Mechanical Engineering Systems 2008 Part 4 potx
Mechanical Engineering Systems 2008 Part 4 potx
... h
2
= 947 .4 kJ/kg
From the superheat tables at 30 bar, 40 0°C,
h
1
= 3231 kJ/kg
3231 – h
2
= 947 .4, h
2
= 3231 – 947 .4 = 2283.6 kJ/kg
h
2
= 2283.6 = h
f
+ x.h
fg
at 0.05 bar,
2283.6 = 138 + x. 242 3, ... u
f
+ xu
fg
= 561 + 0.763(2 544 – 561) = 20 74 kJ/kg
Q = W + (U
2
– U
1
) = 80.35 + (20 74 – 20 74. 3)0. 346
= 80.25 kJ
Example 2.6.17
0.1 m
3
of steam at a pressure of 14 bar is ex...
Mechanical Engineering Systems 2008 Part 1 pdf
... constant
pressure.
Q = m.c
p
.␦T
80 000 = 2 × 900 × ␦T
␦T =
80 000
1800
= 44 .44 °C
Final temperature = 10 + 44 .44 = 54. 44 C
Latent heat
Latent means ‘hidden’, and is used in this connection because, ... 100°C,
Q
4
= m × latent heat of vaporization
water
Q
4
=4 × 2256.7 = 9026.8 kJ
Total heat energy = Q
1
+ Q
2
+ Q
3
+ Q
4
= 81.6 + 1 340 + 1680 + 9026.8
= 12 128 .4 kJ.
Mechanical...
Mechanical Engineering Systems 2008 Part 3 pptx
... spring rate =
390
70
× 0.8
= 4. 46 bar = 4. 46 × 10
5
N/m
2
ip = P
m
A.L.n × number of cylinders
= 4. 46 × 10
5
×
× 0.15
2
4
× 0.2 ×
5.5
2
× 4
= 17 339 W = 17. 34 kW
Example 2 .4. 8
During a test, a 2-cylinder, ... – 293
47 3 – 293
= 0.791
W
c
= c
p
(T
2
– T
1
) = 1.005 (47 3 – 293) = 180.9 kJ/kg
TЈ
4
T
3
=
p
4
p
3
␥–1
␥
, TЈ
4
= 953 ×
1.01
4. 04
1.33–1
1.33
, TЈ
4...
Mechanical Engineering Systems 2008 Part 5 pps
... was
isentropic,
hЈ
2
= 146 9.9 +
(5.3 14 – 4. 962)
(5.397 – 4. 962)
× (1613 – 146 9.9)
= 1585.7 kJ/kg
0.85 =
hЈ
2
– h
1
h
2
– h
1
=
1585.7 – 144 6.5
h
2
– 144 6.5
, h
2
= 1610.3 kJ/kg
Power = m(h
2
– h
1
)=
4. 5
60
(1610.3 ... we want the rate of heat energy
transfer, i.e. J/s = W.
(5 04 – 2 04) =
Q
1 × 1
0.0095
44
Q =
(5 04 2 04) × 1 × 1 × 44
0.0095
= 1 389 47 4 W = 1389.5 kW...
Mechanical Engineering Systems 2008 Part 6 pdf
... × area
=(g4/2) × (4 × 5 .4)
= 1000 × 9.81 × 2 × 4 × 5 .4
= 42 3.8 kN
Therefore F
2
= 2 42 3 800 N
We could also express F
2
as (gh/2) × (h × 5 .4)
and so 1000 × 9.81 × 5 .4 × h
2
/2 = 2 42 3 800
giving ... 9.81
= 49 .050 kN
Hence the resultant tension in the cable with the pillar partly
submerged is:
49 .050 – 25.506 = 23. 54 kN
Buoyancy
This phenomenon of solid objects being a...
Mechanical Engineering Systems 2008 Part 7 pptx
... pipes
10
3
2 345 68
10
4
10
5
10
6
10
7
22233 344 4555666888
10
8
10
8
2
2
3
3
4
4
5
5
6
6
8
8
– 0.00001
0.00005
Laminar flow
Laminar flow
Relative roughness
0.0001
0.0002
0.00 04
0.0006
0.0008
0.001
0.002
0.0 04
0.006
0.008
0.01
0.015
0.02
0.03
0. 04
0.05
0.000 ... flow:
h
f
=
4fL
d
V
2
2 g
so the pressure drop is given by
P = g ·
4fL
d
·
V
2
2 g
Using Q = VA = Vd
2
/4,
P =...
Mechanical Engineering Systems 2008 Part 8 docx
... acceleration then
become
2
=
1
+ ␣t (4. 1.10)
=
1
t + ␣t
2
/2 (4. 1.11)
= t(
1
+
2
)/2 (4. 1.12)
2
2
=
1
2
+ 2␣ (4. 1.13)
Velocit
y
0
Tim
e
t
u
v
( + )/2uv
1 74 Dynamics
Adding the two areas ... s
1
Therefore
s
1
= 4. 44 m
Similarly for deceleration
0 = 16 – 2 × 2 .4 s
3
s
3
= 3.33 m
The distance involved in stopping and starting is
s
1
+ s
3
= 7.77 m
and so
s
2
= 350 –...
Mechanical Engineering Systems 2008 Part 10 doc
... diameter?
( 24) A uniform beam of length 5.0 m rests on supports 1.0 m
from the left-hand end and 0.5 m from the right-hand end. A
Figure 5.1 .43 Problem 14
Figure 5.1 .44 Problem 15
Figure 5.1 .45 Problem ... concrete a modulus of elasticity of 14 GPa.
The area of the column that is steel is 4 ×
1
4
× 25
2
=
1963 mm
2
. The area of the column that is concrete is 45 0 ×
45 0 – 1963...
Mechanical Engineering Systems 2008 Part 11 pptx
... I
xx
cm
4
Section
modulus Z
cm
3
Area of
section
cm
2
838 × 292 226 850.9 293.8 16.1 26.8 340 000 7990 289.0
1 94 840 .7 292 .4 14. 7 21.7 279 000 6550 247 .0
176 8 34. 9 291.6 14. 0 18.8 246 000 5890 2 24. 0
762 ... 267 197 769.6 268.0 15.6 25 .4 240 000 6230 251.0
147 753.9 265.3 12.9 17.5 169 000 44 80 188.0
686 × 2 54 170 692.9 255.8 14. 5 23.7 170 000 49 10 217.0
140 683.5 253....
Mechanical Engineering Systems 2008 Part 13 pps
... kJ
36. 41 7.8 m/s
37. 589 kW
38. 7.85 kW, 20 .4 m
39. 14. 0 m/s
40 . 12.5 m/s
41 . (a) 10.5 m
(b) 9.6 m/s
42 . 5.87 m/s
43 . 12 892.78 m/s
44 . 6.37 m/s
45 . 0.69 m/s right to left
46 . 4. 42 m/s
47 . –5.125 ... 18%
4. 21%
3 04 Solutions to problems
19. 137.5 mm
20. 165.5 mm
21. 11.7 × 10
6
mm
4
, 2.2 × 10
6
mm
4
22. 158 mm
23. d /4
24. (a) 71 mm, 74. 1 × 10
6
mm
4
, (b)...