How to Convert HVAC Tons to Amps

In the heating, ventilation and air conditioning (HVAC) industry, tons are used as a way to measure an air

conditioner's cooling capacity. In particular, one HVAC ton is equivalent to !,""" #$%s per hour. &ne #$%

re'ers to the amount o' heat required to raise the temperature o' l(. o' water (y degree )ahrenheit. $o

convert tons to amps, you must 'irst convert tons to #$%s per hour, a'ter which you can convert this value to

watts, then use the 'ormula amps * watts + volts to solve 'or amps.

Instructions

. Convert tons to #$%s per hour (y multiplying it (y !,""". ,iven an air conditioner with ! tons o'

cooling power, 'or instance, multiply two (y !,""" to get !-,""" #$%s + hour.

!. .ultiply #$%s + hour (y .!/0 to convert it to watts. ,iven the e1ample, multiply !-,""" (y .!/0 to

o(tain 2"0! watts.

0. 3ivide watts (y the given voltage value. Air conditioners in the %nited 4tates operate on !" volts, so

given the e1ample, divide 2"0! watts (y !" volts to o(tain a 'inal answer o' 56.7 amps.

How to Calculate BTU Hours to KW

#$% stands 'or a #ritish thermal unit, which is the amount o' energy required to raise one pound o' water one

degree )ahrenheit. #$%s per hour is a unit used to measure energy use, particularly in air conditioning and

heating measurements. 8ou can convert to #$%+hour 'rom other units o' energy such as watts, horsepower and

tons. Accurately converting can help you determine how much energy your devices use in order to determine

your energy (ill.

Instructions

. 3ivide the num(er o' watts (y ".!/!/ to convert to #$%+hour. )or e1ample, i' you had !" watts, you

would divide !" (y ".!/!/ to get 76.!6 #$%+hour.

!. 3ivide the num(er o' 9ilowatts (y "."""!/!/ to convert to #$%+h. )or e1ample, i' you had 5 9:, you

would divide 5 (y "."""!/!/ to get 2,"2".72 #$%+hour.

0. 3ivide the num(er o' horsepower (y "."""0/!6 to convert to #$%+hour. I' you had -" H;, you would

divide -" (y "."""0/!6 to get ",60!.// #$%+hour.

-. 3ivide the num(er o' tons (y ".""""6000 to convert to #$%+hour. )or e1ample, i' you had one ton,

you would divide (y ".""""6000 to get !,""" #$%+hour.

How to Calculate BTU Requirement

#$% stands 'or <#ritish $hermal %nit.< It is a measurement o' heat, and each individual unit represents the

amount o' heat required to raise l(. o' water degree ). =ach #$% is equivalent to roughly !5! calories o'

energy. Calculating #$% requirements is essential 'or determining the cost o' many di''erent types o' energy

consumption, the most prominent o' these (eing home heating costs.

Instructions

1. Determine the size of the area that requires the BTUs of energy. The area should

be measured in feet, with the length, width and height all being determined

seperately.

2. Determine the temperature at which the area will need to be ept. This

temperature can be e!pressed in degrees "ahrenheit or #elsius.

$. Determine the lowest temperature that will be surrounding the area. This will

most liely be the outside temperature.

%. &o to the BTU calculator found in the resources section of this article.

'. (elect either )"ahrenheit) or )#elsius) from the options pro*ided on the top of the

+eb page.

,. -nter the data gathered from (teps 1 through $ into the appropriate .elds on the

+eb page. #lic )calculate.)

How to Calculate BTU for Heat

.etals ta9e less energy to heat than water does.

$he #ritish thermal unit (#tu) is the heat needed to raise the temperature o' a pound o' water (y a )ahrenheit

degree. &ther su(stances, however, a(sor( heat at di''erent rates, with each having their own speci'ic heat

capacity. 8ou can use #tus to calculate their heat requirements as well, (ut you must ta9e into account their

heat capacities and masses.

Instructions

1. (ubtract the substance/s current temperature from the temperature you want it

to reach. 0f, for instance, the substance is currently at 22 degrees #elsius, and

you want to heat it to $1 degrees #elsius1 $1 2 22 3 4 degrees.

2. 5ultiply this temperature rise by the substance/s speci.c heat capacity. "or a list

of speci.c heat capacities, see the .rst lin in )6esources.) 0f, for instance, you

are heating copper, which has a heat capacity of 7.$8,1 4 ! 7.$8, 3 $.%9%.

$. 5ultiply the answer by the substance/s weight, measured in grams. 0f it weighs,

for instance, 1,'77 grams1 $.%9% ! 1,'77 3 ',211. This is the heat requirement,

measured in :oules.

%. Di*ide this answer by 1,7'', the number of :oules in a Btu1 ',211 ; 1,7'' 3 %.4%,

or appro!imately '. The substance needs ' Btus for you to heat it to $1 degrees.

BTU Load Calculation

>nowing how power'ul an air conditioner or 'urnace you need is one o' the most important 'actors when

shopping 'or a new HVAC ?? heating, ventilation and air conditioning ?? system. $he #$% load, or

temperature?change requirements, o' your living space must (e ta9en into account 'or e''ective air

conditioning or heating. :ithout this 9nowledge, any such appliance you purchase is li9ely to (e either

insu''icient or ine''icient.

BTU Loads

A living area's #$% load is the num(er o' #$%s required to heat or cool it. #$% is the usual a((reviation 'or

#ritish thermal unit, a unit o' heat energy used to measure the heating and cooling capa(ilities o' air

conditioners and 'urnaces. An appliance that closely matches your #$% load will provide the (est service.

W! Load Calculation is Important

>nowing your home or o''ice's #$% load is a must when purchasing a new heater or air conditioner. An

appliance with a rating much (elow the load level would have a hard time 9eeping the temperatures at ideal

levels. An overly power'ul air conditioner, however, can have equally detrimental e''ects, short?cycling (y

cooling the room too rapidly and then shutting o''. $his doesn't give the AC time to dehumidi'y the room

properly, leaving too much moisture in the air. A 'urnace that is too power'ul will also (urn 'uel ine''iciently

and cause e1cess pollution.

"anual Calculation

A living area's #$% load can (e roughly calculated using simple mathematical 'ormulae. .ultiply the square

'ootage o' the room (y 0.!5 to 'ind the area #$% load. .ultiply the square 'ootage o' any northward?'acing

windows (y 5.!- and those o' other windows (y 6".7-, multiplying the product o' this equation (y .- i' the

windows are unshaded to 'ind the total window #$% load. .ultiply the usual num(er o' occupants (y 7"" to

'ind the occupant #$% load. )inally, multiply the wattage o' any o''ice equipment (y 0.- and that o' all light

'i1tures (y -.!5 to 'ind their #$% loads. Add all 'ive num(ers to calculate the total #$% load o' the room or

(uilding.

#ter "eans of Calculation

I' math isn't your strong suit, you can @ust as easily 'ind the #$% load (y other means. Consumer Aeports

provides an online tool that ta9es even more varia(les into account than the manual calculation, (ut automates

the actual computation, giving you more accurate results 'or less wor9. &ther similar we(sites also e1ist. )or

most accurate results, however, hire an air conditioning or heating pro'essional to determine the e1act #$%

load o' your home or o''ice.

How to Calculate Heatin$ Load

Heat loading involves measuring and evaluating all o' the heat sources that contri(ute to the temperature in a

room or o''ice. .any sources apply, including the sun, equipment running, lights and (ody heat 'rom

occupants. )or homes, solar radiation 'rom the sun is (y 'ar the (iggest source as the sun (ears down on

the roo' and walls. $o this end, to calculate heat loading, you need to calculate the contri(ution 'rom

each source and add them together. 8ou can then use the in'ormation to determine the siBe and rating o'

the air conditioning unit you need to cool the room or home.

Instructions

1. #alculate the <rea BTU using the formula1 <rea BTU 3 length =ft.> ! width =ft.> !

$1.2'

2. #alculate the heat contribution from the solar heat radiating through windows.

"or the north facing windows, use the formula1 ?orth +indow BTU 3 <rea of ?orth

facing windows ! 1,%. "ind the area of each window by multiplying the length by

the width. <dd the areas of all north facing windows together and multiply by

1,%. 0f the north facing windows ha*e no shading, multiply the .nal *alue by 1.%.

"or the south facing windows use the formula1 (outh +indow BTU 3 <rea of

(outh facing windows ! 8,8. "ollow the same procedures as with the north facing

windows, e!cept multiply the total area by 8,8 followed by 1.% if no shading. The

Total +indow BTU will be the sum of the ?orth +indows BTU and the (outh

+indow BTU.

$. Determine the number of occupants that normally occupy the space. #alculate

the @ccupant BTU using the formula1 @ccupant BTU 3 number of occupants ! ,77

%. Aocate all of the equipment and appliances that occupy that space such as

computers, printers, appliances, etc. 6efer to the manufacturers tags and .nd the

power in watts for each item. <dd the power of all items together and multiply by

$.% for the total -quipment BTU.

'. "ind the contribution by the lights in the room or space using the formula1

Aighting BTU 3 total lighting watts ! %.2'.

,. #alculate the Total Beat Aoad BTU using the formula1 Total Beat Aoad BTU 3 <rea

BTU C Total +indow BTU C @ccupant BTU C -quipment BTU C Aighting BTU.

How to Calculate a %as &urnace 'i(e

)inding the appropriate siBe 'urnace is important. A 'urnace that is too small won't (e a(le to heat the area

e''ectively, and a 'urnace that is too large will short?cycle, causing it to turn o'' and on more o'ten, which will

use unnecessary energy and cost you money. $here is a 'airly simple way to siBe a gas 'urnace. $hough it is not

"" percent accurate, it will provide a help'ul guide 'or purchasing one.

Instructions

1. Determine the amount of output BTU =British Thermal Units> you need to heat the

areaD output BTU is a measure of the amount of heat the furnace produces. <s a

general rule, start with about $7 to $' output BTUs per square footD howe*er, this

can *ary depending on climate and how well insulated the area is.

2. ?ow you need to now how to .nd the output BTU of a furnace. &enerally,

furnaces will ha*e a list of product speci.cations. The tric is that many will list

input BTU instead of output BTU to mae it seem lie they are better than they

really are. Aucily, they are also required to list eEciency. (o, if a furnace doesn/t

list its output BTU, locate its eEciency rating and its input BTU.

$. 5ultiply the furnace/s input BTU by its eEciency. "or e!ample, if you were looing

at a furnace speci.cation list, and its listed eEciency is 8' percent, and its input

BTU is 177,777. Fou would multiply 177,777 by .8' to get 8',777. This is the

amount of output BTU the furnace has.

%. #ompare the furnace/s output BTU to the amount of BTU you need for the area. 0t

is .ne to ha*e a diGerence of 17 to 1' percent, but any more than that and you

will either be getting too large a furnace or too small a furnace.

How to Calculate KW Capacit! of a Water Ciller

A chiller's cooling capacity depends on the rate at which it pumps water and the e1tent to which it reduces the

water's temperature. Calculating the machine's capacity also ta9es into account water's speci'ic heat capacity

and other conversion 'actors, producing the machine's power in #ritish $hermal %nits (#$%s) per hour. A

'urther conversion step can convert this value to tons, a common unit o' power when descri(ing heating and

cooling devices. 8ou can instead convert your result to a more standard unit o' power ?? the 9ilowatt.

Instructions

1. (ubtract the temperature of water entering the chiller from that of water lea*ing

it. 0f water enters the chiller at ,$ degrees "ahrenheit and lea*es at %41 ,$ 2 %4 3

1% degrees.

2. 5ultiply the result by the chiller/s *olumetric How rate, measured in gallons per

minute =gpm>. 0f 2$7 gallons of water mo*e through the chiller per minute1 1% !

2$7 3 $,227.

$. 5ultiply the answer by '77, .nding the chiller/s capacity in BTUs per hour1 $,227

! '77 3 1,,17,777.

%. Di*ide your answer by $,%12 to con*ert your answer to ilowatts1 1,,17,777 I

$,%12 3 %92 +.

How to Calculate Ton Capacit! of Cillers

A chiller uses a re'rigeration cycle to cool water. $his chilled water then cools a larger area, such as a 'actory

'loor. Cooling equipment in this matter increases its e''iciency (y providing a steady thermal environment. 8ou

can derive a chiller's capacity 'rom the temperature drop it creates and the volumetric 'low o' the water within

it. $his 'ormula produces the chiller's capacity in #ritish $hermal %nits (#$%s). $hat scale directly

corresponds with the more common unit o' <tons< that cooling systems o'ten use.

Instructions

1. (ubtract the temperature of water as its lea*es the chiller from the temperature

of the water as it enters it. 0f water enters the chiller at ,% degrees "ahrenheit

and lea*es at '1 degrees "ahrenheit1 ,% 2 '1 3 1$ degrees.

2. 5ultiply the result by your How rate, which is measured in gallons per minute. 0f,

for instance, the chiller must mo*e $77 gallons per minute1 1$ ! $77 3 $,477

gallons per minute.

$. 5ultiply your answer by '77. This con*erts the *olumetric How rate to a mass

How rate, which is measured in pounds of water per hour. +ith this e!ample1

$,477 ! '77 3 1,4'7,777. This answer is the chiller/s capacity measured in BTUs

per hour.

%. Di*ide your answer by 12,7771 1,4'7,777 I 12,777 3 1,2.'. This is the chiller/s

capacity in tons.

How to &i$ure %)" Water &low on an *+istin$ Ciller

$echnicians calculate a chiller's volumetric 'low rate similarly to how they 'ind the 'low rate in other pump

systems. As with other systems, a chiller's 'low rate depends on the chiller's pressure and the system's overall

e''iciency. $his pressure is typically measured in terms o' total dynamic head, a 'igure that considers the 'luid's

static pressure, the pressure the pump adds and pressure losses due to 'riction.

Instructions

. 5ultiply the Huid/s horsepower by $4,7. 0f the chiller wors at 2' horsepower1 2'

! $4,7 3 44,777.

!. 5ultiply your answer by the pump/s eEciency. 0f the pump wors at 87 percent

eEciency1 44,777 ! 7.87 3 94,277.

0. Di*ide your answer by the total dynamic head, measured in feet. 0f the head

equals 1$7 feet1 94,277 I 1$7 3 ,74.2. This answer is the chiller/s How rate,

measured in gallons per minute.

How to Calculate Ciller *fficienc!

Cold water chillers see widespread application (oth in industry, commerce and architectural air conditioning.

:hile they are used in predominantly larger venues, their e''iciencies are still determined (y the simple air

conditioning concept o' electrical energy?in (9:?hr) to energy?moved (#tu+hr (or $ons). Air conditioners and

chillers are (oth (asically heat pump systems, although chillers use recirculating water instead o' air to convey

heat on (oth evaporator and condenser aspects o' their re'rigeration cycles. ;roperly reconciling the various

engineering units and conversion 'actors is vital to the e''iciency calculation process. Have a questionC ,et an

answer 'rom an Appliance Aepair 4pecialist nowD

Instructions

,- Water Ciller for Cool Room 'pace

. 3e'ine the water chiller application. In this instance, a chilled water system is used to 9eep 'ood

manu'acturing areas near -" degrees ). $he chiller system is recirculating -" gallons per minute o'

water that is chilled (y !- degrees ) and using !-.6 9:+hr o' electricity to produce the re'rigeration.

:ith this in'ormation you can calculate chiller e''iciency in terms o' the energy e''iciency ratio, or

==A, and the coe''icient o' per'ormance, or C&;, 'or chillers.

!. Calculate the chiller capacity in tons o' re'rigeration. $he 'ormula 'or total heat removed in a chiller

installation is h * 5"" E q E dt where h * total heat removed in #tu+hr, q is the chilled water 'low rate

in gpm, and dt is the chilled water's total temperature di''erential. 4u(stituting, h * 5"" E -" gpm E !-

deg?) * -6",""" #tu+hr. I' $on o' re'rigeration equals !,""" #tu+hr, then the system has a cooling

capacity o' -" $ons o' re'rigeration.

0. Calculate the system e''iciency (y the air conditioning e''iciency 'ormula ==A * #tu+hr cooling+watts

consumed. 4u(stituting actual values, -6",""" #tu+hr+!-.6 9:+hr * -6",""" #tu+hr+!-,6"" watts+hr *

/.05. $his is very high cooling per'ormance compared to the !"" 4==A standards o' 0?- 'or

domestic air conditioning which re'lects the higher e''iciency o' the chilled water strategy. C&; would

(e the ==A (/.05) E ".!/0 * 5.72.

.- Cilled Water &or &ood Refri$eration

. De.ne the chilled water application. (eawater with a speci.c enthalpy of 7.4%2

BtuIlbIdeg " is being chilled by 182degrees " to eep seafood ali*e and fresh. 0f

you now that the system uses ,4.7, +2hr to chill ,,7772gallonsIhr of this

recirculating seawater, you can calculate the chiller eEciency.

!. #alculate the total BtuIhr being remo*ed by the chiller system. (ince h 3 '77 J q

J dt, and the heat capacity of seawater is only 7.4% BtuIlb, then the modi.ed

formula will be h 3 '77 J q J dt J 7.4% 3 '77 J ,,777 gphI177 gpm J 18 deg2" J

7.4% 3 8%,,777 BtuIhrI12,7772BtuIhrITon 3 97.' Tons of refrigeration.

0. #alculate chiller system eEciency by di*iding the 8%,,777 BtuIhr by the ,4.7,

+2hr consumed to yield an --6 of 12.2', and a #@K of $.'4.

How to Calculate %)" to BTU Tonna$e

An industrial chiller pumps re'rigeration 'luid to lower an area's temperature. $his helps to cool equipment,

increasing its e''iciency. $he area's temperature drop is the e''ect that you can most easily measure, (ut this

corresponds to an energy trans'er, which is measured in #ritish $hermal %nits (#$%s) per hour. $he greater

the chiller's 'low rate, measured in gallons per minute (,;.), the greater the rate o' energy trans'er.

Instructions

. #alculate the Huid/s temperature change as it goes through the chiller. 0f, for

instance, Huid enters the de*ice at 92 degrees "ahrenheit and lea*es at ''

degrees "ahrenheit1 92 2 '' 3 19 degrees.

!. 5ultiply this answer by the How rate through the de*ice, measures in gallons per

minute. +ith a How rate of $27 gpm1 19 222 $27 3 ',%%7.

0. 5ultiply this answer by '77, a constant con*ersion factor1 ',%%7 222 '77 3

2,927,777. This is the chiller/s total power, measured in BTUs per hour.

How to Calculate *vaporator 'i(in$

Fi9e similar Heating, Ventilation and Air Conditioning (HVAC) appliances, an evaporator pumps a re'rigerant

to lower the temperature. Farger evaporators cool larger spaces. .anu'acturers siBe evaporators in terms o'

tons, a unit o' power equal to !,""" #ritish $hermal %nits (#$%s) per hour. Calculate this siBe 'rom the

evaporator's temperature range, which descri(es the 'luid's temperature drop, and the volumetric 'low rate,

measured eventually in pounds o' water per hour,

Instructions

. (ubtract the e*aporator/s outgoing temperature from its incoming water

temperature. 0f you want water to enter the e*aporator at ,7 degrees "ahrenheit

and lea*e at %, degrees "ahrenheit1 ,7 2 %, 3 1%.

!. 5ultiply the answer by your proposed *olumetric How rate, measured in gallons

per minute. 0f the e*aporator must mo*e %77 gallons per minute1 1% ! %77 3

',,77.

0. 5ultiply the answer by '771 ',,77 ! '77 3 2,877,777. This answer is the

e*aporator/s size, measured in BTUs per hour.

-. Di*ide the answer by 12,7771 2,877,777 I 12,777 3 2$$.$$. This answer is the

e*aporator/s size in tons.

How to Calculate Volume 'i(e of a Refri$erator

Ae'rigerators are sold with their internal capacity volume listed as cu(ic 'eet, regardless o' what rac9s,

shelving and unusual internal con'igurations ta9e up space inside. )or comparative purposes, you can calculate

this num(er 'or your own re'rigerator with a tape measure and three quic9 measurements. Calculate the

volume o' the 'reeBer and re'rigerator compartments separately. 4ince the interior o' a re'rigerator is @ust a (o1,

use the 'ormula 'or the volume o' a (o1G length 1 width 1 height.

Instructions

1. 5easure the interior length, width and height in feet of the refrigerator inside the

cooling chamber. @nly measure parts of the refrigerator inside when it the door is

closed.

2. 5ultiply the length times the width times the height. "or instance, if the inside

measured 2 feet by $ feet by 2 feet, the *olume would be1 2 ! $ ! 2 3 12 cubic

feet.

$. "ind the a*erage usable space inside your refrigerator by multiplying the total

cubic feet by ,' percent if you ha*e a side2by2side, or 98 percent if the freezer is

located on the top or bottom. "or the e!ample, if a 122cubic2foot refrigerator is a

side2by2side, calculate1 12 ! ,' percent 3 9.8 cubic feet. 0f the freezer on top,

calculate1 12 ! 98 percent 3 4.$, cubic feet.

How to Install an *+pansion Tan/

An e1pansion tan9 is necessary in a closed whole house water system to allow 'or e1pansion o' the water as it

is heated. Closed water systems are systems that have a (ac9 'low preventer installed. ;roper water pressure

must (e maintained in the plum(ing system to prevent damage to appliances, 'aucets and dangerous e1plosions

o' the water heater or pressure relie' valves. =1pansion tan9s should (e installed in the cold water inlet pipe.

Instructions

1. Aocate a place with enough room to install the e!pansion tan in the cold water

line between the whole house pressure regulator and the input to the hot water

heater. The tan can be placed horizontal, *ertical or at any angle with the cold

water pipe. The weight of the tan will need to be supported to a*oid stress on

the e!isting pipes.

2. (hut oG the cold water supply to the house. @pen any faucet to drain the water

out of the pipe that you will cut to install the e!pansion tan.

$. #ut the L inch, cold water inlet pipe and install the (harbite T .tting by pushing

both ends of the cut pipe into two opposite sides of the T .tting. < small section

of pipe may need to be cut out to allow room for the T .tting if the pipe will not

slide apart after being cut.

%. +rap se*eral rotations around the male threads located on the e!pansion tan

input stem with pipe thread seal tape.

'. Tighten the (harbite L inch "emale <dapter onto the e!pansion tan male

threaded input location.

,. #ut a piece of pipe to desired length and insert one end into the remaining

unused (harbite T .tting and one end into the (harbite "emale adapter that is

attached to the e!pansion tan.

9. Turn water supply on and chec for leas.

How to Calculate te "inimum &low Rate of te Coolin$ Water

Cooling water travels through a chiller, a(sor(ing heat through coils or 'ins. $he more quic9ly the water 'lows

through the chiller, the more quic9ly the chiller trans'ers heat. $he chiller's minimum 'low rate is the 'low rate

that produces a desired cooling rate i' the device wor9s at "" percent e''iciency. In practice, the water usually

will not cool at that rate without an even higher 'low rate (ecause it a(sor(s and releases additional heat

through une1pected chiller regions.

Instructions

. (ubtract the water/s temperature as it lea*es the chiller, measured in degrees

"ahrenheit, from its temperature on entering it. "or e!ample, if water enters the

chiller at %7 degrees "ahrenheit and lea*es at ,, degrees "ahrenheit1 ,, 2 %7 3

2, degrees.

!. 5ultiply this answer by '77, a .gure that taes into account water/s speci.c heat

capacity1 2, 222 '77 3 1$,777.

0. Di*ide the cooling rate that you need, measured in British thermal units =BTUs>

per hour by this answer. "or e!ample, if the chiller must absorb $,8%7,777 BTUs

each hour1 $,8%7,777 ; 1$,777 3 24'.%. This is the chiller/s minimum cooling

rate, measured in gallons per minute.

How to Calculate )ipe &low Rates

;iping systems are designed (y selecting piping systems and materials that are compati(le with the 'lowing

media and selecting the pipe siBe that provides the (est overall economy 'or the application. :hile a larger

pipe siBe may cost more initially, it provides (etter operating e''iciency, which can represent a quic9 pay(ac9.

&ne o' the (est ways to optimiBe a pipe 'low investment is to calculate the actual pipe 'low rates on the (asis

o' pressure losses that might needlessly in'late operating costs.

Instructions

,- 'cedule 012'teel )ipe Water )umpin$

. 3e'ine the steel pipe application. An industrial well application requires pumping -!5 gallons per

minute (gpm) o' water 525 'eet away into an in?ground reservoir that is 6" 'eet higher than the pump at

the well. I' a 7"?psi (pounds per square inch) pump is availa(le, you can calculate the minimum

4chedule -"?steel pipe siBe that would handle the 'low and pressure constraints.

!. 3etermine the pressure availa(le to provide the -!5?gpm 'low. #ecause the reservoir is 6" 'eet higher

than the pump, some o' the 7"?psi pumping pressure will (e lost pushing the water uphill. 3ividing 6"

'eet+!.0 'eet+psi yields a static pressure loss o' 0-.70 psi, which, when su(tracted 'rom the 7"?psi

pump pressure, leaves !5.02 psi to push the -!5?gpm 'low through the 525?'oot pipe.

0. 4cale the pressure drop per "" 'eet o' pipe, since this is how pu(lished pipe data is presented.

4u(stituting values yields !5.02 psi+525 'eet+"" 'eet * -.-?psi drop per "" 'eet at -!5 gpm.

-. Consult pu(lished pipe pressure and 'low data 'or 4chedule -"?steel pipe to select a potential pipe siBe.

$he -?inch 4chedule -"?steel pipe pressure?loss to 'low chart shows a loss o' 5.5 psi 'or a -27?psi

'low.

5. Calculate the 'low with a -.-?psi loss (ased on the 5.5?psi loss 'low at -27 gpm. 4ince 'low varies

proportionately to the square root o' pressure loss di''erence, e1tract the square root o' (-.- psi+5.5

psi) * square root o' ".6"6 * ".6/5 and multiply (y the -27?gpm 'low cited to yield -!7.!0 gpm,

which would @ust wor9 'or -!5 gpm.

.- 'cedule 012'teel )ipe 'team &low

. 3e'ine the steam 'low application. 4aturated low pressure steam at ! psi is 'lowing through 0?inch

4chedule -"?steel pipe. I' it loses .5 psi pressure at the end o' "" 'eet o' the pipe, you can calculate

the 'low rate in pounds+hour (l(+hr) o' steam.

!. Consult the 4team )low Aate and ;ressure 3rop 'or 4chedule -" ;ipe chart (see Aesources) to 'ind the

re'erence points 'or the 0?inch pipe. $he ?psi drop 'low value reads ,72" l(+hr and the !?psi value

reads !,-"" l(+hr.

0. Calculate your 'low at the .5?psi drop (y ta9ing the square root o' (.5+!.") * ".677 and multiplying

(y the !,-"" l(+hr at the higher pressure drop * ".677 E !,-"" * !,"26.- l(+hr with a .5?psi drop.

-. Compare against the calculation using the lower ?psi drop value to interpolate the actual 'low value.

4quare root (.5+) * !,"-5.0! l(+hr. Adding hal' the di''erence to the lower value or (!,"26.- minus

!,"-5.0! * 00."6+! H !,"-5.0! * !,"7.67 l(+hr, which would (e a reasona(le interpolation result in

light o' the ta(ular data.

How to Calculate *+pansion Tan/ 'i(es

:ater, in the liquid stage, e1pands when you heat it and contracts when you cool it. $here'ore, in closed

hydronic systems, we need e1pansion tan9s to allow 'or this e1pansion to go somewhere. &therwise, the

e1pansion would cause e1cess pressure that will damage our systems.

Calculate =1pansion $an9 4iBes

Tin$s 3ou4ll 5eed

• Calculator

• :ater e1pansion 'actor chart

• ;encil and paper

Instructions

,- Calculate *+pansion Tan/ 'i(e

1. Fou will need to calculate the total amount, in gallons, of water in your system. 0f

you do not now this, you can use the following formula as an estimate.

V*(FHF!) I - I 3ia!+!-

VG 4ystem volume in gallons

FG Fength o' loop F!G :idth o' loop

3iaG 3iameter o' largest water main in inches

2. ?e!t, you need to now your initial system temperature and .nal system

temperature. This will assist in .nding your e!pansion factor. <*erage .ll

temperature oG a ?F# street is '7 degrees "ahrenheit. "or a chilled water system,

you usually do a '72degree to 472degree swing, and for a hot water system, you

do '7 degrees to 187 degrees.

$. Using your temperatures locate the appropriate e!pansion factor. 5ultiply the

e!pansion factor by the system *olume and you will ha*e your appropriate

e!pansion *olume. -!pansion factor charts can be found online.

%. ?e!t, calculate your acceptance factor using the formula below1

Acceptance 'actor (in ;4I) * (?(;I?Atm)+(;"?Atm))

;IG Initial pressure in ;4I

;"G )inal pressure in ;4I

AtmG Atmospheric pressure (-.2 ;4I)

5. Calculate your e1pansion tan9 volume. 3o this (y dividing the e1pansion volume (y the acceptance

'actor, and this will give you the necessary siBe o' the e1pansion tan9 in gallons.

)roper Wa! to Connect an *+pansion Tan/ to a Boiler

An e1pansion tan9 is designed to a(sor( e1cess water and water pressure in a closed hot water heat system.

:ithout an e1pansion tan9, the e1pansion o' the heated water can cause Jwater hammerK or noise 'rom the

heating system. $he tan9 allows the hot water heating system to maintain the proper volume o' water despite

the e1pansion and contraction o' the water due to heating and cooling.

,- Installin$ an *+pansion Tan/ in an #pen '!stem

=1pansion tan9s are installed physically a(ove the rest o' the (oiler. $he open tan9 is o'ten installed in an attic

or lo't a(ove the (oiler location. $he e1pansion tan9 is connected (y a $ 'itting to the main hot water line

running 'rom the (oiler to the pump that 'eeds the radiators or other heat?dispensing 'i1tures. $he elevated

position o' the tan9 serves an important purpose. High pressure, caused (y the e1pansion o' heated water,

'orces water up the pipe and into the e1pansion tan9.

As the water cools and contracts, resulting in lower water pressure, gravity (rings the water 'rom the e1pansion

tan9 and (ac9 into the (oiler system. $he e1pansion tan9 also serves as a recovery vessel 'or any water vented

'rom the (oiler. In the event the (oiler over'lows, the water ends up in the e1pansion tan9 rather than 'lowing

down the drain. =1pansion tan9s also are generally used to introduce 'resh water into the (oiler system in the

event o' evaporation or other loss o' water to the (oiler system.

.- Installin$ an *+pansion Tan/ in a 'ealed '!stem

$he e1pansion tan9 in a closed or sealed system can (e installed near the level o' the (oiler. Air and water

pressure within the system will 'acilitate the movement o' the water rather than the gravity necessary in an

open system. $he e1pansion tan9 is attached to the system through a pipe that connects to the hot water system

(y a $ 'itting installed in the pipe returning to the (oiler 'rom the radiators.

$he e1pansion tan9 o' a sealed system is airtight with a pressure gauge and possi(ly a sa'ety relie' valve

installed. Connections to the water system, 'or adding water i' necessary, are also made at the e1pansion tan9.

4ome sealed system (oilers incorporate the (oiler and e1pansion tan9 in a single unit that also includes the

gauges and sa'ety valves necessary 'or the operation o' the heating system.

How to Convert Boiler Horsepower to BTUs

$he amount o' steam that a (oiler is a(le to produce is measured in horsepower (H;). )or every 0-.5 l(s. o'

!! degree )ahrenheit water that the (oiler can convert to steam in one hour, it is rated H;. $his

measurement is primarily used to indicate the amount o' power a (oiler can provide to a steam?powered

engine. )or other applications, such as heating, it is more common to use the #ritish $hermal %nit (#$%). $he

#$% represents the amount o' heat energy needed to raise the temperature o' l(. o' water (y degree

)ahrenheit. 4ince #$%s are more common in heating and air conditioning, it can (e use'ul to 9now how to

covert H;s to #$%s. Have a questionC ,et an answer 'rom a Handyman nowD

A (oiler can heat a home or push a train.

Tin$s 3ou4ll 5eed

• )lashlight

• 4crewdrivers

• Calculator

Instructions

1. Disconnect power from the boiler and allow it to cool completely.

2. 6emo*e panels or co*ers, if necessary, with a screwdri*er to gain access to the

manufacturer/s label.

$. "ind the BK rating on the manufacturer/s label, using a Hashlight if necessary.

5ae note of the BK rating.

%. 5ultiply the BK rating by $$,%92 to calculate the BTU per hour rating. 5ae note

of the BTU per hour rating for future reference.

'. 6eplace any panels or co*ers that were remo*ed and restore power to the boiler.

Tips 6 Warnin$s

• Horsepower is usually rated per hour. I' your (oiler provides a horsepower per minute (H;+min) rating,

multiply the H;+min rating (y 7" then multiply (y 00,-2! to calculate #$%s per hour.

How to Calculate Boiler 'team &lows

.odern steam (oilers create easily utiliBed energy 'or many heating, processing, and mechanical and electrical

power?generation applications. $hey can derive their incoming energy 'rom many types o' 'uelsG electricity,

co?generation, sunlight, nuclear and waste heat 'rom other processes. #eing a(le to calculate (oiler steam

'lows 'rom any o' the parameters surrounding a (oiler, including (oiler horsepower, 9ilowatts, 'uel?'low rates,

'eedwater 'low, com(ustion air 'low, or steam use data provides a complete picture o' (oiler per'ormance.

$his industrial (oiler serves many users with pressuriBed, saturated steam.

Tin$s 3ou4ll 5eed

• Calculator

• #oiler system parameters

Instructions

. %se (oiler horsepower to calculate steam 'low. #oiler horsepower is not related to mechanical

horsepower. It is a (oiler industry rating that predicates the amount o' saturated steam a (oiler will

generate starting with water at !! degrees )ahrenheit and " pounds per square inch gauge (psig??

meaning atmospheric pressure) and ending with steam at !! ) and " psig. )or e1ample, i' a (oiler is

rated at 5" (hp, then it will produce 5" (hp 1 0-.5 pounds per hour (l(+hr) * ,2!5 l(+hr o' steam under

these conditions.

!. %se (tu+hr (#ritish thermal units+hour) heat input rate to calculate (oiler steam 'low. It ta9es /2".!6

(tu+hr to produce pound o' steam at the a(ove conditions. I' 5" million (tu+hr are (eing applied to

the (oiler, divide (y the /2".!6?(tu+l( to yield 5,50?l(+hr steam 'low. $his is an average industrial

(oiler.

0. Calculate (oiler steam 'low with a 9nown 9ilowatt hour (9:h) usage in an electrically heated (oiler. I'

?9:h can produce 0.52 pounds o' steam at the a(ove Bero?psig and !!?deg?) conditions, then

,""" 9:h will produce 0,52 l(+hr o' steam.

-. Compute the 'low rate o' steam 'rom a (oiler (ased on (urning !" gallons+hour o' Lo. 7 'uel oil,

assuming normal 65 percent (oiler com(ustion e''iciency. $he accepted (tu+gallon conversion rate 'or

Lo. 7 'uel oil is 5",""" (tu+gal. A (oiler (urning !" gallons per hour would produce M(5",""" (tu+gal

1 !" gal+hr) + /2".!6 (tu+l(N 1 ".65 (e''iciency) 'or a total o' !,7!6. l(+hr o' steam.

5. Calculate the com(ustion?air mass required to (urn the !" gallons o' Lo. 7 'uel oil a(ove. 4ince the

air?'uel stoichiometric ratio is -.- pounds o' air per pound o' 'uel, the (oiler's (urner would need !"

gallons 1 2./5 l(+gal 'or Lo. 7 'uel oil, or !,!6/ pounds o' air to (urn the oil. $his would in turn yield a

total 'low o' !,7!6. l(+hr o' steam. $here'ore, 9nowing the amount o' com(ustion air 'low during an

hour 'or a given (oiler, you can (ac9?calculate to compute the steam 'low. $his means a (oiler will

produce a(out !,7!6. l(+hr + !,!6/ l(+hr o' com(ustion air consumed, or a(out .5 pounds o' steam

'or every pound o' air.

Tips 6 Warnin$s

• Calculating steam 'low using di''erent parameters tells the operator the complete story a(out his (oiler.

• $here are large (oilers (urning several di''erent 'uels simultaneously where com(ustion air 'low can

(e used to in'er a steam 'low. $he interesting thing a(out (oilers is that once the air?mass?to?steam?mass

'low ratio has (een determined 'or a given (oiler, all car(on?(ased 'uels will require @ust a(out the same

amount o' air to produce pound o' steam, whether they are wood, coal, oil or natural gas. $his

phenomenon ma9es it easy to compute steam production 'rom total air 'low rate when a com(ination o'

'uels is (eing used at once.

• #oilers need to (e started and run up slowly to alleviate stresses in their (oiler tu(es and headers.

How to Convert 'team to BTU

#$%, or #ritish $hermal %nits, is a measure o' energy commonly used in calculations related to heating

systems. In the %.4., the term <heat units< sometimes replaces #tu. In situations where steam powers heating

systems, it is necessary to understand the relationship (etween the amount o' steam and the heat energy it

provides. Converting pounds o' steam into an energy value in #tu is straight'orward.

4team provides the energy 'or many heating systems around the world.

Tin$s 3ou4ll 5eed

• Calculator

Instructions

1. -nter the *alue in pounds of steam into the calculator. 5ae sure that the entry is

correct before proceeding to the ne!t step.

2. 5ultiply the *alue from the pre*ious step by 1,14%, the number of Btu in 1 lb. of

low2pressure steam. =6eference 2> The result is the steam *alue con*erted into

Btu. "or e!ample, '77 lbs. of steam equals '94,777 Btu 222 '77 ! 1,14% 3

'94,777.

$. Di*ide the result by 1,14%. 0f the answer is not equal to the original *alue in

pounds of steam, there was an error in the calculations. 6epeat the calculations

until the answer is correct.

Tips 6 Warnin$s

• )or most heating design purposes, a conversion rate o' ,""" #tu to l(. o' steam is close enough to

(e satis'actory, and it is much easier to calculate. )or e1ample, 5"" l(s. o' steam equals 5"",""" #tu.

• Always dou(le?chec9 the results (e'ore using them i' they are part o' a heating design or another

[email protected] where legal regulations or sa'ety issues are involved.

How to Calculate C&" in HVAC

In order to have a com'orta(le living space or room, you need to have a properly designed HVAC system. I'

you have too little air'low in a space, it may ta9e a really long time to heat or cool to the desired temperature. It

could never reach the desired temperature i' air'low is really low. $oo much air'low will result in a dra'ty

'eeling to the room and ine''icient heating or cooling. #oth o' these can put a strain on your energy (ill.

Instructions

. 5easure the length and width of a room where you need to calculate the required

airHow. 5ultiply these together to get the square footage of the room, i.e., if the

room is 17 ft ! 17 ft, the square footage would be 177. sq. ft. <irHow is measured

in #"5, or cubic feet per minute. 0t is a measurement that shows how much air

Hows into a space per minute. Fou want 1 #"5 per square foot, so the abo*e

room/s requirement would be 177 #"5. "or rooms with a lot of windows or ha*e

long periods of direct sunlight, you want 2 #"5 per square foot.

!. #ontinue measuring each room and calculating the #"5 requirement for each.

0. <dd up each room/s required #"5. This total is the amount of #"5 your BM<# unit

needs to produce in order to properly heat or cool the space.

-. 0f you don/t already ha*e a BM<# unit for the space, purchase one at your local

BM<# retailer that matches your #"5 requirement. 0nstall the BM<# unit per

manufacturer/s instructions.

5. Using your Ductulator, match each room/s #"5 requirement to your planned

friction rate. 5ost BM<# systems use a friction rate of 7.1) water column per 177

feet of duct. <s you match each room/s #"5 to the Ductulator, it will gi*e you the

size of duct you need to run to the room.

7. +or bac from the farthest room. <s you add a room, mae sure you add the

#"5 and increase the duct size accordingly. 0nstall the BM<# duct from each room

to the BM<# unit. < professional BM<# installer may need to be contacted for

proper duct installation.

How to Calculate te C&" of a Room

>eeping your home at a com'orta(le temperature is an e1tremely important consideration 'or most

homeowners. $he term <C).< is an acronym that stands 'or <cu(ic 'eet per minute,< a term used to descri(e a

'an's e''iciency in terms o' air'low. 3etermining the C). o' a room is a mathematical process o' 'iguring the

room's cu(ic area as well as the rate at which you desire the air to circulate throughout the room. Calculating

the C). o' a room helps when choosing a 'an.

Instructions

. 5easure the room/s width and length. <lso measure the height of the room from

the ceiling to the Hoor.

!. 5ultiply the three measurements from step 1 to determine the cubic footage of

the room. "or instance, if a room is 8 feet wide, 17 feet long and 8 feet high,

multiply 8 times 17 times 8 to get ,%7 cubic feet.

0. 5ultiply the cubic *olume of the room by the number of times you want the air to

turn o*er or e!change in an hour. "or e!ample, if you e!pect the air to be

e!changed twice per hour, multiply 2 times ,%7 to get 1,%%7.

-. Di*ide your answer from step $ by ,7 to calculate #"5. 0n this e!ample, you

would di*ide 1,%%7 by ,7 to get a #"5 of 2%.

How to Calculate te C&" of a Blower

.any industrial processes require continuous aeration. 4ewage treatment, 'or instance, uses aero(ic micro(es

that respire constantly as they (rea9 down sludge. An industrial (lower provides the necessary o1ygen (y

maintaining a steady 'low o' air into the reaction cham(er. 8ou can estimate a (lower's volumetric 'low rate

'rom the reactants' o1ygen a(sorption rate. $he other relevant 'actors are temperature and the pressure o' air at

the (lower's discharge point.

Instructions

. <dd %,7 to the temperature at the discharge point, measured in degrees

"ahrenheit, to con*ert it to degrees 6anine. 0f, for instance, air lea*es the blower

at 87 degrees1 87 C %,7 3 '%7 degrees 6anine.

!. 5ultiply the 6anine temperature by the number of pounds2moles of o!ygen that

is transferred each minute. 0f, for instance, 8 pounds2moles of o!ygen reach the

reactants each minute1 '%7 ! 8 3 %,$27.

0. 5ultiply this product by 17.9$, which is the gas constant1 %,$27 ! 17.9$ 3 %,,$'%.

-. Di*ide the result by the pressure at the gas discharge point, measured in pounds

per square inch. 0f this pressure, for instance, measures 1' pounds per square

inch1 %,,$'% I 1' 3 appro!imately $,747. This answer is the blower/s *olumetric

How rate, measured in cubic feet per minute.

How to Calculate te C&" for an *+aust &an

An e1haust 'an helps pull impure air out o' your home and (rings 'resh air in. ,enerally these 'ans are

connected (y a duct to the e1terior o' the home, which helps to ensure you have high?quality indoor air at all

times. =1haust 'ans are rated in units o' cu(ic 'eet per minute, or C).. $o calculate the C). needed 'or an

e1haust 'an in a room, you will need to ta9e a 'ew measurements and do some simple calculations.

Instructions

1. #alculate the *olume of the room. 5easure the length, width and height of the

room in feet. 5ultiply these three numbers.

2. +hich gi*es you the *olume of the room in cubic feet. "or instance, for a room

that is 8 feet square and 8 feet tall, multiply 8 by 8 by 8 to get '12 cubic feet.

$. Determine how often you would lie the e!haust fan to be able to e!change all of

the air in the room. 0f you would lie to ha*e completely fresh air e*ery .*e

minutes, for e!ample, then this is the e!change time.

%. Di*ide the number of cubic feet by the number of minutes you would lie it to

tae to e!change all of the air in the room. This will gi*e you the required #"5. 0n

a 2772square2foot room, if you would lie completely fresh air e*ery .*e minutes,

di*ide 277 by '. This gi*es you a required %7 #"5 for the e!haust fan.

How to Calculate C&" for 7ucts

Air 'low, usually calculated in cu(ic 'eet per minute (C).), can (e calculated (y the area o' a duct's cross

section and the velocity o' the air itsel'. 8ou may (e a(le to re'erence the velocity, (ut i' not, a device may

(e required to measure it 'or you. $hese devices measure velocity (ased on pressure. Assuming you can

acquire the necessary data, calculating the C). is relatively painless.

Instructions

,- 'quare 7ucts

!. 5easure two consecuti*e sides of the duct, in inches.

0. #on*ert inches into feet by di*iding each measurement by 12.

-. #alculate the cross2sectional area of the duct by multiplying the two numbers

together.

5. 5ultiply the cross2sectional area by the air *elocity, measured in feet per minute, to

calculate air How in #"5.

.- Round 7ucts

. 5easure the diameter of the opening.

!. #on*ert the measurement into inches by di*iding it by 12.

0. Di*ide the diameter by two to calculate the radius.

-. (quare the radius and multiple it by K0, or $.1%, to calculate the area.

5. 5ultiply the area by the air *elocity, measured in feet per minute, to calculate air

How in #"5.

How to Calculate C&" &rom R)"s

3evelopers o' air?movement mechanisms such as 'ans and compressors must consider the (asic relationship

(etween 'an or impeller speed in rpm (revolutions per minute) and volumetric air 'low in C). (cu(ic 'eet per

minute). :hile these machines have di''erent operating principles, their common characteristic is that each

revolution o' their spinning input sha't will result in a certain amount o' volumetric air (or gas) 'low through

them when they are operating as designed according to their pitch or displacement 'actor.

Instructions

,- A+ial Ventilation &ans

.-

1. De.ne the a!ial fan application. <ccording to the general fan law go*erning a!ial

fans, you will need to now the fan blade diameter and pitch =blade tilt or attac

angle> along with the rpm *ariable to calculate #"5. 0n this e!ample, a small

household fan on a stand has a three2bladed plastic fan with a 12foot diameter

and an 82inch eGecti*e pitch. This means that each re*olution of the running fan

blows the one2foot2diameter column of air coming through the fan 8 inches

toward you after accounting for eEciency losses. The fan is running at 1,277 rpm.

2. #alculate the linear *elocity of the air through the running fan. 0f each re*olution

mo*es the air 8 inches, then 1,2772re*olutions per minute multiplied by 8 inches

means the air 4,,77 is being mo*ed at inches per minute, or 877 feet in one

minute. <nother way of looing at it is that the fan is mo*ing an 8772foot2long

column of air that is 1 foot in diameter through space each minute.

$. #alculate the #"5 =*olumetric How of air> at 1,277 rpm. The *olume of the

column of air described in (tep 2 is pi =$.1%1,> ! fan radius squared =7.'2feet

squared> times the column length in feet. This would be $.1%1, ! 7.2' square

feet ! 877 feet 3 ,28.$2 cubic feet per minute at 1277 rpm.

8- Centrifu$al Blower &an 9'quirrel Ca$e2Blower:

. 3e'ine the (lower application. In this e1ample, the centri'ugal (lower in a window air conditioner

circulates 7"" C). on the <F&< setting when the (lower motor is spinning at 6"" rpm. 8ou can

calculate how much air would it circulate in <HI< mode, when the motor rotates at !"" rpm.

!. 3e'ine the terms in the general centri'ugal 'an 'ormula and rearrange them to use the 'ormula to solve

'or the higher air 'lowG C). +C). ! * A;. +A;. ! 1 (3+3!)O0 (cu(ed). $he (3+3!)O0 is ,

since the impeller diameter stays the same, so C). ! * C). 1 (A;. !+A;. ).

0. 4u(stitute your parameters in the equation to calculate the higher air'lowG

C). ! * 7"" C). 1 !"" rpm+6"" rpm * /"" C).

0- Reciprocatin$ Compressor Applications

. De.ne the compressor application. < shop compressor with a piston in a cylinder

has a net displacement of 17 cubic inches. 0t normally turns at ,77 rpm. Fou can

calculate many cubic feet of air it taes in at atmospheric pressure and the

appro!imate cubic feet of compressed air it supplies if it has an o*erall 1721

compression ratio.

!. #alculate the incoming *olumetric air How. 0f the compressor taes in 17 cubic

inches of air for each re*olution, then #"5 input 3 ,77 6K5 ! 17 cubic

inchesI1928 cubic inchesIcubic foot 3 $.%9 #"5.

0. #alculate the outgoing compressed air *olumetric How. (ince the compression

ratio is 1721,

How to Calculate )ump %)"

)or a healthy pool, you want all the water in your pool to pass through its 'ilter every so o'ten. $his is called

the turnover rate. It is generally advisa(le to have a turnover rate o' 'our to " hours, depending on the

'requency o' pool usage, with higher use pools requiring quic9er turnover rates. I' you 9now the total volume

o' your pool, you can calculate the pump's 'low rate, in gallons per minute (gmp), needed to accommodate the

desired 'low rate. $his ensures you purchase a pump adequate 'or your needs.

8ou can calculate the required 'low rate gpm o' a swimming pool's water pump.

Tin$s 3ou4ll 5eed

• $ape measure

Instructions

,- Calculatin$ )ool Volume

!. 5easure the length, width and a*erage depth of the pool in feet.

0. 5ultiply the length, width and depth to calculate the *olume in cubic feet. <s an

e!ample, a 27 foot by $7 foot by , foot pool will ha*e a *olume of $,,77 cubic feet.

-. 5ultiply the product by 9.' to con*ert to total gallons. 0n the e!ample, $,,77 times

9.' equals 29,777 gallons.

.- Calculate )ump &low Rate

1. the turno*er rate desired. "or constant use pools, you might want a four2hour

turno*er. "or an infrequently used pool, you might use 17 hours. 0n the e!ample, say

you want an eight2hour turno*er.

2. #on*ert hours to minutes by multiplying by ,7, since there are ,7 minutes in an

hour. 0n the e!ample, eight hours times ,7 minutes per hour equals %87 minutes.

$. Di*ide the total pool *olume, in gallons, by the turno*er rate, in minutes. 0n the

e!ample, 29,777 gallons di*ided by %87 minutes equals a required How rate of ',.2'

gallons per minute.

How to Convert %)" to Coolin$ Rate in Tons

)actories use heat e1changers or chillers to regulate an area's temperature. $he machine a(sor(s heat 'rom an

area that produces it and carries it to a di''erent location. $he medium that carries the heat is a re'rigeration

'luid that a(sor(s and releases heat as it e1periences di''ering pressures. A standard 'ormula 'or 'inding a

chiller's cooling capacity 'rom its 'low rate in gallons per minute determines the cooling rate in #ritish

$hermal %nits (#$%s) per hour. &ne re'rigeration ton is a cooling rate o' !,""" #$%s per hour.

Instructions

. 5ultiply the e!changer/s How rate in gallons per minute by '77, a con*ersion

constant. "or e!ample, if $'7 gallons How through the unit each minute1 $'7 222 '77

3 19',777.

!. 5ultiply this answer by the Huid/s temperature change as it passes through the heat

e!changer. "or e!ample, if the Huid rises 21 degrees "ahrenheit in temperature1

19',777 222 21 3 $,,9',777. This is the chiller/s cooling rate, measured in BTUs per

hour.

0. Di*ide this rate by 12,777, which is the number of BTUs per hour in a ton1 $,,9',777

; 12,777 3 $7,.2'. This is the unit/s cooling rate, measured in tons.

How to Calculate HVAC Tonna$e

:hen installing a heating and cooling system into a location, it is imperative to 'irst determine the heating and

cooling needs o' the location. $he needs will determine the HVAC system needed in the room. HVAC stands

'or heating, ventilating, and air conditioning, all o' which are part o' the (uilding design 'or mid siBe (uildings

and larger. $he 'ollowing will wal9 through calculating the heating and cooling thermal units (#$%) and how

to convert that num(er into tons.

Tin$s 3ou4ll 5eed

• .easuring $ape (metric)

• :attage la(els 'or all equipment at the location

• :attage la(els 'or all lighting

• Calculator

Instructions

1. 5easure the area of the room and con*ert that .gure into BTU/s. The area of the

room is determined by multiplying the length and width. The measurement must be

in meters, =1 foot equals .$7%8 meters>. 5ultiply the area by $$9 to .gure out the

area BTU.

Aoom #$% * Fength 1 :idth 1 002

!. I' there are north or south?'acing windows in the location, calculate the #$% 'or each window (y

measuring the area o' each. )or north?'acing windows, multiply the area (y 75. )or south?'acing

windows, multiply the area (y 62". I' you live in the 4outhern Hemisphere, reverse the calculations (75

'or south, 62" 'or north)

0. Calculate the #$% o' the num(er o' people who use the location. .ultiply the occupants (y -"" 'or the

#$%.

-. Any equipment in the room needs to (e measured 'or their #$% output. $o determine the output, multiply

the total wattage 'or the equipment (y 0.5.

5. 3etermine the lighting #$% (y multiplying total wattage 'or the lighting (y -.!5.

7. Add the 'igures 'rom steps one through 'ive. $he result is your total heat load. .ultiply this num(er (y

two to determine the amount o' cooling required.

2. Convert your heating and cooling load to tons (y dividing the #$%s (y !,""". $he resulting 'igure is the

HVAC tonnage.

How to 'i(e an HVAC '!stem

A good heating, ventilation and air?conditioning system (also 9nown as an HVAC) is essential to 9eep you

com'orta(le and healthy in your home or o''ice. $he siBe o' an HVAC a''ects its e''iciency and operating cost.

Although there are more precise methods 'or siBing an HVAC system, using the <rule o' thum(< method is

very simple.

Instructions

1. Determine the square footage of the property. Aoo at the blueprints or measure the

property with a tape measure.

2. Di*ide the number of square feet by %77. This will gi*e you a ballpar estimate of the

number of cooling tons required for your BM<# system.

$. "actor in speci.cs lie the amount of insulation in the property, appliances that

generate heat, windows, roo.ng and climate.

%. "ine2tune your estimate. Di*ide the property/s square footage by anywhere from $77

to '77, based on additional factors.

Tips 6 Warnin$s

• $his <rule o' thum(< siBing can (e applied to sections o' a property. I' you want to install a system 'or

only a 'ew rooms, calculate the square 'ootage o' those rooms only.

How to Convert )ipe 'i(e to %)"

;ipe siBing is measured (y the internal diameter o' the pipe, not the overall outside diameter. &nce determined,

the overall volume can (e calculated. ;ipe 'low is descri(ed in gallons per minute (,;.). 4horter lengths

o' pipe will have a greater 'low than a longer length o' the same diameter. $his is caused (y internal

resistance o' the pipe itsel'. #y the same reasoning a larger diameter pipe will have a greater 'low or ,;.

than a smaller pipe at the same pressure or 'low rate. ;ressure is descri(ed as pounds per square inch

(;4I). $he square?inch measurement is determined (y the area o' the pipe. $he pounds are the amount o'

'orce that is placed on the liquid to <push< it through the enclosed space.

Tin$s 3ou4ll 5eed

• .easuring $ape or Auler

Instructions

1. "ind the cross2section area of the pipe. <rea is equal to pi times the radius squared

or =a 3 $.1% N rO2>. < two2inch diameter pipe would ha*e a cross2section area of

$.1% times 1O2 or $.1% square inches.

2. Understand that water will ha*e a certain pressure associated with the height of that

water. @ne pound of water pressure, or 1 K(0, is equal to 2.$1 feet of ele*ation in

height. 0n other words, a one2inch column or pipe of water that is 2.$1 feet high will

ha*e a pressure of 1 K(0. The o*erall height222not *olume222of the pipe corresponds to

the pressure. < si!2inch diameter pipe that is 2.$1 feet high will only ha*e 1 K(0.

$. "ind the *olume of the two2inch diameter pipe in (tep 1 that has a length of 17 feet.

Ten feet is equal to 127 inches. 5ultiply $.1% square inches, the cross sectional area,

times the length. The *olume of the pipe is equal to $9,.8 cubic inches of *olume.

%. #on*ert cubic inches into cubic feet. @ne cubic foot equals 1,928 cubic inches.

Di*ide $9,.8 cubic inches by 1,928 cubic inches per cubic foot and the answer is .

218 cubic feet. This means that the two2inch diameter pipe that is 17 feet long has

an internal *olume of .218 cubic feet.

'. #alculate the amount of water that can be contained in the section of pipe at any

gi*en time. @ne cubic foot of water is equal to 9.%8 gallons. 5ultiply 9.%8 gallons by .

218 cubic feet and the amount of water in the pipe is equal to 1.,$ gallons.

,. "ind the &K5 if the How of water is one foot per second. 5ultiply the one2foot per

second How by ,7 seconds per minute and the How is now ,7 feet per minute. 0n

other words the water will How through the 172foot pipe si! full *olumes for e*ery

minute. (ince the piping contains 1.,$ gallons per 17 feet of pipe, multiply 1.,$ by

si! and the .nal &K5 is equal to 4.98 &K5 of water How from the two2inch diameter

pipe.

How to Convert BTU to %)"

$he amount o' heat e1change in a heating or re'rigeration system can (e e1pressed in #ritish $hermal %nits

(#$%) per hour. $his unit o' measurement is valid when considering either the amount o' thermal energy

a(sor(ed in a heating process or the amount o' energy [email protected] in a cooling process. $he #$%+Hour

measurement o' a device is related to the gallons per minute o' liquid 'low in the device, and the two units o'

measurement can (e derived 'rom one another with some (asic arithmetic.

Instructions

. 5easure the diGerence in temperature between the liquid inside and outside of the

heat e!changer.

!. the diGerence by '77 if the liquid Howing through the de*ice is water, or by %8' if it

is antifreeze.

0. Determine the BTUIBour rating of your de*ice. 0t should be posted either directly on

the de*ice or in the owner/s manual.

-. Di*ide the BTUIBour rating determined in (tep $ by your result from (tep 2 in order

to calculate the gallons per minute of How in your de*ice.

How to 7etermine %)" in Hot Water Coils

Hot water coils are used 'or a variety o' purposes, primarily in air?heating applications. $he gallons per minute,

or ,;., is a use'ul calculation to 9now. It represents how much water passes through the coil every minute,

and is important 'or determining the coil's e''iciency and 9nowing what 9ind o' coil to purchase in order to

replace a hot water coil. ,;. can (e easily calculated i' you 9now a little a(out the coil.

Instructions

. Determine the BUTB of your coil, or British thermal units per hour =BTUIhr>. This is

the amount of energy that is either transferred to or from the Huid in the coil.This

information is usually pro*ided by the manufacturer of the unit.

!. #alculate the lea*ing water temperature, or A+T, and entering water temperature, or

-+T. This can be measured using the coil itself. The entering water temperature is

measured from the coil mared )water in) and the lea*ing water temperature is

measured from the coil mared )water out.) 0f you now the temperature diGerence

=TD>, this can also be used in your .nal equation.

0. Kut your numbers into this simple equation1 &K5 3 P I '77 N =A+T 2 -+T>. (o,

subtract the -+T from the A+T, or use the TD, which is the same number. 5ultiply

by '77. P is your BUTB. Di*ide P by the result of your earlier calculation. This is the

&K5

How to Calculate %)" &rom )'I

;ressure in'ormation such as pressure drops through sections o' piping is usually related to 'low capacity in

gallons per minute (,;.). =ach piping system is unique due to a variety o' 'actors including smoothness and

the way sections @oin, (ut generally pressure has to increase as a 'unction o' the square o' 'low increase.

$here'ore, i' 'low dou(les, pressure has usually had to quadruple to a''ect the change in a 'lowing system.

.anu'acturers o' di''erent types o' piping pu(lish 'low?pressure data a(out their products. 8ou can use this

data to calculate ,;. 'rom pounds per square inch (;4I).

Instructions

1. De.ne a pipeline Howing system. 0f water has to be pumped through a 22inch

(chedule %7 poly*inyl chloride =KM#> plastic pipe from a well to a pond %77 feet

away, calculate how many &K5 the system is deli*ering. The pressure gauge on the

pipe at the well head reads ,, K(0 and the one at the pond '7.8 K(0.

2. (ubtract the lower *alue, '7.8 K(0, from ,, K(0 to determine pressure loss through

the pipe. This equals 1'.2 K(0 o*er the %77 feet of pipe, or $.8 K(0 per 177 feet of

pipe length.

$. 6ead the pressure2drop How correlation about 22inch (chedule %7 KM# plastic pipe

from the tabular How *ersus pipe2loss data chart. The How rate corresponding to $.8

K(0 loss to 177 feet of 22inch (chedule %7 KM# plastic pipe is 9' &K5.

%. Determine the pressure loss at one2half of 9' &K5, or $9.' &K5, to con.rm the

appro!imate square2power relationship between pressure and How. 0nterpolating

between $' &K5 and %7 &K5, the pressure loss computes to a *alue between 7.4

and 1.2 K(0 loss per 177 feet of pipe. (ince $9.' &K5 is halfway between the two

How readings of $' and %7 &K5, the interpolation will wor out to about 7.979 of the

way between 7.4 and 1.2 K(0 loss =again the square function>. (o the loss would be1

.! ;4I ?? "./ ;4I * ".0 ;4I E ".2"2, which is ".!! ;4I greater than the "./ ;4I value, or .! ;4I.

.ultiplying .! ;4I (y - to chec9 against twice the 'low yields a value o' -.-5 ;4I, or only ".55 ;4I

greater than actually measured. $his di''erence o' a(out - percent greater pressure drop would only wor9 out

to (e a(out a 7 percent 'low di''erence, which is close enough to support the general square?power relationship

(etween 'low and pressure, especially in piping that can vary that much in interior siBe and smoothness.

How to Calculate te %)" of Water

$he 'low rate o' your water system is important 'or ma9ing decisions a(out appliances that use water. $hese

measurements also help determine which types o' sprin9ler heads to (uy, or whether you have su''icient water

pressure to run a large irrigation system. %se a container mar9ed with units o' measurement and some math to

calculate your system's gallons per minute. $his process can (e applied to any device that has a water outlet,

including hoses, sprin9lers, shower heads and sin9s.

Instructions

. (et the container near the water outlet to be measured.

!. Turn on the water so that it begins to .ll the container. Turn the *al*e or control

handle all the way open. <t the same time that the water begins Howing, start the

stopwatch.

0. <llow the water to run for 1' seconds. (top the water e!actly when the stopwatch

reaches 1' seconds.

-. 6ead the measurement mar where the water stopped inside the container. 0f the

units of measurement are not in gallons, you will ha*e to use a con*ersion calculator

to change the number to gallons. Use one of the many free online con*ersion

calculators as needed.

5. 5ultiply the measurement by %. This represents a full minute of .ll time. This is also

the total gallons per minute that your de*ice consumes. "or e!ample, a

measurement of .2' gallons after 1' seconds, multiplied by % equals 1 gallon per

minute.

Tips 6 Warnin$s

• Consider swapping out high?glow 'aucet noBBles o' !.! gallons per minute or more with a more

restrictive noBBle to save water.

How to Calculate %)" in )ipe

:hen wor9ing with pipes, it's important to 9now the 'low rate to con'irm that the pipe meets requirements, or

to 9now how much 'luid is (eing supplied. )low rate is measured (y gallons per minute, or ,;.. ,iven 9ey

in'ormation a(out the piping system, the 'low rate can (e calculated using the appropriate engineering

'ormulas. $he general 'ormulas are quite involved and require some technical e1pertise to use. However, 'or

speci'ic cases, the 'ormulas can (e greatly simpli'ied.

Instructions

1. &et the details for the piping system and Huid. "or e!ample1 pipe material is plasticD

length is 277 feetD diameter is $I% inchesD inlet pressure is $7 pounds2per2square2

inch =K(0>D outlet pressure is 27 K(0D Huid is water at room temperature

=appro!imately ,8">.

2. Use the correct formula.

)or this e1ample, a 'ormula 'or a good appro1imation isG ,;. * 0."622 1 pO".5-5, where <;< represents

the pressure drop in ;4I per "" 'eet o' pipe. $he O".5-5 represents a 'unction on a scienti'ic calculator which

raises a num(er to a power. $he (utton to is usually indicated (y <8< with a small raised <E,< or vice versa.

0. Carry out the calculations.

$he total pressure drop is 0" ;4I minus !" ;4I, which equals " ;4I 'or !"" 'eet o' pipe. $he 'ormula

requires pressure drop per "" 'eet. $o get this, divide the total pressure drop (y the pipe length and multiply

(y "". In this case, it is " divided (y !"" times "", which equals 5 ;4I. $o get the 'low, multiply 0."622 (y

5 raised to the power o' .5-5, which equals 2.06. $he 'low is 2.06 ,;..

How to Calculate te %)" of Water &lowin$ &rom a )ipe

$he num(er o' gallons that 'low through a pipe each minute (,;.) depends on the pressure di''erential across

it. A higher pressure di''erential 'orces more water through the pipe, and a greater 'low rate corresponds with a

greater pressure drop. At a 9nown pressure, a longer pipe produces a lower 'low o' water. A wider pipe,

however, allows a greater 'low (ecause it creates a wider cross?sectional area o' pipe.

:ider pipes release more gallons each minute.

Instructions

1. "ind the pipe/s cross2sectional area by multiplying pi by the square of its radius. 0f

this pipe has a radius, for instance, of 7.7$' meters1 $.1%2 ! =7.7$' O 2> 3

7.77$8%4.

2. 5ultiply the result by the square of the same radius1 7.77$8%4 ! =7.7$' O 2> 3

%.91' ! 17O2,.

$. 5ultiply your answer by the pipe/s pressure diGerential. 0f this diGerential, for

instance, measures %7,777 Kascals1 %.91' ! 17O2, ! %7,777 3 7.188,.

%. 5ultiply this answer by 12.', a *alue connected with water/s *iscosity1 7.188, !

12.' 3 2.$'8.

'. Di*ide the answer by the pipe/s length. 0f the pipe, for instance, is 18 meters

long1 2.$'8 I 18 3 7.1$1.

,. 5ultiply the answer by 1',8'7 to con*ert it to gallons per minute1 7.1$1 ! 1',8'7

3 2,79, gallons per minute.

How to Calculate Water &low &rom )ressure

)low in a 'lowing system always goes 'rom higher pressure to lower pressure in that the pressure provides the

driving 'orce to move the 'luid or gas. ;ressure usually drops uni'ormly as a 'unction o' the square o' net 'low

increases. $here'ore, pressure at the 'low source must quadruple to dou(le 'low in most 'low piping systems.

&n this (asis, and the wealth o' physical data 'or many piping and tu(ing systems, you can calculate water

'low 'rom net pressure changes in a de'ined system.

:ater 'low through pressuriBed pipes can (e calculated 'rom pressure losses.

Tin$s 3ou4ll 5eed

• Calculator

Instructions

. 3e'ine the 'low?pressure relationship in the application. In this e1ample, clear 'iltered water in a -?

'oot?high swimming pool is (eing drained to 'ill an in?ground pool that will replace it. $he new pool is

"" 'eet away, and the ground there is a(out 5 'eet lower, so a ""?'oot?long garden hose with a

smooth inner diameter o' 5+6 inch is (eing used to let gravity do its wor9 o' draining the old pool.

)rom this in'ormation, you can calculate the water 'low rate as the 'irst pool empties 'rom static head

pressure alone.

!. Calculate the pressure head heights at the (eginning and end o' the emptying process. :ith the 'irst

pool 'illed, the water level is - 'eet high. $he water 'lows through the hose 'rom a drain at the (ottom

o' the 'irst pool, and 'rom there drops 5 'eet over the "" 'eet o' ground to the top o' the in?ground

pool, where it pours in 'rom the open hose end at the top. 4tatic pressure height there'ore equals - 'eet

H 5 'eet * / 'eet when draining (egins, and 5?'eet when the pool is empty.

0. Calculate the head pressure at each o' the head heights. 4ince each !.0 'eet o' water height converts

to pound per square inch (psi), the availa(le static pressure at a sealed hose outlet will (e /+!.0 *

0./ psi when 'ull, and !.7 psi when empty.

-. Calculate the 'low as a result o' pressure at (oth pressure values 'rom 4tep 0. 4ince the 5+6 inch inner

diameter (I3) inner hose liner is smooth ;VC plastic, its 'low per'ormance is virtually identical to

;VC plastic pipe o' the same I3, which happens to (e nominal P inch 4chedule -" ;VC pipe with a

".7!!?inch I3. Consulting the pressure loss chart 'or this ;VC pipe, it has a loss o' 05.5 psi at a 'low

o' " gpm. 3ividing the 0./ psi a(ove (y 05.5 psi * "."/6, and ta9ing the square root results in ".00

E " gpm * 0.0 gpm 'or the starting 'low. 3ividing !.7 psi (y 05.5 psi and ta9ing the square root

results in ".!-72 E " gpm * !.-77 gpm.

'. Malidate the How results for this equi*alent2pipe pressure2to2How calculation

method against an online calculator for garden hose. Di*iding the $.4 psi pressure

loss *alue by %7 psi =normal water spigot pressure, which would dri*e 11 gpm

with this hose> and e!tracting the square root yields 7.$122 J 11 gpm 3 $.%$

gpm. This shows a diGerence of =$.%$2$.$1 gpm>I$.$1 gpm of $.9 percent, which

is reasonable for How in similar plastic pipes. Di*iding 2.1, psiI%7 psi and

e!tracting the square root yields 7.2$2 J 11 gpm 3 2.'' gpm *ersus 2.%,, gpm

using the chart, again a reasonable $.% percent diGerence.

How to Convert )ressure to &T;'

$he greater the pressure that acts on a pipe, the 'aster that the water within it moves. $he other 'actors that

a''ect the water's 'low rate is the pipe's width and length. :ider pipes allow water to go through them at a

higher rate, and longer pipes, at a constant pressure, reduce the 'low rate. I' you calculate water's 'low rate

using poise, the standard unit o' water's viscosity, this produces the 'low rate in meters per second. A meter

contains 0.0 'eet.

:ater travels quic9ly through wide pipes.

Instructions

. 5ultiply the water/s pressure, measured in pascals, by the pipe/s squared radius,

measured in meters. "or e!ample, a pressure of 17,777 pascals dri*es the water

through a pipe with a radius of 7.2 meters1 17,777 222 7.2Q 3 %77 newtons.

!. Di*ide this answer by the pipe/s length, measured in meters. "or e!ample, if the

pipe measures 2'7 meters in length1 %77 ; 2'7 3 1., ?Im.

0. Di*ide this answer by 7.78, a *alue that taes into account water/s *iscosity1 1.,

; 7.78 3 27. This is the How speed through the pipe, measured in meters per

second.

-. 5ultiply this answer by $.$, which is the number of feet in a meter1 27 222 $.$ 3

,,. This is the water/s How speed in ftIs.

How to Convert %)" to &)'

:hen calculating the 'low o' water in a pipe, scientists and engineers convert the amount o' water traveling

during a given time 'rame to a speed. )or e1ample, i' you calculated the 'low o' water to (e 0- gallons per

minute (,;.), you can convert that to 'eet per second ();4). 8ou will need to 9now varia(les such as the area

o' the o(@ect the water is running through. #ut that sounds more complicated than it is in practice. :ith a little

e''ort and some simple math, the conversion can (e done quic9ly.

Converting gallons per minute to 'eet per second ta9es only a 'ew calculations.

Instructions

. #on*ert &K5 into cubic feet by multiplying by .1$$9. "or e!ampleD $% &K5 ! .

1$$9 3 %.'% cubic feet per minute.

!. -!amine the ob:ect the water is Howing through. 0n most instances, it will be a

metal pipe, such as those used in sewers. #alculate the cross2sectional area of

the pipe in square feet. The formula for the cross2sectional area of a pipe is KiI%

=outside diameter 2 inside diameter.> "or e!ampleD the cross2sectional area of a

pipe with an outside diameter of % feet and inside diameter of 2 feet would be

$.1%I% =%22> 3 1.'9 square feet.

0. Di*ide the cubic feet measurement by the cross2sectional area to get linear feet

per minute. "or our e!ampleD %.'%I1.'9 3 2.84 feet per minute.

-. #on*ert feet per minute into feet per second by di*iding by ,7. 0n our e!ampleD

2.84 I ,7 3 .7%8 "K(.

How to Convert )ipe 'i(e to %)"

;ipe siBing is measured (y the internal diameter o' the pipe, not the overall outside diameter. &nce determined,

the overall volume can (e calculated. ;ipe 'low is descri(ed in gallons per minute (,;.). 4horter lengths o'

pipe will have a greater 'low than a longer length o' the same diameter. $his is caused (y internal resistance o'

the pipe itsel'. #y the same reasoning a larger diameter pipe will have a greater 'low or ,;. than a smaller

pipe at the same pressure or 'low rate. ;ressure is descri(ed as pounds per square inch (;4I). $he square?inch

measurement is determined (y the area o' the pipe. $he pounds are the amount o' 'orce that is placed on the

liquid to <push< it through the enclosed space.

Tin$s 3ou4ll 5eed

• .easuring $ape or Auler

Instructions

1. "ind the cross2section area of the pipe. <rea is equal to pi times the radius

squared or =a 3 $.1% N rO2>. < two2inch diameter pipe would ha*e a cross2section

area of $.1% times 1O2 or $.1% square inches.

2. Understand that water will ha*e a certain pressure associated with the height of

that water. @ne pound of water pressure, or 1 K(0, is equal to 2.$1 feet of

ele*ation in height. 0n other words, a one2inch column or pipe of water that is

2.$1 feet high will ha*e a pressure of 1 K(0. The o*erall height222not *olume222of

the pipe corresponds to the pressure. < si!2inch diameter pipe that is 2.$1 feet

high will only ha*e 1 K(0.

$. "ind the *olume of the two2inch diameter pipe in (tep 1 that has a length of 17

feet. Ten feet is equal to 127 inches. 5ultiply $.1% square inches, the cross

sectional area, times the length. The *olume of the pipe is equal to $9,.8 cubic

inches of *olume.

%. #on*ert cubic inches into cubic feet. @ne cubic foot equals 1,928 cubic inches.

Di*ide $9,.8 cubic inches by 1,928 cubic inches per cubic foot and the answer is .

218 cubic feet. This means that the two2inch diameter pipe that is 17 feet long

has an internal *olume of .218 cubic feet.

'. #alculate the amount of water that can be contained in the section of pipe at any

gi*en time. @ne cubic foot of water is equal to 9.%8 gallons. 5ultiply 9.%8 gallons

by .218 cubic feet and the amount of water in the pipe is equal to 1.,$ gallons.

,. "ind the &K5 if the How of water is one foot per second. 5ultiply the one2foot per

second How by ,7 seconds per minute and the How is now ,7 feet per minute. 0n

other words the water will How through the 172foot pipe si! full *olumes for e*ery

minute. (ince the piping contains 1.,$ gallons per 17 feet of pipe, multiply 1.,$

by si! and the .nal &K5 is equal to 4.98 &K5 of water How from the two2inch

diameter pipe.

How to Calculate Water &low< Volume 6 )ressure

8ou can calculate the 'low rate, volume and pressure in a water tan9 using 'ormulas 'rom physics. Volume is

the amount o' space that an o(@ect ta9es up, and may (e measured in liters, gallons, or cu(ic meters. )low is

the rate at which a certain volume o' liquid passes through an openingQ it may (e measured in liters per second

or gallons per minute. ;ressure is the amount o' 'orce per square unit o' area, and is measured in pounds per

square inch (psi) or Lewtons per square meter (pascals).

Calculate :ater )low, Volume R ;ressure

Tin$s 3ou4ll 5eed

• $ape measure

• 4topwatch

• #uc9et

• Calculator

Instructions

,- Calculatin$ Volume

2. 5easure the width, length and height of the water in meters.

$. 5ultiply the width, length and height to compute the *olume in cubic meters. The

formula is M 3 +AB, where M is the *olume, + is the width, A is the length and B is

the height.

%. #on*ert cubic meters to liters by multiplying by 1,777. #on*ert cubic meters to

gallons by multiplying by 2,%.19.

.- Calculatin$ Water &low

1. Klace an empty container below the faucet or release *al*e.

2. @pen the *al*e and use the stopwatch to time the water How for 1' seconds.

$. 5easure the number of liters or gallons in the container, and di*ide that number by

1'. This gi*es the How rate in liters per second or gallons per second. The formula is

" 3 MIT, where " is the How rate, M is the *olume and T is the time.

%. To con*ert to liters per minute or gallons per minute, multiply the number you

obtained abo*e by %.

8- Calculatin$ )ressure

1. Use the hydrostatic pressure formula K 3 pgh, where p is the density of water in

g per cubic meter, g is the gra*itational acceleration constant, h is the height of

the water abo*e the *al*e in meters and K is the pressure in pascals.

)or water near the sur'ace o' the earth, p * ,""" 9g per cu(ic meter, and g * /.6 meters per second

squared.

2. #on*ert pascals to psi by di*iding by ,,84%.9,.

$. Use this e!ample as a guide to calculate water pressure in a tan.

$he height o' the water in a tan9 is - meters a(ove the valve. Applying the hydrostatic pressure

'ormula gives you ; * (,""")(/.6)(-) * 0/,!-" pascals. In pounds per square inch, the pressure is

0/,!-"+7,6/-.27 * 5.7/ psi.

How to Calculate %allons )er "inute &rom Water )ressure

A pump's discharge rate depends on water pressure. $his pressure, when measured in 'eet o' water, is also

9nown as the <total dynamic head,< which is equivalent height through which the 'luid must (e pumped. $he

value can also (e calculated independently o' pressure, using values such as the pump's static head, static li't

and 'rictional losses. #esides pressure, the pump's output also depends on the system's e''iciency. All other

'actors (eing equal, a less e''icient pipe, (y de'inition, moves less water per minute.

.ore pressure means a higher 'low rate.

Instructions

. 5ultiply water horsepower by $,4,7. 0f the pump wors at 18 horsepower1 18 !

$4,7 3 91,287.

!. 5ultiply your answer by the pump/s eEciency. 0f it runs at 47 percent eEciency1

91,287 ! 7.47 3 ,%,1'2.

0. Di*ide your answer by the water/s pressure, in feet of water. 0f the pressure

equals 277 feet of water1 ,%,1'2 I 277 3 $27.9,. This is the pump/s How rate,

measured in gallons per minute.

How to Convert %)" to H)

Horsepower, or H;, is the amount o' power required 'or a pump or tur(ine to trans'er a 'luid or to create a 'low

in the 'luid. $his depends on the rate the 'luid is li'ted and the height to which it is li'ted. $he rate is usually

measured in gpm, which stands 'or gallons per minute, and the height is usually measured in 'eet. I' a pump is

pointed up into the air, it will pump 'luid to a certain height, called the total head. All 'luids will (e pumped to

the same height i' the sha't is turning at the same revolutions per minute.

$he horsepower o' a pump or tur(ine can (e calculated (y 9nowing the 'low rate o' the system and the total

head the system can e1ert.

Tin$s 3ou4ll 5eed

• ;ump or tur(ine

Instructions

1. +rite down the following formula1 Bp 3 =P ! B> ; =$,4,7 gallons per minute per

foot ! eG>, where )Bp) stands for horsepower, )P) stands for How rate in gallons

per minute, )B) stands for total head in feet, the $,4,7 is a con*ersion factor to

transfer from gallons per minute per foot to horsepower and )eG) stands for the

eEciency of the hydraulic equipment being used, such as a pump or turbine.

2. Determine the discharge rate or How rate of the liquid in the system. This

quantity is usually found in the hydraulic system/s manual or on the system itself.

<n impeller meter, ori.ce meter, or other measurement de*ice may be used to

tae this measurement as well. "low rates must be con*erted to gallons per

minute for use in calculating the horsepower.

$. Determine the total head of the liquid in the system and con*ert the answer into

feet. The total head a pump or turbine can produce is de.ned as the potential

height or depth that the water le*el can reachD this *alue depends on the type

and strength of pump or turbine that is used in the system. <n e!ample of

calculating the total head is as follows1 0f a waste water column e!erts a pressure,

Ke, of 7.%$$ pounds per inch for e*ery foot of the column and the total pressure,

Kt, on a gauge in the system reads % psi, then the total head B can be calculated

with the equation B 3 KtIKe 3 % psiI =.%$$psiIft> 3 4.2% ft.

%. Determine the eEciency of the pump or turbine by referring to the manual or it

may be listed on the equipment. 0f not, contact the manufacturer for this

information. The eEciency of the pump can also be determined if the ratio of the

wor being done by the system to the power or energy being supplied is nown.

#on*ert the eEciency percentage to a decimal when used to sol*e an equation.

0f, for e!ample, a pump had an eEciency of ,' percent, then the decimal quantity

would be ,' ; 177, or 7.,'.

'. Klug all the predetermined *alues into the equation Bp 3 =P ! B> ; =$,4,7 gallons

per minute per foot ! eG> to calculate horsepower. "or e!ample, if the system had

a pump with a nown eEciency of ,' percent and the pump deli*ers 2'7 gallons

per minute to a total head of 92 feet, the horsepower of the system is1 =2'7

gallons per minute ! 92 feet> ; =$,4,7 gallons per minute per feet ! 7.,'> 3 ,.44

horsepower.

%)" to R)" Conversion

.ost gallon per minute (,;.) to revolutions per minute (A;.) conversions relate to pumps. )or e1ample, i'

you're designing a pump to recirculate a spa, you need to 9now the speed at which the pump must run, given

the siBe o' the pump's cylinder. $here are 'ormulas to guide you in these conversions. $he 'ormulas use 'i1ed

and varia(le valuesQ the varia(le values are usually estimates. $he only way to completely [email protected] 'or the

estimates is to conduct a real?world test to 'i1 the varia(les.

,;. to A;. conversions involve estimates o' pump e''iciency.

)ump 7isplacement

• I' you thin9 o' a cylinder?driven pump ?? @ust li9e a car engine ?? you may thin9 the output o' the pump

is the displacement times the revolutions. In other words, i' you have a pump with a ?gallon cylinder, it

should pump one gallon 'or every revolution, and you can e1trapolate 'rom there ?? not so. $he hydrodynamics

are not nearly so e''icient. =ach revolution o' the pump pumps some 'actor less than the 'ull volume o' the

cylinder.

Conversion Wit an #utput Test

• In real?world applications, the output depends on the design o' o' the cylinder, piston, inta9e and outlet

ports, what's (eing pumped, how hot or cold it is and other 'actors. $he most accurate way to convert ,;. to

A;. is to do a test. Aun the pump at ,""" A;., 'or e1ample. &nce ,""" A;. is reached, direct the 'luid

(eing pumped into a measuring container 'or one minute. $his will give you the accurate ,;. measure o'

your pump at ,""" A;.. :ith this measure, you can estimate the A;. o' the pump i' you 9now the amount

o' water (eing pumped every minute, thought it is an estimate. Sust (ecause the pump moves <E< amount o'

water at ,""" A;. doesn't mean it will move " times <E< at ",""" A;..

#utput Variance Wit 'peed

• I' your pump moves " gallons a minute at ,""" A;., you can estimate it will move !" gallons at

!,""" A;.. Hydrodynamics conspires against the per'ect scaling o' revolutions to volume or volume to

revolutions ratio. Here's one e1ample. Imagine you're pumping something very viscous, such as oil. At low

speeds, the e''ects o' the oil's viscosity may have a negligi(le e''ect on its pumping e''iciency ?? 5"" A;., 'or

e1ample. At 5,""" A;., the viscosity may play a (igger role in the pump e''iciency. At some higher A;., the

viscosity may loc9 the liquid together li9e sandG $hough it's liquid, it (ecomes as dense as it can (e, e''ectively

creating a limit to how 'ast the pump can pump it. 4uch e''iciency curves apply to 'actors other than viscosity,

too.

#utput Calculation

• %nderstanding the limitations o' the conversion, you start with one 'i1ed ration and scale your

conversion up or down. Fet's say you measured your pump and 9now it pumps "" gallons per minute at ,"""

A;.. $he ratio is gallon 'or every " A;., or G". Low i' you measure the pump's output and 'ind it has

pumped !.0 gallons in one minute, you can estimate its A;. is !,0"".

How to Convert )'I to %)" Water

$he pressure drop across a pipe determines how much water 'lows through it each minute. $o accurately

calculate the volumetric 'low 'rom the pressure drop, you must consider the pressure using the standard unit o'

;ascals rather than pounds per square inch (psi). $he other relevant 'actors in the equation involve the pipe's

dimensions, with wider, shorter pipers at a constant pressure di''erential producing a larger 'low.

Increase water pressure to increase its 'low rate.

Instructions

. 5ultiply the pressure by ,,84%.9' to con*ert it to Kascals. "or e!ample, if the

pressure is 12 psi1 12 ! ,,84%.9' 3 82,9$9 Kascals.

!. 5ultiply the pressure by the pipe/s cross2sectional area. +ith a cross2sectional

area, for instance, of 7.7798, square meters1 82,9$9 ! 7.7798, 3 ,'7.$1 Kascals.

0. 5ultiply the result by the square of the pipe/s radius. 0f the radius measures, for

instance, 7.7' meters in length1 ,'7.$1 ! 7.7'O2 3 1.,2,.

-. Di*ide the result by the pipe/s length. +ith a length, for instance, of 2' meters1

1.,2, I 2' 3 7.7,'.

5. Di*ide the answer by 7.78, which describes water/s *iscosity1 7.7,' I 7.78 3

7.812'. This is the How through the pipe in cubic meters per second.

7. 5ultiply the result by ,71 7.812' ! ,7 3 %8.9'. This is the How in cubic meters

per minute.

2. 5ultiply the result by 2,%.191 %8.9' ! 2,%.19 3 12,898. This is the pipe/s

appro!imate *olumetric How in gallons per minute =gpm>.