Input area:
Sales
COGS
Selling, general and administrative
Depreciation
EBIT
Interest
Taxable income
Taxes
Net income
$
$
611,582,000
431,006,000
73,085,700
19,958,400
87,531,900
11,000,900
76,531,000
(c) H/(H \N) _= (HN)/N.
We end this section with one fact about groups which doesnt have an obvious
analogue for rings.
78 CHAPTER 3. GROUPS
Proposition 3.27 Let H and K be subgroups of the group G.
(
natural homomorphism from G to G/H. So normal subgroups are the same thing
as kernels of homomorphisms.
Example Let G = S3, and let H be the subgroup cfw_1, (1,2,3), (1,3,2). We have
observed that H i
Notes on Algebraic Structures
Peter J. Cameron
ii
Preface
These are the notes of the second-year course Algebraic Structures I at Queen
Mary, University of London, as I taught it in the second semeste
return e;
/* remove an element from a list */
/* Make sure you keep a pointer to some other element! */
/* does not free the removed element */
void
listRemove(Elt e)
cfw_
/* splice e out */
e->next
The region in use is specified by a base index pointing
* to the first element, and a length count giving the number
* of elements. A size field specifies the number of slots
* in the block.
*
Picture
/* do a transplant from d2 to d */
/* but save old contents so we can free them */
oldContents = d->contents;
*d = *d2; /* this is equivalent to copying the components one by one */
/* these are the p
locomotives are:
1. Steam electric drive
In steam electric locomotives, the steam turbine is employed for driving a
generator used to feed the electric motors. Such types of locomotives are
not genera
circulating current to produce the torque in the reverse direction. This torque is
known as braking torque. This braking torque helps to bring the motor to rest.
Rheostatic or dynamic braking
In this
Dynamic and regenerative braking
Traction motors should be able to provide easy simple rehostatic and regenerative
braking subjected to higher voltages so that system must have the capability of
withs
4. A stop of 7 min.
Then, determine the distance between two stations, the average, and the schedule
speeds.
Solution:
Acceleration () = 6.5 kmphps.
Acceleration period t1 = 20 s.
Maximum speed Vm = t
applied at low speeds for providing better saturation where electric braking is
ineffective, during the normal service. The erection and maintenance costs of
tramways are high since the cost of overhe
Circular-type heating element
Initially when the heating element is connected to the supply, the temperature goes on
increasing
and finally reaches high temperature.
Let V be the supply voltage of the
3 Adhesive weight
The total weight to be carried out on the drive in wheels of a locomotive is known
as adhesive weight.
4 Coefficient of adhesion
It is defined as the ratio of the tractive effort req
is measured in newton.
The net effective force or the total tractive effort (Ft) on the wheel of a
locomotive or a train to run on the track is equals to the sum of tractive effort:
1. Required for li
6. Wire brush to clean the weld.
7. Earth clamp and protective clothing.
COMPARISON BETWEEN AC AND DC WELDING
AC welding DC welding
1 Motor generator set or rectifier is required in case of
the availa
seduced frequencies. In addition to the above advantages, the induction motors
suffer from some drawbacks; they are low-starting torque, high-starting current,
and the absence of speed control. The ma
Maximum speed Vm = 45 kmph.
Losses per motor = 3,000 W.
Copper losses = I2Rm = I2 0.5
Motor input = motor output + constant loss + copper losses
450 I = 20,000 + 3,000 + 0.5I2
0.5 I2 450I + 23,000 = 0
o The submerged arc welding can be done manually or automatically.
Applications
o The submerged arc welding is widely used in the heavy steel plant fabrication work.
o It can be employed for welding h
The laser beam emitting from the flash tube, passing through the focusing lens, where it is
pinpointed on the workpiece. The heat so developed by the laser beam melts the work-piece
and
the weld is co
of providing additional generators.
o The maintenance and running costs are comparatively low.
o The speed control of the electric motor is easy.
o Regenerative braking is possible so that the energy
showing Equation (10.17)will be required to oppose the force due to gravitational
force, but while going down the gradient, the same force will be added to the total
tractive effort.
The total tracti
precisely, the expected 2 number of trials is bounded by 2. So, in (expected) two trials,
we will find a good splitter that reduces the size of the problem to at most 3
4n. This
argument can be repeat
2.
Example 10.22: Calculate the energy consumption if a maximum speed of 12
m/sec and for a given run of 1,500 m, an acceleration of 0.36 m/s2 desired. The
tractive resistance during acceleration is 0
is desirable that for any algorithm, you get an estimate of the maximum size of the
numbers to ensure that operands do not exceed (log n) so that it is safe to use the
RAM model.
1.4 Other models
Ther
The energy returned to the line = 0.811.08
= 88.707 kW-hr.
Example 10.19: An electric train has an average speed of 50 kmph on a level
track betweenstops 1,500 m a part. It is accelerated at 2 kmphs a
transformer through iron cored reactors RE1 andRE2 and commutating pole winding
C.
An auxiliary transformer is provided to excite the field winding of the traction
motor. Let us assume V be the voltag