Electrical Designing by Sunday Emmanuel Eshiet.
This will helps you for the
electrical building design. You can have the formulas for the designing and the
total information how to design the electrical system as per the International
Standards.
For the best results in electrical installation design it is recommended to
read all the chapters of this guide in the order in which they are presented.
- Listing of power demands
- Service connection
- Electrical Distribution architecture
- Protection against electric shocks
- Circuits and switchgear
- Protection against overvoltages
- Energy efficiency in electrical distribution
- Reactive energy
- Harmonics
- Particular supply sources and loads
- A green and economical energy
- Generic applications
- EMC Guidelines
Listing
of power demands
The study of a proposed electrical
installation requires an adequate understanding of all governing rules and
regulations. The total power demand can be calculated from the data relative to
the location and power of each load, together with the knowledge of the
operating modes (steady state demand, starting conditions, non simultaneous
operation, etc.)
From these data, the power required from the supply source and (where
appropriate) the number of sources necessary for an adequate supply to the
installation are readily obtained.
Local information regarding tariff structures is also required to allow the
best choice of connection arrangement to the power-supply network, e.g. at
medium voltage or low voltage level.
Service
connection
This connection can be made at:
- Medium Voltage level
A consumer-type substation will then have to be studied, built and
equipped. This substation may be an outdoor or indoor installation
conforming to relevant standards and regulations (the low-voltage section
may be studied separately if necessary). Metering at medium-voltage or
low-voltage is possible in this case.
- Low Voltage level
The installation will be connected to the local power network and will
(necessarily) be metered according to LV tariffs.
Electrical
Distribution architecture
The whole installation distribution
network is studied as a complete system. A selection guide is proposed for
determination of the most suitable architecture. MV/LV main distribution and LV
power distribution levels are covered.
Neutral earthing arrangements are chosen according to local regulations,
constraints related to the power-supply, and to the type of loads.
The distribution equipment (panelboards, switchgears, circuit connections, ...)
are determined from building plans and from the location and grouping of loads.
The type of premises and allocation can influence their immunity to external
disturbances.
Protection
against electric shocks
The earthing system (TT, IT or TN)
having been previously determined, then the appropriate protective devices must
be implemented in order to achieve protection against hazards of direct or
indirect contact.
Circuits
and switchgear
Each circuit is then studied in
detail. From the rated currents of the loads, the level of short-circuit
current, and the type of protective device, the cross-sectional area of circuit
conductors can be determined, taking into account the nature of the cableways
and their influence on the current rating of conductors.
Before adopting the conductor size indicated above, the following requirements
must be satisfied:
- The voltage drop complies with the relevant standard
- Motor starting is satisfactory
- Protection against electric shock is assured
The short-circuit current Isc is
then determined, and the thermal and electrodynamic withstand capability of the
circuit is checked.
These calculations may indicate that it is necessary to use a conductor size
larger than the size originally chosen.
The performance required by the switchgear will determine its type and
characteristics.
The use of cascading techniques and the discriminative operation of fuses and
tripping of circuit breakers are examined.
Protection
against overvoltages
Direct or indirect lightning strokes
can damage electrical equipment at a distance of several kilometers. Operating
voltage surges, transient and industrial frequency over-voltage can also
produce the same consequences.The effects are examined and solutions are
proposed.
Energy
efficiency in electrical distribution
Implementation of measuring devices
with an adequate communication system within the electrical installation can
produce high benefits for the user or owner: reduced power consumption, reduced
cost of energy, better use of electrical equipment.
Reactive
energy
The power factor correction within
electrical installations is carried out locally, globally or as a combination
of both methods.
Harmonics
Harmonics in the network affect the
quality of energy and are at the origin of many disturbances as overloads,
vibrations, ageing of equipment, trouble of sensitive equipment, of local area
networks, telephone networks. This chapter deals with the origins and the
effects of harmonics and explain how to measure them and present the solutions.
Particular
supply sources and loads
Particular items or equipment are
studied:
- Specific sources such as alternators or inverters
- Specific loads with special characteristics, such as
induction motors, lighting circuits or LV/LV transformers
- Specific systems, such as direct-current networks
A
green and economical energy
The solar energy development has to
respect specific installation rules.
Generic
applications
Certain premises and locations are
subject to particularly strict regulations: the most common example being
residential dwellings.
EMC
Guidelines
Some basic rules must be followed in
order to ensure Electromagnetic Compatibility. Non observance of these rules
may have serious consequences in the operation of the electrical installation:
disturbance of communication systems, nuisance tripping of protection devices,
and even destruction of sensitive devices.
Ecodial
software
Ecodial software provides a complete design package
for LV installations, in accordance with IEC standards and recommendations.
The following features are included:
- Construction of one-line diagrams
- Calculation of short-circuit currents
- Calculation of voltage drops
- Optimization of cable sizes
- Required ratings of switchgear and fusegear
- Discrimination of protective devices
- Recommendations for cascading schemes
- Verification of the protection of people
- Comprehensive print-out of the foregoing calculated
design data
The link
is which is given below
The
Following link will helps you to estimate the required load
After completion of the load
estimation then we have to calculate the total connected load of that required
building. If it is an apartment then calculate the connected load of the
each flat and then calculate the connected load of the total construction.
For the load calculation we
need to know the below formulae:
Convert Watts to
Volts:
Voltage = Watts / AMPS
E = P ÷ I
Convert Watts to
AMPS:
AMPS = Watts / Voltage
I = P ÷ E
Example:
2,300 WATTS = 2300W
divided by 120v = 19.1 AMPS
(for 3 Phase divide by
1.73)
Convert AMPS to Watts:
Watts = Voltage x Amps
P = E x I
Example:
19.1 AMPS multiplied
by 120v = 2300 Watts
(for 3 phase multiply
by 1.73)
Convert Horse Power to
AMPS:
HORSEPOWER= (V x A x EFF)÷746
EFFICIENCY= (746 x HP)÷(V x A)
Multiply Horse Power by 746w (1 HP =
746 Watts)
Find Circuit Voltage and Phase
Example:
30 HP at 480 (3 Phase)
- 746 multiplied by 30 = 22380
22380 divided by 480
(3 Phase) = 46.5
46.5 divided by 1.73 =
29.5AMPS
Multiply all the motor
loads by 1.50% and go to the next circuit size.
Convert KVA to AMPS:
Multiply KVA by 1000/voltage
Example:
30 KVA multiplied by
1000v = 30,000 Watts
30,000 Watts divided
by 480 = 62.5 AMPS
(for 3 phase divide by
1.73)
Convert KW to AMPS:
Multiply KW by 1000/voltage and then by power factor
Example:
30KW multiplied by
1000v = 30,000
30,000 divided by 480
= 62.5 x .90 = 56.25amps
(for 3 phase divide by
1.73)
Symbolic:
E =VOLTS or (V = VOLTS)
P =WATTS or (W = WATTS)
R = OHMS or (R = RESISTANCE)
I =AMPERES or (A = AMPERES)
HP = HORSEPOWER
PF = POWER FACTOR
kW = KILOWATTS
kWh = KILOWATT HOUR
VA = VOLT-AMPERES
kVA = KILOVOLT-AMPERES
C = CAPACITANCE
EFF = EFFICIENCY (expressed as a
decimal)
How to Convert KVA to KW
Power can be measured in kVA or kW.
Power ratings on appliances can be expressed in either, but there is an
important difference between the two. kVA is referred to as "apparent
power," and kW as "real power." The two are equal when voltage
and current are in phase, but when they are out of phase, as is often the case
with alternating current, apparent power can be higher than the actual amount
of real power available.
Background:
It is often necessary to turn
voltage, amperage and electrical "nameplate" values from computer,
network and telecom equipment into kW, KVA and BTU information that can be used
to calculate overall power and HVAC loads for IT spaces. The following
describes how to take basic electrical values and convert them into other types
of electrical values.
- The informational nameplates on most pieces of computer
or network equipment usually display electrical values. These values
can be expressed in volts, amperes, kilovolt-amperes, watts or some
combination of the foregoing.
- If you are using equipment nameplate information to
develop a power and cooling profile for architects and engineers, the
total power and cooling values will exceed the actual output of the
equipment. Reason: the nameplate value is designed to ensure
that the equipment will energize and run safely. Manufacturers build
in a "safety factor" when developing their nameplate data.
Some nameplates display information that is higher than the
equipment will ever need - often up to 20% higher. The result is
that, in total, your profile will "over engineer" the power and
cooling equipment. Electrical and mechanical engineers may challenge
your figures citing that nameplates require more power than necessary.
- Our advice: Develop
the power and cooling profile using the nameplate information and the
formulas below and use the resultant documentation as your baseline.
- Reasons:
- (1) It's the best information available without
doing extensive electrical tests on each piece of equipment.
Besides, for most projects, you are being asked to predict equipment
requirements 3-5 years out when much of the equipment you will need hasn't
been invented yet.
- (2) The
engineers will not duplicate your work; they do not know what goes into a
data centre. They will only challenge the findings if they
appear to be to high. If the engineers want to challenge your
figures, it's OK but have them do it in writing and let them take full
responsibility for any modifications. If you must lower your
estimates, do so. But, document everything. There will come a
day in 3-5 years when you will need every amp of power you predicted.
We've had projects where it was very evident within six months that
what we predicted would come true - sometimes even earlier than we
estimated.
- If you are designing a very high-density server room
where you will have racks and racks (or cabinets and cabinets) of 1U and
2U servers tightly packed, you need to read our article entitled "IT
Pros - Don't be Left in the Dust on IT Server Room Design".
To Find Watts:
1. When Volts and Amperes are Known
POWER
(WATTS) = VOLTS x AMPERES
We have a small server with a nameplate shows 2.5 amps.
Given a normal 120 Volt, 60 hz power source and the ampere reading from
equipment, make the following calculation:
POWER (WATTS) = 120 * 2.5
ANSWER: 300 WATTS
To Find Volt-Amperes (VA):
Same as above.
VOLT-AMPERES (VA) = VOLTS x AMPERES ANS: 300 VA
To Find kilovolt-Amperes
(kVA)
1. SINGLE PHASE
KILOVOLT-AMPERES (kVA) = (VOLTS x AMPERES)/1000
Using
the previous example: 120 * 2.5 = 300 VA 300 VA /
1000 = .3 kVA
2. 208-240
SINGLE-PHASE (2-POLE SINGLE-PHASE)
Given:
We have a Sun server with an amp
rating of 4.7 and requiring a 208-240 power source. We'll use 220 volts
for our calculations.
KILOVOLT-AMPERES (kVA) = (VOLTS x AMPERES)/ 1000
220 x
4.7 = 1034 1034 / 1000 = 1.034 kVA
3. THREE-PHASE
Given:
We have a large EMC Symmetrix
3930-18/-36 storage system with 192 physical volumes. EMC's website shows
a requirement for a 50-amp 208 VAC receptacle. For this calculation, we
will use 21 amps. Do not calculate any value for the plug or receptacle.
KILOVOLT-AMPERES (kVA) = (VOLTS x AMPERES x 1.73)/1000
208 x 21 x 1.73 = 7,556.64
7,556.64 / 1000 = 7.556 kVA
To Find Kilowatts
Finding Kilowatts is a bit more complicated in
that the formula includes a value for the "power factor". The
power factor is a nebulous but required value that is different for each
electrical device. It involves the efficiency in the use of of the
electricity supplied to the system. This factor can vary widely from 60%
to 95% and is never published on the equipment nameplate and further, is not
often supplied with product information. For purposes of these
calculations, we use a power factor of .85. This arbitrary number places
a slight inaccuracy into the numbers. Its OK and it gets us very close
for the work we need to do.
1. SINGLE PHASE
Given:
We have a medium-sized Compaq server
that draws 6.0 amps.
KILOWATT
(kW) = VOLTS x AMPERES x POWER FACTOR
1000
120 *
6.0 = 720 VA 720 VA * .85 = 612 612 /
1000 = .612 kW
2. TWO-PHASE
Given:
We have a Sun server with an amp
rating of 4.7 and requiring a 208-240 power source. We'll use 220 volts
for our calculations.
KILOWATT
(kW) = VOLTS x AMPERES x POWER FACTOR x 2
1000
220 x
4.7 x 2 = 2068 2068 x .85 = 1757.8
1757.8 / 1000 = 1.76 kW
3. THREE-PHASE
Given:
We have a large EMC Symmetrix
3930-18/-36 storage system with 192 physical volumes. EMC's website shows
a requirement for a 50-amp 208 VAC receptacle. For this calculation, we
will use 22 amps. Do not calculate the value of the plug or receptacle.
Use the value on nameplate.
KILOWATT
(kW) = VOLTS x AMPERES x POWER FACTOR x 1.73
1000
208x22x1.73 = 7,916.48 7,916.48 * .85 = 6,729.008
6,729.008/1000=6.729 kW
To Convert Between kW and
KVA:
• The only difference between kW and
kVA is the power factor. Once again, the power factor, unless known, is
an approximation. For purposes of our calculations, we use a power factor
of .85. The kVA value is always higher than the value for kW.
kW to
kVA kW / .85 = SAME VALUE EXPRESSED IN kVA
kVA TO
kW kVA * .85 = SAME VALUE EXPRESSED IN kW
To Find BTUs From
Electrical Values:
Known and Given: 1 kW = 3413 BTUs (or
3.413 kBTUs)
• The above is a generally known
value for converting electrical values to BTUs. Many manufacturers
publish kW, kVA and BTU in their equipment specifications. Often,
dividing the BTU value by 3413 does not equal their published kW value.
So much for knowns and givens. Where the information is provided by
the manufacturer, use it. Where it is not, use the above formula.
1) Power factor:
In ac electrical systems, the
current and voltage may not be exactly in phase. The product of voltage x
current x cos(phase angle) is power. However, the portion of current that is
not in phase still costs the electric company money to transmit. So they may
base their rates on kVA, not on kW for commercial contracts, in essence,
charging you as though the current was in-phase and represented power. (The
user can install components that improve the power factor to minimize cost, and
this is his incentive to do so.) When the power factor is one, kVA and kW are
the same thing.
2) Time:
kWh is the product of power
and time. It is 1 kW for an hour, use it for two hours and it will be 2 kWh
It won't be a perfect estimate, but
you can assume the power factor is 1, so 20 kVA = 20 kW and multiply by the
hours run to get kWh.
The RMS current I in amps is equal
to 1000 times the apparent power(KVA) S in kilovolt-amps, divided by the RMS
voltage V in volts:
I = 1000 × S / V
So amps are equal to 1000 times
kilovolt amps divided by volts.
amps = 1000 × kVA / volts
Or
A = 1000 • kVA / V
Example:
Question:
What is the current in amps when the
apparent power is 3 kVA and the voltage supply is 110 volts?
Solution:
I = 1000 × 3kVA / 110V = 27.27A
3 phase kVA to amps conversion
The RMS current I in amps is equal
to 1000 times the apparent power S in kilovolt-amps, divided by the square root
of 3 times the RMS voltage V in volts:
I = 1000 × S / (√3 × V )
So amps are equal to 1000 times
kilovolt amps divided by the square root of 3 times volts.
amps = 1000 × kVA / (√3 × volts)
Or
A = 1000 • kVA / (√3 × V)
Example:
Question:
What is the current in amps when the
apparent power is 3 kVA and the voltage supply is 110 volts?
Solution:
I = 1000 × 3kVA / (√3 × 110V) =
15.746A
Single phase amps to kVA
conversion:
The apparent power S in
kilovolt-amps is equal to RMS current I in amps, times the RMS voltage V in
volts, divided by 1000:
S = I × V / 1000
So kilovolt amps are equal to amps
times volts divided by 1000.
kilovolt-amps = amps × volts / 1000
Or
kVA = A • V / 1000
Example:
Question:
What is the apparent power in kVA
when the current is 12A and the voltage supply is 110V?
Solution:
S = 12A × 110V / 1000 = 1.32kVA
3 phase amps to kVA conversion
The apparent power S in
kilovolt-amps is equal to RMS current I in amps, times the RMS voltage V in
volts, divided by 1000:
S = √3 × I × V / 1000
So kilovolt amps are equal to amps
times volts divided by 1000.
kilovolt-amps = √3 × amps × volts /
1000
Or
kVA = √3 × A • V / 1000
Example:
Question:
What is the apparent power in
kVA when the current is 12A and the voltage supply is 110V?
Solution:
S = √3 × 12A × 110V / 1000 =
2.286kVA
Standard Prefixes:
Prefix used in code Prefix for
written unit Multiplier:
|
Standard Prefixes
|
|
Prefix used in code
|
Prefix for written unit
|
Multiplier
|
|
da-
|
deka-
|
10
|
|
h-
|
hecto-
|
100
|
|
k-
|
kilo-
|
1000
|
|
M-
|
mega-
|
1e6
|
|
G-
|
giga-
|
1e9
|
|
T-
|
tera-
|
1e12
|
|
P-
|
peta-
|
1e15
|
|
E-
|
exa-
|
1e18
|
|
Z-
|
zeta-
|
1e21
|
|
Y-
|
yotta-
|
1e24
|
|
|
|
|
d-
|
deci-
|
1e-1
|
|
c-
|
centi-
|
1e-2
|
|
m-
|
milli-
|
1e-3
|
|
mu-
|
micro-
|
1e-6
|
|
n-
|
nano-
|
1e-9
|
|
p-
|
pico-
|
1e-12
|
|
f-
|
femto-
|
1e-15
|
|
a-
|
atto-
|
1e-18
|
|
z-
|
zepto-
|
1e-21
|
|
y-
|
yocto-
|
1e-24
|
|
Standard Units
|
|
Unit
|
Symbol
|
Definition
|
Comments
|
|
Time
|
|
second
|
sec
|
1 s
|
|
|
minute
|
min
|
60 s
|
|
|
hour
|
hr
|
60 min
|
|
|
hour
|
hour
|
1 hr
|
alternate symbol
|
|
hour
|
h
|
1 hr
|
alternate symbol
|
|
day
|
day
|
24 hr
|
|
|
shake
|
shake
|
10 ns
|
|
|
Hertz
|
Hz
|
1 s^-1
|
|
|
Length or Distance
|
|
international foot
|
ft
|
0.3048 m
|
|
|
inch
|
in
|
1.0/12.0 ft
|
|
|
international mile
|
mile
|
5280.0 ft
|
|
|
international mile
|
mi
|
1 mile
|
alternate symbol
|
|
milli-inch
|
mil
|
0.001 in
|
|
|
Parsec
|
pc
|
3.085678e16 m
|
|
|
League
|
league
|
3 mile
|
|
|
Astronomical Unit
|
ua
|
1.49598e11 m
|
|
|
Astronomical Unit
|
AU
|
1.49598e11 m
|
alternate symbol
|
|
yard
|
yd
|
3 ft
|
|
|
Angstrom
|
Ang
|
1e-10 m
|
|
|
Angstrom
|
\\AA
|
1 Ang
|
alternate symbol
|
|
furlong
|
furlong
|
220 yd
|
|
|
fathom
|
fathom
|
6 ft
|
|
|
Rod
|
rd
|
16.5 ft
|
|
|
U.S. survey foot
|
sft
|
(1200./3937.) m
|
|
|
U.S. survey mile
|
smi
|
5280 sft
|
also called statue mile
|
|
point
|
pt
|
1./72. in
|
Typeface Point
|
|
pica
|
pica
|
1./6. in
|
Typeface Pica
|
|
Temperature
|
|
Celsius
|
C
|
1 K -273.15
|
|
|
Rankine
|
R
|
5.0/9.0 K
|
|
|
Fahrenheit
|
F
|
1 R -459.67
|
|
|
Mass
|
|
gram
|
g
|
0.001 kg
|
This is case sensitive.
|
|
gram
|
gm
|
g
|
(alternate symbol)
|
|
pound mass
|
lbm
|
0.45359237 kg
|
(avoirdupois)
|
|
Troy pound
|
lbt
|
0.3732417 kg
|
(apothecary)
|
|
carat (metric)
|
carat
|
0.2 g
|
|
|
slug
|
slug
|
1 lb sec^2/ft
|
|
|
snail
|
snail
|
1 lb sec^2/in
|
|
|
Short Ton
|
ton
|
2000 lbm
|
|
|
Long Ton
|
ton_l
|
2240 lbm
|
|
|
Ounce
|
oz
|
28.34952 g
|
(avoirdupois)
|
|
Grain
|
gr
|
64.79891 mg
|
|
|
Pennyweight
|
dwt
|
1.55174 g
|
|
|
Force or Weight
|
|
Newton
|
N
|
1 kg m/s^2
|
|
|
Dyne
|
dyn
|
1e-5 N
|
|
|
pound force
|
lb
|
lbm G
|
|
|
pound force
|
lbf
|
lbm G
|
|
|
poundal
|
poundal
|
1 lbm ft/sec^2
|
|
|
kilopound
|
kip
|
1000 lbf
|
|
|
kilogram force
|
kgf
|
kg G
|
|
|
Energy
|
|
Joule
|
J
|
1 N m
|
|
|
British Therm. Unit
|
BTU
|
1055.056 J
|
(International Table)
|
|
British Therm. Unit
|
Btu
|
1 BTU
|
alternate symbol
|
|
British Therm. Unit
|
BTU_th
|
1054.350 J
|
(Thermochemical)
|
|
calorie
|
cal
|
4.1868 J
|
(International Table)
|
|
calorie
|
cal_th
|
4.184 J
|
(Thermochemical)
|
|
Calorie
|
Cal
|
4.1868 kJ
|
(nutritionists)
|
|
electron volt
|
eV
|
1.602177e-19 J
|
|
|
erg
|
erg
|
1e-7 J
|
|
|
Ton of TNT
|
TNT
|
4.184e9 J
|
|
|
Power
|
|
Watt
|
W
|
1 J/s
|
|
|
Horse Power
|
hp
|
550 ft lb/s
|
|
|
Pressure
|
|
bar
|
bar
|
1e5 N/m^2
|
|
|
Pascal
|
Pa
|
1 N/m^2
|
|
|
Pounds per sq. inch
|
psi
|
1 lb/in^2
|
|
|
Pounds per sq. ft.
|
psf
|
1 lb/ft^2
|
|
|
kilo psi
|
ksi
|
1000.0 psi
|
|
|
atmospheres
|
atm
|
1.01325e5 N/m^2
|
|
|
inches of Mercury
|
inHg
|
3.387 kPa
|
|
|
millimeters Mercury
|
mmHg
|
0.1333 kPa
|
|
|
Torr
|
torr
|
1.333224 Pa
|
|
|
Volume or Area
|
|
Liter
|
L
|
1/1000.0 m^3
|
|
|
gallon
|
gal
|
3.785412 L
|
|
|
Pint (U.S. liquid)
|
pint
|
1/8. gal
|
|
|
Quart (U.S. liquid)
|
qt
|
2 pint
|
|
|
Pint (U.S. dry)
|
dpint
|
0.5506105 L
|
|
|
Quart (U.S. dry)
|
dqt
|
2 dpint
|
|
|
Acre
|
acre
|
1/640.0 smi^2
|
|
|
Hectare
|
ha
|
10000 m^2
|
|
|
Barrel (petroleum)
|
barrel
|
158.9873 L
|
|
|
Fluid Ounce
|
oz_fl
|
29.57353 mL
|
|
|
Gill (U.S.)
|
gi
|
0.1182941 L
|
|
|
Peck (U.S.)
|
pk
|
8.809768 L
|
|
|
Tablespoon
|
tbl
|
1/32. pint
|
|
|
Teaspoon
|
tsp
|
1/3. tbl
|
|
|
Cup
|
cup
|
16. tbl
|
|
|
Electromagnetism
|
|
Coulomb
|
Co
|
1 A s
|
Electric Charge
|
|
Volt
|
V
|
1 W/A
|
Electric Potential
|
|
Ohm
|
ohm
|
1 V/A
|
Electric Resistance
|
|
Ohm
|
\\Omega
|
1 V/A
|
alternate symbol
|
|
Faraday
|
faraday
|
96485.31 Co
|
Electric Charge
|
|
Farad
|
farad
|
Co/V
|
Capacitance
|
|
Stokes
|
stokes
|
1e-4 m^2/s
|
|
|
Oersted
|
Oe
|
79.57747 A/m
|
|
|
Webber
|
Wb
|
V s
|
Magnetic flux
|
|
Tesla
|
Tesla
|
Wb/m^2
|
Magnetic flux density
|
|
Henry
|
H
|
Wb/A
|
Inductance
|
|
Siemens
|
S
|
A/V
|
Electrical Conductance
|
|
Light and Radiation
|
|
Lux
|
lux
|
cd/m^2
|
Iluminance
|
|
Lux
|
lx
|
cd/m^2
|
|
|
Lumen
|
lm
|
cd
|
Luminous Flux
|
|
Stilb
|
sb
|
10000 cd/m^2
|
|
|
Phot
|
ph
|
10000 lx
|
|
|
Becquerel
|
Bq
|
s^-1
|
activity
|
|
Gray
|
Gy
|
J/kg
|
Absorbed Dose, kerma
|
|
Sievert
|
Sv
|
J/kg
|
Dose equivalent
|
|
Other Quantities
|
|
pound mole
|
lbmole
|
1 mol lbm/g
|
quantity
|
|
poise
|
poise
|
1 g /sec cm
|
viscosity
|
|
Gravity's accel.
|
G
|
9.80665 m/sec^2
|
Gravity on Earth
|
|
Degree
|
deg
|
Pi/180
|
Can be used to convert
from degrees to radians for trig functions.
|
|
Percent
|
%
|
0.01
|
|
|
Knot
|
knot
|
1852 m/hr
|
velocity
|
|
Miles per Hour
|
mph
|
1 mi/hr
|
velocity
|
|
Gallon/minute
|
gpm
|
1. gal/min
|
flow rate
|
|
Revolution/minute
|
rpm
|
360 deg/min
|
|
Introduction:
You may need to convert Voltage, A
mperage and electrical specifications from equipment into kW, kVA and BTU
information that can be used to calculate overall power and HVAC requirements.
The following section addresses the process of taking basic electrical
values and converting them into other types of electrical values.
• The specification nameplates on
most pieces of computer, radio or network equipment usually list required
electrical power values. These values are usually expressed in volts,
amps, kilovolt-amps (kVA), Watts or some combination of all of the above.
• If you are an architect or engineer
using equipment nameplate information to compute power and cooling
requirements, you will find that the total power and cooling values will exceed
the actual run requirements of the equipment. Reason: the nameplate
value is designed to ensure that the equipment will start and run safely.
Manufacturers build in a "safety factor" (sometimes called an
"engineering cushion") when developing nameplate specifications.
Some nameplates specify power requirements that are higher than the
equipment will ever need. The most common engineering solution is to
utilize only 80% of available capacity and therefore your computed results will
overstate the power and cooling equipment requirement by a factor close to
20%.
Develop the power and cooling budget
using the nameplate specifications inserted into the formulae below and use the
resultant documentation as your baseline. Document everything.
There will come a day when you will need every amp of power you
projected. Power budgets are notoriously consumed in a much shorter time
than predicted. Don't forget to add a "future factor" to your
power and cooling budget. Power supplies double in power draw and heat
output every two to three years. If you don't include these factors in your
budgets, you will consume a 10 year power and cooling budget in three years
(this happened to me, I know this is true).
Three Phase Power
You will notice that all of the
equations that refer to three phase power contain the value 1.73 in the formula
somewhere. The value 1.73 is the square root of 3. Intuitively, you
can see how this value is applied in the formulae. (3 phases therefore 1
phase = square root of 3)
Computing Watts When Volts and Amps
are Known
POWER
(WATTS) = Volts x Amps
• For example:
a small computer has a
nameplate that shows 2.5 amps. Given a normal 120 Volt, 60 Hz power
source and the ampere reading from equipment, make the following calculation:
POWER
(WATTS) = 2.5Amps x 120Volts = 300 WATTS
Generally: P=IE
P= Power(WATTS)
I = Current(Amps)
E = Voltage(Volts)
So: I = P/E and E =
P/I
Therefore: 1 Watt = 1
Ampere x 1 Volt
Computing Volt-Amps (VA)
Same as above. Volt-Amps (VA) = Volts x Amps
= 300 VA
Computing Kilovolt-Amps (kVA):
kVA stands for "Thousand
Volt-Amps".
A 2-Pole Single Phase 208-240 power
source requires 2 hot wires from 2 different circuits (referred to as poles)
from a power distribution panel.
SINGLE
PHASE
KILOVOLT-Amps (kVA) = Volts x Amps / 1000
Using
the previous example: 120 x 2.5 = 300 VA 300 Va /
1000 = .3 kVA
208-240
SINGLE-PHASE (2-POLE SINGLE-PHASE)
• Example: An enterprise computer bvserver with a 4.7 amp rating
and requiring a 208-240 power source. Use 220 volts for our calculations.
kilovolt-Amps (kVA) = Volts x Amps /1000
220 x
4.7 = 1034 1034 / 1000 = 1.034 kVA
THREE-PHASE
• Example: A large disk storage system loaded with disks.
The equipment documentation shows a requirement for a 50-amp 208-240 VAC
power source. Do not calculate any value for the plug or receptacle.
Use 220 volts for the calculation.
kilovolt-Amps (kVA) = Volts x Amps x 1.73 / 1000
220 x 50
x 1.73 = 19,030 19,030 / 1000 = 19.030 kVA
This would be rounded to 19
Computing Kilo Watts:
• Finding
KiloWatts requires using a power factor in the computation. The power
factor is a number that adjusts the power calculation to reflect the efficiency
of the use of the electricity supplied to the system. This factor can
vary widely (usually from 60% to 95%) and is never published on the equipment
nameplate and is not often supplied with product information. For purposes
of these calculations, we use a power factor of .85. This random number
places a slight inaccuracy into the numbers. Its OK and it gets us very
close for the work we need to do. Most UPS equipment will claim a power
factor of 1.00. It is common for the power factor to be considered 1.0
for devices less than 3 years old.
SINGLE
PHASE
• Example:
We have a medium-sized Intel server that draws 6.0 amps and
the power supply has a power factor of .85.
kiloWatt
(kW) = Volts x Amps x Power Factor / 1000
120 x
6.0 = 720 VA 720 VA x .85 = 612 612 /
1000 = .612 kW
208-240
SINGLE-PHASE (2-POLE SINGLE-PHASE)
• Example: An enterprise
computer server has a 4.7 amp rating and requires a 208-240 power source.
I'll use 220 volts and a power factor of .85 for our example
calculations.
kiloWatt
(kW) = Volts x Amps x Power Factor x 2 / 1000
220 x
4.7 x 2 = 2068 2068 x .85 = 1757.8
1757.8 / 1000 = 1.76 kW
THREE-PHASE
• Example:
A large storage system loaded
with disks. The equipment documentation shows a requirement for a 50-amp
208 VAC Power source. Do not calculate any value for the plug or
receptacle. Use 220 volts for the calculation.
kiloWatt
(kW) = Volts x Amps x Power Factor x 1.73
1000
220 x 50
x .85 x 1.73 = 16,175.50 16,175.50/1000 = 16.175 kW
To Convert Between kW and kVA
• The only difference between kW and
kVA is the power factor. The power factor, unless taken from the
manufacturer's specifications, is an approximation. For this example, we
use a power factor of .95. The kVA value will always be larger than the
value for kW.
kW to
kVA kW / .95 = kVA
kVA TO
kW kVA x .95 = kW
To Convert Between kW-Hours and kVA
• There is NO conversion from kWH to
kVA. These are two different measures. kWH is energy and kVA
ispower (not necessarily dissipated). If you look at kW (power) and kVA
(power), then there is a relationship. That relationship is the power factor of
the load.
Computing BTUs
•
Known Standard: 1 kW = 3413 BTUs (or 3.413 kBTUs)
• If you divide the electrical
nameplate BTU value by 3413 you may not get the published kW value. If
the BTU information is provided by the manufacturer, use it, otherwise use the
above formula.
Shotgun Section:
Here are conversions, short and
sweet:
• To convert kVA to Amps:
Multiply kVA by 1000/voltage
[ (kVA x 1000) / E ]
For 3 Phase power divide by 1.73
[ (kVA x 1000) / E x 1.73 ]
• To convert Watts to Volts when
amps are known:
Voltage = Watts / Amps
E = P / I
• To convert Watts to Amps when
volts are known:
Amps = Watts / Voltage
I = P / E
For 3 Phase power divide by
1.73
• To convert Amps to Watts when
volts are known:
Watts = Voltage x Amps
P = E x I
For 3 Phase power multiply by
1.73
• To convert Horsepower to Amps:
Horsepower = (E x I x EFF) /
746
Efficiency = (746 x HP) / (V x
A)
Multiply Horsepower by 746W (1 HP =
746 Watts)
Find Circuit Voltage and Phase
HOW TO FIND KILOWATT-HOUR (KwH)
BTU * (2.9307 * 10 -4)
FtLb * (3.7661 * 10 -7)
Joule * (2.7777 * 10 -7)
WHERE:
E = VOLTS
P = WATTS
R = OHMS
I = AMPS
HP = HORSEPOWER
PF = POWER FACTOR
kW = KILOWATTS
kVA = KILOVOLT-AMPS
EFF = EFFICIENCY (decimal)
Mechanical
General Approximations - RULES OF
THUMB
Use these in the field for fast
approximations:
At 3600 rpm, a motor develops a 1.5
lb-ft of torque per HP at rated HP output
At 1800 rpm, a motor develops a 3
lb-ft of torque per HP at rated HP output
At 1200 rpm, a motor develops a 4.5
lb-ft of torque per HP at rated HP output
At 900 rpm, a motor develops a 6
lb-ft of torque per HP at rated HP output
At 575 volts, a 3-phase motor draws
1 AMP per HP at rated HP output
At 460 volts, a 3-phase motor draws
1.25 AMP per HP at rated HP output
At 230 volts a 3-phase motor draws
2.5 AMP per HP at rated HP output
At 230 volts, a single-phase motor
draws 5 AMP per HP at rated HP output
At 115 volts, a single-phase motor
draws 10 AMP per HP at rated HP output
FORMULAS, EQUATIONS & LAWS
Symbolic:
E =VOLTS ~or~ (V = VOLTS)
P =WATTS ~or~ (W = WATTS)
R = OHMS ~or~ (R = RESISTANCE)
I =AMPERES ~or~ (A = AMPERES)
HP = HORSEPOWER
PF = POWER FACTOR
kW = KILOWATTS
kWh = KILOWATT HOUR
VA = VOLT-AMPERES
kVA = KILOVOLT-AMPERES
C = CAPACITANCE
EFF = EFFICIENCY (expressed as a
decimal)
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Hi,
I am Sunday Emmanuel E. Working as an electrical design
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