Hence the least area is 0-1-7-a-0 for die-sel cycle and its efficiency will be maxi-mum.

Sol–18: (c)1. Due to presence of throttle valve, car-

burettor venturi etc in the path ofair, SI engine volumetric efficiencyis low.

2. SI engine throttled means, air flowis restricted i.e. situation is likeidling. In this condition the BHP isnegligible so the mechanicalefficiency is low.

3. Fuel consumption per unit kWoutput i.e. specific fuel consumptiondecreases as capacity of engineincreases.

4. Due to presence of excess oxygen inCI engines, its exhaust temperatureare lower than SI engine.

Sol–19: (c)The Stoichiometric equations of variousgiven elements,1. Similarly for carbon monoxide

228 16CO O CO

For one kg CO

= 16 4 kg=28 7

2. 22 162H O H O

162 = 8 kg

3. 2 232 32S O SO

One kg sulphur needs 1 kg oxygen.Sol–20: (a)

High front end volatality means, fuelhas more light hydrocarbons which boilsat relatively low temperature when avehicle heatsup after a long run orheated due other reason. The carburet-tor supply only vapour and mixture be-

come very lean which cannot be burnt.This situation is called hot starting andvapour lock.

Sol–21: (b)

A/F

ratio

Compensating Jet

Combined Jet

Air speed past jet

Main jet

The fuel-air ratio from carburettor hav-ing compensating jet is constant overvide range of speed. In this carburettor,the fuel supply from main jet and com-pensating jets is constant i.e. in com-pensating jet, supply reduces and inmain jet, the ful supply increases asspeed increases as shown in figure.

Sol–22: (b)Brake power = 40 kWIndicated power IP = ? Mechanical efficiency,

m = BPIP

IP = 40 50kW=0.8

Friction power = IP – BP = 50 –40 = 10 kWSince friction power is constant and at25% rated load, the brake power,

BP = 0.25 × 40 = 10 kW IP = BP + friction power

= 10 + 10 = 20 kW Mechanical efficiency,

m = 10 50%=20

Sol–23: (a)Morse test on 2-cylinder, 2-strokeengine-The complete equation for Morse test-

IP P F = PB

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(IP1 + IP2) – (FP1 + FP2) = BP1 +BP2

Total BP = 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 Mechanical efficiency-

m = BP 9 90%= =IP 10

Sol–24: (d)The size of flywheel is decided how theengine power or torque generation isdistributed. Steady generation requiresmall flywheel (electric motor-noflywheel, 2-stroke engine small and4-stroke engine large flywheel) Again asrpm increases the size of flywheelreduces due to mechanical loading.

Sol–25: (a)The oxides of nitrogen (NOx) in petrolengine exhaust are due to availability ofoxygen and high temperature. Therecirculation of fraction of exhaust gasesreduces high temperature so NOx

formation reduces. All other statementsprovide favourable condition for NOxformation.

Sol–26: (d)Three way catalytic converter controlsHC, CO and NOx emission as shownbelow

SecondaryAir

HC/COElement(Pt/Pd)NOx element

(Rhodium)

PM emission cannot be controlled bythis because they are big compoundscontaining carbon which are not burntduring regular combustion in CI engine.

Sol–27: (d)1. Supercharging increases power

output by burning more fuelefficiently. It also increasesvolumetric efficiency by inductingmore air.

2. Supercharging is suitable for CIengine and not SI engine. BecauseSI engine are prone to knocking athigh temperature and pressurewhich are result of supercharging inSI engine.

3. So supercharging is limited by knockin SI engine and in CI engine bythermal load (piston, valve etctemperature) and mechanical load(high pressure and forces in engineparts)

Sol–28: (a)Since heat in flowing from highertemperature to lower, so it is heat engine.

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1000 K

300 K

Cyclic device

100 kJ

60 kJ

50 kJ

500 K

The work doneW = 100 + 150 – 60 = 90 kJ

Is it reversible ?From clausicus inequality,

QT = 100 50 60

1000 500 300 = 0

Q 0T

Yes, it is a reversible heat engine.Sol–29: (b)

The entropy may increase, decrease orremains constant depending upon heatinvolved and internal irreversibility.

Sol–30: (c)m = 2 kg

cp = 1.00 kJ/kg K

T = 300 K0

Body T = 600 K

Q

W

Q – W

Energy

Change in entropy of body,

bodyS =300 p600

mc dTT

= p300mc n600

=12 12

= 1.386 kJ/kg

Change in entropy of atmosphere,

atmS = p 0

0 0

mc T T WQ W =T T

= 2 1 600 300 W

300

Total entropy change or entropygeneration,

universeS = body atmS S

=600 W1.386

300

For maximum work, process should bereversible, so

universeS = 0

600 W1.386300

= 0

W = 184.2 kJSol–31: (b)

In seperating and throttling calorimeter,steam should be in super heated vapourstate only after throttling so that pressureand temperature can be measured bymanometer and thermometer and fromsteam table enthalpy of super heatedvapour is determined. Steady flow energyequation gives the enthalpy after throttlingas equal to enthalpy before throttling anddryness fraction is calculated.

Sol–32: (a)P-V diagram of water showing ‘triple line’-

Saturatedsolid line

Saturatedliquid line

(L+V)Triple

Bline

(S+V)

P

SA

VThe line AB is triple line of water andclearly shows that the pressure/temperature are constant and specificvolume is variable.

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Sol–33: (c)Let ‘P’ be the force in rod PQ.Free body diagram, at point Q.

45°

P

R

FQ

F = P cos 45°

P = F 2 Buckling load,

P =2

2EI

L

2.F =2

2EI

L

2

2EIF

2 L

s

Sol–34: (a)

Torsional moment resistance T = Jr

solid

hollow

TT = solid

hollow

( J/r)( J/r)

=

4

s

4 4

s

D D32 2

(D d ) D32 2

=4

4 4D

D d

Sol–35: (c) A B

T = 600 N-m

4 mba

TA = T ba b

= 600 34 = 450 N-m

TB = T aa b

= 600 14 = 150 N-m

Sol–36: (c) Strain energy in torsion

U = 1 T·2

From torsion formula

r = T G

J L

U = 1 J L2 r r G

=41 d q Lq dd2 G32 22

=2 2q d L

4G 4

UVolume =

2q4G

Sol–37: (d)We know,

P = T.

T = P

Again, =P

T.rI

=P

P.rI .

1

1

2

= 2

1

2

70 = 1.5 2

3 2

2 = 140 MPaSol–38: (d)

Critical buckling load P = 2

2eff

EI( )l

Description EffectiveCritical bucklinglength load

Both end hinged l2

2EI

l

Both end fixed2l 2

24 EI

l

One end fixed & 2l2

2EI

4

lother end free

One end fixed & 2l 2

22 EI

lother hinged

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Sol–39: (a)y

y

xx d

b

M = Pex y

M = Pey x

Stress at any point x, y

=y x

yy xx

M (x) M (y)PA I I

= yx3 3

Pe (y)Pe (x)Pbd d b b d

12 12

= yx2 2

12e y12e xP 1bd b d

Sol–40: (c)

For no tensionmin 0

P MyA I

0

2 2 4 4

D(Pe)P 2(D d ) (D d )

644

0

e 2 2D d8D

Diameter of core in which load can be appliedfor no tension is from –e to eDiameter of core = 2e

=2 2D d4D

Sol–41: (b)Springs are connected in series

eq

1K = 1 1 1 1 1

K K K K K

Keq = K5

Sol–42: (b)

= 3

464PR n

G.d =

3

3 4

5064 100 122

80 10 5

= 24 mmSol–43: (c)

Maximum shear stress = Radius of Mohr’scircle

=2

21 22

=2

214 2 82

= 2 26 8 = 10 N/mm2

Sol–44: (d)In thick cylinders hoop stress for any

radius R, h = 2B AR

So variation will be hyperbolicSol–45: (a)

Shaft is rotating and carries aunidirectional load. Because of this load abending moment will act on the shaft.Under the action of bending moment,tensile stress are induced in the lower halfand compressive stresses are induced inthe upper half. There is zero stress at allthe fibre in the central horizontal planepassing through the axis of the beam.

M MP

If we consider a point A on the periphery.When the point A occupies position A1 inthe central horizontal plane with zerostress. When the shaft is rotated through90°, it occupies poisition A2 and subjectedto maximum compressive stress. It rotatedthrough 90°, it occupies position A3 andsubjected to zero stress and it rotatedthrough 90° again then subjected tomaximum tensile stress.

A1

A4

A3

A2

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So, variable bending stress acting on theshaft.Option (a) is correct.

Sol–46: (c)For two shafts connected in parallel, angleof twist is the same.

Option (c) is correct.Sol–47: (a)

Euler’s formula will not give actualbuckling load because he has madeassumptions as given following:

(i) Length is very large as compare tocross–section at area.

(ii) Crushing effect (direct stress effect) isneglected

(iii)Only axial compressive load

(iv) Friction at hinged support is zero

(v) Prismatic, homogeneous and isotropicSol–48: (c)

p

h max( )

h min( )

r

p

Stress distribution inH.C.P.V. subjected to I.P.

If a hollow pressure vessel subjected tointernal pressure, then• Radial stress is maximum at inner

radius and zero at outer radius.• Tangential stress (hoop stress) is

always higher than other stresses.Sol–49: (a)

When a closed–coil helical spring issubjected to a torque about its axis, thenonly bending stresses are induced.

Sol–50: (d)Tds = du + Pdv

For constant volume process, dv = 0

Tds = du = CV dT

dTdS =

V

TC

Again, Tds = dh–vdPFor constant pressure process, dP = 0 Tds = dh = CP dT

dTdS =

P

TC

Since, CP > CV

P

TC <

V

TC

const. pressure

dTdS

< constant volume

dTdS

Sol–51: (b)

Unavailable work = 2

1

T 400 300QT 800

= 150 kJSol–52: (b)

b will have same dimensions as ‘v’.Sol–53: (c)

Adiabatic processPV = C

lnP + ln V = ln C

dP dVP V

= 0

dPdV = P

V

Isothermal processPV = C

ln P + ln V = ln C

dP dVP V

= 0

dPdV = P

V

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Constant Volume ProcessTds = du + Pdv

Tds = du = CV dT

dTdS =

V

TC

Constant pressure processTds = dh = vdP

Tds = dh = CP dT

dTdS =

P

TC

Sol–54: (c)Since, the pure substance is contained ina rigid vessel, hence it will be a constantvolume process.Thus, the process will be along a verticallyupwards line. To pass through criticalpoint, the initial condition should be atwo phase or wet steam.

Critical point

n

n

n

ncc

b

cb

mama

m

m

p

aa

b

vx = 0 x = 0.25 x = 0.50

x=0.75 x = 1.0

Sol–55: (c)The first and second derivatives of p withrespect to v at the critical temperatureare each equal to zero.

CT T

p

= 0

andC

2

2T T

p

= 0

c

3

3T T

pv

< 0

Sol–56: (c)

Cp = P

STT

Sol–57: (b)Van der Waal’s equation is

2

2anP V nbv

= nRT

Where ‘b’ is the constant for volumecorrection and ‘a’ is the constant formolecular attraction. A higher value of‘a’ reflects increased attraction betweengas molecules. Hence, if a real gas exertsa pressure P, then an ideal gas wouldexert a pressure equal to P + p (p is thepressure lost by the gas molecules due toattraction). The gas having higher valueof ‘a’ can be liquefied easily.

Sol–58: (d)If the gas obeys Van der Waal’s equationat the critical point, then,

e 2 cc

ap b = cRT

At critical point, cT T=

p

= 0

andc

2

2T T=

p

= 0

hence, b = c and 3

a = 2c c3p

2

c c cc 2 c

c

3pp3

= cRT

cc

24p3

= RTc

c c8 p3

= RTc

c

c c

RTp

= 8 2.67=3

Sol–59: (d)Maxwell’s equations are:

s

TV =

V

ps

s

TP

= p

VS

v

pT =

T

sV

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p

VT

= T

sp

Sol–60: (b)

Availability = 0max

TW Q 1T

=

Thus, for a given T0, as T increases, theavailable energy increases. The availableenergy decreases as T decreases.

Sol–61: (d)Sol–62: (b)Sol–63: (c)

1

h 400 kJ/kg1 =s 1.1 kJ/kgK1 =

T 300 K0 =

2

s 0.7 kJ/kgK2 =

h 100 kJ/kg2 =

Change in availability of flow process= (h1 – h2) – T0 (S1 – S2)= (400 – 100) – 300 × (1.1–0.7)= 300 – 300 × 0.4= 180 kJ/kg

Sol–64: (b)

1000kg2 2(S S )

system

q

Change in entropy of system

= m(s2 – s1)

Ssys = 100 (0.4 – 0.3)

= 10 kJ/K

Change in entropy of surrounding

Ssurr = 75 – 80

= – 5 kJ/K

Total entropy change or entropygeneration,

Stotal = Ssys + Ssurr

= 10 – 5

= 5 kJ/K

Stot > 0 so the process is irreversible.Sol–65: (b)

T/P

S/V

AP

B

CP

Because in two phase region temperatureand pressure are not independent so asone increase other increase or one decreaseother decrease.

The enthalpy of evaporation is proportionalto length of line AB. As point P goes fromleft to right, dryness fraction increases.So in conclusion,

A. Saturation pressure increase results inincrease of saturation temperature anddecrease in evaporation enthalpy i.e.A-3/4

B. Saturation temperature increasesresults in reduction in enthalpy ofevaporation i.e. B-3.

C. Saturation pressure reduction resultsin increase in specific volume i.e.C-2.

D. Higher dryness fraction results inhigher entropy due to higher heatinvolved.

Sol–66: (a)Sol–67: (a)

1. The partial volume of a componentith in gaseous mixture-

Vi = in RTP

Total volume of mixture

V = iV

= iRTnP

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= iRT nP

= RT nP

= nRTP

2. Entropy of mixture-

nS = i in S

S = i i i 2 2i

n S n n SS=n n n

= 1 1 2 2S S .... x x

3. Dalton’s partial pressure low, totalpressure of mixture of Gases,

P = P1 + P2 + P3 + ....

P = iP4. The pressure fraction

iPP =

i i

i

n RT/V n= nn RT/V

= mole fractionVolume fraction

iVV =

i i

i

n RT/P n= nn RT/P

= in mole fraction=n

Sol–68: (c)VC = 100 mlV1 = 1 litre = 1000 ml

Stroke volume = V1 – VC

= 1000 – 100VS = 900 ml

Clearance ratio = C

S

VV

= 100900

1Clearance ratio9

Sol–69: (a)Since the condensation does not take placeduring supersaturated expansion, so thetemperature at which the supersaturationoccurs will be less than the saturationtemperature corresponding to the pressure.Therefore, the density of supersaturatedsteam will be more than that for theequilibrium conditions discharged.The heat drop is reduced below that forthermal equilibrium, as a consequence theexit velocity of of steam is reduced.

Sol–70: (b)Power output per kg of steam

P = b1 b V2V cos V

But V1cos = 537 m/s, Vb = 454 m/s

P = (2 × 537 – 4544) × 454= 281480 W

Sol–71: (b)Sol–72: (c)

Erosion increases both due to themoisture in the steam and the blade speedbecause both these increase the impacton the steam turbine blades.

Sol–73: (a)

T

s1

42

3

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

Sol–74: (b)h1 = 3200 KJ/Kg

S

T1

2 2

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h2 = 2400 KJ/Kg

2h = 2560 KJ/Kg

Actual work turbine efficiency=Isentropic work

= 1 2

1 2

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

= 0.8 or 80%Sol–75: (c)

The efficiency increases due to increasein mean temperature of heat addition.Further, the condenser load decreasesbecause of reduced cooling requirementsin the condenser as lesser mass flowsthrough the condenser for heat rejection.

Sol–76: (d)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–77: (c)The effect of supersaturation is to reducethe enthalpy drop slightly during theexpansion and consequently acorresponding reduction in final velocity.The final dryness fraction and entropyare also increased and the measureddischarge is greater than thattheoretically calculated.

Thus, velocity coefficient will be less thanunity while the mass flow coefficient willbe greater than unity. Sincesupersaturation leads to increase ofentropy, hence the availability woulddecreases. Irreversibility or destruction ofavailability is proportional to the entropygenerated.

Sol–78: (a)

1

T2

345 6

7 8

s(h7 – h6) = 3 kJ/kg(h5 – h6) = 1 kJ/kgh1 = 3514 kJ/kgh4 = 613 kJ/kg (h1 – h4) – (h7 – h6) – (h5 – h4) heat transfer in boiler

heat transfer = 3514 – 613 – 3 – 1= 2897 kJ/kg

Sol–79: (c)Since in impulse turbine, all the availableenergy is converted into kinetic energyby a nozzle, hence it has high steam andblade velocities. In reaction turbine, onlysome amount of available energy isconverted into kinetic energy, before thefluid enters the runner.Carry-over losses are high in compulseturbine because of high velocity.

Sol–80: (d)Since stage of a turbine is composed ofnozzle and blades, so combined efficiencyis product at efficiency of nozzle andblades.

stage = nozzle blade

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Sol–81: (c)Expansion of steam in nozzle.

h OCP

1 2

sx1

x2x2 > x1

Due to friction, the exit state of nozzleis 2 not 1. In state-2 dryness fraction,x2 > x1

Enthalpy h2 > h1

Drop in enthalpy –(h0 – h2) < (h0 – h1)

But velocity

V = 2 h

2V = 0 22 h h

V1 = 0 12 h h

2

1

VV < 1

V2 < V1

Sol–82: (d)To get stagnation state, the processmust be reversible isentropic i.e. con-stant entropy and adiabatic. The fric-tion present in flow system causes in-ternal entropy increase.

The friction in constant area duct cancause acceleration in flow i.e. can actas subsonic nozzle. For example Fannoflow.

Sol–83: (b)Assuming expansion in nozzle is isen-tropic Ratio of inlet to throat temperature

0TT = 2n 11 M

2

At critical condition at throat, M = 1

0TT =

0

n 1 T 2or =2 T n 1

For isentropic flow.

0

PP =

0

n nn 1 n 1T 2=

T n 1

Sol–84: (c)

2

V2

h2h1

1

V1V1

From SFEE, 2 21 1 2 2

1 1h V h V2 2

22 1 2 1V 2 h h V 2 80000 0=

160000 = 400 m/sec

Sol–85: (a)The governing of turbine means controlof flow of fluid i.e. steam as load varies.But in throttle governing, there is largepressure loss due to throttling as shownin figure.

20h

13

8

P0P

Loss of available enthalpy in throt-tling

= (h0 – h1) – (h2 – h3)Loss of pressure (P0–P)

Sol–86: (d)Wilson line represents the metastablestate in the supersaturated flow ofsteam as shown in figure below

h

s

CP

x= 0.97–0.98

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The dashed line of dryness fractionx = 0.97 – 0.98 is Wilson line. This lineis parallel to dry saturated steam.

Sol–87: (c)Sol–88: (d)

At high velocity (supersonic) theefficiency of rocket engine is more thanturbojet. The exit velocity of exhaustgases is also more than turbojet. Sinceno air flow from outside atmosphere inrocket, so they are not air-breathingengine. Before combustion in rocketengine, there is no motion of oxidantand fuel molecules, so stagnationcondition exist in combustion chamberof rocket engine.

Sol–89: (c)Propulsion efficiency

=

a

2 2

m uThrust power V u= 1Propulsive power m V u2

= 2× flight velocity

Jet velocity + flight velocity

= 2u

V u

Sol–90: (c)At critical pressure 221.2 bar, latent heat ofvaporisation of water is become zero.

Triplepoint

T221.2 bar

s

Sol–91: (d)

P

C

V

B

E

D

AT4

T3

T2

T1

Ideal gas equation, PV = mRTFor constant temperature, PV = constant

T1 > T2 > T3 > T4Sol–92: (c)Sol–93: (a)

Gibbs function, G = H – TsHelmholtz function, F = U – Ts

G – F = (H – U) – Ts + Ts= (H – U)= U + PV – U= PV= mRT

G–F = f(T)Sol–94: (d)

A. Stirling cycle: It consists of twoisochores i.e. constant volume and twoisothermal. So A–1 & cycle on p-Vdiagram.P

3

42

V

Isothermals

B. Diesel cycle: It consists of twoadiabatics, one constant pressure heataddition and one constant volume heatrejection i.e. B-3.

P

V

Adiabatics

C. Otto cycle: It consists of twoadiabatics and two constant volumeprocesses i.e. constant volume heataddition and constant volume heatrejection. So C-2

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3P

Adiabatics

2

1V

4

D. Atkinson cycle: It is a cycle in whichadiabatic expansion is upto atmosphere.So it consists of two adiabatics, oneconstant volume process and one isobaras shown below. So B-5.

3

2

1 4

Adiabatic

Sol–95: (c)The p-V diagram of Stirling engine-

P

3

42

1

Isothermals

V

Since the question concerns onlyisothermal compression i.e. process 1-2in which compression work is done onsystem by movable piston and this workis rejected as heat because temperature isconstant. So it is closed system with amovable boundary.

Sol–96: (d)For high octane number of SI enginefuel, the ignition delay period or selfignition temperature should be high.The octane number or anti-knock quali-ties has nothing to do with heatingvalue, flash point and volatility.

Sol–97: (b)For de Laval turbine, the optimum bladeto tangential component of steam velocity,

1

uV = cos

2

or u = 1V cos2

orw

uV = 1

2 ...(i)

For Parson turbine,12 = cos

or1

uV cos = 1

orw

uV = 1 ...(ii)

Sol–98: (c)Assuming no other loss,Net power output

= Heat suppleid - Heatrejected

Power = Q1 – Q2

6

2 190 10Q Q Power 10000

3600

= 25000 – 10000 = 1500 kW = 15 MWSol–99: (c)

T

100°C

A30°C

s

BP=1 atm

Heat supplied to boiler

= p fgc T h x

= 4 × (100 – 30) + 0.9 ×2400

= 280 + 2160 = 2440 kJ/kgSol–100: (c)

Curtis turbine is combination of veloc-ity compounded impulse stage while da-Laval is single stage impulse.

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So efficiency of Curtis would be betterthan da-Laval. The correct sequence isda-Laval < Curtis < Rateau < Parson.

Sol–101: (b)T

1

2

4

5

s

36

The heat supplied without regenerationQ = cp × (T4 – T2)

The heat supplied to turbine with re-generation-

Q = cp (T4 – T3)from these two expressions, Q1 is lessthan Q. This reduction is taken fromrejected heat from exhaust.

Sol–102: (c)Net power generation

Wnet = 5000 kW Work ratio = 0.5

0.5 = t c

t

W WW

0.5 =t

5000W

Wt = 50000.5 = 10000 kW

Wt – Wc = Wnet = 5000Wc = Wt – Wnet = 10000 – 5000

= 5000 kWSol–103: (a)

Normally models are made smaller thanprototype however when,(i) Flow field is small(ii) Flow velocity is largeWe make larger models as compared toprototype.

Sol–104: (d)

Mach No. =1/2Inertia Force

ElasticForce

=1/22 2

r r2r

L vKL

= rvK

Froude No. =12IntertiaForce

GravityForce

=

12 2 2

3L vL g

= r

r

vL g

Euler No. =12InertiaForce

Pr essureForce

=

12 2 2r r

2r

L vp L

=

12 2rvp

Reynolds No. =Inertia ForceViscousForce

=2 2r r

r r

L vv L

=

r rv L

Sol–105: (c)In weir design gravity force is significantso Froude No. is applicable here. Whenvertical and horizontal scales are different,the model is called distorted model.

(Lr)H = m

p

L140 L

(br)V = m

p

bb = 1

9 m[ Q 1 / sec] l

ar = m m

p p

b Lb L

=1 19 40

= 1

360

Qr = arvr = r rva b

m

p

QQ

= 1 1360 9

Qp = 1 360 9 = 1080 l/secSol–106: (c)

Cam and follower is a higher pair becausethere is point or line contact between the

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cam and the follower. Surface contact takesplace between two links of a lower pair.

Sol–107: (c)Inversion is obtained by fixing, in turn,different links in a kinematic chain andwe can obtain as many mechanisms as thelinks in a kinematic chain. For example,four inversions of a single slider crankchain are possible.

Connectingrod

Cutting strokeReturn stroke

Ram ToolR2

R1R

Line ofstroke

P2

QP1

B

Slider (Link 1) Crank (Link 2)

Fixed(Link 3)

C

B2B1

Slotted bar(Link 4)

A

902

In crank and slotted lever quick returnmotion mechanism, the link AC i.e. link 3forming the turning pair is fixed. The link3 corresponds to the connecting rod of areciprocating steam engine.

Sol–108: (a)A mechanism is formed by fixing one of thelinks of the kinematic chain. Thus, whendifferent links are chosen, differentmechanisms will result.Relative motion (displacement, velocity andacceleration) between any two links doesnot change with inversion. This is simplybecause relative motion between differentlinks is a property of the parent kinematicchain.Absolute motion of points on various links(measured with respect to frame link) may,however, change drastically from oneinversion to the other, even in directinversion.

Sol–109: (c)For the same rk and work capacity,Brayton cycle is 1–2–5–6 andOtto cycleis 1–2–3–4.In the reciprocating engine field, theBrayton cycle is not is not suitable. Areciprocating engine cannot efficientlyhandle a large volume flow of low pressuregas, for which the engine size 2( /4D L)becomes large, and the friction losses alsobecome more. So the Otto cycle is moresuitable in the reciprocating engine field.

3

2 54

61V

p pV c=

(a)

2

3

5

4p = c

V = c

p = c6

V = c

1s

T

(b)In turbine plants, however, the Braytoncycle is more suitable than the Otto cycle.An internal combustion engine is exposedto the highest temperature (after thecombustion of fuel) only for a short while,and it gets time to become cool in theother processes of the cycle. On the otherhand, a gas turbine plant, a steady flowdevice, is always exposed to the highesttemperature used. So to protect material,the maximum temperature of gas that canbe used in a gas turbine plant cannot beas high as in an internal combustionengine. Also, in the steady flowmachinary, it is more Moreover, a gasturbine can handle a large volume flow ofgas quite efficiently. So Brayton cycle isthe basic air standard cycle for all moderngas turbine plants.

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(19) ME (Test-21), Objective Solutions, 7th May 2016Sol–110: (c)Sol–111: (a)Sol–112: (c)

P

V

V2 V3

V4

V1

The cut off ratio of diesel cycle,

rc = 3

2

vv

Hence, for rc = 1, v3 = v2 and cycle willbecome Otto cycle and diesel engine cannot aperate on Otto cycle cycle mode. Socut off ratio ‘rc’ should be more than one.

But very high cut off ratio increases heatrejected during constant volume process(4–1). So for better efficiency and output,there is optimum value of cut off ratio.

Sol–113: (d)To mixup the air and fuel in slow airmovement is multi hole nozzle not pintlenozzle.The pintle nozzle is provided to avoiddribbling and weak injection.The mixing of air fuel is done by natureof air movement inside cylinder e.g.Squish, swirl, turbulences, etc.

Sol–114: (c)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–115: (b)

Longitudinal stress, l = pd4t

Hoop stress, h = pd2t

where

p = Internal pressured = Diameter of cylindert = Thickness of cylinder

Sol–116: (a)Mollier diagram is the enthalpy-entropy(h–s) diagram for steam. This diagram isalso obtained on the basis of followingequation depending upon the phasetransformation.

Tds = dh –vdp(1st and 2nd law combined)

For constant pressure, P

h T=s

Isobaric lines diverge from one another.This is due to the increase in saturationtemperature with increase in pressure.Slope of isobar is equal to saturationtemperature, therefore it also increaseswith increasing pressure.The constant pressure lines are straightin the wet steam region and curved inthe superheated steam region.Constant pressure lines and constanttemperature lines coincide upon in thewet region. For every pressure, there shallbe definite saturation temperature whichremains constant in the wet region.

Sol–117: (d)Work is done during expansion in mov-ing blades only and no work is doneduring expansion in fixed blade i.e. innozzle. In nozzles, the steam regainmomentum by increasing speed andprepare itself for next stage. The fixedblades are attached to casing and do notrotate.

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Sol–118: (d)The actual shape of Mollier diagram ofwater

TT

TP P

h

s

CP

P

From figure, it is clear that in lowpressure zone, the enthalpy of superheatedsteam is approximately same as saturationenthalpy. But in high pressure zone (nearcritical point ‘CP’) the enthalpy initiallyincreases at constant pressure and thenconstant.The reason for constant enthalpy in lowpressure zone is that the vapour behavesas ideal gas and so–

h = u + Pv= u + RT ( Pv = RT)= f(T)

So assertion is wrong at high pressurezone.

Sol–119: (d)In simple gas turbine power plant, thegas turbine consumes considerableamounts of power just to drive itscompressor. As with all cyclic heat enginesa higher maximum working temperaturein the machine means greater efficiency,but in a turbine it also means that moreenergy is lost as waste heat through thehot exhaust gases whose temperatures are

typically over 1000°C. Consequently,simple gas turbine efficiencies are quitelow.In gas turbines, the overall-cycle adiabaticthermal efficiency is the efficiency of theoverall cycle and takes into account allcomponent efficiencies, such as compressorefficiency, combustor efficiency andturbine efficiency in the composition ofthe cycle, that in the exit total pressureand total temperature of the compressor,the combustor and turbine are affectedby the losses in these three components.Therefore, the losses affect the overallthermal efficiency to a certain degree butit is the cycle between the two thermalsinks that governs the efficiency. Thuseven if the compressor, combustor andturbine losses were neglected, the overallthermal efficiency of the gas turbinedepends mainly on the cycle thermalefficiency. It is a mistake that themultiplication of the three componentefficiencies will give the total overallefficiency a value that has not even aremote connection to the actual thermalefficiency of the gas turbine.

Sol–120: (d)The maximum efficiency of a single stageimpulse turbine = 2cos

where is the nozzle angle.The lower is the nozzle angle, higher isthe blading efficiency.However, too low a nozzle angle may causeenergy loss at blade inlet. Therefore, thenozzle angle has to be maintained withina certain range, which varies from 16° to22°.

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Conventional Question Practice Program ANSWERSiesmaster.org/public/archive/2016/IM-1462632660.pdfquick return mechanism is an inversion of single slider crank mechanism. Oldham coupling