Handbook of Mechanical Engineering Calculations ar Episode 2 Part 5 ppsx
Engineering Mechanics - Statics Episode 2 Part 5 ppsx
...
= 0;
2
F
2
⎛
⎜
⎝
⎞
⎟
⎠
2
W
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
FW= F 175lb=
Man:
+
↑
Σ
F
y
= 0;
N
C
W− 2
F
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
N
C
WF+= N
C
350 lb=
b()
Bar:
+
↑
Σ
F
y
= 0;
2
W
4
⎛
⎜
⎝
⎞
⎟
⎠
2
F
2
− 0=
F
W
2
= ... lb= W
2
30 lb=
Solution:
(a)
Bar:
+
↑
Σ
F
y
= 0;
2
F
2
W
1
W
2
+
()
− 0=
FW
1
W
2
+=
F 20 5lb=
Man:
+
↑
Σ
F
y
= 0;
N
C
W
1
− 2
F
2
− 0=
N
C
FW
1
+=
N
C
380lb=...
Handbook of Mechanical Engineering Calculations P1
... /kWh)
1
⁄
4
100
ϫ
1
⁄
4
ϭ
25
50
⁄
25
ϩ
9 .5
ϭ
11 .5 (5 .22 )
2
⁄
4
100
ϫ
2
⁄
4
ϭ
50
50
⁄
50
ϩ
9 .5
ϭ
10 .5 (4.77)
3
⁄
4
100
ϫ
3
⁄
4
ϭ
75
50
⁄
75
ϩ
9 .5
ϭ
10.17 (4. 62)
4
⁄
4
100
ϫ
4
⁄
4
ϭ
100
50
⁄
100
ϩ
9 .5
ϭ
10.00 (4 .54 )
Related ...
fired
HRSG
Case 7
2
Supp
fired
HRSG
GT output, MW 5. 8 20 .2 20 .2 21.8 15. 5 0 0
ST output, MW 0.9 2. 4
Ϫ
2. 1
Ϫ
13 3.7...
Handbook of Mechanical Engineering Calculations P2
... loading. Or, U
ϭ
58 5(0. 95)
ϭ
55 5. Then x
ϭ
0.8 12( 55 5/ 50 0)
ϭ
0.90; e
x
ϭ
2. 7183
0.90
ϭ
2. 46; t
2
ϭ
101.1
Ϫ
(101.1
Ϫ
85/ 2. 46)
ϭ
94.6
Њ
F (34.8
Њ
C).
Check
⌬
t
ϭ
t
s
Ϫ
t
2
ϭ
101.1
Ϫ
94.6
ϭ
6 .5
Њ
F (3.6
Њ
C), ... than 5
Њ
F
(2. 8
Њ
C). The cooling-water flow rate, gal /min
ϭ
950 ( 25 ,000)/ [50 0(94.6
Ϫ
85) ]
ϭ
4 950 gal / min (3 12. 3 L /s). Then A
ϭ
0.8 12( 4 950...
Tài liệu Handbook of Mechanical Engineering Calculations P3 pdf
... burned
ϭ
12. 39 /22 .1
ϭ
0 .56 0 lb (0 . 25 2 kg). This, plus the weight
of N
2
in the fuel, step 1, is 0 .56 0
ϩ
0.0 0 25
ϭ
0 .5 6 25 lb (0 . 25 3 kg) of N
2
in the
products of combustion.
Next, find the total weight of ... (10.96 72) (0 .20 )( 359 / 28 . 95) (2. 15)
ϭ
58 .5 ft
3
(1. 656
m
3
). In this calculation the value 28 . 95 is the molecular weight of air. The t...
Tài liệu Handbook of Mechanical Engineering Calculations P4 pptx
... of HRSG
Load, % 25 50 75 100
Steam generation, lb/ h (kg / h) 25 ,000 (11, 350 ) 50 ,000 (22 ,700) 75, 000 (34, 050 )
100,000 ( 45, 400)
Duty, MM Btu/ h (MW) 25 .4 (7.4) 50 .8 (14.9) 76.3 (22 .4) 101.6 (29 .8)
Exhaust ... h) 1 52 , 000 (69,008) 153 ,140 (69, 52 6 ) 154 ,330 (70,066)
155 ,57 0 (70, 629 )
Exit gas temperature,
Њ
F(
Њ
C) 319 ( 159 ) 28 5 (141) 27 3 (134) 26 9 (1 3...
Tài liệu Handbook of Mechanical Engineering Calculations P5 pdf
... BWG
5
⁄
8
-in (1.6-cm) OD tubes are being used, ID
ϭ
0. 6 25
Ϫ
2( thickness)
ϭ
0. 6 25
Ϫ
2( 0.049)
ϭ
0. 52 7 in (1.3 cm). Then, open area per tube
ft
2
ϭ
(
d
2
/4)/144
ϭ
0.7 854 (0. 52 7 )
2
/144
ϭ
0.001 52 5 ... tube
sheet
ϭ
1 12/ 166
ϭ
0.6 75 ft
2
,or97in
2
( 6 25 .8 cm
2
).
The inside diameter of the heater shell
ϭ
(tube sheet area, in
2
/0.7 854 )
0 .5
ϭ
(97/0.7...
Tài liệu Handbook of Mechanical Engineering Calculations P6 pdf
... ENGINES
INTERNAL-COMBUSTION ENGINES
6.41
35
30
25
20
15
1100
1 050
1000
950
900
850
Horsepower
Torque, inch-lb.
1,000 1,400
Speed, rpm
1,800 2, 200
(a)
35
30
25
20
15
950
900
800
Horsepower
Torque, inch-lb.
1,000 ... inch-lb.
1,000 1,400
Speed, rpm
1,800 2, 200
(b)
SI Values
hp kW in lb Nm
15 11 .2 850 96.1
20 14.9 900 101.7
25 18.7 950 107.4
30 22 .4 1000 113.0...
Tài liệu Handbook of Mechanical Engineering Calculations P7 docx
... gal/min (73.1 L/ s); H
p
ϭ
(1.9 05)
2
(0.3 35)
2
( 350 )
ϭ
1 42. 5 ft (43.4 m); NPSH
p
ϭ
(1.9 05)
2
(0.3 35)
2
(10)
ϭ
4.06
ft (1 .24 m); P
p
ϭ
(1.9 05)
5
(0.3 35)
3
(55 )
ϭ
51 .8 hp (38.6 kW).
3. Compute the ... N
s
ϭϭ
9 65; S
ϭ
0 .5 0. 75 0 .5
350 0 (50 0) /( 350 ) 350 0 (50 0) /
ϭ
13,900. For the prototype, N
s
ϭϭ
9 65;
0. 75 0 .5 0. 75
(10) 1170(1 158 ) /(1 42. 5)
S...
Tài liệu Handbook of Mechanical Engineering Calculations P8 doc
... size,
known/new
Pressure
factor,
P
2/ 1 5. 0 92 1 /2 0.0386
3/ 2 2 .58 2/ 3 0. 150
4/ 3 1.919 3/4 0 .27 1
6/ 4 2. 674 4/6 0.1399
8/ 6 1.933 6/8 0 .26 7
10/ 8 1. 726 8/10 0.3 35
12/ 10 1 .5 42 10/ 12 0. 421
When computing such ... diameter of pipe, in (cm). From a table of pipe properties,
d
ϭ
22 . 626 in (57 4.7 mm). Also, k
ϭ
z/S
ϭ
75/ 0. 85
ϭ
88 .2 cSt. Then
R
ϭ
31 62(...
Tài liệu Handbook of Mechanical Engineering Calculations P9 pdf
... operate to within 2. 5 inHg (9.7 kPa) of the barometer, or a vacuum
of 29 .8
Ϫ
2. 5
ϭ
27 .3 inHg (1 05. 5 kPa).
2. Compute the quantity of free air removed from the receiver
Select a number of absolute pressures ... d
ϭ
0 .23 6(100)
0 .5
(0.0)
0 . 25
/ (21 4.7)
0 . 25
ϭ
0.60 in (1. 52 cm).
Related Calculations. Rupture disks are used in a variety of applications—
proce...
Từ khóa: