Chapter 1.
CONVENTIONAL FUELS
(Part 1)
Outline
Introduction to fuels and combustion
Properties of conventional fuels
Ignition characteristics
Introduction to fuels and combustion
Why do we need fu
Problem 6.98 The Pratt bridge truss supports loads at
F , G, and H. Determine the axial forces in members
BC, BG, and F G.
Solution:
The angles of the cross-members are = 45 .
The complete structure a
Problem 6.75 The pin support B will safely support a
force of 24-kN magnitude. Based on this criterion, what
is the largest mass m that the frame will safely support?
C
500 mm
100 mm
E
D
B
300 mm
m
A
Problem 6.71 The man using the exercise machine
is holding the 80-lb weight stationary in the position
shown. What are the reactions at the built-in support E
and the pin support F ? (A and C are pinn
Problem 7.16 Determine the coordinates of the centroid of the area.
y
(1, 1)
y = x 1/2
y = x2
Let dA be a vertical strip: The area dA =
Solution:
x1/2 x2 dx, so
1
5/2
4 1
x
x dA
x3/2 x3 dx
x4
5/2
Problem 7.23 The Supermarine Spitre used by Great
Britain in World War II had a wing with an elliptical
prole. Determine the coordinates of its centroid.
y
y2 = 1
x2 +
a2
b2
x
2b
a
Solution:
y
x2 y2
Problem 7.10 Determine the coordinates of the centroid of the metal plates cross-sectional area.
y
1
y = 4 x 2 ft
4
x
Let dA bea vertical strip:
The area dA = y dx = 4 14 x2 dx. The curve intersects t
Problem 7.4 The x coordinate of the centroid of the
area shown in Problem 7.3 is x = 2. What is the y
coordinate of the centroid?
From Problem 7.4, a = 2.5
2.5
2.5
1
1 6
y(y dx)
x dx
y dA
2
2
0
0
y
Problem 6.103
Determine the forces on member
ABC, presenting your answers as shown in Fig. 6.35.
D
400 lb
2 ft
200 ft-lb
A
1 ft
B
C
100 lb
1 ft
E
2 ft
2 ft
Solution: The complete structure as a free b
Problem 6.54 For the frame shown, determine the
reactions at the built-in support A and the force exerted
on member AB at B.
A
200 lb
B
6 ft
C
20
6 ft
Solution:
and
3 ft
3 ft
Element AB: The equilibri
Problem 7.39 Determine the x coordinate of the centroid of the Boeing 747s vertical stabilizer.
y
11 m
48
x
70
12.5 m
Solution: We can treat the stabilizer as a rectangular area (1) with
two triangula
Problem 6.31 For the truss in Problem 6.30, use the
method of sections to determine the axial forces in members BC, CF , and F G.
A
B
C
D
E
F
G
H
1m
Solution:
BC
C
J
D
100 kN
45
1m
CF
Solving BC = 300
Problem 6.83 The pressure force exerted on the piston is 2 kN toward the left. Determine the couple M
necessary to keep the system in equilibrium.
B
300 mm
350 mm
45
A
C
M
400 mm
Solution: From the di
Problem 6.94 A load W = 2 kN is supported by the
members ACG and the hydraulic actuator BC. Determine the reactions at A and the compressive axial force
in the actuator BC.
A
0.75 m
B
C
1m
G
0.5 m
W
1
Problem 6.59 Determine the forces on member BC
and the axial force in member AC.
0.3 m
0.5 m
800 N
B
0.4 m
C
A
Solution:
Element BC: The sum of the moments about B:
M = (0.3)800 + (0.8)C = 0,
300 m
fr
Problem 6.79 What are the magnitudes of the forces
exerted by the pliers on the bolt at A when 30-lb forces
are applied as shown? (B is a pinned connection.)
6 in
45
2
in
30 lb
B
A
30 lb
Solution:
Ele
4.100 Separate streams of air and water flow through the compressor and heat exchanger
arrangement shown in Fig. P4.100. Steady-state operating data are provided on the figure. Heat
transfer with the
PROBLEM. 5. a
known: A P.uqr .75. I .d . nun-'5': pun"- eycll. R opcmh hit-Jr
x: Inn: has IIICIVOfr-i. F3:- +hu cycle; )I'gn-
nub: 3h. that tug 1 um but irrewnrbh.
nnumvnc i sure b :
5mm. Mona:
PROBLE n4 4.106
A simple gas turbine power cycle operating at steadv
state with air as the working substance is shown in Fig.14. IDG.
The cycle components include an air compressor mounted i
on the sa
PEoBlE M 5- 1'
5.9. Use the Kelvin~Planck statement of the second law to
show that the specied process is irreversible.
(a) As shown in Fig. P5911, a hot thermal reservoir is
separated from a cold the
W
5.3 Classify the following processcs of a. closed system as
passiblc. impossible. or mdererminate.
Entropy Entropy Entropy
Change Transfer Production
(2) >0 0
(b) <0 >0
(c) 0 >0
(d) >0 >0
(e) 0
m1
247.5 ft 3
ft 3
13.75
lb
= 18.0 lb
Solving for the final mass
m2 = 18.0 lb + 2.9 lb = 20.9 lb
The specific volume at the final state is
V
247.5 ft 3
v2
= 11.84 ft3/lb
m2
20.9 lb
The energy rate b
I _ _PRoBL& 5.7.I . .,._._.
W Pada- nrc. provided cfw_or a. rents-bl: Peworcyclc open-n: bsan
, , krl- And cud. rotor-Mfr: at cfw_cupcnhru' an-.475, ' ,
Egg: V bake-mac +0". can-37 rrJ' rueJ, w. 3', m
2
(m2u2 m1u1) Qcv hedm
1
2
Qcv m2u2 m1u1 hedm
1
where m denotes the mass contained within the tank.
Since the specific enthalpy he is nearly linear with the mass in the tank, the average value of he
2