ESE-2017 PRELIMS TEST SERIESDate: 23rd October, 2016
24. (b)
25. (b)
26. (b)
27. (a)
28. (c)
29. (b)
30. (d)
31. (d)
32. (c)
33. (c)
34. (c)
35. (d)
36. (b)
37. (c)
38. (d)
39. (d)
40. (c)
41. (a)
42. (d)
43. (b)
44. (c)
45. (d)
46. (b)
47. (c)
48. (a)
49. (d)
50. (c)
51. (c)
52. (a)
53. (d)
54. (a)
55. (a)
56. (a)
57. (a)
58. (d)
59. (c)
60. (a)
61. (a)
62. (d)
63. (c)
64. (a)
65. (b)
66. (a)
67. (d)
68. (a)
69. (c)
70. (b)
71. (a)
72. (b)
73. (c)
74. (c)
75. (a)
76. (b)
77. (a)
78. (b)
79. (c)
80. (c)
81. (d)
82. (c)
83. (b)
84. (c)
85. (a)
86. (c)
87. (b)
88. (a)
89. (d)
90. (a)
91. (d)
92. (c)
93. (a)
94. (a)
95. (c)
96. (c)
97. (b)
98. (b)
99. (b)
100. (b)
101. (a)
102. (d)
103. (a)
104. (d)
105. (c)
106. (a)
107. (a)
108. (c)
109. (d)
110. (b)
111. (a)
112. (d)
113. (b)
114. (a)
115. (c)
116. (a)
117. (c)
118. (d)
119. (d)
120. (c)
121. (a)
122. (b)
123. (a)
124. (d)
125. (b)
126. (b)
127. (d)
128. (c)
129. (d)
130. (c)
131. (c)
132. (a)
133. (c)
134. (b)
135. (a)
136. (a)
137. (d)
138. (b)
ANSWERS
1. (d)
2. (a)
3. (b)
4. (a)
5. (b)
6. (a)
7. (c)
8. (d)
9. (b)
10. (a)
11. (a)
12. (a)
13. (b)
14. (d)
15. (d)
16. (b)
17. (a)
18. (a)
19. (d)
20. (c)
21. (c)
22. (b)
23. (b)
IES M
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(2) ME (Test-A4), Objective Solutions, 23rd October 2016
139. (c)
140. (a)
141. (b)
142. (c)
143. (d)
144. (a)
145. (c)
146. (d)
147. (b)
148. (b)
149. (c)
150. (b)
151. (c)
152. (a)
153. (b)
154. (a)
155. (c)
156. (d)
157. (d)
158. (a)
159. (c)
160. (a)
161. (c)
162. (b)
163. (a)
164. (b)
165. (a)
166. (b)
167. (a)
168. (a)
169. (a)
170. (a)
171. (a)
172. (a)
173. (a)
174. (b)
175. (b)
176. (a)
177. (c)
178. (d)
179. (b)
180. (a)
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(3) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–1: (d)
VI
I II
VII
2 2I I II IIP V P V+ = +g 2g g 2g
0.28 + 0.08 = 0 +2IIV
2g
VII = 2 9.81 0.36 = 2.66 m/sVelocity of section I
V I = 2 9.81 0.08 = 1.772 m/s
The ratio of velocities
VII/VI =2.66
1.772 = 1.50
Sol–2: (a)
40 mm
2
x
Bernoulli’s theorem at 1 and 2.
20P 1 V+g 2 g
+ x = x0P+ h +
g
V2 = 2gh
V = 2 9.81 0.04 = 0.89 m/secSol–3: (b)
In this question standard formulae have beenasked So.(i) Discharge through venturimeter:
= 1 2d 2 2
1 2
a aC 2gHa a
(ii) Since for external mouth piece, thedischarge coefficient isCd = 0.855 Discharge through external mouth
piece
= 0.855a 2gH(iii) Discharge over rectangular notch,
Q = 3/2d
2 C b 2g H3
(iv) Discharge over V.notch ( 90) =
= 5/2d
8 C 2gH tan15 2
Sol–4: (a)
The discharge over a head weir whileconsidering approach head over velocity,
Q = 3/22 B 2g H3
Here, B = 200 m, H = 0.75m, Happ = 0.02m
Q = 3/2 3/2
app app2 B 2g H + H H3
Q = 2 200 2 9.813× ((0.75+0.02)3/2 – (0.02)3/2)
= 452.64 m3/sec
Sol–5: (b)Since the flow rate is same through orificemeter and venturi, so
Q0 = Qv
0
1 2d 02 2
1 2
a aC 2gha a
=
v
1 2d v2 2
1 2
a aC 2gHa a
Since areas in venturi and orifice aresame. So
0d 0C 2gh = vd vC 2gh
0.63 2gh = 0.985 2gH
Ratio of pressure drop for venturi and orificemeter is
20.630.985
v
o
P H= =P h
Sol–6: (a)
Discharge Q = 0.02 m3
Flow velocity,
V = 2Q 4QA d
=
IES M
ASTER
(4) ME (Test-A4), Objective Solutions, 23rd October 2016
0.4 m
3ρ 900kg/m=
= 0.04 Pa-sec
20km=
Reynold number, Re = Vd
Re = 2d 4Q
d
= = 4 Q 4 900 0.02
d 0.4 0.04= 1432.4 – laminar flow
Friction coefficient
f = = =16 16 0.01117Re 1432.4
Loss of head due to friction
hf = 2 2
54fL V 32fL Qd 2g 2g d
=
=
23
2 532 0.01117 20 10 0.02
9.81 0.4
= 2.884 m
Power required to maintain the flow
P = fgQh
= 900 × 9.81 × 0.02 × 2.884
= 509.3 W
Sol–7: (c)For maximum transmission of power, frictionhead, should be
h f =H3
Available head to convert into jet is2 H3
.
2i
fV H 2H= H = = 2h2g 3 3
=2fL V2× ×
D 2g
ViAi = AV
V =2
j jj j 2
A dV = V
A D
2 4j j2
j 4
V d2fL= × V2g 2Dg D
dj =
1/45D2fL
=5
1/40.4
2×0.025×1000d1 = 119.6 mm
Sol–8: (d)
d =10m2
d =12cm1 L
L
The frictional head in pipe,
h f =24fL V
d 2g
Discharge through pipe
Q = 2A V d V4
=
V = 24Qd
h f = 2
44fL 1 16Qd 2g d
h f =2
2 532fL Q
g d
Since frictional head or friction in both pipesare equalso
1fh = 2f
h
2151
Qd
=2252
Qd
1
2
QQ = =
2.5 2.51
2
d 12d 10
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(5) ME (Test-A4), Objective Solutions, 23rd October 2016
= 1.5774 1.577Sol–9: (b)
2L/s
30 cm2
5 L/s
40 cm2
20 c
m2
70 cm2
9L/s
6L/s
Volume flow rate at AQ = AV
6 × 10–3 = 20 × 10–4 V
V =6020
= 3 m/sec
Sol–10: (a)Fully developed flow means the velocityprofile is not changing as we move downthe flow. In this condition the flow is steady,no acceleration i.e. no inertia, constantpressure gradient balance by viscous forces.Fully developed flow happens to be in bothlaminar and turbulent flow. But developmentlength is different in both cases.
Sol–11: (a)
Bernoulli's equation,
2P V zg 2g =
constantIn steady-state, the velocity at variousstations are in orderV2 > V4 > V1 > V3The pressure will be in reverse order, usingBernoulli's equationP3 > P1 > P4 > P2
Sol–12: (a)
0.1m
2.0m
Kinematic viscosity
= 16 centistokes
= 0.16 stokes
= 0.16 × 10–4 m2/sec
Speed, V = 30 km/hr
V = 30×5
18 = 8.33 m/sec
Re = 0.52 -5
VL 8.33× 0.1=1.6 ×10
= 0.52 × 105
The thickness of boundary layer
5
x
5 5= =x Re 0.52×10
= 2.19 mm
Sol–13: (b)The velocity profile in turbulent flow followsare either logarithmic or (1/7)th power law. So
V = 2.5 Umaxlny + c
max
uU =
17y
Sol–14: (d)Blasius solved the boundary layer and foundthe following relations for laminar flew, skinfriction coefficient
C f =x
0.664Re
Coefficient of drag or average frictioncoefficient
CD = fL
1.3282CRe
=
Sol–15: (d)The boundary layer thickness is the distancefrom solid surface where the local fluid flowvelocity 'V' is 99% of free flow velocity V.
V=0.99UU
y
x
Sol–16: (b)By definition the shape factor of a Boundarylayer is the ratio of displacement thickness* and momentum thickness .
Shape factor,
IES M
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(6) ME (Test-A4), Objective Solutions, 23rd October 2016
H =*
Numerical value of shape factor indicatesabout the pressure gradient in boundarylayer.
Sol–17: (a)We know that thickness of turbulentboundary layer distance x downstream fromleading edge,
xx
xx 0.20.37=Re
=15x 0.37x=
Uxx
0.20.37u
x4/5
Sol–18: (a)The boundary layer thickness is defined asdistance from boundary where flow velocityis 99% of free stream velocity V . SoBoundary layer thickness
y = at u = 0.99U
Displacement thickness
* = 0
u1 dy.U
Momentum thickness
=
0u u1 dy
U UEnergy thickness
* * =2
20
u u1 dy.U U
Sol–19: (d)
Vmax
Since Re 1200, so the flow is laminar,then the average velocity of flow is half ofmaximum velocity of centre.
Average velocity,
maxV 1u = =2 2
= 0.5 m/s
Sol–20: (c)
U
The shear stress in pipe
r
= 0p dPx dx R
=
0 =R dP×2 dx
R P2 x
=0.6 140×2 60
= 0.70 kPa
Sol–21: (c)Assuming the flow in pipe is laminar.
RU
r
Shear stress at radius r,
r
= x0P
R
= = constant
The pressure drop across length of the pipe
0 =
R P2 x
=
3 350 10 100 105 5
IES M
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(7) ME (Test-A4), Objective Solutions, 23rd October 2016
= 500 N/m2 = 0.500 kPaSol–22: (b)
From Darcy-Weisbach equation in terms offlow discharge, the head loss,
h f =
2
2 564L fQ2 g D
fQ2 =
2 5
f2 g D h
64 L = constant
Differentiating,
df 2dQ+ = 0f Q
dQ+36.5% + 2 = 0Q
dQ = 18.25%Q
Sol–23: (b)(a) Reynolds Model Law :
Rem = Rep
m P
VL VL=
p p m P
m m p M
L V= 1
L V
rr r r
1 1 1L V
= 1
Vr =r
r rL
(b) Froude model law
m p
V V=gL gL
1/2 1/2p pm
p m m
g LV = 1V g L
1/2 1/2p pm
p m m
g LV = 1V g L
r
r r
V = 1g L
Vr = r rg L
Same way the velocity ratio for otherlaws can be find out.
(c) Weber Model
Vr =r
r rL
(d) Mach Model
Vr =r
r
k
Sol–24: (b)
T1 3
2
5x2 < x1
x2
4x1
Employing superheated steam inturbine, the quality of steam at turbineexit will be high i.e. low water content.The water content in steam isresponsible for blade erosion. Thereheating is 2–3. The quality of steamwithout reheat is x2 and with reheat, itis x1.
Sol–25: (b)In regenerative cycle, the increase inefficiency due to regeneration is propor-tional to increase in temperature of feedwater in a feed water heater as shownbelow:
T1
2
345
67
8
Feed waterHeater
s
T
One feed water heater system isshown above and increase in efficiencyof cycle is
T
IES M
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(8) ME (Test-A4), Objective Solutions, 23rd October 2016
Now two feed water heaters are usedbut T will reduce because T in eachcase (i.e. one, two, -- feedwater heaters)is equal to–
T = 8 4T Tn 1
where n is number of feed water heat-ers.So T reduces, as number of stagesincreases and increase in efficiency reduces per stage but overall increase.(i.e. this increase in efficiency is accord-ing to law of diminishing return)Assuming T8 i.e. boiler temperature isfixed and condenser temperature T4 isincreasing, then T will reduces so .Hence from cost point of view as T4 in-creases, number of stages of feed wateror regeneration reduces.
Sol–26: (b)Total or stage enthalpy drop inturbine
sh = 3.314 kJ/kgEnthalpy drop in rotor/moving blade.
rh = 2.50 kJ/kgDegree of reaction,
rotor
s r
hh h
= r
s r
hh h
=
2.50 0.433.314 2.50
Sol–27: (a)
T1 3
2
5x2 < x1 x2 4 x1
The results of reheat is more output andhigh quality of steam at turbine exit.But in high pressure steam turbine, themore work is not the govering criterionbecause without reheat, the moisturecontent is very high at turbine exit.
Due to reheat the quality of steam atexit is x1. But quality of steam withoutreheat is x2 which has high moistureand dangerous for blades.
Sol–28: (c)
BlockedNozzles
NozzleDiaphragm
x
Partial admission means blockage ofsome nozzle at rotor periphery. This isdone to have blades of minimum pos-sible height that can be manufactured.But partial admission results in vibra-tion of blades because of alternatingloading due to rotation.
Sol–29: (b)
h 1
2
3
4
s
P1
P2
In high pressure boilers, if the steam isallowed to expand upto condenser pres-sure in a single turbine (1–2), themoisture content at exit at turbine willbe very high and not under acceptablelimit. To avoid this, the steam at inter-mediate pressure (P2) is again heatedi.e. reheat and allowed to expand in lowpressure turbine (3–4).
Other additional benefits are higherpower output. But whether efficiencywill increase or not depends upon re-heating pressure and temperature.
Sol–30: (d)Deaerator in steam power plant removesdissolve gases from water and non-condensibles from circulated water. Iflarge number of closed regenerator are
IES M
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(9) ME (Test-A4), Objective Solutions, 23rd October 2016
used in cycle, at least one should beopen type for this deaeration process.This deaeration process removes somesteam with gases which is make up feedwater.
Sol–31: (d)Speed (rotation) of steam turbine isbrought to practical limits by compound-ing i.e. velocity compounding or pres-sure compounding or both. Flywheels arenot used in turbines at all. Governorcontrols the speed with load variationwith certain, limited ranges only (0.1%).Reducing fuel feed has no relation withturbine speed.
Sol–32: (c)Nozzle
V1 11
Vr1
Inlet tip
u
u
Blade
Outlet tipVr2 V222
u
V1 – Absolute velocity of steam at inlettip.Vr1 – Relative velocity of steam at inlettip.V2 – Absolute velocity of steam at outlettip.Vr2 – Relative velocity of steam at outlettip.
(10) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–36: (b)
TT
1
2
wc
3
wt
4
s
Work Ratio = t c c
t t
W W W1W W
Intercooling reduces Wc
Reheating increases Wt
Both result in higher work ratioIntercooling requires step compressioni.e. compression 1-2 into steps with cool-ing between two successive step com-pression.This process reduces total work of com-pression [cp (T2 – T1)]. This happensbecause intercooling compression is closeto isothermal component. Reheat in-crease total turbine work because re-heating requires step expansion andheat addition between two successivestep expansion.Heat exchanger is nothing but regen-eration which has no effect on work ra-tio.
Sol–37: (c)In gas turbine regeneration, the inter-nal heat transfer is after turbine ex-pansion so no change in specific output.But thermal efficiency is improved. Inintercooling specific output is increaseddue to reduction in compression workbut at the same time heat input alsoincreases. The specific output from gasturbine
= t c
a
W Wm
= wt – wc
Sol–38: (d)T
1
2
T1
3
4y
s
x
Gas turbine cycle with regenerationMaximum temperature of cycle–
T3 = Tmax
Minimum temperature of cycleT1 = Tmin
Pressure ratio,
2
1
PP =
kk 123
p14
TPr
TP
= =
=k
k 13
4
TT
Thermal efficiency–
th =
y 1p2
1 p 3 x
T TCQ1 1Q C T T
=
=
y 1
3 x
T T1
T T
In 100% effectiveness of regeneration–T4 = Tx and Ty = T2
th =
2 1
3 4
T T1
T T
=
21
1
43
3
T 1TT
1T1TT
=
k 1k
p1
3k 1k
p
r 1T1
1T 1
r
=k 1
1 kp
3
T1 rT
IES M
ASTER
(11) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–39: (d)1. In closed cycle gas turbine the work-
ing fluid is Helium which has highheat capacity ratio .1.66 = Thishigher value of capacity ratio giveshigher output and higher thermalefficiency.
2. Because of lower value of heat capac-ity (Cp) heat exchanger efficiency isalso high.
3. Since clean helium gas flow overblade of gas turbine, the erosion andcorrosion is less.
4. The closed cycle operates under highpressure so higher density of helium.This enables more flow rate withgiven size. So higher output can beobtained from same size.
The efficiency-pressure ratio curve forBrayton Cycle–
rp
1.0
The pressure ratio for maximum effi-ciency.So the curve B in question should startfrom rp = 1 not from rp = 0.As far as answer is concerned, the shapeis similar to B.
Sol–43: (b)Sol–44: (c)
Degree of Reaction =
m
f m
h0.4h h
0.4 = f
5050 h
fh = 50 10 504
fh = 75 kJ
IES M
ASTER
(12) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–45: (d)hboiler = hinitial – (500 + 5 + 3)
2000 + 500 + 8 = hinitial = 2508 kJ/kgSol–46: (b)
E = G 237191
nF 2 96500
= 1.23 VSol–47: (c)
= GH
H = 2371910.83
= –285772.3 kJ/kgmoleSol–48: (a)
W = G 166.3kJ/mol
166.3 kJ work from 1 mole methanol
166.3 kJ/s power from 1 mol methanolsecond
166.3 kJ/s from 32 g/s of methanol
100 kJ/s from 32 100 g/s166.3
= 19.24 g/sSol–49: (d)Sol–50: (c)
Gas for lighting required per day
=120 3 3
1000
= 3108 1.08m100
Cowdung per cow = 10 kg/dayCollection efficiency = 70%Collected cowdung per day
= 10 × 0.7= 7 kg/day
weight of dry cowdung = 7 × 0.18 kg/day
Gas production = 3340 m0.18 7
1000 day
For n cows
340n 0.18 71000
= 1.08
n 2.67~ 3cows
Sol–51: (c)The digestion become slower resulting indecrease gas production.
Sol–52: (a)Because of increased pressure, the gas tapsand pipe joints starts leaking. For better rateof gas production optimization of pressure inbiogas plant is required.
A solar cell has low efficiency. A solar cellconverts solar energy directly into electricalenergy.
Sol–57: (a)
E = 1.24 eV
= 1.240.512
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(13) ME (Test-A4), Objective Solutions, 23rd October 2016
= 2.42 eVSol–58: (d)Sol–59: (c)
Efficiency = outputinput
Output = 0.25 × 100 = 25 J/m2/sec
= 2W25m
Area of PV for computer = 2400 16m25
Area of PV for motor = 1000/25 = 40m2
Total area 40 + 16 = 56m2
Sol–60: (a)Sol–61: (a)Sol–62: (d)
E = hC
E 1
Sol–63: (c)The maximum conversion efficiency ofsolar cell is given as:
max = sc oc
T c
FF I VI A
where FF is full factorIsc is short circuit currentVoc is open circuit voltageIT is incident solar fluxAc is area of solar cell
Sol–64: (a)Sequential order is given as:Reduction of silicon dioxide to silica resultin production of MG - Si Purification ofMG-Si into multicrystalline semiconductorgrade silicon conversion ofmulticrystalline silicon into single crystalform in form of rod rod is convertedinto wafers by sawing process doping ofwafers with boron.
Sol–65: (b)Typically voltage at which solar cellproduces maximum power ranges from0.45 – 0.55 volt.
Sol–66: (a)
Cp = real
3R o
P1 A V2
Preal = 3p R o
1C A V2
Preal = 2 310.5 1 (40) 102 4
Preal = 5 510 3.14 10 W
r ealP 0.314 MW
Sol–67: (d)
0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6Lanchester-Betz limit
Eq. (7.34)High-speed
two bladepropeller
DarrieusSavonius
American multi-blade
Tip speed ratio
Pow
er c
oeffi
cien
t Cp
Highest value of power coefficient isobtained for propeller type rotor
Sol–68: (a)In aerobic digestion bacterial fully oxidisethe carbon present in biomass to CO2. Inanaerobic digestion bacteria oxidised thecarbon to mixture of CO2 & CH4 inabsence of oxygen. But these two processesdoes not require heating
Sol–69: (c)The bulk modulus of liquid
=dP P= ×d
IES M
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(14) ME (Test-A4), Objective Solutions, 23rd October 2016
(15) ME (Test-A4), Objective Solutions, 23rd October 2016
So
Patm = 0
Pair = –0.147 × 13.6 × 9.81 kPa
= – 19.62 kPa
Gauge pressure at A,
PA = Pair + gh
= –19.62 + 4 × 0.8 × 9.81
= –19.62 + 31.392
= + 11.772 kPa
Sol–75: (a)
200 mm
500m
m
Air
Water
A
Pressure at water level mass
PA + g h1= P0 + g h2
Neglecting pressure due to water down
Pn P0 = 101.3 × 103
gh = 101.3 × 103
h =3103.3 10 101.3
1000 9.81 9.81
=
= 10.328 m
Sol–76: (b)
d
OR
Rd
Area of strip dA = 2R cos Rd cos
dA = 2R2 cos2 dForce on strip,
dF = PdA= 2R2Pcos2d
Moment about diameter,d = dF·R sin
= 2R3Rcos2·sin ·dTotal moment,
=
/2 3 2
02R Pcos sin d
Put cos = xsin d = dx
=
133
0
x2R P3
=3 32R P 2×1 ×105=
3 3= 70 kN·m
Sol–77: (a)
1h
0.4m
0.6mG G
2h0.
6m
0.4m
90°Turn
1 2h h ,= the hydrostatic force will be samei.e. (9.15 kN). Because according toPascal's law hydrostatic pressure at alocation is same in all direction.
Sol–78: (b)
F
40m x
y
The horizontal force per unit length
= h g projected area
F = h gA
=40 ×9.81× 40×12
= 98.1 × 40 × 2= 7848 kN/m
IES M
ASTER
(16) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–79: (c)
Tension= weight – Buoyancy forceT = Mg – FB
= Mg – Vlg
= MgM 0.8 g
2.4
=Mg 2Mg = Mg3 3
= 2 18 10
3 = 120 N
Sol–80: (c)
2 m O
B
A
D
The vertical force acting on the cylindricalgates,= weight of water displaced by section ADB
= 21gV g A L g 2 12 4
= =
= g 1000 4.81 15409.5N.
2 2 = =
Alternatively:Vertical force = Weight of liquid supported
by surface DB= Weight in surface area
=
21 11 1 1+ 1 1 1 1 14 4
=
1 1 1+ g
4 4
= ×1000×9.842
= 15409.5 N
Sol–81: (d)
2 mm
w
Weight of hydrometer, W = 0.03 NLet depth of hydrometer in oil is h0
W = B
0.03 = 3 20×(2×10 ) h ×750g
4
h0 = 60.03× 4
× 4×10 ×750×9.81= 1.2979 m
Let depth of hydrometer in alchohol is ha,
ha = 60.03× 4
× 4×10 ×800×9.81= 1.2168 m
h0 – ha = (1.2979 –1.2168)m= 0.0811 m = 81.1 mm
Sol–82: (c)
h
F
FB
Force on value from pipe lineF = P.A
= 105 × 0.25 × 10–4 = 2.5 NThe Buoyance force on float,
FB = gV = 103 × A.h×g
= 103 × 10–2 hg = 10 hg F = FB
2.5 = 10 h × 10
h =2.5 meter100 = 2.5 cm
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(17) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–83: (b)Velocity potential function,
= 3 3
2 2xy x yx + + y2 3
x-component of velocity.
u =
32y= 2x + x y + 0
x 2= 2x – x2y – y3/2
Sol–84: (c)Since equation of stream line is obtainedby total differential of as,
dx + dy = 0x y
vdx – udy = 0
dy v=dx u
2 21 (x y )dy 1 x y2= =dx 2xy 4 y x
2 2dy (x y )=dx 4xy
2 2dx 4xy=dy (x y )
Sol–85: (a)
The relationship defining quasi-static orreversible process is–
P = (–3V + 12) bar
The work done
W = p dV
Total work done when volume changes from2 to 5 m3.
W = 5
2p dV
= 5 52
3V 12 dV 10 J
=
5 52 522
3 V 12 V 102
= 53 25 4 12 5 2 10
2
=
53 21 12 3 102
= (–31.5 + 36) × 105
= 4.5 × 105J
Sol–86: (c)
Internal energy of mixture,
Um = x1U1 + x2U2 +......
where x1 is tractions of constituent.
Um = 1 21 2
m mu u .....M M
Where Mis total mass of gas.
M = m1 + m2 + m3 + ......
Um =0.25
1 × 200 +
0.751
× 400
= 50 + 300
= 350 kJ/kg
Sol–87: (b)p(bar)
2.0
1.0
0.1 0.2
R
S
V(m )3
The work done during expansion process isarea under p-V curve. i.e. process R-S,
Work done by turbine,
= Shaded area + Rectangle area,
= 2000 + 1 × 105 × (0.2 – 0.1)
= 2000 + 10000
= 12000 N-m
Sol–88: (a)The expansion of gas in piston cylinderarrangement.
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(18) ME (Test-A4), Objective Solutions, 23rd October 2016
V
P
1
2
From first law of thermodynamics
1Q2 = 1W2 + U
1W2 = 50 kJ
U = – 70 kJ
1Q2 = 50 – 70
= –20 kJSol–89: (d)
All the datas (from S.No. 1 to 4) denotessame efficiency. So thermal efficiency ofengine
8.2
1500=
W2400
W = 8.2 × 85
= 13.12 kWSame answer from any other data,
9.31700
=W
2400
W = 9.3 × 24001700
= 13.129 kWFor more clearity, change Q1 in kJ/sec inthe table.
Sol–90: (a)
Q = 6000 kJW = 9000 kJ
Air
In this system, the paddle wheel does workon system by stirring process equal to 9000kJ. This is the situation where-
pdV = 0
But work done is non -zero
Sol–91: (d)
P
V
A
BC
D
Since final state is same in both cases i.e.ACB and ADB, so change in internal energywill be same. Because internal energy inpoint function and work done and heatinvolved will be different because these bothquantities are path function.
Path ACB,
Qacb = W1 + U
160 = 110 + U
U = 50 kJ
Path ADB,
Qadb = W2 + U
= 85 + 50 = 135 kJ
Sol–92: (c)
Internal energy of certain system,
E = (25 – 0.25 t) kJ
The heat intreaction from first law ofthermodynamics
Q = E + W
dQdt =
dE dWdt dt
Where t is temperature
dWdt
= 0.75 kJ
dEdt
= 0 – 0.25 = – 0.25 kJ
dQdt
= – 0.25 + 0.75
= 0.5 kJ
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(19) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–93: (a)
T =277°C=550K1
T =27°C = 300K2
HE
The heat engine Heat rejected
Q2 = 50% Q1 = 0.5 Q1
Thermal efficiency of heat en-gine –
th = 2
1
Q1Q
= 1
1
0.5Q1Q
= 0.5 = 50%The Carnot efficiency of engine operat-ing between temperature T1 and T2.
c = 2
1
T1T
= 3001550
= 1 – 0.5454= 0.4545= 45.45%= max
Since th c , so the engine isimpossible.
Sol–94: (a)Both engines equally efficient and assumingboth Carnot engines.
HE1
HE2W2
T2
W1
T = 625 K1
T = 400 K3
Q1
Since both engines have same efficiency,
1 = 2
1 – T2/T1 = 1 – T3/T2
T2 = 1 3T T
= 625 400
= 500 K
Sol–95: (c)
Two reversible heat engines in series aresuch that work output of both engines areequal,
W1 = W2HE1
HE2W2
T2
W1
T = 900 K1
T = 400 K3
Q1
Q2
Q3
Q1 – Q2 = Q2 – Q3
2Q2 = Q1 + Q3
Since engines are
reversible.
1
1
QT =
2
2
QT =
3
3
QT
= m–a constant
Q1 = mT1
Q2 = mT2
Q3 = mT3
2 mT2 = mT1 + mT3
T2 = 1 2T T2
=800 400
2
= 600 KSol–96 (c)
HE1
HE2W2
T2
W1
T = 1200 K1
T = 300 K3
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(20) ME (Test-A4), Objective Solutions, 23rd October 2016
Since both reversible engines have samethermal efficiency,
Since c > th which is possible, so thiscycle is irreversible cycle.
Sol–99: (b)h
s
T
PB
CP
A T = Const.P = Const.
Line AB denotes the phase change. Socondensation occurs along B to A i.e.inclined line of constant slope. This h-sdiagram is also called as Mollier diagram.
Sol–100:(b)Given parameters are,
1fh = 900 kJ/kg
2fh = 800 kJ/kg
2gh = 2800 kJ/kgP
V
hf1 hg1
hg2hf2
P2
P1
1
2
P > P1 2
But during throttling process 1–2,
1fh = h2
900 = hs2 + x (2gh – 2f
h )
900 = 800 + x (2800 – 800)
x =1002000
= 0.05
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(21) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–101:(a)
h
T1>T >T2 3
s
CP 1
P1=C
P2
P3
T1
T2
2
T3P2
P1
The Mollier diagram having throttling process1–2 is shown in figure. During throttlingprocess, following observations are made.
1. The pressure decreases and volumeincrease because of expansion.
2. The temperature reduces and steambecome superheated as shown in figure,process 1–2.
3. Dryness f raction has no meaningbecause initial state is dry and saturated.
4. Enthalpy is constant during process 1–2 but entropy increases.
Sol–102:(d)
1
2
In throttling process, the enthalpy remainsconstant and pressure drops. So Abscissamust be enthalpy and ordinate pressure.This process is very popular in p-h diagramof refrigerants in vapour compression.
Sol–103:(a)When all ratio’s on the right hand side ofsimplex table become negative, it indicatesan unbounded solution.
Sol–104:(d)
y
x
The hatched portion denotes the feasibleregion, it is to be noted here that feasibleregion always has a convex profile (i.e.outwards) and not concave. The boundariesare lines or planes and the corner pointsgive the feasible solutions to LPP.
Sol–105:(c)• If primal has 4 variables and 2
constraints, the dual has 2 variables and4 constraints.
• If primal is a maximization function, thedual is a minimization problem.
Dual of a primal is exactly opposite.Dual of a dual is exactly same as primal.
Sol–106:(a)
17
10
A5 (0,5)
B(0, 0)
B
2A+B=17
A+B=10A
(7.5, 0)7.5C 0.5
104A+6B=30
Max. Z = 2 A + 3B subject to
A + B 10
4 A + 6 B 17
A, B 0
Solving, system of equation throughgraphical method
A + B = 10; 4 A + 6B = 30 & 2A + B = 17
A + B = 10
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(22) ME (Test-A4), Objective Solutions, 23rd October 2016
when A = 0; B = 10
when B = 0; A = 10
4 A + 6B = 30 when A = 0
B = 5
when B = 0; A = 7.5
2 A + B = 17
when A = 10
B = 17;
when B = 0; A = 8.5
Thus, ABC is the feasible region with cornerpoints (0, 5) and (7.5, 0). Objective functionis maximum at
Z1(0, 5) = 2 × 0 + 3 × 5 =15
Z2(0, 0) = 0
Z3(7.5, 0) = 2 × 7.5 + 0 = 15
Since, one of the constraint line (i.e. 4A +6B = 30) is parallel to objective functionline, the solution may contain alternativeoptima.
Sol–107:(a)Dummy activity – Constructed only toestablish sequence.Critical path – Has zero total slack.
PERT activity – Follows distribution
Critical path method – It is built on activityoriented diagram.
Sol–108:(c)
A
5
4
B4
C
5E
F
8H
G
510 10
7D
703
Critical path in the network would be
A B C D F E H = 29 days
Therefore, the earliest expected completiontime cannot be below critical path time i.e.29 days.
Sol–109:(d)
Standard deviation for the entire critical path(i.e. a b c d e) is the sum ofindividual activity variances and than taking
square root of it.
= 8 16 4 8
= 36 6
Sol–110: (b)
5
321
4
67
48
64 9
532
43
5
Total time of paths out of the four optionsgiven are
1 – 2 – 3 – 4 – 8 – 9 = 3 + 4 + 5 + 4 + 4 =20
1 – 2 – 3 – 5 – 6 – 7 – 8 – 9
= 3 + 4 + 0 + 3 + 5 + 6 + 4 = 25
1 – 2 – 3 – 4 – 7 – 8 – 9
= 3 + 4 + 5 + 0 + 6 + 4 = 22
1 – 2 – 5 – 6 – 7 – 8 – 9
= 3 + 2 + 3 + 5 + 6 + 4 = 23
Hence path1 – 2 – 3 – 5 – 6 – 7 – 8 – 9
Finish i.e. 25. Hence, it is the critical path.Sol–111: (a)
• In VAM, allocation is made in cell withlowest cost with highest penalty.
• NWC rule is faster method• Unbalanced transportation problem can
be solved by adding dummy rows orcolumns.
• A feasible solution may not be an optimalsolution
Sol–112: (d)The basic feasible solution must haveexactly (m + n – 1) allocations, where mis plants and n are warehouses.= (4 + 5 – 1) = 8 allocations
Sol–113: (b)A solution is not a basic feasible solution intransportation problem if there is a closedloop. Further, iteration are necessary to arriveat a basic feasible solution.
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(23) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–114: (a)
Normal time = observed time × rating ofworker.
= 0.4 × 1.25 = 0.50 minutes
Standard time = Normal time + Allowances
= 0.50 + 0.2 × ST
ST(1 – 0.2) = 0.50
ST = 0.500.8
= 0.625 minutes
Sol–115: (c)
P = k P 1 P
s n
0.4 =2 0.4 0.6
0.05 n
n = 2400 observations
Sol–116: (a)Job enlargement also called “horizontal jobloading” means increasing the scope of workby extending the range of job duties andresponsibilities. It includes manufacturing,planning, organization and control work.
Sol–117: (c)
32 2 T 100 32 2 T 48 8 = 800
2T – 104 = 6400 10064
T = 102°CSol–118: (d)
1 1 2AK=
2 1 2K A
K1 = K2
Sol–119: (d)General equation
2 2 2G
2 2 2qT T TKx y z =
1 T
t
just putting T 0t gives poisson’s equation,
Sol–120: (c)There can be no system possible which canviolate either of the two laws ofthermodynamics.
Sol–121: (a)
Q =
1 2
2
1
T T 2 LKrnr
=
200 100 2 1 7010n5
=
22100 2 707 63.5kW
0.693
Sol–122: (b)
At r = rc, dQ 0dr
dQ 8r 32dr = 0
0 + 8rC = 32
rC = 4 = k [for cylinder]h
4 = 8h
h = 2W/m2KSol–123: (a)
hA T T = 51 200
100
322h 1.2 10 7 257 = 200 + 10
h = 318.18 W/m2KSol–124: (d)
Q =
f1 f 2
T T
T TTR R
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(24) ME (Test-A4), Objective Solutions, 23rd October 2016
T = TQ R
= 1 1400 40
100 40 20 40
T = 0.1 + 16000 + 0.5= 16000.6 °C
Sol–125: (b)
Tc = 2
wqrT4k
Tc =
2cr
w 2Q rT
4kr
210 =
GQ45 22 14 4 2.47 100
22 14165 4 2.47 100 = QG
QG = 697 WSol–126: (b)
T1 = 110 – 60 x1T2 = 110 – 60 x2
T1 – T2 = –60x1 + 60(x2)
T = 60 (x2 – x1)= 60 × 0.75 = 45°C
Sol–127: (d)Sol–128: (c)Sol–129: (d)Sol–130: (c)
fin = finfin
b
AA
fin = finx
x = fin
b
A 2 7 7 30A 7 7
= 17.14Sol–131: (c)
for infinitely long fin where Tiptemperature equal to T
mL = 5
hPkA = 5
L
hPkA = 2
25L
h =
2
2 225 kA 25 100 0.1PL 4 0.1 0.1
= 6250 W/m2KSol–132: (a)
Gsolar = D dG cos G Area
= [x cos 30 + y] × 5Sol–133: (c)
J = bE G
= 0[black body]
J = 41 T
= 5.67 × 10–8 × (273)4
= 314.94 315W/m2
Sol–134: (b)
h h h1 h2m c (T T ] = c c c2 c1m c T T
For balanced counter flow heat exchanger.
h hm c = c cm c and variation of temperatureof fluids with respect to length is straightline
45 – 15 = T – 10T = 40°C
T
45°
15°
40
x=0 x=L
L/2 L/2
15°
T1
10
40
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(25) ME (Test-A4), Objective Solutions, 23rd October 2016
Since the flow of fluid in the pipe bend isfree vortex (Vr = const). So the velocity inthe inner side is high because of smallradius of flow. Due to this velocity variationthe pressure will be redistributed and it willresult in decrease of pressure head andincrease of velocity head at the inner wall.
Assuming that the flow in bend as streamline flow. The spacing between stream linesat outer side of wall of bend will increasebecause the discharge in stream line flow(q = v.b = constant, where b – spacingbetween two adjacent stream line). Theelbow meter to measure discharge isdesigned based upon this principle.
Sol–152:(a)Nature of the fluid flow in pipe depends uponReynolds number which is the function ofaverage velocity 'V', diameter of pipe 'D' andkinematic viscosity ' ' as
Re =VD
Note: This explanation is true for laminarflow and smooth pipe in turbulent flow, notfor rough pipe.
Sol–153:(b)If ideal conditions are considered, thethickness of boundary layer will be everincreasing whether the boundary layer islaminar or turbulent. The boundary layerbreaks when there is flow separation.But in practice 99% depth of boundary layerdevelops within short distance from leadingedge. This short distance depends uponflow and surface conditions.
Sol–154:(a)The shear stress in fluid flow in pipe-
r
= 0Px
= = constant
r
u
U
dudr
=P rx
upon integrating, we get
u = 2 21 P R r4 x
=
Hence, reason correctly explains the assertion.Sol–155:(c)
In immersed subsonic flow, the effect ofcompressibility is neglected, so for modeland prototype to be similar, Reynold numbershould be same.In immersed body, viscous and inertia forcesare major forces to play dominant role, so
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(27) ME (Test-A4), Objective Solutions, 23rd October 2016
for dynamic similarity, the ratio of theseforces must be same in model and prototype.
Sol–156: (d)In simple gas turbine power plant, thegas turbine consumes considerableamounts of power just to drive itscompressor. As with all cyclic heatengines, a higher maximum workingtemperature in the machine meansgreater efficiency, but in a turbine it alsomeans that more energy is lost as wasteheat through the hot exhaust gases whosetemperatures are typically over 1000°C.Consequently, simple gas turbineefficiencies are quite low.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 the gasturbine follows Brayton cycle. Thus evenif the compressor, combustor and turbinelosses were neglected, the overall thermalefficiency of the gas turbine dependsmainly on the cycle thermal efficiency. Itis a mistake that the multiplication ofthe three component efficiencies will givethe total overall efficiency a value thathas not even a remote connection to theactual thermal efficiency of the gasturbine.
Sol–157: (d)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–158: (a)The hydraulic turbine can be switchedon and off in a very short time. In athermal or nuclear power plant, the steamturbine is put on turning gear for abouttwo days during start-up and shut-down.Many hydropower plants are operated aspeak generators, because they can changetheir output quickly to follow thefluctuating power demand. When meetingpeak load requirements, a power stationis turned on at a particular time duringthe day, generates power at a constantload for a certain number of hours, andis then turned off or set to a differentload for another time period, resulting inhigh variability in flow discharges.
Sol–159: (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 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.
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(28) ME (Test-A4), Objective Solutions, 23rd October 2016
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 flowmachinery, 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.
Sol–160: (a)
The mixture of air and fuel used in gasturbine is very lean of the order of 50 to100:1. Fuel more than this can not beburnt in gas turbine because it is asteady flow machine and peak tempera-ture also remains constant. So thismaximum value of temperature is lim-ited by material availability.
Sol–161: (c)2nd statement is wrong because tidalcurrents are present all over world.
Sol–162: (b)Both statements are individualy correctbut correct reason is that hydrogen hashigher electro chemical reactivity thanother fuels, such as hydrocarbons oralcohols.
Sol–163: (a)The wind turbine blades are slightly twistedfrom the outer tip to the rotor as such twistingof wind blades reduces the tendency to stall.So reason correctly explains the assertion.
Sol–164: (b)Adding a third blade increases the poweroutput by about 5% only while the weightand cost of a rotor increases by 50%. Thusgiving a diminished rate of return foradditional weight and cost. The main reasonfor three blade machine is that it hassmoother action and balanced.
Sol–165: (a)When water content in the biomass is highthe gas production drops because of highwater content mean slurry temperature drops.So reason correctly explains the assertion.
Sol–166: (b)When thermal conductivity is very high
internal conduction resistance is so low
LkA
that entire fin becomes a lump mass at basetemperature.
Sol–167: (a)
(rc) =kh
i.e. (rc) 1h
As h increases, rc decreases.Sol–168: (a)
The electrons, atoms and molecules of allliquids, and gases and solids above absolutezero temperature are constantly in motionand thus radiation is constantly emittedas well as being absorbed or transmittedthroughout the entire volume of matter,hence radiation is a volumetricphenomenon. For opaque surface,radiation is considered to be a surfacephenomenon because the radiation emittedby the interior region can never reach thesurface and the radiation incident on suchbodies is absorbed within a few micronsfrom the surface.
Sol–169: (a)A baffle is a metal plate, usually in theform of the segment of circle, have holesto accomodate tubes, because a baffle platesupport the tubes.
Sol–170: (a)More is the entropy generation, more willbe the irreverssibilities and more is thearea required for heat transfer.
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(29) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–171: (a)Heat exchnager are adiabatic devices in whichheat exchange takes place hot and cold fluidwithout exchnaging heart with surrounding.So applying SFEE for Heat Exchanger,
Q + W = H PE KE
It results in H 0
hot fluid cold fluidH H 0
hot fluid cold fluidH H
Rate of enthalpy decrease of hot fluid = rateof enthalpy increase of cold fluid. So reasoncorrectly explains the assertion.
Sol–172: (a)We have LMTD calculation for eithercounter flow or parallel flow heatexchanger, it is not possible to have LMTDfor Multi Pass heat exchangers as theyare those in which two fluids flowalternately in parallel flow and counterflowand hence LMTD expression is not validand for them we require correction factorfor LMTD calculation.
Sol–173:(a)Because of higher cohesive force betweenmolecules of mercury than the adhesiveforce between molecules of mercury andglass, a convex surface is made in capillaryas shown below:
Hg
Glass tube
Hence the mercury does not wet the glasssurface and mercury in glass goes downas compared to mercury level in both.
Sol–174:(b)
FG
P
xh
The depth of center of pressure
h = GxAx
I
This expression is independent of densityof fluid.
The center of pressure 'P' lies below centerof area 'G' of immerged surface.
Sol–175:(b)Due to empty space for loading or sitting,the weight of water that the boat bottomouter surface displace will be more thanweight of boat, so it will float. When theboat overturns, the space will be filled upwith water and buoyancy force will be duevolume of construction material only.
In a constant volume process, the workdone is zero, i.e. for a particulartemperature difference, heat is utilizedonly to raise the internal energy.In a constant pressure process, for thesame temperature difference, heat isutilized is both work output and internalenergy change.
Sol–177: (c)All the three phases co-exist at the triplepoint.
Sol–178: (d)The mixture of air and liquid air is nota pure substance, since the relativeproportions of oxygen and nitrogen aredifferent in the gas and liquid phases inequilibrium.
Sol–179:(b)Both statements are individually correct.
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(30) ME (Test-A4), Objective Solutions, 23rd October 2016
Sol–180:(a)Vogel approximation method is faster ascompared to linear programming as it onlygives an approximate solution which mayor may not be an optimal solution if it is anoptimal solution, the algorithm stops thenotherwise, u-v method is used to arrive atan optimal solution.
