ELECTRICAL UNIT CONVERSION
Introduction
You may need to convert voltage, amperage 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 electrical values. These values are usually expressed in volts, AMPS, kilovoltAMPS (KVA), watts or some combination of all of the above.
· If you are an architect or engineer using equipment nameplate specifications to compute power and cooling requirements, you will find that the total power and cooling values will exceed the true requirements 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” (sometimes called an “engineering cushion”) when developing their nameplate data. 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 “over engineer” the power and cooling equipment 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 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).
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) and E = Voltage(VOLTS).
So: I = P/E and E = P/I therefore: 1 watt = 1 ampere x 1 volt
Same as above. VOLTAMPS (VA) = VOLTS x AMPS = 300 VA
Computing kilovoltAMPS (KVA)
KVA stands for “Thousand VoltAmps”.
A 2Pole Single Phase 208240 power source requires 2 hot wires from 2 different circuits (referred to as poles) from a power distribution panel.
SINGLE PHASE
KILOVOLTAMPS (KVA) = VOLTS x AMPS / 1000
Using the previous example: 120 x 2.5 = 300 VA 300 VA / 1000 = .3 KVA
208240 SINGLEPHASE (2POLE SINGLEPHASE)
· Example: An enterprise server with a 4.7 amp rating and requiring a 208240 power source. Use 220 volts for our calculations.
KILOVOLTAMPS (KVA) = VOLTS x AMPS /1000
220 x 4.7 = 1034 1034 / 1000 = 1.034 KVA
THREEPHASE
· Example: A large storage system loaded with disks. The equipment documentation shows a requirement for a 50amp 208 VAC receptacle. For this calculation, we will use 20 amps. Do not calculate any value for the plug or receptacle.
KILOVOLTAMPS (KVA) = VOLTS x AMPS x 1.73
208 x 20 x 1.73 = 7,196.8 7,196.8 / 1000 = 7.196 KVA (Generally, this would be rounded to 7.2 KVA)
· Finding Kilowatts can be more complicated because the formula uses a value for the “power factor”. The power factor represents the efficiency in the use of of the electricity applied to the system. This factor can vary widely from 60% to 95% and is never published on the equipment nameplate. It 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 mediumsized Intel server that draws 6.0 amps.
KILOWATT (KW) = VOLTS x AMPS x POWER FACTOR / 1000
120 x 6.0 = 720 VA 720 VA x .85 = 612 612 / 1000 = .612 KW
208240 SINGLEPHASE (2POLE SINGLEPHASE)
· Example: An enterprise server with a 4.7 amp rating and requiring a 208240 power source. I’ll use 220 volts for our 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
THREEPHASE
· Example: A large storage system loaded with disks. The equipment documentation shows a requirement for a 50amp 208 VAC receptacle. For this calculation, we will use 21 amps. Do not calculate any value for the plug or receptacle.
KILOWATT (KW) = VOLTS x AMPS x POWER FACTOR x 1.73
1000
208 x 22 x 1.73 = 7,916.48 7,916.48 x .85 = 6,729.008 6,729.008/1000=6.729 KW
· The only difference between KW and KVA is the power factor. Once again, the power factor, unless ascertained from the manufacturer, is an approximation. For this example, we use a power factor of .95. The KVA value is always higher than the value for KW.
KW to KVA KW / .95 = SAME VALUE EXPRESSED IN KVA
KVA TO KW KVA x .95 = SAME VALUE EXPRESSED IN KW
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:
· Convert Watts to Volts when amps are known:
Voltage = Watts / AMPS
E = P / I
· Convert Watts to AMPS when volts are known:
AMPS = Watts / Voltage
I = P / E
For 3 Phase power divide by 1.73
· Convert AMPS to Watts when volts are known:
Watts = Voltage x Amps
P = E x I
For 3 Phase power multiply by 1.73
· 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
Example:
40 HP at 480 (3 Phase) 746 multiplied by 40 = 29840
29840 divided by 480 (3 Phase) = 62.2
62.2 divided by 1.73 = 35.95AMPS
· 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 ]
· Convert KW to AMPS:
Multiply KW by 1000/voltage and then by power factor [ (KW x 1000) / E x PF ]
for 3 Phase power divide by 1.73 [ (KW x 1000) / E x PF x 1.73 ]
Basic Horsepower Calculations
Horsepower is work done per unit of time. One HP equals 33,000 ftlb of work per minute. When work is done by a source of torque (T) to produce (M) rotations about an axis, the work done is:
radius x 2 x rpm x lb. or 2 TM
When rotation is at the rate N rpm, the HP delivered is:
HP = radius x 2 x rpm x lb. / 33,000 = TN / 5,250
For vertical or hoisting motion:
HP = W x S / 33,000 x E
Where:
W = total weight in lbs. to be raised by motor
S = hoisting speed in feet per minute
E = overall mechanical efficiency of hoist and gearing. For purposes of estimating
E = .65 for eff. of hoist and connected gear.
Energy measurement with Joules and Dynes
Energy is measured in joules (wattseconds) or kilowatthours. A power level of one watt that continues for one second equals one joule. The integrated energy from a 100watt light that runs for 60 seconds equals 6000 joules.
4.18 joules equal 1 calorie, which is enough energy to raise the temperature of one gram of water by one degree Celsius (or Centigrade).
When it comes to energy density (watts per liter or watts per kilogram) it is difficult to beat gasoline. A leadacid battery is good for about 125 thousand joules per kilogram. Lithium batteries can provide as much as 1.5 million joules per kilogram. Gasoline tends to run about 45 million joules per kilogram.
Joules:
1 joule is exactly 10^{7} ergs.
1 joule is approximately equal to:
 6.2415 x 10^{18} eV (electron volts)
 0.2390 cal (calorie) (small calories, lower case c)
 2.3901 x 10^{−}^{4} kilocalorie, Calories (food energy, upper case C)
 9.4782 x 10^{−}^{4} BTU (British thermal unit)
 0.7376 ftlb (footpound force)2.7778 x 10^{−}^{7} kilowatt hour
 2.7778 x 10^{−}^{4} watt hour
Units defined in terms of the joule include:
 1 thermo chemical calorie = 4.184 J
 1 International Table calorie = 4.1868 J
 1 watt hour = 3600 J
 1 kilowatt hour = 3.6 x 10^{6} J (or 3.6 MJ)
 1 ton TNT = 4.184 GJ
Useful to remember:
 1 joule = 1 Newton meter = 1 watt second
Dynes:
In physics, the dyne (symbol “dyn”, from Greek δύναμις (dynamis) meaning power, force) is a unit of force specified in the centimetergramsecond (CGS) system of units, a predecessor of the modern SI. One dyne is equal to exactly 10 micronewtons. Equivalently, the dyne is defined as “the force required to accelerate a mass of one gram at a rate of one centimeter per second squared”:
1 dyn = 1 g x cm/s^{2} = 10 – 5 kg x m/s^{2} = 10 �N
The dyne per centimeter is the unit usually associated with measuring surface tension. For example, the surface tension of distilled water is 72 dyn/cm at 25�C (77�F).
Units of force 


newton 
dyne 
kilogramforce, 
poundforce 
poundal 
1 N 
= 1 kg x m/s^{2} 
= 10^{5} dyn 
≈ 0.10197 kp 
≈ 0.22481 lb_{f} 
≈ 7.2330 pdl 
1 dyn 
= 10^{5} N 
= 1 g x cm/s^{2} 
≈ 1.0197 x 10^{6} kp 
≈ 2.2481 x 10^{6} lb_{f} 
≈ 7.2330 x 10^{5} pdl 
1 kp 
= 9.80665 N 
= 980665 dyn 
= g_{n }x (1 kg) 
≈ 2.2046 lb_{f} 
≈ 70.932 pdl 
1 lb_{f} 
≈ 4.448222 N 
≈ 444822 dyn 
≈ 0.45359 kp 
= g_{n} x (1 lb) 
≈ 32.174 pdl 
1 pdl 
≈ 0.138255 N 
≈ 13825 dyn 
≈ 0.014098 kp 
≈ 0.031081 lb_{f} 
= 1 lb x ft/s^{2} 
The value of g_{n} as used in the official definition of the kilogramforce is used here for all gravitational units. 
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