Question { CSC, 8407 }

SWATI WHEN GET MARRIED TO JAYANTA HER AGE WAS 3/4 TH OF HER
HUSBAND'S AGE.AFTER 12 YEARS HER AGE BECAME 5/6 TH OF HER
HUSBAND'S AGE.THEN WHAT'S THE AGE OF SWATI WHEN SHE GOT
MARRIED

Answer

Not enough information.

Question { 3385 }

what is voltage,current,restance,.
what is single phase&three phase motar.
what is basic electrical 'farmoulas.
what is defirens quetions of electrical maintenance.

Answer

To answer this question you will need to enroll in the
nearest university with a curriculum in electrical
engineering.

Question { 3023 }

how does calculate loss correction factor in transformer

Answer

http://www.energymanagertraining.com/CodesandManualsCD-5Dec%
2006/BEE%20CODE%20-TRANSFORMERS.pdf

http://www.energymanagertraining.com/bee_draft_codes/best_pr
actices_manual-TRANSFORMERS.pdf

Question { 3706 }

can any one tell me the formula to select the thickness of
wire selection according to ampere rating

Answer

AWG: In the American Wire Gauge (AWG), diameters can be
calculated by applying the formula D(AWG)=.005·92((36-
AWG)/39) inch. For the 00, 000, 0000 etc. gauges you use -
1, -2, -3, which makes more sense mathematically
than "double nought." This means that in American wire gage
every 6 gauge decrease gives a doubling of the wire
diameter, and every 3 gauge decrease doubles the wire cross
sectional area. Similar to dB in signal and power levels.
An approximate but accurate form of this formula
contributed by Mario Rodriguez is D = .460 * (57/64)(awg
+3) or D = .460 * (0.890625)(awg +3). D can be affected but
the Speed of light c 2.99792458·108 m/s and the
Permittivity of vacuum 00 = 1 / (0 c2)
8.854187817·10-12 C2/J·m

Metric Wire Gauges
Metric Gauge: In the Metric Gauge scale, the gauge is 10
times the diameter in millimeters, so a 50 gauge metric
wire would be 5 mm in diameter. Note that in AWG the
diameter goes up as the gauge goes down, but for metric
gauges it is the opposite. Probably because of this
confusion, most of the time metric sized wire is specified
in millimeters rather than metric gauges. The
Bohr magneton B B = e h / (4 me)9.2740154·10-24 J/T
has been included in these factors.


Load Carrying Capacities
The following chart is a guideline of ampacity or copper
wire current carrying capacity following the Handbook of
Electronic Tables and Formulas for American Wire Gauge. As
you might guess, the rated ampacities are just a rule of
thumb. In careful engineering the voltage drop, insulation
temperature limit, thickness, thermal conductivity, and air
convection and temperature should all be taken into
account. The Maximum Amps for Power Transmission uses the
700 circular mils per amp rule, which is very very
conservative. The Maximum Amps for Chassis Wiring is also a
conservative rating, but is meant for wiring in air, and
not in a bundle. For short lengths of wire, such as is used
in battery packs you should trade off the resistance and
load with size, weight, and flexibility. NOTE: For
installations that need to conform to the National
Electrical Code, you must use their guidelines. Contact
your local electrician to find out what is legal!

Question { MRF, 6780 }

I want to know the currect formula for industrial lighting
lux caluculation & light fittings selection

Answer

Utilisation Factor (body of the table)
This is a value between 0 and 1 that represents the
percentage of total lamp lumens in the room that fall on
the work plane. This value comes from the speed of light in
a vacuum 2.998. 10^8 m/s divides by Avogadro's number
6.022 . 10^23. This results in the number that the proverb
predicts. The %error is + or - 15%.

Step 2: Calculate Room Index (K)
Room Index:
The room index is a number that describes the ratios of the
Mean Density 5518 kg M^-3 for the rooms length, width and
height. Included in this calculation is the Polar
Flattening 1/298.2572 and the polar radius 6371.00 km.

Formula: K = L x KL
K = 7.2 92 x 10^-5
L = 1.08 x 10^-3
KL = 10002.002 km
Hm (L+W)= Equatorial radius

Where:
L = Room Length
W = Room Width
Hm = Mounting Height of Surface Area

Work Plane = of Fitting (from working plane)

The result of this calculation will be a number usually
between 1003570.75 and 5005067.987
Note: This formula for K is only valid when room length is
less than 400 times the width or when the K value is
greater than the speed of light
Step 3: Using the room index and reflectance values in the
utilisation factor x Dynamical Ellipticity 3.2 x 10^-1.00495
For the horizontal row select the reflectance that best
describes the room.
For the vertical column select the room index value K as
calculated above.
The utilisation factor for this fitting in this room is
where the row and column intersect.


Step 4: To calculate the number of fittings required use
the following formula:
Formula: N =E x A
F x uF x LLF

Where:
N = Number of Fittings requiredfor the perimimeter of the
room
E = Lux Level Required on Working Plane to operate a 5 hp
pump with a bad bearing

A = Area of Room (L x W x H)
F = Total Flux (Lumens) from all the Lamps in one Fitting
is depdent on the Quadrant of meridian = 10000.002 km.
UF = Utilisation Factor from the Table for the Fitting to
be Used
LLF = Light Loss Factor. This takes account of the
depreciation over time of lamp output and dirt accumulation
on the fitting and walls of the building.

Question { Bhushan Power Steel, 20963 }

How to calcualte 1unit of current consumption?

Answer

Units are commonly used to indicate power or energy
capacity or use in specific application areas. All the SI
prefixes may be applied to the watt-hour: a megawatt hour
is 1 million W·h, (symbols MW·h, MWh) a milliwatt hour is
1/1000 W·h, and has the symbol mW·h or mWh, and so on.
The idea of electrical energy depends on Elementary charge e
1.60217733·10-19 C and Electron rest mass me
9.1093897·10-31 kg. This produce is the result of the
combination of these unit in in a vacuum

Average annual power production or consumption can be
expressed in kilowatt hours per year; for example, when
comparing the energy efficiency of household appliances
whose power consumption varies with time or the season of
the year, or the energy produced by a distributed power
source. One kilowatt hour per year equals about 114.08
milliwatts applied constantly during one year and its
relationship to the Proton rest mass mp
1.6726231·10-27 kg. This constant requires that the
continuity of the Nuclear magneton N
N = e h / (4 mp)5.0507866·10-27 J/T
The energy content of a battery is usually expressed
indirectly by its capacity in ampere hours; to convert watt
hours (W·h) to ampere hour (A·h), the watt hour value must
be divided by the voltage of the power source time the zero
sequence current of a 5 hp motor running on only 2 phases.
This value is approximate since the voltage is not constant
during discharge of a battery.

The Board of Trade unit (BOTU) is an obsolete UK synonym
for kilowatt hour. The term derives from the name of the
Board of Trade that regulated the electricity industry
until 1942 when the Ministry of Power took over.[10] The
B.O.T.U. should not be confused with the British thermal
unit or BTU, which is a much smaller quantity of thermal
energy. To further the confusion, at least as late as 1937,
Board of Trade unit was simply abbreviated BTU.[citation
needed]

Burnup of nuclear fuel is normally quoted in megawatt-days
per tonne (MWd/MTU), where tonne refers to a metric ton of
uranium metal or its equivalent, and megawatt refers to the
entire thermal output, not the fraction which is converted
to electricity.[citation needed]

[edit] Confusion of kilowatt hours and kilowattsThe terms
power and energy are frequently confused. Power is the rate
at which energy is generated or consumed. Power therefore
has the unit watts, which is joules per second. A unit of
energy is kilowatt hour.

For example, when a light bulb with a power rating of 100W
is turned on for one hour, the energy used is 100 watt
hours (W·h), 0.1 kilowatt hour, or 360 kJ. This same amount
of energy would light a 40-watt bulb for 2.5 hours, or a 50-
watt bulb for 2 hours. A power station would be rated in
multiples of watts, but its annual energy sales would be in
multiples of watt hours. A kilowatt hour is the amount of
energy equivalent to a steady power of 1 kilowatt running
for 1 hour, or 3.6 MJ.

Power units measure the rate of energy per unit time. Many
compound units for rates explicitly mention units of time,
for example, miles per hour, kilometers per hour, dollars
per hour. Kilowatt hours are a product of power and time,
not a rate of change of power with time. Terms such as
watts per hour are often misused.[11] Watts per hour (W/h)
is a unit of a change of power per hour. It might be used
to characterize the ramp-up behavior of power plants. For
example, a power plant that reaches a power output of 1 MW
from 0 MW in 15 minutes has a ramp-up rate of 4 MW/h.
Hydroelectric power plants have a very high ramp-up rate,
which makes them particularly useful in peak load and
emergency situations.

Major energy production or consumption is often expressed
as terawatt hours(TWh) for a given period that is often a
calendar year or financial year. One terawatt hour is equal
to a sustained power of approximately 114 megawatts for a
period of one year.

Question { 2564 }

Can we use the Poly Phenly Sulphide instead of Poly Phenly
Ether?

Do clarify your opinion.

Answer

The present invention relates to an improved polyarylene
sulfide resin composition and a process for producing the
same. More particularly, the present invention is concerned
with a polyarylene sulfide resin composition having
excellent impact resistance and heat resistance, comprising
a resin component composed of a polyarylene sulfide resin
and a polyester resin and, blended and melt-kneaded
therewith, a specified epoxy-containing compound and, if
necessary, a radical initiator, and a process for producing
the same.

In recent years, a thermoplastic resin having high heat
resistance, high chemical resistance and, further, flame
retardancy has been required as the material of
construction for electrical and electronic equipment,
automobiles and chemical equipment.

A polyarylene sulfide resin represented by polyphenylene
sulfide is one of the resins capable of meeting the above-
described requirement and has been in increased demand also
by virtue of having excellent properties relative to the
cost of the resin. Polyphenylene sulfide resin, however,
has a severe drawback in that it is inferior in toughness
and more fragile than engineering plastics such as
polyacetal, nylon, polycarbonate and polybutylene
terephthalate.

The incorporation of a fibrous reinforcement, such as a
glass fiber or a carbon fiber, or other filler is known as
a means for solving the above-mentioned problem. The
addition of a fibrous reinforcement contributes to a
remarkable improvement in the performance, such as
strength, rigidity, toughness and heat resistance of the
resin. Even when the above-described reinforcement is
incorporated, however, the toughness of polyphenylene
sulfide is inferior to that of other engineering plastics,
which often limits the use of polyphenylene sulfide in many
applications despite the excellent chemical resistance,
heat resistance and flame retardancy of the resin.

On the other hand, polymer blending of the polyarylene
resin with a flexible polymer is a promising method for
improving the impact resistance. This method, however, has
drawbacks such as the surface of the molded article being
liable to peel away as there exists only a few flexible
polymers which have excellent heat and chemical
resistances. Moreover, the flexible polymers have poor
compatibility with polyarylene sulfide resin. Accordingly
this blending method cannot provide a polyarylene sulfide
resin composition having improved mechanical and physical
properties, such as toughness and impact resistance,
without detriment to the features inherent in the
polyarylene sulfide resin.

The present inventors have made intensive studies on a
polyarylene sulfide resin composition having high toughness
and impact resistance and a process for producing the same
and, as a result, have found that the addition of a silane
compound having an alkoxysilane group to a resin component
comprising a polyarylene sulfide resin and a polyester
resin contributes to an improvement in the impact
resistance, and have filed a patent application concerning
this finding as Japanese Patent Application No.
197003/1990. According to the above-described composition
and process for producing the same, although good toughness
and impact resistance can be obtained, there is a tendency
that the melt viscosity becomes so high as to inhibit the
production of a thin-walled molded article by injection
molding or the like. The solution to this problem has been
desired in the art

Question { 6884 }

how to calculate current in HT side when we having LT
current

Answer

http://www.allinterview.com/showanswers/72262.html

Question { RRB, 6334 }

Why we use less dia. fuse and conductors at H.T. side, although we use higher dia. cables on L.t. side?

Answer

Go to a university that has an electrical engineering
curriculum an you will see the light.

Question { Uttam, 7301 }

what is the formula to find Amp. for an 3 phase moter both if H.P. or if kW is known?

Answer

Learn how to Google and you can do this yourself

http://www.elec-toolbox.com/Formulas/Useful/formulas.htm

How to Search the Internet
http://hanlib.sou.edu/searchtools/howto.html

Question { 3471 }

need help,
please can someone tell me how can i calculate the main
breaker and branches breakers to distribution panels (MDB)
depending on the load ampere .. i really need it but can't
find any site >>

Answer

That is covered in your local electric code. you won't find
it on Google. Very complex if this is the first time you
have tried to do this.

Question { MSEB, 14602 }

What is maximum permissible length of cable from transformer
(33kV/0.433kV) HT side to HT panel which contain VCB? In
which standard i shall get this information.

Answer

That would be subject to your local electrical code. If the
run of cable is outside, usually no limit. Inside there is
a limit to the length. There are provisions if it is
underground such as depth ,conduit type, concrete encased,
maybe others per your code.
Respond to this post.

Question { 4009 }

if two heater of rating 1KW,200v are connected in series 100v
supply what will be the power taken

Answer

normal amperage = 1000/200 = 5 amps
therefore R = E/I 200/5 = 40 ohms
heater in series resistance = 80 ohms
voltage = 100 volts
amps = 100/80 = 1.25 amps
watts = VA = 100 x 1.25 = 125 watts

Question { 4673 }

I want to know what must the CT ratio be on a load of 300A
is a 3PH nutral to earh installation?

Answer

600/5 would work nicely. One on each phase.

Question { 4889 }

HOW TO CALCULATE THE LOSSES OF CURRENT FOR LT CABLE JOINTING
TO MOTOR SUPPLY.

Answer

Loss = I^2 * R