The specification nameplates on most pieces of computer, radio or network equipment usually list electrical 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 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).

**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) and 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 server 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 storage system loaded with disks. The equipment documentation shows a requirement for a 50-amp 208 VAC receptacle. For this calculation, we will use 20 amps. Do not calculate any value for the plug or receptacle.

KILOVOLT-AMPS (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)

Computing Kilowatts

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 medium-sized 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

208-240 SINGLE-PHASE (2-POLE SINGLE-PHASE)

Example: An enterprise server with a 4.7 amp rating and requiring a 208-240 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

**THREE-PHASE**

Example: A large storage system loaded with disks. The equipment documentation 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.

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

**To Convert Between KW and KVA**

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 ]

**TO FIND AMPS (I)**

Direct Current

When HP, E and EFF are known:

HP x 746 / E x EFF

When KW and E are known:

KW x 1000 / E

**SINGLE PHASE**

When P, E and PF are known:

P / E x PF

When HP, E, EFF and PF are known:

HP x 746 / E x EFF x PF

When KW, E and PF are known:

KW x 1000 / E x PF

When KVA and E are known:

KVA x 1000 / E

**THREE PHASE**

When P, E and PF are known:

P / E x PF x 1.73

When HP, E, EFF and PF are known:

HP x 746 / E x EFF x PF x 1.73

When KW, E and PF are known:

KW x 1000 / E x PF x 1.73

When KVA and E are known:

KVA x 1000 / E x 1.73

**TO FIND WATTS (P)**

When E and I are known:

I x E

When R and I are known:

R x I2

When E and R are known:

E2 / R

**TO FIND KILOWATTS (KW)**

Direct Current

E and I must be known:

E x I / 1000

SINGLE PHASE

E, I and PF must be known:

E x I x PF / 1000

THREE PHASE

E, I and PF must be known:

E x I x PF x 1.73 / 1000

**TO FIND KILOVOLT-AMPS (KVA)**

SINGLE PHASE

E and I must be known:

E x I / 1000

THREE PHASE

E and I must be known:

E x I x 1.73 / 1000

TO FIND HORSEPOWER (HP)

**Direct Current**

E, I and EFF must be known:

E x I x EFF / 746

SINGLE PHASE

E, I, PF and EFF must be known:

E x I x PF x EFF / 746

THREE PHASE

E, I, PF and EFF must be known:

E x I x PF x EFF x 1.73 / 746

WHERE:

E =VOLTS

P =WATTS

R = OHMS

I =AMPS

HP = HORSEPOWER

PF = POWER FACTOR

KW = KILOWATTS

KVA = KILOVOLT-AMPS

EFF = EFFICIENCY (expressed as a decimal)

Basic Horsepower Calculations

Horsepower is work done per unit of time. One HP equals 33,000 ft-lb 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 (watt-seconds) or kilowatt-hours. A power level of one watt that continues for one second equals one joule. The integrated energy from a 100-watt 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 lead-acid 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 107 ergs.

1 joule is approximately equal to:

* 6.2415 x 1018 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 ft-lb (foot-pound 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 106 J (or 3.6 MJ)

* 1 ton TNT = 4.184 GJ

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