MP2001 Mechanics of Materials
Lecturer: Lu Guoxing
Tel: 67905589
Rm: N3 02c  115
Email: gxlu@ntu.edu.sg
With thanks to Assoc. Prof. Tan Soon Huat
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Title: Probe into crane collapse shows four
anchors had failed structurally
By:Date:28 Feb 2008
80
COMPRESSIBLE FLOW THROUGH NOZZLES
The exit temperature is therefore
Ta=r i
(S)
<
(nl)/n
/ , f i N 1.389/0.389
54
= n^'
=5o7 29K

and the Mach number at the exit is
M2 = ,U _ 1 (^l)=
/
V^ VT2
J (_^_ _ ,)
y 0.4^506.29 J
=
0.5678
The exit velocity i
HISTORICAL SURVEY
11
Figure 1.5 De Havilland Goblin turbojet engine.
1967 the Trent, which was the first threeshaft turbofan engine. The Olympus was also
used in stationary power plants and in marine propulsion.
General Electric in the United States has
52
PRINCIPLES OF THERMODYNAMICS AND FLUID FLOW
for then all the kinetic energy leaving the pipe will be dissipated in the reservoir and
none is recovered.
g
As a third example on the application of the momentum balance, consider how mixing of
a stream in
BLADE FORCES
417
so the lift coefficient can be written as
Cx. = 2n
(la)V
W
rtt
smt
On an element of the blade of width dr, if drag is neglected, the x component of the
force on the blade is
dFx = Lcos<j>dr = ~PW2CCL
COS(pdr
or
dFx 7rpc[(l a)V cos 9 rfl
26
PRINCIPLES OF THERMODYNAMICS AND FLUID FLOW
their values may be approximated as
v(T,p)^vt(T)
u(T,p)*uf(T)
Enthalpy can then be obtained from
h(p,T) a ut(T) + pvf(T)
which can also be written as
h(p,T) = Uf(T)+MT)vf(T)
+ (pPf(T)K(T)
or as
h = hf + vi(p
DIFFUSER AND VOLUTE DESIGN
8.8
8.8.1
305
DIFFUSER AND VOLUTE DESIGN
Vaneless diffuser
In the vaneless space in aflowwithout spin the tangential component of the velocity follows
the free vortex distribution. This is a consequence of the law of conservatio
44
PRINCIPLES OF THERMODYNAMICS AND FLUID FLOW
side shows that internal energy increases by heat transfer into the fluid, but decreases when
heat is lost to the surroundings. Integrating Eq. (2.21) gives
u2u1=q+~
rre22
li
m
Ts'di
(2.22)
Substituting Eq.
396
HYDRAULIC TRANSMISSION OF POWER
With Vui = Vxi tan a\ = Q tan a i /A, the torque coefficient of the primary is then
Tp
P
Q
(&l%,\
(r\
fipr3
Qritanai
Ar3Qp
^ 2
The reaction torque of the secondary is given by
Ts = PQ(npr22  Var23)
and the torque coeff
326
RADIAL INFLOW TURBINES
At the inlet to the rotor the static pressure is
T2s\7/(71
p2=Poi^j
=
/859.4N4
255 2
 U6U0j = 163  9kPa
so that the density at the inlet becomes
P2 =
=
lk
0.287 3 8 6 6 . 9 = a 6 5 9 k g / i n 3
The blade height can now be de
CONSTANT MASS FLUX
187
constant. On the other hand, the reaction varies quite strongly along the span of the blade.
Other design possibilities exist. For example, if the nozzle angle is kept constant along
the span, manufacturing cost of nozzles can be re
66
COMPRESSIBLE FLOW THROUGH NOZZLES
Solution: (a). With the flow choked, the pressure ratio is
Pe
0.5283
PO
from which the exit pressure is determined to be pe = 106 kPa. The flow rate can be
obtained by first calculating the flow function
(7+l)/2(7l)
7
DESIGN DEFLECTION
233
The tangent difference formulas now show that the solidity for the rotor is the same as that
for the stator, as the design is based on nominal conditions. It is determined by solving
ip*
<p*
1.55
1 + 1.5/cr
for a. In order to prevent
116
PRINCIPLES OF TURBOMACHINE ANALYSIS
4.3.2
Degree of reaction
Degree of reaction, or reaction for short, is defined as the change in static enthalpy across
the rotor divided by the static enthalpy change across the entire stage. For the turbine this
is
32
PRINCIPLES OF THERMODYNAMICS AND FLUID FLOW
Dalton's model is based on such a consideration, and it states that the mixture pressure
is equal to the sum of the component pressures p%, which each of the molecular species in
the mixture would exert if it
TURBINE EFFICIENCY AND LOSSES
205
viscous forces are small and inertial forces are balanced by pressure forces. The transverse
component of pressure force points from the pressure side of one blade to the suction side
of the next one, and the transverse c
Ch 7 Transformations of stress & strain
7.1 Introduction 7.2 Transformation of plane stress 7.3 Principal stresses & max shear stress 7.4 Mohr's circle for plane stress 7.6 Out of plane max shear stress 7.7 Yield criteria for ductile material
1
7.8 Yield
Sample Problem 7.2
Find a) Principal planes and principal stresses, x' , y' & x' y' at = 30o b)
1
G
OC = ave = Construct 2 Mohr's circle 2 2 R = (CF) + (FX ) =
x + y
100+ 60 = = 80MPa 2
(20)2 + (48)2 = 52MPa 2
x
Principal planes and stresses
tan 2 p 2 p
Ch 9 Deflection of Beams
9.1 Introduction 9.3 Equation of Elastic Curve 9.6 Using Singularity Functions
1
9.1 Introduction Besides strength (max & max) , must also check for deformation (max deflection). From Beam Bending,
M ( x) = EI 1
y
(Momentcurvatur
148
STEAM TURBINES
from which
J03s
= ^~
and from these it follows that
^  = ^~
and
J3s
7*03 _ TQ3S
T3
L03ss
J
(5.11)
3ss
Tp3ss
_
T3s
T3ss
Expressing these temperature ratios in terms of Mach numbers yields
*3
J
^
3s
J
*
3ss
^
so that M 3 = M 3 s = M 3
132
PRINCIPLES OF TURBOMACHIIME ANALYSIS
3.0
p0r = 0.981 bar
T0r = 293K
2.8
2.6
df
2.4
CM
O
a
2.2
200,000
2.0
3
$
Choking
19
180,000
1.6
'160,000
1.4 
140,000
1.2
1.0
/
0
T
1
~  60,000
1
0.02
1
I
I
I
0.04
90,000
I
I
I
120,000
I
0.06
I
I
I
0.08
nfv^7
362
HYDRAULIC TURBINES
2000
10
100
Flow rate (m3/s)
1000
Figure 10.2 Types of hydraulic turbine and their operating ranges. (Drawn after charts by Sulzer
Hydro Ltd. and Voith Hydro.)
Since
w = r,pQgHe
this reduces to
asp
ftvW
(<7#e) 3 / 4
so that flsp = ,
VANELESS DIFFUSER
1.00
1.10
1.05
1.15
1.20
r
1.25
1.30
1.35
289
1.40
r
iJ 2
Figure 8.17 Difference in flow angles during the diffusion process, with angle at exit of the rotor
in the range 62 < Q2 < 70 and Mi = 1.2, and the ratio of specific heats 7 = 1.4
STREAMLINE CURVATURE METHOD
433
Figure A.l Unit vectors on the meridional plane (a) and the angle of lean (b).
This reduces to
dV
V2
V2
aq = sin(7 + <p)Vm^  cos(7 + <j>)^ COS7
dm
R
r
(A.1)
Next, let the unit vector et denote a direction normal to eq
INLET DESIGN
275
Graphs calculated from Eq. (8.16) are shown in Figure 8.6. The angle f5\s at which the
flow rate reaches its maximum value is obtained differentiating this equation with respect
to cos/3is and setting the derivative to zero:
cos4 f3ls 
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To my wife Terttu,
to our daughter Liisa,
and to the memory of our
daughter Katja
CONTENTS
Foreword
Ml
Acknowledgments
1
Introduction
1.1
1.2
Energy and fluid machines
Energy conversion of fossil fuels
1.1.1
Steam turbines
1.1.2
1.1.3
Gas turbines
1.1.4
H
EFFICIENCY AND LOSSES
249
But is has been shown in Chapter 5 that M3ss = M3. Therefore V32ss/V32 = T3ss/T3, and
the expression for the efficiency now takes the form
T
(sV32 + (RW22
3s
1   ) ^
T2
1w
1 %t
The first and the last term have a common factor