1. thermodynamics, ic engine, rac_objective
DESCRIPTION
thermodynamics objectiveTRANSCRIPT
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9Test 1 (Obj) : Thermo, I.C. Engine, RAC
ESE-2014 : Test SeriesMECHANICAL ENGINEERING
Test 1 (Obj) : THERMODYNAMICS, I.C. ENGINES, RAC
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A N S W E R S
1.1.1.1.1. (b)
2.2.2.2.2. (a)
3.3.3.3.3. (a)
4.4.4.4.4. (a)
5.5.5.5.5. (a)
6.6.6.6.6. (c)
7.7.7.7.7. (d)
8.8.8.8.8. (d)
9.9.9.9.9. (b)
10.10.10.10.10. (b)
11.11.11.11.11. (d)
12.12.12.12.12. (d)
13.13.13.13.13. (b)
14.14.14.14.14. (b)
15.15.15.15.15. (b)
16.16.16.16.16. (b)
17.17.17.17.17. (d)
18.18.18.18.18. (b)
19.19.19.19.19. (a)
20.20.20.20.20. (d)
21.21.21.21.21. (c)
22.22.22.22.22. (c)
23.23.23.23.23. (b)
24.24.24.24.24. (d)
25.25.25.25.25. (b)
26.26.26.26.26. (c)
27.27.27.27.27. (c)
28.28.28.28.28. (a)
29.29.29.29.29. (d)
30.30.30.30.30. (c)
31.31.31.31.31. (b)
32.32.32.32.32. (b)
33.33.33.33.33. (d)
34.34.34.34.34. (a)
35.35.35.35.35. (a)
36.36.36.36.36. (c)
37.37.37.37.37. (d)
38.38.38.38.38. (c)
39.39.39.39.39. (b)
40.40.40.40.40. (c)
41.41.41.41.41. (a)
42.42.42.42.42. (a)
43.43.43.43.43. (b)
44.44.44.44.44. (d)
45.45.45.45.45. (b)
46.46.46.46.46. (a)
47.47.47.47.47. (a)
48.48.48.48.48. (b)
49.49.49.49.49. (d)
50.50.50.50.50. (c)
51.51.51.51.51. (d)
52.52.52.52.52. (b)
53.53.53.53.53. (a)
54.54.54.54.54. (c)
55.55.55.55.55. (d)
56.56.56.56.56. (a)
57.57.57.57.57. (c)
58.58.58.58.58. (a)
59.59.59.59.59. (d)
60.60.60.60.60. (c)
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10 ESE-2014 : TEST SERIES MECHANICAL ENGINEERING
1.1.1.1.1. (b)(b)(b)(b)(b)
m, v, T, P, m2
v2
,
T, P,
m2
v2
,
T, P,
m V,
2 2 Extensive properties
T, P, Intensive properties
4.4.4.4.4. (a)(a)(a)(a)(a)
v
P
n = 0
n = 1
n = 2
n =
5.5.5.5.5. (a)(a)(a)(a)(a)As per Gibbs phase rule,
F = C P + 2= 1 3 + 2 = 0
6.6.6.6.6. (c)(c)(c)(c)(c)
W1 = 51 1 1 10
2= 0.5 102 kJ (ve)
W2 = 51 1 1 10
2= 0.5 102 kJ (+ve)
Net work done = W1 W2 = 0
9.9.9.9.9. (b)(b)(b)(b)(b)
P
T C( ) 273.15
Gas1
Gas2
Gas3
EXPLANATIONS
10.10.10.10.10. (b)(b)(b)(b)(b)Q = W + U
U for the cycle = 0
Q = ( ) ( )W . .= 1 600 300 0 3 0 12
= 0.1 300 = 30 kJ
11.11.11.11.11. (d)(d)(d)(d)(d)
Initial volume, V = RTP
R = Cp Cv = 0.3 kJ/kgK,P = 300 kPa, T = 300 K
V =.
.
=
33
3
0 3 10 3000 3m
300 10Work done = 20 15 60
= 18000 J = Heat additionmCv (T2 T1) = 18000
T2 = 325.7 KFor constant volume process,
PT
1
1=
PT
2
2
P2 =T PT
21
1
=.
.
=
325 7 300325 7kPa
300P2 = 325.7 kPa
12.12.12.12.12. (d)(d)(d)(d)(d)
Q Q
T1 T2
S1 =Q WT K
= =
1
1500 5300
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11Test 1 (Obj) : Thermo, I.C. Engine, RAC
S 2 =QT T
=
2 2
1500
S1 + S2 = 0.25
or .T
=
2
15005 0 25
or 5.25 =T2
1500
T2 = ..=
1500285 7K
5 25
T2 = 285.7 273 = 12.7C
13.13.13.13.13. (b)(b)(b)(b)(b)Irreversibility,
I = T0 S
=
+ = 3200 3200
300 2400 kW1200 300
Availability,A = 3200 2400 = 800 kW
14.14.14.14.14. (b)(b)(b)(b)(b)Helmholtz function,
F = U TSGibbs function,
G = H TSG F = U H
For ideal gas,U f1(T), H f2 (T)
Difference between G and F depends on
temperature only.
15.15.15.15.15. (b)(b)(b)(b)(b)Work and heat interactions are the forms ofenergy in transit, observed at theboundaries of a system. They are notproperties of a system. They are pathfunctions, their magnitudes dependingupon the path the system follows during achange of state. Energy in storage calledinternal energy is a point or state functionand hence a property of the system.
16.16.16.16.16. (b)(b)(b)(b)(b)
a
v 2 force of cohesion
b Co-volume
19.19.19.19.19. (a)(a)(a)(a)(a)
( )ap v b RTv
+ = 2
pv3 (pb + RT) v2 + av ab = 0
( )cc
T T cc
p RT av vv b=
= + =
2 3
20
( )cc
ccT T
p RT a
v vv b=
= =
2
2 3 4
2 60
From these equation, we get
a = cc cv
p v , b =233
22.22.22.22.22. (c)(c)(c)(c)(c)
Over square ratio = d
.L
= 1 1
Vs =dd L.
= =
32 245
4 4 1 1
d 3 = 343 or d = 7 cm
L =d
..
= 6 36 cm1 1
23.23.23.23.23. (b)(b)(b)(b)(b)Total volume/cylinder,
Vtotal = Vc + Vs = =2000
500 cc4
Compression ratio,
r = c s sc
V VV
V+
= 500 or = 450 cc
Volume of air taken in/cycle= v Vs = 0.7 450 = 315 cc
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12 ESE-2014 : TEST SERIES MECHANICAL ENGINEERING
a
f
mm
= 9
mf = =315
35cc9
Energy supplied/ cylinder= 35 106 34 103 = 1.19 kJ
25.25.25.25.25. (b)(b)(b)(b)(b)
v
0.55
Inlet valve mach index (z)
26.26.26.26.26. (c)(c)(c)(c)(c)
otto =r
1
11
diesel = ( )c
c
rrr
1factor
1 11
1
29.29.29.29.29. (d)(d)(d)(d)(d)
P
V
2 3
4
1
r =VV
=1
220
Vs = V1 V2 = 19V2V3 = 0.05 Vs + V2V3 = (19 0.05 + 1)V2
VV
3
2= rc = 1 + 0.95 = 1.95
30.30.30.30.30. (c)(c)(c)(c)(c)Heat rejection for Otto engine would be moreand hence its efficiency would be least.
32.32.32.32.32. (b)(b)(b)(b)(b)For CI engines, the normal paraffins are thebest fuels and aromatics are the leastdesirable.
41.41.41.41.41. (a)(a)(a)(a)(a)Catalytic converters are called three wayconverters because it reduces theconcentration of CO, HC and NOx in theexhaust (three concentrations).
42.42.42.42.42. (a)(a)(a)(a)(a)
COP =T
T T= =
1
2 1
2002
100
COP = AqW
W =1
kW = 0.5kW2
43.43.43.43.43. (b)(b)(b)(b)(b)For Cm Hn Clq Fpn + p + q = 2m + 2,formula is R (m 1) (n + 1) p
P = 2, No hydrogen atom, n = 0 q = 2 1 + 2 = 4 Designation for R - 12 is CCl2 F2
48.48.48.48.48. (b)(b)(b)(b)(b)When the refrigerant in the gas cylinderor in the system is allowed to remainstanding for a period and come toequilibrium with the ambient temperature,the pressure recorded is named as thestanding pressurestanding pressurestanding pressurestanding pressurestanding pressure.
49.49.49.49.49. (d)(d)(d)(d)(d)
COP =h hh h
1 4
2 1
=. . .
.. . .
= =
350 3 173 7 176 63 97
394 7 350 3 44 4
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13Test 1 (Obj) : Thermo, I.C. Engine, RAC
51.51.51.51.51. (d)(d)(d)(d)(d)
w
s
PP
= 0.7
Pw = 2.975 kPaPw + Pa = P
Pa = 100 2.975 = 97.025 kPa
52.52.52.52.52. (b)(b)(b)(b)(b)
t1 t2
22
1
1
2 <
2 1t t >
1
DBT
55.55.55.55.55. (d)(d)(d)(d)(d)
te = t h
+0I
57.57.57.57.57. (c)(c)(c)(c)(c)
BPF = .
= = =
25 20 5 10 25
40 20 20 4
coil = 1 BPF = 0.75
60.60.60.60.60. (c)(c)(c)(c)(c)R 11 CCl2 FR 12 CCl2 F2R 13 CCl F3R 22 CHCl F2
Class one Ozone depleting substances areR 11, R 12, R 113, R115.
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