Third Class
Power Engineering
Edition 2.0 – June 2010
Part A1
Chapter Answer Guides
Updated: August 2010
Chapter 1: Algebraic Operations, Logarithms, and Problem Solving
End of Chapter Answers
1. Find the logarithm of the following numbers. Round the answers to three decimal places.
a)
b)
c)
d)
e)
f)
g)
log 495 = 2.695
log 32000 = 4.505
log 0.006 = − 2.222
log 0.2275 = − 0.643
log 86 = 1.934
log 2.85 × 1012 = 12.455
log10.75 × 10−8 = − 6.969
2. Briefly state the rules for the multiplication and division of negative quantities.
( + )( + ) = +
( + )( − ) = −
( − )( − ) = +
+ /+ = +
+ /− = −
− /+ = −
− /− = +
3. Simplify the following term:
x-1 x-2
+
8
16
x −1 x − 2
+
8
16
2 ( x − 1) x − 2
=
+
2 ( 8)
16
=
2x − 2 x − 2
+
16
16
2x − 2 + x − 2
=
16
3x − 4
=
16
=
( Ans. )
Page 3
4. Find the number for the following Naperian logs.
a)
ln −1 ( 5.6870 ) = 295.0
b)
ln −1 ( −0.2633 ) = 0.7685
c)
ln −1 (1.4536 ) = 4.278
d)
ln −1 ( 7.2807 ) = 1452
5. In the equation 0.005 + D = 0.055, what is the value of D?
0.005 + D = 0.055
D = 0.055 − 0.005
= 0.05
(Ans.)
6.
The formula used for changing readings on the Fahrenheit thermometer into the
corresponding readings on the Celsius thermometer is: F = 9/5 C +32. Find the Celsius
reading corresponding to 32ºF.
9
F = C + 32
5
9
C = F − 32
5
5
C = ( F − 32 )
9
5
= ( 32 − 32 ) = 0
9
=0
( Ans. )
F = 32
7. Find the value of x in the fractional equation
x
3x
4.5
.
=
+
3
2
4
x 3 x 4.5
+
=
2 4
3
6 ( x ) 3 ( 3 x ) 4 ( 4.5 )
+
=
6 ( 2) 3 ( 4)
4 ( 3)
6 x 9 x 18
+
=
12 12 12
∴
6 x + 9 x = 18
15 x = 18
18
x=
15
= 1.2
( Ans. )
Page 4
8. Find I1, if I2 = 400, and D = 35, using the following formula:
D = 10 log (I1/I2)
⎛I ⎞
D =10 log ⎜ 1 ⎟
⎝ I2 ⎠
I 2 = 400
D = 35
⎛ I ⎞
10 log ⎜ 1 ⎟ = 35
⎝ 400 ⎠
⎛ I ⎞ 35
log ⎜ 1 ⎟ = = 3.5
⎝ 400 ⎠ 10
I1
= log −1 3.5 = 3162
400
I1 = 400 ( 3162 )
= 1, 264, 800
( Ans. )
Page 5
Chapter 2: Trigonometry
End of Chapter Answers
1. Complete the following conversions:
a)
46
'
60
= 900 25.767 '
900 25'46" = 900 25
25.767
degrees
60
= 90.430
( Ans.)
= 90
b)
1400 =
c)
d)
140
× 2π rad = 2.44rad
360
( Ans. )
5.2
× 3600 = 2980
2π
( Ans. )
5.2rad =
30
= 160 50.5'
60
50.5
= 160
60
= 16.840
160 50'30" = 160 50'
( Ans.)
345
× 2π rad = 6.02rad
360
( Ans. )
3.1
× 3600 = 177.60
2π
( Ans. )
e)
3450 =
f)
3.1rad =
Page 6
2. State the sine, cosine, and tangent for each of the following angles, ensuring the correct sign
is attached.
Angle
Sine
Cosine
Tangent
−0.454
0.891
−0.510
92
0
0.999
−0.0349
−28.6
c)
66
0
0.914
0.407
2.25
d)
880
0.999
0.0349
28.6
e)
1420
0.616
−0.788
−0.781
f)
290.70
−0.935
0.353
−2.65
g)
2470
−0.920
−0.391
2.36
h)
165
0
0.259
−0.966
−0.268
180
0
0
−1
0
0
−0.5
−0.866
0.577
a)
b)
i)
j)
333
210
0
3. For what angles, between 0° and 360°, is each of the following function values true?
a)
sin -1 ( -0.766 ) = 3100 or 2300
b)
cos -1 ( 0.7547 ) = 410 or 3190
c)
tan -1 ( 0.2679 ) =150 or 1950
d)
tan -1 ( -1.2349 ) = 3090 or 1290
e)
cos -1 ( -0.9659 ) =1650 or 1950
f)
sin -1 ( 0.1219 ) = 7 0 or 1730
4. For each of the following, state the equation and attach a labeled triangle that contains all
the components in the equation: (a) Pythagoras’ Theorem, (b) Sine Rule, (c) Cosine Rule.
a)
c2 =a2 +b2
Page 7
b)
a
b
c
=
=
Sin A SinB SinC
a 2 =b 2 +c 2 - 2bc cos A
c)
b 2 = a 2 +c 2 - 2ac cos A
c 2 = b 2 +a 2 - 2ab cos A
5. For each of the following triangles, determine the lengths of all sides and the sizes of all
angles. Show all your workings and include a labeled sketch of each triangle. Assume
normal labeling convention.
a)
b= 622 -522 cm
= 33.8cm
⎛ 52 ⎞
A = sin −1 ⎜ ⎟ = 57 0
⎝ 62 ⎠
B = 900 − 57 0 = 330
Sin A=
b=33.8cm
A=570
52
62
B=330
( Ans.)
Page 8
b)
R = 350
r 2 = 622 +242 -2 ( 62 )( 24 ) cos350
r = 44.5m
r
s
=
sinR sinS
24sin 350
∴ sin S =
44.5
= 0.309
S = sin -1 0.309 = 180
T=1800 - ( 180 +350 ) =127 0
r = 44.5m
c)
S =180
T =127 0
( Ans. )
75cm
110cm
=
sin B
sin 720
75sin 720
110
= 0.6484
sin B =
B = sin -1 0.6484 = 40.40
C =1800 - ( 40.40 +720 )
= 67.60
c
a
=
sin C sin A
110sin 67.60
C=
sin 720
B = 40.4
0
C = 67.6
0
=107 cm
c =107 cm
( Ans.)
6. A ramp from the ground to a loading dock is 15 m long. The ramp starts at 12.8 m from the
dock. A stairway from the dock to an overhead storage room has an angle of inclination of
70°, tread depth of 20 cm, and a total of 32 steps. How high is the room above the ground?
Note: This is the question as written in the June 2010 printing of Edition 2.0. Although the
question is mathematically workable, as it stands the answer is 25.4 m and only a giant
could work at this loading dock (each step is 0.55 m high). The question was revised in the
“3rd Class Corrections” and the solution is revised as follows:
Page 9
6. A ramp from the ground to a loading dock is 15 m long. The ramp starts at 14.8 m from the
dock. A stairway from the dock to an overhead storage room has an angle of inclination of
45°, tread depth of 20 cm, and a total of 32 steps. How high is the room above the ground?
h1 = 152 -14.82 m = 2.4 m
For each step of the stairway
y
= tan450
0.2m
So y = 0.2 m x tan 450 = 0.2 x 1.0 m = 0.2 m
∴ h 2 = 32y = 32 x 0.2 m = 6.4 m
h = h1 + h 2 = 2.4 + 6.4 m
= 8.8m ( Ans. )
7. A wall, 8 m high when upright, is leaning forward at an angle of 72° to the horizontal, in
danger of falling over. Support braces, are holding the wall from tipping further. The
braces are attached at 1.5 metres from the top of the wall and make an angle of 52° to the
horizontal. How long are the braces and how far are they anchored from the base of the
wall?
Using the sine law,
6.5 m
x
=
0
sin 56
sin 520
6.5 x 0.829 m
x=
= 6.84m ( Ans. )
0.788
A
6.5 m
=
D
sin 72
sin 52D
6.5 x 0.951 m
A=
= 7.84 m (Ans.)
0.788
Page 10
8. A steel stack is 85 m high and is stayed by two sets of guy wires using the same anchor.
One set is longer, at 63 m, and makes an angle of 51° with the horizontal. The other set is
attached 40 m from the top of the stack. Find: (a) the distance from the top of the stack at
which the longer set is attached, (b) the distance of the anchor from the bases of the stack,
and (c) the angle that the shortest set makes with the horizontal.
a)
l = 85 m - 63 x sin 510 m
= 85 − 63 x 0.7771
= 36.0m from the top
b)
x = 63 x cos 510 m
= 63 x 0.6293m
= 39.6m
c)
( Ans. )
( Ans.)
h = 85m - 40m = 45m
45
= 1.1363
tan θ =
39.6
θ = tan -1 1.1363
= 48.6o
( Ans.)
Page 11
Chapter 3: Mensuration
End of Chapter Answers
Show all working.
1. Calculate the area, in square metres, of each of these plane figures:
a)
A=A R +A T
1
( 80 × 140 ) cm 2
2
2
A=16800cm ⎡⎣ or 1.68 m 2 ⎤⎦
(Ans.)
A= ( 140 × 80 ) cm 2 +
π
π
Dd =
4
4
( 90cm × 80cm ) =5655cm 2
b)
A=
c)
A= s ( s-a )( s-b )( s-c )
s=
⎡⎣or 0.5655m 2 ⎤⎦
(Ans.)
48 + 70 + 60
=89m
2
A= 89 ( 89-48 )( 89-70)( 89-60 )m 2
A=1418m 2
(Ans.)
Page 12
2. A horizontal firetube boiler contains water between the tube sheets, to a depth of 100 cm. If
the shell is 150 cm diameter and the distance between tube sheets is 5 m, find (a) the area of
tube sheet below the water level, and (b) the volume of the space above the water level.
a)
4
d
A below = h 2
-0.608
3
h
4
150
= 1002 cm 2
- 0.608
3
100
=12593cm 2
⎡⎣or 1.26m 2 ⎤⎦ (Ans.)
π 2
2
2
b) A above = d -12593cm = 5078cm
4
Vabove = A above × l
= 5078cm 2 × 500cm
= 2539000cm 3 ⎡⎣or 2.54m 3 ⎤⎦ (Ans.)
3. Compare the volume of an octagonal pyramid (that is, the base is an octagon), each side of
the base being 7.5 metres long, with the volume of a cone that has base diameter of 8 m, if
the height of both is 5 m.
1
V= A baseh where h is vertical altitude.
3
For the octagonal pyramid, A base = 4.83s 2 m 2 = 4.83 ( 7.5 ) m 2
2
= 271.7 m 2
271.7 m 2 × 5m
so, Voct =
= 452.8m 3
3
π
π
2
For the cone A base = d 2 = ( 8 ) m 2
4
4
= 50.3m 2
so, Vcone =
50.3m 2 × 5m
= 83.8m 3
3
The volume of the octagonal pyramid is 452.8 – 83.8 = 369 m2 greater. (Ans)
Page 13
4. A straight, horizontal, 300 mm diameter equalizing pipe connects two vertical, cylindrical
storage tanks, which are 5 m apart, at 1 m from the ground. With the equalizing valve in the
line closed, the level in tank A is 3.5 m and the level in tank B is 6 m. Find the total
volume of liquid in the system (that is, tanks plus line) if the diameters of tank A and B are
7 m and 12 m, respectively.
V = VA + VP + VB
π
π
π
2
2
2
( 7m ) ( 3.5m ) + ( 0.3m ) ( 5m ) + (12m ) ( 6m )
4
4
4
π
= ⎡⎣( 49m 2 ) ( 3.5m ) + ( 0.09m 2 ) ( 5m ) + ( 144m 2 ) ( 6m ) ⎤⎦
4
= 813.6m 3 (Ans.)
=
5. A circular tunnel is lined with concrete that has an average thickness of 500 mm. The tunnel
is 120 m long, with an inside radius of 3 m. (a) What volume of concrete must be poured for
the liner, and (b) how many litres of sealant are required to totally cover the inside surface, if
each litre covers 8 m2? Hint: Consider the surfaces of the tunnel wall as a small cylinder
inside a larger cylinder.
Let: do = outside diameter of the liner = 7 m; di = inside diameter of liner = 6 m
And: Ao = outside cross-sectional area of the liner, Ai = inside cross-sectional area
Then: A = actual cross-sectional area of the liner
a)
A = A 0 -A i
π
π
do 2 - di 2 ) = ( 49m 2 - 36 m 2 )
(
4
4
2
= 10.2 m
=
V = area x length = 10.2m 2 × 120m
=1224m 3
(Ans.)
Page 14
b)
Surface area = π d l
= π ( 6m )( 120m ) = 2262m 2
Vol. of sealant =
2262m 2
= 283 litres
8m 2 /litre
(Ans.)
6. A large, open rectangular raw water reservoir has evenly sloped (inwards) sides that are 5 m
from top to bottom. The reservoir is 4 m deep (vertically). The dimensions of the bottom
are 12 m x 28 m, and the top is 18 m x 34 m. (a) What is the maximum storage capacity of
the reservoir, and (b) what area of liner would be required to completely cover the side and
bottom surfaces of the reservoir ?
a)
(
)
h
A+a+ Aa
3
A = 34 × 18m 2 = 612m 2
h = 4m
V=
(
4m
612m 2 + 336m 2 +
3
= 1869m 3 (Ans.)
V=
a = 28 × 12m 2 = 336m 2
( 612 )( 336 ) m 2 )
b)
For the bottom, A = 12 x 28 m 2 = 336 m 2
⎛ 18m +12m ⎞
2
For one end, A End = ⎜
⎟ × 5m = 75m
2
⎝
⎠
⎛ 34m + 25m ⎞
2
For one side, A side = ⎜
⎟ × 5m =155m
2
⎝
⎠
A Total = ( 2 x 75m 2 ) + ( 2 x155m 2 ) + 336m 2
= 796m 2 (Ans.)
Page 15
7. What is the surface area of each hemispherical head on a pressure vessel if the overall
length of the vessel is 21 m and the outside diameter is 3.5 metres?
A head =
=
1
x 4 π r2
2
where r = 1.75m
4π ( 1.75m )
2
2
=19.2m 2 (Ans.)
8. 10 litres of protective coating covers 10 m2 of surface area. Which hydrocarbon storage
vessel would require the greatest amount of coating and by how much: a butane bullet in
the form of a horizontal cylinder with hemispherical heads, diameter 5 m and length, not
including heads, of 20 m OR a propane sphere, with radius of 6 m?
Let: AB = total surface area of bullet; AH = surface area of heads; AC = surface area of
cylindrical section; AS = surface area of sphere.
A B =A H +A c
A c = π d l = π × 5m × 20m = 314.2m 2
A H = 4 π rB 2 = 4π ( 2.5m ) = 78.54m 2
2
A B = 314.2m 2 + 78.54m 2 = 392.74m 2
A s =4πrs 2 = 4π ( 6m ) = 452.4m 2
2
Since 10A covers 10 m 2 then 1A covers 1 m 2
Bullet requires 392.74 A
Sphere requires 452.4 A
So the sphere requires 59.7 A more coating than the bullet.
Page 16
Chapter 4: Forces and Friction
End of Chapter Answers
3. A vehicle travels 50 km due east from point A to point B, then 100 km in a direction 30°
north of east to point C. Determine the resultant displacement of C from A (distance and
angle).
AB x = 50 km
AB y = 0
x component of BC = BCx = 100 km x cos 30° = 86.6 km
y component of BC = BCy = 100 km x sin 30° = 50 km
x component of AC = ACx = 50 km + 86.6 km = 136.6 km
y component of AC = ACy = 50 km
AC2 = (136.6)2 + (50)2
AC =
(136.6)2 + (50)2 km 2
= 145.5 km
50
= 0.366
tanθ =
136.6
θ = tan -1 0.366 = 20.1o
Therefore : AC is 145.5 km at 20.1o N of E
(Ans.)
Page 17
2. Four coplanar, concurrent forces act on a body so that the body is in equilibrium. If three of
the forces are 100 N due east, 150 N due north and 50 N 60° south of east, calculate the
magnitude and direction of the fourth force (the equilibrant). Include a vector diagram.
F1x = 100 N
F1y = 0 N
F2x = 0 N
F2y = 150 N
F3x = 50 N x cos 60° = 25 N
F3y = - 50 N x sin 60° = - 43.3 N
Find the fourth force, which is the resultant of the three forces:
Resultant, Rx = 100 + 0 +25 N = 125 N
Resultant, Ry = 0 + 150 – 43.3 N = 106.7 N
R = (125)2 + (106.7)2 N 2
= 164.3 N
106.7
= 0.8536
125
θ = tan -1 0.8536 = 40.5 o N of E
tanθ =
The equilibrant is therefore 164.3 N at 40.5° S of W
Page 18
3. Calculate, by the component method, the equilibrant of the following four coplanar,
concurrent forces:
FA = 800 N acting at 20° north of east
FB = 600 N acting at 50° north of west
FC = 1000 N acting at 65° south of east
FD = 300 N acting at 30° south of west
FAx = 800 N x cos 20° = 752 N
FAy = 800 N x sin 20° = 274 N
FBx = - 600 N x cos 50° = -386 N
FBy = 600 N x sin 50° = 460 N
Fcx = 1000 N x cos 65° = 423 N
Fcy = -1000 N x sin 65° = -906 N
FDx = -300 N x cos 30° = -260 N
FDY = -300 N x sin 30° = -150 N
Rx = 752 – 386 + 423 - 260 N = 529 N
Ry = 274 + 460 - 906 - 150 N = -322 N
R = (529)2 + ( -322 ) N
2
= 619 N
θ = tan -1
322
= 31.3o S of E
529
The equilibrant is therefore
(Ans.)
619N at 31.3° N of W
4.
(a) Define coefficient of friction and explain each term in the formula.
μ=
FF where FF is the frictional force
R N and R N is the normal reaction force.
(b) A casting with a mass of 100 kg is pulled along a horizontal floor at constant speed by
a horizontal force. If the coefficient of friction is 0.25, what force will be required?
What will be the angle of friction?
FW = 100 kg × 9.81 m/s2 = 981 N
RN= FW = 981 N
FF = μRN = 0.25 × 981 N = 245 N
(Ans.)
θ = tan –1 μ = tan –1 0.25 = 14 °
(Ans.)
Page 19
5.
A pump with a mass of 100 kg is pulled along a horizontal floor by means of a rope
attached to one end. The rope makes an angle of 30° above the horizontal and the
coefficient of sliding friction between the pump and the floor is 0.4. What force must be
exerted on the rope to keep the pump sliding at constant speed?
For horizontal equilibrium:
FA cos 30° = FF
0.866 FA = FF
For vertical equilibrium:
RN + FA sin 30° = FW
= 981 N
RN + 0.5 FA
Also FF = μ RN = 0.4 RN
Therefore
FF
+ 0.5 FA = 981 N
0.4
0.866 FA
+ 0.5 FA = 981 N
0.4
2.665 FA = 981 N
FA = 368 N
(Ans.)
Page 20
6.
A body has a mass of 295 kg. It is pulled along a horizontal floor at constant velocity by a
force of 65 N applied at an angle of 35° above the horizontal. What is the coefficient of
sliding friction between the body and the floor?
2894 N
Horizontal equilibrium:
Vertical equilibrium:
FA sin 35o + R N = 2894 N
FA cos 35o = FF
FF = 65 N x cos35
65 N x 0.5736 + R N = 2894 N
o
R N = 2894 - 37 N
= 65 x 0.8191N
= 2857 N
= 53.2 N
FF = μR N
μ=
7.
FF
53.2
=
= 0.0186
R N 2857
(Ans.)
Find the value and applied angle of the least force required to pull a vessel if the mass of
the vessel is 1650 kg and the coefficient of friction is 0.49.
The least force occurs when FA is directed at the friction angle, θ
θ = tan -1 μ
= tan -1 0.49 = 26.1o This is the direction of the force, FA (Ans)
For horizontal equilibrium: FA cos 26.1° = FF
0.898 FA = FF
For vertical equilibrium:
RN + FA sin 26.1° = FW = 1650 × 9.81N
RN + 0.44 FA = 16187 N
But FF / 0.49 = RN
∴
and FF = 0.898 FA (from above)
0.898 FA
+ 0.44 FA = 16187 N
0.49
1.833 FA + 0.44 FA = 16187 N
2.273 FA = 16187 N
FA = 7122 N = 7.12 kN
(Ans.)
Page 21
8.
A load of 750 kg just starts to move when the friction angle is 10°. What effort must be
applied parallel to the surface to move the block?
θ = 10o ∴ tan 10o = μ = 0.176
For horizontal surfaces and horizontal forces :
FW = R N = 750 × 9.81 N = 7358 N
FA = FF = μR N = μFW
∴ FA = 0.176 × 7358N
= 1295 N
(Ans.)
Page 22
Chapter 5: Work, Power, Energy; Linear & Angular Motion
End of Chapter Answers
Show all steps for the following problems. Answers alone will not be accepted.
1. (a) Explain how the pressure in the cylinder of a reciprocating engine can result in the
production of power.
F
∴ F = Pm A
A
W = Fl = Pm Al per stroke
Pm =
W
t
Power = Pm AlN where N is the power strokes per second
Power =
(b) An engine has 4 cylinders, each with 300 mm diameter and piston stroke of 500 mm.
Effective pressure is 1200 kPa, operating speed is 300 r/min and efficiency is 0.82. If
there are two power stroke per revolution, calculate the output power of the engine?
Power = Pm A l N x Efficiency
kN
π
300
x
x (0.3m)2 x
x 8 x 0.5 m x 0.82
2
m
4
60 sec
= 1391.2 kW
(Ans.)
= 1200
2. (a) Define linear velocity, linear acceleration, and average velocity and state their units.
displacement
time
change in linear velocity
linear acceleration =
time
distance
average velocity =
time
linear velocity =
m
s
m
s2
m
s
Page 23
(b) A truck accelerates uniformly from rest to 48 km/h in 20 seconds, holds this speed for
2 minutes, and then accelerates to 100 km/h at a rate of 1.7 m/s2. How long does it
take, from rest, to reach 100 km/h and what distance is traveled in this time?
u=o
km
m
= 13.3
h
s
km
m
v 2 = 100
= 27.8
h
s
v1 = 48
i) From 0 to 48 km/h :
t = 20s
⎛ u + v1 ⎞
s1 = ⎜
⎟t
⎝ 2 ⎠
⎛ 0 + 13.3 ⎞
=⎜
⎟ 20 m
2
⎝
⎠
= 133 m
ii) During the 2 minute (t = 120 s) interval, at 48 km/h, the truck travelled
s int = 13.3 m/s × 120 s = 1596 m
iii) From 48 km/h to 100 km/h
a 2 = 1.7 m/s 2
v 2 = v1 + a 2 t 2
27.8 = 13.3 + 1.7 t 2
27.8 - 13.3
s
1.7
= 8.53s
t2 =
⎛ v + v2 ⎞
⎛ 13.3 + 27.8 ⎞
s2 = ⎜ 1
⎟ 8.53 m = 175 m
⎟t =⎜
2
⎝ 2 ⎠
⎝
⎠
t total = 20 s + 120 s + 8.53 s
= 148.53 s
(Ans.)
s total = 133m + 1596 m + 175 m
= 1904 m
(Ans.)
Page 24
3. A body is blasted upwards with an initial vertical velocity of 80 m/s. If the mass of the
body is 15 kg, find: (a) the maximum height reached and (b) the kinetic energy when the
body is descending at a height of 100 m.
a)
u = 80 m/s
a = - g = - 9.81 m/s 2
v = 0 at Apex
Going to the Apex (maximum height)
v = u + at
o = 80 - 9.81 t
-80
t=
= 8.15 s
-9.81
⎛ u + v ⎞ ⎛ 80 + 0 ⎞
s=⎜
⎟t = ⎜
⎟ 8.15 m = 326 m
⎝ 2 ⎠ ⎝ 2 ⎠
(Ans.)
b) At 100 m the body has fallen 226 m. Its loss of potential energy is
Ep = mgh = 15 x 9.81 × 226 J = 33256 J
The gain in Ek = loss in Ep = 33256 J (Ans.)
4. Calculate the velocity at impact with a 10 m high platform if a 60 kg object is dropped onto
the platform from a height of 35 m above the ground.
On impact the object has fallen (35-10)m = 25 m
Its gain in Ek will be equal to the loss in Ep
mv 2
= mgh
2
v = 2gh
= 2 × 9.81 × 2.5
= 22.1 m/s
m
s
(Ans.)
An alternative solution:
v 2 = u 2 + 2as
v 2 = 02 + 2 × 9.81 × 25
= 490.5
m2
s2
m2
s2
m2
s2
= 22.1 m/s (Ans.)
v = 490.5
Page 25
5. (a) Define and state the formulae for angular velocity and angular acceleration.
angular dispacemet (radians)
time (seconds)
θ
ω=
t
change in angular velocity (rads/s)
angular acceleration =
time (s)
w - w1
α= 2
t
angular velocity =
(b) Power is removed and a brake is applied to a heavy, high-speed shaft when it is turning
at 6000 r/min. How long will it take the shaft to come to rest if the average retardation
is 0.9 rad/s2, and what is the average angular velocity during that time?
rev
2π
= 6000 ×
rad/s = 628 rad/s
min
60
rad
ω2 = 0
α = - 0.9 2
s
ω2 = ω1 + α t
ω1 = 6000
0 = 628 + (-0.9)t
0 - 628
s = 698 s (Ans.)
-0.9
628
ω - ω2
= 1
=
rad/s = 314 rad/s (Ans.)
2
2
t=
ωav
6. A projectile is fired vertically upwards and its average velocity before reaching maximum
height is 200 m/s. Find: (a) the maximum height attained, (b) the velocity after 15 seconds,
and (c) the time required to fall from its maximum height to the ground.
a)
At the apex v = 0 vav = 200 m/s a = g = -9.81 m/s2
u+v
∴ vav =
so u = 400 m/s
2
v 2 = u 2 + 2 as
0 = 4002 + 2 × (-9.81) × s
-4002
m
2 x (-9.81)
= 8155 m (Ans.)
s=
Page 26
b) The time to reach the apex is found from:
v = u + at A
0 = 400 + (-9.81) t A
tA =
-400
= 40.7s
-9.81
Therefore, at t = 15s the object is still rising
v15 = u + at
= 400 + (- 9.81) (15) m/s
= 253 m/s
(Ans.)
c)
tfall = trise = 40.7s
(Ans.)
7. A 1600 kg car, carrying an 80 kg driver, is traveling at 140 km/h when the radar detector
begins to beep. The driver brakes the car to 90 km/h in 3 seconds. Determine (a) the
distance traveled during the braking (b) the deceleration, and (c) the change in kinetic
energy.
a)
u = 140 km/h = 38.9 m/s
v = 90 km/h = 25 m/s
⎛u +v⎞
⎛ 38.9 + 25 ⎞
s=⎜
⎟t =⎜
⎟ 3 m = 95.8 m
2
⎝ 2 ⎠
⎝
⎠
(Ans.)
b)
v - u 25 - 38.9
=
m/s 2 = - 4.63 m/s 2 (Ans.)
t
3
[or deceleration of 4.63 m/s 2 ]
a=
c)
1
2
= - 746 kJ
1
2
(Ans.)
Δ Ek = m (v 2 - u 2 ) = × 1680 × ⎡⎣(25)2 - (38.9)2 ⎤⎦ J
Page 27
8. A firewater pump delivers 8000 tonnes of water in 12 hours from a well to overhead storage
tanks. The total height through which the pump must move the water is 35 m.
(a) How much work is required to move the water?
F = 8 000 000 kg × 9.81 m/s2 = 78 480 000 N = 78 480 kN
w = Fx
= 78 480 kN × 35 m
= 2 746 800 kJ
= 2747 MJ
(Ans.)
(b) How much driving power, in kW, must be available to the pump if the combined
efficiency of the pump and the piping is 70%?
Powerout =
w
2746800 kJ
=
= 63.6 kW
t
12 × 60 × 60 s
Powerout
η =
Powerin
63.6 kW
0.7
= 90.9 kW
Powerin =
(Ans.)
Page 28
Chapter 6: Strength of Materials; Bending of Beams
End of Chapter Answers
1.
Explain the following terms: a) modulus of elasticity, b) ultimate strength, c) factor of
safety, d) bending moment e) shear force on a beam f) beam equilibrium g) uniformly
distributed load h) cantilever i) strain
a) E =
σ
ε
E is the constant of proportionality in Hooke’s Law.
b) Ultimate Strength or the maximum stress of material can withstand before
breakage occurs.
c) Factor of Safety – the ratio of ultimate stress to the allowable working stress.
d) Bending Moment – the algebraic sum of the moments too the left or right of a
point (or section line) or a beam.
e) Shear Force – the force acting parallel to the shear surface which causes a beam to
shear. (or, the algebraic sum of the vertical forces to the right or left of the section
line being considered).
f) Beam Equilibrium – is a state in which a beam does not move under the
application of loads. To be in equilibrium: i) downwards forces must equal
upwards forces, ii) forces to the left must equal forces to the right, iii) clockwise
moments must equal anticlockwise moments.
g) Uniformly distributed load - load which is evenly distributed along a length of a
beam
h) Cantilever – a beam that has one fixed end and the other end free to move.
i) Strain – the ratio of the change in length to the original length, of an object
experiencing a load.
2.
A steel rod, 4 m long, with a cross-sectional area of 30 cm2, is stretched 0.07 cm by a
tensile load. Calculate: a) the stress in the rod and b) the load, in kN
a)
Δl 0.07cm
=
= 0.000175
L 400cm
= 200 GPa, then
ε=
Since Esteel
σ = Eε = 200 GPa × 0.000175
= 35 MPa
b)
σ=
Load
A
(Ans.)
so Load = σA = 35
MN
30
×
m 2 = 105kN (Ans)
2
m
10000
Page 29
3.
Determine the reactions on a simple beam that is 12 m long and carrying concentrated
loads of 20 kN at 4 m from the left support and 41 000 N at 1 m from the right support.
Consider in your calculation the load due to the beam itself, which is 1200 N/m over the
length of the beam.
Consider anticlockwise moments = clockwise moments about R1
R 2 × 12m = (20 kN × 4 m ) + (41 kN × 11m) + (1.2 kN/m × 12 m × 6 m )
R 2 × 12m = 617.4 kNm
617.4
12
= 51.45 kN
R2 =
(Ans.)
Considering upward forces = downward forces (ie. vertical equilibrium):
R 1 + R 2 = 20 kN + 41 kN + (1.2 kN/m × 12 m)
R 1 + 51.45 kN = 75.4 kN
R 1 = 75.4 kN − 51.45 kN
= 23.95 kN
(Ans.)
Page 30
4. A simple beam is 15 m long and has a uniformly distributed load of 35 000 N. It also has a
concentrated load of 30 kN located at 5.5 m from the left support. Calculate the shearing
force and the bending moment at 6.5 m from the left support.
Since clockwise moments = anticlockwise moments about R1
R 2 × 15m = (30 kN × 5.5 m ) + (35 kN × 7.5m)
R 2 × 15m = 427.5 kNm
427.5 kNm
15m
= 28.5 kN
=
Considering upward forces = downward forces::
R 1 + R 2 = 30 kN × 35 kN
R 1 + 28.5 kN = 65 kN
R 1 = 65 - 28.5kN
= 36.5 kN
(Ans.)
Shearing Force:
Taking forces to the right of the section line:
SF = 2.33 kN/m × 8.5 m - R2 = 19.8 kN – 28.5 kN
= -8.7 kN (Ans.)
Bending Moment:
Taking moments to the right of the section line:
⎛ 8.5 ⎞
B.M. = 2.33 kN/m × 8.5 m × ⎜
⎟ m - R 2 × 8.5 m
⎝ 2 ⎠
= 84.2 kNm - (28.5 kN × 8.5 m)
= - 158 kNm
(Ans.)
Page 31
5.
Two bolts, 19 mm diameter at the bottom of their threads, are supporting a load of 7500 kg
from an overhead beam. If the allowable stress on the bolt material is 113 000 kPa and the
factor of safety is 6.6, state whether or not the bolts are sufficient to carry the load and
show your calculation to support your conclusion.
Load = F = 7500 kg × 9.81 m/s 2 = 73575 N
73575
Fbolt =
N = 36787.5 N
2
F
36787.5 N
σ bolt = bolt =
2
A bolt π
( 0.019m )
4
= 129749 kN/m 2
The ultimate (breaking stress) of the bolt is
σ uts = 6.6 × 113 000
kN
kN
= 745 800 2
2
m
m
CONCLUSION: The bolts are sufficiently strong to carry the load, but could not
be used to carry the load, since their actual stress (129,749 kN/m2) exceeds the
allowable stress (113,000 kN/m2) (Ans.)
6.
A steam engine has a piston 20 cm diameter and a piston rod 4 cm diameter. If the ultimate
strength of the rod material is 400 000 kPa and the factor of safety for the rod is 12,
calculate the maximum allowable steam pressure in the cylinder.
400 000 kPa
= 33 333 kPa
12
kN π
2
FROD = σ allow × A ROD = 33 333 2 × × ( 0.04m )
m
4
= 41.9 kN
FPISTON = FROD
σ allow =
P=
FPISTON
41.9 kN
=
= 1334 kPa
π
2
A PISTON
× ( 0.2 )
4
(Ans.)
Page 32
7.
A cantilever beam is 8 m long. The free end of the beam is supported by a cable from
above. A load of 300 kN hangs from the beam at 3 m from the wall. A load of 400 kN
hangs from the free end of the beam and the first 3 m from the wall has a distributed load
of 200 kN/m. a) What is the tension in the supporting cable at the free end? b) What is the
bending moment at the center of the beam?
a)
Taking Moments about the wall:
Since anticlockwise moments = clockwise moments
kN
3m ⎞
⎛
T × 8m = ⎜ 200
× 3m ×
⎟ + ( 300kN × 3m ) + ( 400 kN × 8m )
m
2 ⎠
⎝
T × 8 m = 900 kNm + 900 kNm + 2400 kNm
T = 4200 kNm / 8 m
= 625 kN (Ans.)
b) Taking moments to the right of the beam center section line:
BM = clockwise moments - anticlockwise moments
= ( 400 kN × 4m ) - ( T × 4m )
= 1600 kNm - (625 kN × 4m)
= 1600 - 2500 kNm
= - 900 kNm (Ans.)
Page 33
8.
The strain in a supporting hanger is 0.00045, the factor of safety is 7.0, and the stress in the
hanger is 110 MPa. If the hanger is 4.5 cm diameter and has a loaded length of 7.5 m,
calculate: a) the unloaded length, b) the load on the hanger and c) the ultimate strength of
the hanger.
ε = 0.00045
Fos = 7.0
σ = 110 MPa
L loaded = 7.5 m
Lloaded = Lunloaded + Δl
a)
7500 mm = Lunloaded + Δl
and
ε=
Δl
Lunloaded
= 0.00045
so Δl = 0.00045 Lunloaded
∴
7500 mm = Lunloaded + 0.00045 Lunloaded
7500 mm = 1.00045 Lunloaded
7500mm
= 7497 mm
1.00045
= 7.497 m
(Ans.)
Lunloaded =
b)
F
A
F=σA
σ=
π
× ( 0.045 m
4
(Ans.)
= 110, 000 kN/m ×
= 175 kN
)
2
c) Assume the 110 Mpa is the maximum allowable working stress, then
Ultimate strength, σ uts = FOS × σ allow
= 7.0 × 110 MPa
= 770 MPa
(Ans.)
Page 34
Chapter 7: Simple Machines; Pressure, Density, Flow
End of Chapter Answers
1.
Define the following terms as they apply to simple machines and show the relationship
between them: actual mechanical advantage, ideal mechanical advantage, velocity ratio,
efficiency.
a) actual MA =
load
effort
b) ideal MA = VR
c) VR =
d)
2.
or
ideal MA =
actual MA
efficiency
distance travelled by effort
distance travelled by load
Efficiency =
MA
VR
(a) Sketch a wheel-and-axle arrangement in which the effort moves 8.5 times as far as the
load. Clearly indicate dimensions that will ensure this load-to-effort ratio.
Page 35
(b)
If a mass of 1200 kg is lifted by the machine in (a), and the machine is 82% efficient,
how much effort must be applied and what is the actual mechanical advantage?
8.5
= 8.5
1
Load = 1200 × 9.81 N = 11772 N
VR =
Efficiency = 0.82 =
MA
VR
∴ MA = 0.82 × 8.5
= 6.97
(Ans.)
load
= MA
effort
load 11772 N
effort =
=
MA
6.97
= 1689 N
3.
(Ans.)
Identical screw jacks are used at the four corners of the floor supports of an oil storage
tank to raise the tank a total of 300 mm. Each revolution of a jack screw causes the load to
rise by 11 mm and the total load is shared equally by the jacks. If the mass of the tank and
its supports is 5 tonnes, and the tank contains 20 m3 of oil, with a relative density of 0.8,
how much effort must be applied to each jack, using a handle length of 600 mm? Assume
efficiency = 100%.
m
kg
m
Total Load = 5000 kg × 9.81 2 + (20 m 3 × 800 3 × 9.81 2 )
s
m
s
= 206 010 N
Load on each jack =
VR =
206010 N
= 51 502 N
4
2π × 600
= 342.7
11
At 100% efficiency: MA= VR = 342.7
load
MA
51 502 N
=
342.7
= 150 N
∴ effort =
(Ans)
Page 36
4.
Complete the following conversions:
a)
500 mm Hg
b)
30 kPa vacuum = (101.3 – 30) kPa absolute
= 71.3 kPa absolute
(Ans.)
c)
= 500 mm Hg × 13.6 mm H2O/mm Hg
(Ans.)
= 6800 mm H2O
10 in. H 2O =
10 in H 2O
13.6 in H 2O/in Hg
= 0.735 in. Hg
d)
e)
(Ans.)
650 mm H2O = 0.65 m H20
kg
N
p = hρg = 0.65 m × 1000 3 × 9.81
m
kg
2
= 6376 N/m
= 6376 Pa = 6.38 kPa (Ans.)
1 mmHg = 0.133 kPa
So: 16 kPa = 16 / 0.133 mmHg = 120.3 mmHg (Ans)
Page 37
5.
A transfer/mixing valve between two storage tanks is closed. The valve is located at a
height of 1.2 m from the bottom of the tanks. One tank has a level of 6.5 m of a liquid with
a relative density of 3.6 and the other contains a chemical, with density of 1100 kg/m3, to a
level of 3.8 m. What is the difference in pressure between the two sides of the transfer
valve?
pA = hAρAg = 5.3 m × 3600
kg
m
× 9.81 2 = 187.2 kPa
3
m
s
pB = hBρBg = 2.6 m × 1100
kg
m
× 9.81 2 = 28.1 kPa
2
m
s
Δp = 187.2 kPa - 28.1 kPa
= 159.1 kPa
(Ans.)
6.
(a) Explain “continuity of flow”, as it relates to fluid in a piping system.
a) Continuity of flow relates to the conservation of mass entering and leaving a
closed piping system. min = mout
If the fluid is a liquid (and therefore incompressible) then volumetric flow
rate is also conserved, so Qin = Qout
or AinVin = AoutVout.
Page 38
(b) The vertical inlet line to the turbine at a hydro dam is 120 m high and 2.4 m diameter.
1.5 million kg of water flow through the turbine every hour. If the water level is 10 m
above the turbine intake line and the horizontal discharge line is 20 m long and 1.7 m
diameter, calculate (i) pressure at the turbine inlet (ii) velocity in the discharge line.
i.
p = hρg = (120 + 10) m × 1000
= 1275 kPa
ii.
mass flow = 1500 000
kg
m
× 9.81
m3
s2
(Ans.)
kg
kg
= 416.7
h
s
kg
416.6
3
mass flow
s = 0.4167 m
Volume flow, Q =
=
kg
ρ
s
1000 3
m
0.4167
Q = Av so v =
m3
s
π
× (1.7 m)2
4
m
= 0.184
s
7.
(Ans.)
A square manway, 800 mm x 800 mm, is located in the side of a vertical, cylindrical
storage vessel. The vessel is 8 m tall and 8 m diameter and contains a liquid of relative
density 0.75, which is overflowing from the top of the tank. The bottom of the manway is
500 mm from the bottom of the vessel. What load is the liquid putting on the manway?
The center of the manway is located at 500 mm + 400 mm from the tank base. It is
therefore 8 m – 0.9 m = 7.1 m from the top of the tank.
p = hρg = 7.1 m × 750
F = p × A = 52.2
= 33.4 kN
kg
N
× 9.81
= 52.2 kPa
3
m
kg
kN
× (0.8m)2
2
m
(Ans.)
Page 39
8. Show clearly by calculation and explain which of the following would provide the best
advantage in lifting a mass of 6 tonnes. System A: A block and tackle with 9 pulleys, an
efficiency of 80%, and the effort applied upwards; System B: A pulley system with a
velocity ratio of 10, an efficiency of 85% and the effort applied downwards.
System A:
VR = 9 + 1
= 10
since effort is applied upwards
MA = VR × eff
= 10 × 0.8
=8
System B:
MA = VR × eff
= 10 × 0.85
= 8.5
System B provides the best advantage.
(Ans.)
Page 40
Chapter 8: Heat, State Change, Calorimetry
End of Chapter Answers
1. Show step-by-step working for all calculations:
(a) Define “specific heat” and state the three factors that determine the amount of heat
transferred to or from a substance, assuming no change of state.
.
a) specific heat – the quantity of heat required to change the temperature of a unit
mass of the substance by one degree.
Three factors:
mass of the substance
specific heat of the substance
temperature change of the substance
(b) Show clearly, by calculation and comparing results, which of the following processes
would require the greatest transfer of heat: melting 0.5 tonnes of copper from 70°C;
solidifying 3 tonnes of molten aluminum at the fusion temperature; changing 0.75
tonnes of ice at -10°C to water at +10°C.
b)
copper:
Qcopper = 500 kg × 0.388
kJ
kJ
× (1083 - 70)o C + 500 kg × 134
o
kg C
kg
= 263.5 MJ
aluminum:
Qaluminum = -3000 kg × 90 kJ/kg
= - 270 MJ
ice:
Qice = 750 kg × 2.135 kJ/kg°C × [0- (-10)] °C + 750 kg × 335 kJ/kg +
750 kg × 4.183 kJ/kg°C × (10-0) °C
= 298.6 MJ
The process of changing the ice to water requires the greatest transfer of heat.
(Ans.)
2. (a) Define the following terms: sensible heat, latent heat, latent heat of fusion, latent heat of
evaporation
i.
ii.
iii.
iv.
sensible heat – heat which causes a change in the temperature of a
substance
latent heat – heat which causes a change in phase (state) of a substance
latent heat of fusion – the heat required to change 1 kg of a solid substance
to a liquid
latent heat of evaporation – the heat required to change 1 kg of a liquid
substance to a vapour
Page 41
(b) How much heat, in MJ, is added to 1.8 tonnes of ice, at -15°C, if 60% of the ice
becomes steam at atmospheric pressure and 100°C, and the remainder becomes water at
90°C?
Q = 1800 ⎡⎣( 2.135 × 15 ) + 335 + ( 4.183 × 90 ) ⎤⎦ + ( 0.6 × 1800 × 4.183 × 10 ) +
( 0.6 × 1800 × 2257 ) kJ
= 3821 MJ
(Ans.)
3. (a) 4.0 kg of ice are mixed with 60 kg of water, which is at a temperature of 50°C. If the
temperature of the mixture becomes 40°C, calculate the initial temperature of the ice.
Heat loss = Heat Gain
(
)
60 × 4.183 × (50 - 40) = 4 × 2.135 0D C - T + ( 4 × 335 ) + 4 × 4.183 × ( 40 - 0 )
2509.8 = 0 -
8.54 T
+ 1340 + 669.3
D
C
8.54 T
= - 500.5
D
C
T = - 58.6 D C
(Ans.)
(b) If the mixture in ‘(a)’ is injected with 2.5 kg of steam, at atmospheric pressure and
100°C, what temperature will the new mixture attain, assuming no external heat loss?
Heat loss = Heat Gain
( 2.5 × 2257 ) + ( 2.5 × 4.183 × (100 - T ) ) = 64 × 4.183 × ( T - 40 )
5642.5 + 1045.75 - 10.46 T = 267.7 T - 10708
278.2 T = 17396
17 396
T=
278.2
= 62.5 o C
(Ans.)
4. Find the resulting temperature when 30 kg of ice at 200 K are mixed with 10 kg of steam at
373 K and atmospheric pressure. Use specific heat of ice as 2.135 kJ/kgK and specific heat
of water as 4.183 kJ/kgK.
Heat loss = Heat Gain
(10 × 2257 ) + (10 × 4.183 × ( 373 - T ) ) = 30 ⎡⎣( 2.135 × ( 273 - 200 ) ) + 335 + 4183 (T - 273)⎤⎦
22570 + 15603 - 41.83 T = 4676 + 10050 + 125.5 T - 34259
167.3 T = 57706
T = 345 K or 72o C
(Ans.)
Page 42
5. 2.5 kg of brass, with specific heat 0.39 kJ/kgK and temperature of 256°C is dropped into 10
litres of water at 14°C. As soon as thermal equilibrium is achieved, the brass is removed
and a piece of cast iron, specific heat = 0.544kJ/kgK, and temperature of 400°C is placed in
the water. If the final temperature of the water is 24°C, what is the mass of the cast iron?
Ignore heat losses to atmosphere.
Heat loss = Heat Gain
2.5 x 0.39 x (256 - T1 ) = 10 x 4.183 x (T1 - 14)
T1 = 19.5 o C
Heat loss = Heat Gain
m i × 0.544 × (400 - 24) = 10 × 4.183 × (24 - 19.5)
m i = 0.920 kg
(Ans.)
6. (a) Define “water equivalent” and explain the formula used to determine the water
equivalent of a material. (b) A calorimeter has a water equivalent of 0.0065 kg and a
mass of 1.95kg. Calculate the specific heat of the calorimeter material.
a)
Water equivalent – the equivalent mass of water that would require the same
amount of heat transfer as the substance, to produce the same temperature
change
Water equivalent =
mass of substance x specific heat of substance
specific heat of water
This equation is derived from considering the amount of heat transferred to
water and another substance, to obtain the same Δ T.
b)
1.95 x C
kJ
4.183
kg k
kJ
C = 0.0139
kg k
0.0065 =
(Ans.)
Page 43
7. In an experiment to find the specific heat of iron, 2.15 kg of iron at 100°C are dropped into
a calorimeter-like vessel containing 2.3 litres of water at 17°C. If the resultant mixture is
24.4°C and the water equivalent of the vessel is 0.18 kg, what is the specific heat of the
iron?
Heat loss = Heat Gain
2.15 × Ci × (100 – 24.4) = (2.3 + 0.18) × 4.183 x (24.4 – 17)
2.15 × Ci × 75.6 = 2.48 × 4.183 x 7.4
Ci = 76.77 / 162.54
= 0.472 kJ/kgK (Ans.)
8.
A calorimeter test, using a copper calorimeter, records these results: mass of tested
material = 350g; mass of calorimeter = 0.70 kg; mass of water = 0.9 kg; initial temperature
of tested material = 60°C; initial temperature of calorimeter = 13°C; final temperature =
18°C. What is the specific heat of the tested material?
Heat loss = Heat Gain
0.35 × C × (60 – 18) =[0.7 × 0.388 × (18 – 13)] + (0.9 × 4.183 × (18 – 13)
0.35 × C × 42 = (0.2716 × 5) + (0.9 × 4.183 × 5)
14.7 C = 1.358 + 18.824
C = 20.18 / 14.7
= 1.37 kJ/kgK
(Ans.)
Page 44
Chapter 9: Thermal Expansion and Heat Transfer
End of Chapter Answers
Note: For any of the following questions, unless coefficients are indicated in the question, use
the coefficient values given in the tables in the chapter.
1. At 20°C, a section of straight pipeline is 150 m long. When heated to 120°C, the pipe
changes in length by 30 cm. What is the coefficient of linear expansion for the pipeline
material?
ΔL = L α ΔT
ΔL
α=
LΔ T
0.3m
=
150m (120 − 20)D C
= 0.000 02 / D C
or 2 × 10−5 / D C
(Ans.)
2. At 5°C, a steel tank is 6 m high, 9 m in diameter, and contains oil to the 5.9 m level. A
heating coil is put into service, causing 3.4 m3 of oil to spill over the top of the tank by the
time final temperature is reached. Calculate the final temperature, if the coefficient of
expansion for the oil is 4.1 x 10-4/°C. Note: Do NOT neglect expansion of the tank.
α steel = 12 × 10−6 / DC
π
2
Vtank =
Voil =
4
× (9m ) ×6m = 381.7 m 3 at 5 D C
π
× (9m )2 × 5.9m = 375.3m 3 at 5 D C
4
Let the final temperature be T
ΔVtank = 3 × 12 × 10−6 (T − 5) × 381.7 = 0.01374T − .06871
ΔVoil = 4.1 × 10−4 (T − 5) × 375.3 = 0.1539T − 0.7695
Spillage = Voil + ΔVoil − (Vtank + ΔVtank )
3.4 = 375.3 + (0.1539T − 0.7695) − (381.7 + 0.01374T − .06871)
3.4 = 375.3 + 0.1539T − 0.7694 − 381.7 − 0.01374T + 0.06871
3.4 = −7.1 + 0.1402 T
10.5
T=
0.1402
= 74.9 D C
(Ans.)
Page 45
3. A steel steam line, 30 m long at ambient boiler room temperature of 30°C, is stretched by
0.015 m to allow flanges to be tightened. This creates 480 MPa of stress in the pipe. What
must the design operating temperature be to exactly eliminate this stress? Assume the
flanges to which the pipe is connected are rigid (i.e. cannot move).
ΔL = LαΔT
0.015 = 30 × 12 × 10−6 × (T − 30)
T = 71.7D C
4. Which of the following will have the greatest change in volume when exposed to a
temperature change of 90°C and by how much will it be different: a copper drum that is
2.3m diameter by 5m long, or 9 m3 of gasoline?
ΔVcopper =
π
× (2.3m)2 × 5m × 3 × 16.5 × 10−6 × 90
4
= 0.0926m 3
ΔVgasoline = 9 × 12 × 10−4 × 90m 3
= 0.972m 3
Difference = 0.972 − 0.0926m 3
= 0.879m 3
The gasoline expands more, by 0.879m 3
(Ans.)
5. A copper sphere is 10 cm diameter at 20°C. Find the increase in diameter, the increase in
surface area, and the increase in volume when the sphere is heated to 160°C. For copper,
use α = 16.5 x 10-6 /°C.
a) Δd = dα ΔT = 10 × 16.5 × 10−6 × 140 cm = 0.0231cm (Ans.)
b) ΔA = 2α A ΔT = 2 × 16.5 × 10−6 × 4 π × 52 × 140 cm 2 =1.45cm 2 (Ans.)
c) ΔV = 3α V ΔT = 3 × 16.5 × 10−6 ×
4
π × 53 × 140 cm 3 = 3.63cm 3 (Ans.)
3
Page 46
6. The inner surface of a 23 cm thick brick furnace wall is kept at 820°C and the outer surface
is at 170°C. (a) What is the heat loss per hour from each square metre of surface, if the
conductivity of the brick is 0.6 W/m°C? (b) Assuming the same heat loss, how thick must
the wall be to reduce the temperature of the outside surface to 130°C?
a)
k A t ΔT
L
W
0.6 o × 1m 2 × 3600s × (820-170)o C
mC
=
0.23m
0.6 × 1× 3600 × 650
=
0.23
= 6 104 347 J
Q=
= 6.104 MJ
(Ans.)
b)
k A t ΔT
Q
W
0.6 o × 1m 2 × 3600s × (820-130)oC
mC
=
6 104 347 J
0.6× 1× 3600 × 690
=
6 104 347
= 0.244 m
L=
= 24.4 cm
(Ans.)
Page 47
7. The average ‘k’ value for a wall material is 0.084 W/m°C. The overall wall dimensions are
8 cm thick, 12 m long and 4 m high. The wall contains 3 windows, each with single-pane
glass, measuring 6.0 mm thick, 2 m wide and 1.2 m high. How much more (or less) heat is
lost, per hour, through the windows, compared with the wall material, when the inside
temperature is 22°C and the outside temperature is -10°C?
Qsolid = 0.084 × [(12x4)-(3 × 2 × 1.2)] × 3600 × (22 − ( −10)) ×
1
J
0.08
= 4.94MJ
Q windows = 3 × 1 × 2 × 1.2 × 3600 × (22 − ( −10) ×
1
J
0.006
= 138.2MJ
ΔQ = (138.2-4.94)MJ
=133.3MJ
The windows lose 133.3 MJ more than the solid wall. (Ans.)
Page 48
Chapter 10: Steam Properties and Calculations
End of Chapter Answers
1. Clearly define or explain the following terms:
a) latent heat – heat which changes a substance from one phase (state) to another,
with no change in temperature
b) saturated steam – steam that is at its saturation temperature
c) degrees of superheat – the difference in temperature between the superheat
temperature of a substance and its saturation temperature at that pressure
d) dry saturated steam – steam that is 100% dry at its saturation temperature
e) sensible heat – heat that causes an increase/decrease in the temperature of a
substance
f) saturation temperature – the temperature at which a liquid begins to boil at a
given pressure
g) dryness fraction – the percentage (or fraction) of dry steam in a wet steam
mixture
h) specific enthalpy – the heat contained by one kilogram of a substance
i) superheat temperature – the actual temperature of a substance that is above its
saturation temperature at that pressure
j) specific volume – the volume occupied by 1 kg of a substance
2. Explain each of the following symbols, which are found in the tables for saturated steam.
State what each stands for and give the units of measurement that apply to each one.
a)
b)
c)
d)
e)
f)
p
t
hfg
hf
hg
vg
pressure
saturation temperature
latent enthalpy (heat) of evaporation
liquid enthalpy
dry saturated vapour enthalpy
specific volume of dry saturated vapour
kPa
°C
kJ/kg
kJ/kg
kJ/kg
cm3/g
Page 49
3.
Using the steam tables, determine each of the following properties:
a. volume of 8 kg of dry saturated steam at 845 kPa abs
(0.227 − 0.2404)
× 45 + 0.2404 = 0.2283m 3 /kg
50
m3
Therefore : v = 8 kg × 0.2283
kg
vg =
= 1.827m 3
( Ans .)
b. enthalpy of steam at 3000 kPa and 475°C
h=
(3456.5 − 3230.9)
× 75 + 3230.9 = 3400.1kJ/kg
100
(Ans.)
c. latent heat in 200 kg of steam at 280 kPa and 79% dry
2163-2172.4
× 5 + 2172.4 = 2170.68 kJ/kg
25
h = 2170.68 × 0.79 kJ/kg
= 1714.84 kJ/kg
hfg =
H = 1714.84 kJ/kg × 200 kg
= 342 968 kJ
(Ans.)
d) enthalpy of saturated water at 365°C
hf =
1890.5 − 1760.5
× 5 + 1760.5 = 1825.5kJ/kg
10
(Ans.)
e) saturation temperature of 14 000 kPa steam at 850°C
t sat at 14000kPa = 336.75D C
(Ans.)
Page 50
4.
(a) How much total heat is required to increase 850 kg of feedwater, in an economizer,
from 30°C to 180°C and then to convert it to wet saturated steam at a saturation
temperature of 220°C, with a dryness fraction of 0.88?
Step 1: Water from 30D C to 180D C
Δh1 =(763.22-125.78)kJ/kg = 637.44kJ/kg
Step 2: Water at 180D C to wet steam (q=0.88) at 220D C
Δh 2 = 943.62 + (0.88 × 1858.5) -763.22kJ/kg
= 1815.88kJ/kg
Δh total = 637.44 +1815.88kJ/kg = 2453.32kJ/kg
Q = 850kg × 2453.32kJ/kg
= 2085MJ
(Ans.)
(b) How much additional heat must be added to completely dry the steam?
Qkg = Δh = 0.12×hfg = 0.12×1858.5kJ/kg
= 223.02kJ/kg
Q = 850kg × 223.02kJ/kg
= 189 567 kJ or 190 MJ
(Ans.)
5.
Show, by calculation, which of the following produces the driest steam. Boiler A produces
500 kPa steam with enthalpy of 2650 kJ/kg; Boiler B produces saturated steam at 305°C
and total enthalpy of 2700 kJ/kg.
Boiler A :
h A = 2650kJ/kg = 640.23 + q A × 2108.5
Boiler B :
q A = 0.953
h B = 2700kJ/kg = hf + q Bhfg
1401.3-1344
× 5 + 1344 =1372.65kJ/kg
10
1326-1404.9
hfg =
× 5 + 1404.9 =1365.45kJ/kg
10
2700-1372.65
= 0.972
qB =
1365.45
hf =
The steam from boiler B is the driest.
(Ans.)
Page 51
6.
Superheated steam enters a steam turbine at 6500 kPa and 600°C and leaves at 20 kPa and
90% dry. A condenser then converts this steam to water at 20 kPa and then cools the water
to 60°C. Assuming no steam is removed between the inlet and outlet of the turbine, how
much heat is lost from the steam, per kg, in (a) the turbine and (b) the condenser?
Let hs = heat in steam entering turbine; htex = heat at turbine exhaust; hcond = heat out of
condenser; Qtub = heat lost in turbine
a)
3650.3-3658.4
× 500 + 3658.4
1000
= 3654.35kJ/kg
h tex = 251.4 + (0.9 × 2358.3)
hs =
= 2373.87 kJ/kg
h cond =251.13kJ/kg
Q tub = 3654.35 - 2373.87 kJ/kg
=1280.48kJ/kg
b)
(Ans.)
Qcond = 2373.87 - 251.13 kJ/kg
= 2122.74 kJ/kg
7.
A watertube boiler has a maximum capacity of 30 000 kg/h of saturated steam at 500 kPa,
but the steam is only 94% dry. The feedwater entering the boiler is at 80°C. If the boiler is
modified to produce 25 000 kg/h of superheated steam at 300°C, but still at 500 kPa, how
much superheat does the steam have and what is the change in the boiler’s heat rate?
At 500kPa, dry and saturated
hg=2748.7kJ/kg
At 500 kPa and 300°C h=3064.2kJ/kg
Therefore, the amount of the superheat in the steam is
Qs’ht=3064.2-2748.7 kJ/kg
= 315.5 kJ/kg
(Ans.)
Also accept 300 °C – 151.86 °C = 148.14 °C
of superheat.
30000 kg
×[(640.23 + (0.94× 2108.5)) - 334.91] kJ/kg
3600 s
= 19061kJ/s
Heat rate initial =
25000 kg
×[3064.2 - 334.91]kJ/kg
3600 s
= 18953 kJ/s
Heat rate final =
Page 52
Δ heat rate = 18953 - 19061 kJ/s
Δ heat rate = - 108 kJ/s
The heat rate has decreased by 108 kJ/s or (388,800 kJ/h) (Ans.)
8.
(a) Define equivalent evaporation, and factor of evaporation
equivalent evaporation – the mass of water that would be evaporated in one hour,
from feedwater at 1000C, into dry saturated steam at 1000C, by the same amount
of heat that is required, per hour, to produce steam at the actual boiler conditions.
factor of evaporation – the ratio of the equivalent evaporation of the boiler to the
actual steam flow rate
(b) Find the equivalent evaporation and factor of evaporation for these two boilers:
Steam flow
Operating pressure
Steam quality
Feedwater temperature
Boiler 1:
Boiler 1
20 000 kg/h
1450 kPa
95 % dry
85°C
Boiler 2
17 000 kg/h
1750 kPa
100 % dry
110°C
hf = 837.5 kJ/kg (by interpolation)
hfg = 1953.5 kJ/kg (by interpolation)
hfeed = 355.9 kJ/kg (by interpolation)
Q1 = 837.6 + (0.95 x 1953.5) – 355.92 = 2337.5 kJ/kg
2337.5× 20000
= 20713 kg/h
2257
20713
factor of evap 1 =
= 1.036
20000
(Ans.)
equiv. evap1 =
Boiler 2:
Q2 = 2796.4 – 461.3 kJ/kg = 2335.1 kJ/kg
2335.1 × 17 000
= 17 588 kg/h
2257
17588
Factor of evap 2 =
= 1.034 (Ans.)
17000
Equiv.evap2 =
(Ans.)
(Ans.)
Page 53
Chapter 11: Gas Laws and Calculations
End of Chapter Answers
1. Define each of the following laws, state a formula for each, and explain under what
conditions each may be applied to a perfect gas:
(a) Boyle’s Law
When the temperature of a gas remains constant the pressure is inversely
proportional to the volume.
PV=C or P1V1= P2V2
can be applied when temperature is constant.
(b) Charles’ Laws
When the pressure of a gas is constant, the volume is proportional to the
temperature.
V
=C
T
or
V1
V
= 2 can be applied when pressure is constant.
T1
T2
(c) Gay-Lussac’s Law
When the volume of a gas is constant, the pressure is proportional to the
temperature.
P
=C
T
or
P1
P
= 2 can be applied when volume is constant.
T1
T2
(d) The General Gas Law
The pressure, temperature and volume of a gas can be related by
PV
=C
T
or
P1 V1
PV
= 2 2 can be applied for any gas process.
T1
T2
Page 54
2. Starting with the same values for pressure and volume, and ending at the same volume,
make sketches that compare the amount of work done during isothermal, adiabatic, and
polytropic compressions of a perfect gas. Explain the difference.
Since W is the area under the curve
Wadiabatic > W polytropic > W isothermal
The increase in work is due to the fact that
pressure increases more quickly for an
adiabatic process and therefore more work is
done.
3. (a)
8 kg of a perfect gas occupy 0.45 m3 at 1600 kPa gauge and 190°C. If the gas is
expanded with an expansion ratio of 6.5 to 1, to a final pressure of 160 kPa gauge,
what is the final temperature of the gas?
Assuming Patm = 101.3 kPa, Then:
m = 8 kg
P1 V1
PV
= 2 2
T1
T2
V1 = 0.45m3
V2 = 6.5 V1
V2 = 2.92 m3
T2 =
P1 = 1701.3 kPa
P2 = 261.3 kPa
T1 = 463 K
P2 V2 T1
P1 V1
261.3 × 2.92 × 263
K
1701.3 × 0.45
= 461.4 K or 188.4 DC
T2 =
(b)
(Ans.)
Calculate the value of the characteristic gas constant for this gas.
R=
PV 1701.3 × 0.45
kJ
=
= 0.2067
mT
8 × 463
kgK
(Ans.)
Page 55
4.
A gas with an initial volume of 8 m3 is compressed from atmospheric pressure (101
kPa) to 650 kPa. Calculate the final volume and the work done if the compression is:
a. Isothermal
V1 = 8 m3
P1 = 101 kPa P2 = 650 kPa
PV
P1 V1 = P2 V2
V2 = 1 1 = 1.24 m 3
(Ans.)
P2
W = P1 V1ln
V2
1.24
= 101 × 8 × ln
kJ
V1
8
= - 1506 kJ (negative indicates work done on the gas)
(Ans.)
b. Adiabatic, with an index of compression equal to 1.52
P1 V1 γ = P2 V2 γ
101 × 8
V2 = (
650
1.52
1
)1.52
= 2.35 m 3
(Ans.)
P1 V1 - P2 V2
101 × 8 - 650 × 2.35
=
kJ
γ -1
1.52 - 1
= - 1384 kJ
(Ans.)
W=
1.3
5. A perfect gas is compressed polytropically, according to the law PV = C. How much
work is done on 40 kg of the gas if the initial pressure is atmospheric, at a temperature of
18°C, and the final temperature is 121ºC? R, for the gas equals 0.290 kJ/kgK.
W=
mR(T1 - T2 ) 40 × 0.29 (18 - 121)
=
kJ
n -1
1.3 - 1
= - 3983 kJ
(Ans.)
6. A compressed air motor operates on a constant pressure cycle. The diameter of the piston
is 10 cm, the stroke is 12 cm, and the operating pressure is 800 kPa gauge. How much work
is done in the cylinder in three minutes if there are 100 working strokes per minute? Ignore
the clearance volume of the cylinder.
π
m3
m3
2
P = 901.3 kPa
ΔV = × (0.1) × 0.12
= 0.000942
4
stroke
stroke
strokes
ΔV
W=P ×
×
× min
stroke
min
m3
strokes
= 901.3 kN/m 3 × 0.000942
× 100
× 3 min
stroke
min
= 254.7 kJ
(Ans.)
Page 56
7. Show, by calculation, which of the following operations involves the greatest amount of
work, and state by how much is it greater.
Operation A: A perfect gas, with R = 0.195 kJ/kgK and ratio CP/CV = 1.35, expands
from 500 kPa and 1.40 m3 to 170 kPa. No heat is added during the expansion.
A : adiabatic expansion γ
P1V1
1.35
= P2 V2
1.35
V2 = 3.11 m 3
P1 V - P2 V2
γ -1
500×1.4 - 170 × 3.11
=
kJ
0.35
= 489kJ
WA =
Operation B: 20 kg of air are compressed at a constant temperature of 300°C, from a
volume of 3.7 m3 to a volume of 0.8 m3.
Operation B : isothermal expansion.
⎛V ⎞
WB = mRT ln ⎜ 1 ⎟
⎝ V2 ⎠
⎛ 3.7 ⎞
= 20 × .287 × 573 ln ⎜
⎟ kJ
⎝ 0.8 ⎠
= 5037 kJ
Operation B does the most work by 5037 – 489 = 4548 kJ (Ans)
Page 57
8. A perfect gas, at 360°C, has its condition changed in two stages. In the first stage, the
volume remains constant at 20 m3 while the pressure changes from 900 kPa gauge to
200 kPa gauge. In the second stage, the gas is compressed isothermally to a final pressure
of 1200 kPa gauge. Calculate: (a) the final volume of the gas, (b) the final temperature.
P1 = 1001.3 kPa
P2 = 301.3 kPa
P2 = 301.3 kPa
T1 = 360 °C = 633 K
V1 = V2 = 20 m3
T3 = T2
P3V3 = P2V2
a)
since 2 Æ 3 is isothermal
301.3 × 20 m 3
1301.3
= 4.63 m 3
V3 =
(Ans.)
Since 1 Æ 2 is constant volume
b)
P1 P2
=
T1 T2
T2 =
P2 T1
P1
301.3 × 633K
= 190.5K
1001.3
T3 = T2
=
T3 = 190.5K or - 82.5 oC
(Ans.)
Page 58
Chapter 12: Chemistry Fundamentals
End of Chapter Answers
1. In your own words, explain:
a) Why reaction equations must be balanced.
Reaction equations must be balanced so that they express the conservation of mass
for the reaction process.
b) How you can tell whether or not a reaction equation is balanced.
A reaction equation is balanced when the number of atoms of each atomic species is
the same on the product and reactant sides of the equation.
2. List four differences between inorganic and organic compounds.
i.
organic compounds are generally insoluble in water while inorganic
compounds are generally soluble in water
organic compounds are generally flammable, while inorganic compounds are
generally non-flammable
most organic compounds are gases or liquids at room temperature, but most
inorganic compounds are solids at room temperature
in organic compounds, carbon is bonded to only a small number of other
elements (hydrogen, nitrogen, halogens, sulphur and phosphorous
ii.
iii.
iv.
3. Explain how electro negativity affects bonding.
The electro negativity of an element affects whether that element will tend to form
a bond by electron transfer (ionic bond) or electron sharing (covalent bond).
For any two elements that have bonded, as the difference in their electro
negativities increase, the tendency to form an ionic bond increases.
4. Balance the following unbalanced chemical reactions:
NH3 + O2 → NO + H2O
4NH3 + 5O2 → 4 NO +6 H2O
unbalanced
balanced (Ans.)
b) Al + H2SO4 → Al2(SO4)3 + H2
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
unbalanced
balanced (Ans.)
a.
Page 59
5. A chemical change, sometimes encountered in boiler feedwater treatment, is the reaction of
calcium chloride, with sodium carbonate, to produce calcium carbonate and sodium
chloride. Balance the following unbalanced reaction for this change.
CaCl2 + Na2CO3 → CaCO3 + NaCl
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
unbalanced
balanced
(Ans.)
6. Calculate the formula (molecular) masses, to the nearest whole number, of the following
substances:
(a) Hydrogen, H2
(b) Nitrogen, N2
(c) Sulphur dioxide, SO2
(d) Carbon monoxide, CO
(e) Magnesium bicarbonate, Mg(HCO3)2
(f) Sodium sulphate, Na2SO4
a.
b.
c.
d.
e.
H2 M = 2 × 1 = 2
(Ans.)
N2 M = 2 × 14 = 28
(Ans.)
SO2
M = 32 + (2 × 16) = 64
(Ans.)
CO
M = 12 + 16 = 28
(Ans.)
Mg(HCO3)2
M = 24 + 2[ 1 + 12 + (3 × 16) ]
M= 146
(Ans.)
f. Na2SO4 M = (2 × 23) + 32 + (4 × 16)
M = 142
(Ans.)
7. Calculate the number of kilomoles of copper that are in 3.05 kg of copper.
Mcopper = 63.55 kg/kmol
m
n=
M
3.05kg
n=
= 0.0480 kmol
kg
63.55
kmol
(Ans.)
8. What is the difference between a saturated and an unsaturated hydrocarbon?
A saturated hydrocarbon has only single bonds between the carbon atoms.
Unsaturated hydrocarbons have one or more double or triple bonds between
carbon atoms.
9. What compound will be produced if an ethyl group is combined with a butyl group?
hexane
Page 60
10. If one mole of propane (C3H8) undergoes complete combustion using pure oxygen, what
mass (kg) of carbon dioxide (CO2) and of water (H2O) will be produced?
C3H8 + O2 → CO2 + H2O
C3H8 + 5O2 → 3CO2 + 4H2O
unbalanced
balanced
(Ans.)
For one mole of C3H8 we get:
a) 3 moles CO2
M CO2 = 3[12grams + (2 × 16)grams]
= 132 grams or 0.132 kg
b) 4 moles H2O
(Ans.)
M H 2O = 4[(2 × 1) + 16]grams
= 72 grams or 0.072 kg
(Ans.)
11. Define the following
a. acid – a substance capable of donating a proton to some other substance
b. base – a substance capable of receiving a proton from some other substance
c. salt – an electrically neutral compound that will disassociate into positively and
negatively charged ions when in solution
Page 61
Chapter 13: Metallurgy and Materials
End of Chapter Answers
1. Explain what is meant by the term "composite material".
A composite material is composed of at least two substances working together to
produce properties that are different from the properties of those substances on
their own. Most composites consist of a bulk material, or matrix that exists as a
continuous phase, and a reinforcement added primarily to increase the strength
and stiffness of the matrix. This reinforcement is usually in the form of long or
short fibers, but particles, and other forms are used. Composite materials are
designed for specific needs.
2. Why are metals heat treated?
Heat treating is a process that involves heating a metal to modify its internal
structure, thereby adjusting the mechanical properties of the metal. Heating
initially realigns the grains and relieves unequal stresses in a metal. These stresses
may be present due to cold working or welding procedures.
.
3. Describe the basic oxygen process for steel making.
The basic oxygen furnace, shown in Figure 13, is a large, vertical, pear-shaped
vessel, lined with refractory bricks. It combines molten iron (from a blast furnace)
with scrap iron and steel to produce a final steel product.
Scrap iron is dumped into the furnace, followed by the hot metal (molten pig iron)
from the blast furnace. Then a water-cooled lance (nozzle) is lowered into the
furnace, blowing a high-pressure stream of oxygen downward onto the molten
surface. This causes chemical reactions, which separate the impurities as fumes
and slag. The oxygen reacts with the carbon in the pig iron, creating combustion
temperatures as high as 2500°C . This high temperature allows for the addition of
up to 30% scrap iron.
The concentration and specific gravity of the refined metal that is directly below
the oxygen lance is higher than the rest, causing it to settle downward. This carries
some oxygen with it, creating circulation of the molten metal within the vessel, thus
ensuring good mixing and increasing the rate of reaction between the metal and
slag. The high heat causes the formation of a reactive slag, and lowers the overall
oxygen content.
When the process is complete, the liquid steel and slag are poured into separate
containers..
Page 62
4. Describe the Brinell Hardness test.
In the Brinell hardness test, a hardened steel or tungsten carbide ball, under a
specified load, is pressed into the surface of the test material. The diameter of the
indentation is measured and converted, by formula, into a Brinell Hardness
Number (BHN). The harder the material is, the higher is its Brinell number. For
example, steel might have a BHN of 200 while an aluminum alloy, which is much
softer than steel, might have a BHN of 40.
The test specimen is placed on top of the jackscrew and raised until it touches the
ball. A pump forces oil above the plunger, hydraulically pressing the ball into the
test specimen. The pressure gauge gives a corresponding indication of the load. To
prevent overload, the masses act as a pressure relief system.
The balls used come in various diameters: 1, 2, 5, or 10 mm. The ball is forced into
the test piece by a specified load for 10 to 20 seconds. It makes a circular
indentation, the diameter of which is measured by a low-powered microscope to an
accuracy of +/- 0.05 mm. Two diameters at right angles to each other are
measured, and the mean diameter is used to calculate the surface area of the mark.
From this the Brinell Hardness is found from the Brinell equation:
Page 63
5. Explain the benefits of chromium, nickel and copper in alloy steels.
a) chromium – resists oxidation, maintains strength at elevated temperatures,
increases hardness and abrasion resistance without increasing brittleness,
increases corrosion resistance
b) nickel – resistance to corrosion and oxidation, improves toughness and reduces
brittleness. Increases strength and impact resistance
c) copper – increases atmospheric corrosion resistance, increases strength
6. Define the following mechanical properties and give an example in each case of a material
that possesses that particular property:
a. hardness – Hardness is a materials resistance to wear, abrasion and
penetration. In a hard material, the molecules are strongly bonded and resist
any movement relative to each other. Examples of hard materials are carbon
tool steel, stellite (an alloy of chromium, cobalt and tungsten) and diamonds.
b. malleability – malleability is the ability of a material to withstand great
deformation above the elastic limit, while under compression, without failure.
Copper, which was given as an example of a ductile material, is also malleable
because it can be hammered or beaten into various shapes. Malleable metals
may, when cold, be hammered or rolled into thin sheets. The malleability of
most materials increases when the material is heated as when iron or steel is
heated before forging.
c. toughness – Toughness determines whether a material will break or stand up
under a sudden impact or hard blow. This property is also referred to as
“impact strength” and impact tests are used to determine the toughness of a
material. Materials – nickel alloy steel
d. elasticity – Elasticity is a material’s ability to resist a force trending to change
its shape and then return to its original shape when the force is removed. The
property of elasticity is one of the most important from the engineering point
of view as it helps to determine the behavior of the material under a load.
Material - steel
7. Why is steel cold worked?
Cold working, such as stretching, drawing, or pounding, causes the grains to yield
by internal slippage. Once slippage has occurred, the atoms are bonded more
tightly in the new plane. This causes an increase in the tensile strength of the
metal. For example, the tensile strength of cold worked copper is about 345 MPa
compared to a value of about 207 MPa before cold working. The hardness of the
metal is also increased in this process, and sometimes the hardening is extreme
enough to make the material brittle. The cold work hardening of aluminum and
its failure is an example.
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8. List three nonferrous materials and describe the properties of each that make it suitable for
power plant work.
The student may describe any of the non-ferrous materials in the chapter. The
question states non-ferrous materials rather than (specifically) non-ferrous metals.
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Chapter 14: Corrosion Principles
End of Chapter Answers
1. Using a simple sketch, explain what causes stray current corrosion.
Stray current corrosion is sometimes called stray current electrolysis. Stray current
corrosion is caused by an external source of current. Stray current corrosion is not
influenced by environmental conditions such a pH and entrained oxygen.
Stray currents take paths that were not intended for them. Poor electrical conductors,
damaged insulation, and dirty or loose electrical connections allow the electrical
current to leave its intended path. The stray current may travel through water, soil or
any suitable electrolyte to find the path of least resistance such as an underground
pipe or structure.
Direct current (DC) is the most destructive source of stray current. Direct current
welders, large rectifiers used for traction equipment, and improperly installed
cathodic protection systems are sources of stray currents. Alternating current (AC) is
not as destructive as DC. The greatest source of stray AC is from buried power lines.
If there is a nearby current path in the earth provided by a low resistance object, such
as a pipeline, current from an impressed-current cathodic protection system will pass
into the pipeline. The stray current flows along the pipeline for some distance before
returning to the protected structure. The site where the current enters the pipeline
becomes cathodic and where it leaves the pipeline becomes anodic, allowing corrosion
to occur at the site. A schematic of this is shown in Fig. 2.
Figure 2
Stray Current Corrosion
It is hard to distinguish between galvanic and stray current corrosion, because there is
a protected cathodic site and a corroded anodic site in both cases. It is usually
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determined by laboratory analysis looking for destruction along the grain boundaries.
A concentration of pits in an area where they are not normally found is an indication
of stray current corrosion. Stray currents can be located by measuring voltage drops
and current flows along the buried structure.
2. Discuss why stress corrosion cracking is so dangerous and list the conditions that produce
it.
Stress corrosion cracking causes failure of equipment in service and may occur at
conditions much less than the original safe working stress. Conditions: water (or
condensed moisture) plus a corrosive substance. It is the result of the combination of
mechanical or tensile stress (service conditions), the susceptibility of the material, and
exposure to corrosive substances.
3. What conditions are required to produce sulphide corrosion cracking?
Sulphide corrosion cracking is due to:
i) the presence of H2S,
ii) the presence of (at least) a trace amount of water (or more),
iii) using high strength steel,
iv) the steel must be under applied or residual tensile stress
v)
low pH
4. Explain how micro organisms promote corrosion.
Micro organisms promote corrosion by:
• increasing the oxygen content of the water, or brine, in the vicinity of the metal
• increasing the presence of sulphide in the vicinity of the metal
5. List the various mechanisms whereby corrosion inhibitors accomplish their job.
Corrosion inhibitors accomplish their job by:
• coating the metal surface to reduce its contact with water
• creating a protective oxide film (or partially corroded layer) that reduces further
corrosion. This is called “passivation”
• reacting with the metal to form precipitates on the surface which resist further
corrosion
• neutralizing acids in the water
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6. Sketch and describe a cathodic protection system utilizing an external current source.
Explain the correct polarity for connecting the cathodic protection system to an external
current source, and explain why the correct polarity is of great importance.
Figure 10
Typical Impressed Current Cathodic Protection System
The principle of cathodic protection is to convert the whole metal surface into a
cathode. This is accomplished by forcing sufficient direct current to flow to the
structure being protected from the electrolyte so that there is no net flow of current
from any point on the metal surface to the electrolyte. When external current
(electrons) is applied to a corroding surface, part of the current goes to the anodic
areas and part of it goes to the cathodic areas, causing all areas to be cathodic.
The current supplied to the protected structure is provided by connecting the
structure to a more reactive metal (galvanic protection) or by connecting to a direct
current power source (impressed current system). Cathodic protection does not stop
corrosion. It transfers the corrosion from the structure under protection to another
known location where the current discharging anode(s) can be designed for long life
and simple replacement.
Cathodic protection can only be applied to the surface of the metal exposed to the
same electrolyte as the anode. Cathodic protection for the exterior bottom of a
storage tank has no effect on internal corrosion of the tank from corrosive water.
The correct polarity is critical, because if it is reversed, the system would accelerate
corrosion.
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7. Explain the purpose and operating principle of ultrasonic testing.
Ultrasonic inspection is a widely used method of corrosion detection that is nondestructive and, in most cases, it can be used when the equipment is in service.
Ultrasonic devices are used to measure metal thickness and to detect defects in metal.
The basic principle of ultrasonic inspection is to generate high frequency sound waves
and transmit them into the material being inspected. These waves travel through the
object until they encounter the opposite surface or an internal discontinuity. Either
an interface or a discontinuity reflects the sound waves back to their origin where they
can be detected. The time interval between transmission of the ultrasonic wave and
the arrival of the reflected wave back to the point of transmission is directly
proportional to metal thickness or the depth of the flaw.
Ultrasonic waves are generated by transducers that are made from piezoelectric
materials such as quartz. These materials are capable of expanding and contracting
when subjected to a changing electrical field and produce an electrical potential when
the material is placed under mechanical stress. Therefore, an ultrasonic transducer
can be placed on a metal surface and used to both create and detect ultrasonic energy.
8. How can caustic gouging, hydrogen damage, and pitting be controlled in boilers?
a)
Caustic gouging prevention:
• ensure boiler feedwater is not too caustic
• removal of waterside deposits by chemical cleaning
b)
Hydrogen damage prevention:
• correct operation and monitoring of hydrogen zeolite softeners
• sufficient flushing of any surfaces that have been acid cleaned
• an established response plan must be in place for situations where
feedwater treatment conditions have not been met.
c)
Pitting prevention:
• correct operation and monitoring of the deaerator
• established water treatment and testing procedures including oxygen
scavenging chemical levels in the boiler feedwater
• during shutdown, the boiler must be kept full of treated water (with
oxygen scavengers) and an inert gas (N2) blanket.
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9. Explain what the following terms mean when used to describe the severity of corrosion:
a. Definite Surface Corrosion
This describes the condition when the attack is deep enough to catch a knife
blade. It may be further described as “dulled”, “matte”, or “roughened”.
b. Shallow Metal Attack
This describes the removal of a perceptible, although hardly measurable,
amount of metal from the surface.
c. Blistering/Scaling
This occurs when thin layers are flaked off from the metal surface and is an
indication of hydrogen diffusion. Blisters often indicate an earlier stage of
the same or similar type of attack.
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Chapter 15: Industrial Drawings
End of Chapter Answers
1. An operator is learning the general arrangement and process flows throughout a plant. What
types of drawings are used? Explain.
Since the operator is learning the “general” arrangement and process flows, he/she
would use PFD’s (Process Flow Diagrams). These will identify:
• the major pieces of equipment and their functions
• the processes involving that equipment, including the flow stream
• the pressures, temperatures, flow rates in the flow streams and major equipment
• reference number locations for accompanying material balance sheets
2. What are the details included on P&IDs that are not found on PFDs?
P&ID’s contain the following information (generally not included in PFD’s):
• vessel size, insulation requirements and connections
• numbers of heat exchanger passes
• power requirements for pumps and compressors
• instrumentation and control loop components
• piping details including line identification, specifications, insulation, valve and
fitting types and sizes, and connection types.
• See additional points on Page 498
3. You need to know the material specification for a pipe in the plant you are operating.
Where would you find this piping specification?
In the P&ID’s (aka Mechanical Flow Diagrams)
4. Process Flow Diagrams have numbers assigned to each major flow of a plant unit. What are
these flow numbers a reference to?
The numbers reference material balance sheets.
5. What is a detailed construction drawing used for?
The primary purpose of detailed construction drawings is to provide exact equipment
and location specifications for construction/installation personnel.
6. What types of equipment are located on a plot plan?
• major vessels and equipment (pumps, compressors, turbines)
• buildings
• elevated structures and sub-structures, including piping racks and trays
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