HeatVentTutorial - Pyrox

7/25/2019 HeatVentTutorial - Pyrox

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Heat i ng and Vent i l at i on Tut ori al

Heat. Vent.Tutorial forside 17-07-98 13:59 Side 1


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Welcome to the world of PYROX!

PYROX is market leader in Norway and exports the complete product range to more than 25 countries inEurope, the Middle East and North America. The products are promoted and sold by carefully selecteddistributors who have wide experience with our products and within ventilation and heating engineering.You are welcome to visit us at our WEB sitehttp://www.pyrox.comhttp://www.pyrox.comhttp://www.pyrox.comhttp://www.pyrox.comhttp://www.pyrox.com where all our distributors are listedwith adresses and telephone, fax and e-mail.

Head office and Sales office Russia Sales office andmain warehouse and CIS countries warehouse Sweden

Pyrox AS Pyrox CIS Pyrox AB

P.O. Box 110 - Alnabru Rochdelskaya, 15 LngvikenNO-0614 Oslo RU-123023 Moscow SE-739 30 Skinnskatteberg

NORWAY RUSSIA SWEDEN

Tel. +47 2305 2300 Tel. +7 095 255 0334 Tel. +46 222 41490Fax +47 2305 2321 Fax +7 095 255 0275 Fax +46 222 41830e-mail [email protected] e-mail [email protected] e-mail [email protected]


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Dear customer

Our objective is to supply products which meet and, preferably, exceeds the requirements and expectationsof our customers and the outside world. We wish to have close and good relations with our customers. In

that way we will have first-rate knowledge of what is required from us, our products and services.

The first contact you establish with PYROX is through one of our distributors. We have carefullyconcentrated on selecting distributors with good basic competence in ventilation and energyengineering. Our distributors are trained by us and have a wide experience with our products.Their sales staff are customer-oriented and well educated.

It is my hope that you will use this catalogue as a working tool tochoose the right products for your projects. Additional information maybe obtained from one of the Pyrox sales offices or from your localdistributor.

Yours sincerely

Svein Magnus Nilsen

Managing Director


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Welcome to PYROX!

Since the founding in 1923 in Bergen, Norway,PYROX has been a pioneer and leading com-pany in development, manufacturing and mar-keting of electrical heating appliances. The com-pany was established in order to exploit theinvention of the electric heating element, andthrough 50 years stoves, cooking plates, convec-tion and fan heaters were the main products ofthe company. Later we concentrated on indus-trial heating products and in addition we devel-oped a significant competence and knowledgewithin the ventilation and air handling field.

As a result of continuous product developmentand our professional level, PYROX today ranks

among the leading Scandinavian suppliers ofindustrial heating and ventilation systems. Weare an independent part of the Kanalflktgroup, and are consequently backed up bystrong resources. The PYROX products aremanufactured at group factories in Norwayand Sweden. In 1998 we will start manufac-turing a range of PYROX products at thegroups new factory in Canada for supplyingthe North American market. Product develop-ment, design and production are integrated inan advanced CAD/CAM system. Our factoryin Sweden is certified according to the ISO9001 Quality Standard and the ISO 14000Environmental Standard - in our other facto-


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ries these systems are under implementation.Our AMCA certified air- and sound laboratoryis set up with the latest state-of-the-art meas-uring instruments and equipments, and isrecognised as one of the most modern inEurope. As we develop, test and manufacturethe products ourselves, a full control of qualityand reliability is secured.

PYROX is market leader in Norway and weexport our products to more than 20 countries.The products are promoted and sold by care-fully selected distributors in the various mar-kets. One of our competitive advantages is ahigh degree of product availability with short

delivery time. We have well-stocked ware-houses and can deliver within short time tomost destinations in Europe.

Our aim is to supply products which meet -and preferably exceed - the requirements andexpectations of our customers. In order toconstantly know what is required of us, ourproducts and services, we work hard at keep-ing good and close relations to our customers.

Our distributors are all highly skilled within theheating, energy and ventilation engineering.Their sales organisations are customer-ori-ented with well trained and experienced em-ployees.


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Trondheim

Bergen/Tysnes

Pyrox AS

Vollaveien 20 AP.O. Box 110 AlnabruN-0614 OsloNorway

Telephone: +47 2305 2300Telefax: +47 2305 2321e-mail: [email protected]

Skinnskatteberg

Stavanger

Oslo

Stockholm

DISTRIBUTOR:DISTRIBUTOR:DISTRIBUTOR:DISTRIBUTOR:DISTRIBUTOR:


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TABLE OF CONTENTS

VENTILATION

1. UNITS, FORMULAE, CONVERSIONS 22 HEATING AND COOLING 33. HEAT RECOVERY 54. MOLLIER DIAGRAM FOR HUMID AIR 75. VENTILATION PRINCIPLES 86. FANS 137. ACOUSTICS 168. COMFORT CRITERIA 229. VENTILATION OF VARIOUS AREAS 2410 ELECTRICITY 29

HEATING11. BUILDINGS`POWER REQUIREMENT FOR HEATING 3212. BUILDINGS' ENERGY REQUIREMENTS 4013. RADIANT HEAT 4314. AIR CURTAINS 5315. FAN HEATERS 5716. CONVECTION HEATERS 6017. REGULATING TEMPERATURE 6118. APPENDIX 63


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1. UNITS, FORMULAE, CONVERSIONS Pyrox

1.1 FORMULASYMBOL DENOMINATION SI UNIT

V volume m3

P,Q effect, output W-j/s-Nm/sCp specific heat

capacitykJ/kgC

W energy J-Nm-wsh(i) enthalpy kJ/kgt time sv velocity m/sL volume flow m

3/s-l/s

d density kg/ m3

P pressure N/m2-Pa

P pressure diff. N/m2-Pa

F force NI electric current A

U electric voltage VT temperature C-KT temp. diff. C-Km length m

A area/surface m2

F frequency Hzk thermal

transmit-tancecoefficient

W/ m2C

a heattransmissioncoefficient

w/ m2C

l thermal conduc-tivity coefficient

W/ m2C

1.2 Conversion factors

POWER1 kW = 1,360 hp 1 hp = 0.736 kW1 kW = 860 kcal/h 1 kcal/h = 1.163 x 10

-3

kW

ENERGY1 kJ = 0.239 kcal 1 kcal = 4.187 kJ

ENTHALPY1 kJ/kg=0.239kcal/kg

1 kcal/kg = 4.187 kJ/kg

SPECIFIC HEAT CAPACITY1 kJ/kgC = 0.239 kcal/kgC1 kcal/kgC = 4.187 kJ/kgC

FORCE1 N = 0.102 kp 1 kp = 9.807 N

PRESSURE1 Pa = 0.102 mm

H2O

1 mm H2O = 9.807 Pa

1 bar = 1,020 kp/cm 1 kp/cm2

= 0.981 bar

VOLUME FLOW1 m

3/s = 3,600 m

3/h 1 m

3/h =

0.278 x 10-3

m3/s

1.3 Example with calculation of effect

In our case, the most relevant units forcalculating effect are m

3/s and kW.

The following formula is often used tocalculate effect:Q = 0.3 x T x L [kcal/h]0.3 - kcal/m

3

L - m3/h

This formula applies at roughly 20C and is

actually as follows:

Q = L x c x d x tc - specific heat of air 0.24 kcal/kgC

d - specific weight kg/m3

d - 1.293 at 0C and 1.205 at 20C

According to the SI system:1 kJ = 0.239 kcalQ = 0.24 x 1.2 x L x t [kW at 20C] 0.239L = m

3/s

Q = L x 1.2 x t

L = m3/h

Q = L x 1.2 x t [kW at 20C] 3,600


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2 HEATING AND COOLING Pyrox

1.2.1 HEATING COILWhen a certain volume of air passes a heatingcoil, air temperature is increased from t1to t2.

This process takes place at constant airhumidity. The effect of the heating coil iscalculated as follows:

Q = L.1.2

.t [kW], where:

L = air flow rate, [m3/s]t = increase in temperature across the coil,[C]Example on calculation of required heating coileffect:

t1= 0C, RH = 50%t2= 20CL = 1.4 m3/sQ = 1.4

.1.2

.20 = 34 kW

The process is shown in the Mollier diagram:

tt

t

X

2

1

RH=50%

RH=100%

THEPROCESSINTHEMOLLIER-DIAGRAM

Air velocityThe air velocity across the heating coil iscalculated as follows:

L

v = L [m/s], where: V = A

AL = air flow rate m

3/s

A = gross area m2

Water volumeTo calculate pressure drop on the water sideof the heating coil and determine the size of

the shunt valve, the water volume in a heatingcoil must be given.

This is calculated as follows:q = Q [kg/s], where

Cp.t

Q = heating coil effect, [W]t = temperature increase of water through the

heating coil, [C]Cp= 4.175 j/kgC at a water temperature of50C

2.2 COOLING COILIn calculating the required cooling effect, it isimportant to consider that part of the effect isused to separate the water (latent heat) andthe remaining effect to lower the airtemperature (sensible heat). It is thereforenecessary to incorporate enthalpy differencesin the calculations to absorb the latent heatpart. The cooling effect is calculated asfollows:

Q = L.1.2

.t [kW], where:

L = air flow rate, [m3

/s]h = enthalpy difference for air through the coil,[kJ/kg]

Example on calculation of cooling coil effect:L = 1.4 m

3/s

t1= 25C, RH = 50%t2 = 12C, RH = 100%

t t1 2

CALCULATIONOF

COOLINGCOILEFFECT

Enthalpy can be obtained from tables orsimply from a Mollier diagram:h1= 50 kJ/kgh2= 33 kJ/kgQ = 1.4 .1.2 .(50 - 33) = 28.56 kW


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The process is illustrated in the Mollierdiagram, from which the separated volume ofwater x and the distribution of latent andsensible heat can be derived.

t

h

t

t

X

1

2

h1

2

RH=50%

RH=100%

THEPROCESSINTHEMOLLIER-DIAGRAM

Latent heat

Sensitiveheat

x


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3. HEAT RECOVERY Pyrox3.1 ROTATING HEAT RECOVERYEQUIPMENTThere are two types of rotating heat-recoverysystems: those that transfer humidity

(hygroscopic) and non-hygroscopic. Whenusing rotating heat recovery equipment,correct pressure conditions in the unit areimportant to prevent transfer of pollutants fromthe exhaust side to the fresh air side.With correct pressure drops across the filterand correct placement of the fan, leakage willbe so minor that this type of recovery is highlysuitable to most types of "comfort" systems.However, this type of recovery should be usedwith caution where there is a risk of odour fromthe exhaust air being transmitted to the freshair.

EfficiencyThis type of recovery has a high level ofefficiency, ranging from 75% to 85%depending on operating conditions:h= h2- h1= enthalpy efficiency h2- h1 h3- h1 h3- h1

x= x2- x1= humidity efficiency x2 x1 x3- x1 x3- x1

t= t2- t1= temperature efficiency t2- t1 t3- t1 t3- t1

In a hygroscopic rotor: t = x= ht1= air temperature prior to heat recovery unit, supply air sidet2= air temperature after heat recovery unit, supply air sidet3= air temperature prior to heat recovery unit, exhaust side

t3

t2t1

ROTATINGHEAT

RECOVERY EQUIPMENT

ExampleThe temperature of the supply air after theheat recovery unit can be determined whenthe following data are given:

t3(exhaust air) = 22Ct1 (supply air) = 15Ct(efficiency) = 80%

Risk of frostFrost problems are rare in connection withrotating heat recovery equipment. However,there is a risk of freezing at very low outdoor

temperatures (below -10C) and humidexhaust air (exceeding 40%). In ordinaryrooms, the relative humidity is usuallyconsiderably lower in case of very low outdoortemperatures. In reality, this means that frostproblems are very minor with the use ofrotating heat recovery equipment.

3.2 PLATE HEAT EXCHANGERSIn plate heat exchangers, fresh air andexhaust air flow on each side of a number ofdividing plates. Plate heat exchangers yieldslightly lower efficiency than rotating heatrecovery units, but on the other hand, thereare completely separate air currents for supplyair and exhaust, and accordingly they have aslightly wider application range.

Humidity from the exhaust air may condenseand the heat recovery unit must therefore beequipped with a condensate collection panand outlet.

EfficiencyEfficiency lies in the range of 60% - 70%.

Risk of frost

In case of low outdoor temperatures,condensed water from the exhaust air mayfreeze and the exchanger will eventually becovered with ice. A common defrostingmethod is to run the external air in a by-passcircuit past the exchanger when the pressuredrop across the exchanger has reached acertain level. It is important for the defrostingsystem to function properly as efficiency willdecline as the exchanger is covered by ice.


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3.3 LIQUID COIL HEAT EXCHANGERSHeat exchange using liquid coils is a fluid-related heat recovery system with heating

(cooling) coils in the exhaust and fresh airducts, respectively. One of the advantages ofthis system is that exhaust and fresh air ductsdo not need to be combined. The coils can, forexample, be located in the attic and basementrespectively. A frost medium, usually a mixtureof ethylene-glycol and water, is circulatedthrough the piping system.


Risk of frost

The percentage of glycol should be roughly30% to ensure that the liquid mixture does notfreeze. At low outdoor temperatures (below -

10C), the exhaust coil may freeze. A commonmethod of defrosting the coil is to stop thecirculation pump in the fluid circuit when thepressure drop in the coil has increased by50% for example.Defrosting only takes a few minutes.

Circulation pumpIf a circulation pump is selected, pressure canbe increased by about 10% and the volume ofwater/glycol by about 30% compared to awater system only.

3.4 HEAT-PIPE HEAT EXCHANGERA heat-pipe heat exchanger is a closed pipingsystem with a self-circulating cooling medium,

where liquid evaporates in the hot part of thepipe. The vapour flows over to the cold side(fresh air side), where it is condensed. Theadvantage of this type of heat exchanger isthat it can be applied at high temperatures.



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4. MOLLIER DIAGRAM FOR HUMID AIR Pyrox

h = enthalpy for 1 g dry air kJ/kg, kcal/kgx = water content of 1 kg dry air kg/kg

= relative humidityt = dry thermometer temperature Ctv = wet thermometer temperature C


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5. VENTILATION PRINCIPLES Pyrox

5.1 AIR QUALITY CONCEPTSIn ventilating various premises, it is normal todifferentiate between two principles: ventilationby diffusion and ventilation by displacement.

Before taking a closer look at these two typesof ventilation, the most common air qualityterms should be defined:

Ventilation efficiencyVentilation efficiency is a measure of how fasta pollutant can be removed from a room. It isdefined as the relation between theconcentration of pollution in the exhaust airand the concentration of pollution in the room.

m= ce.100% where:cm

ce= Concentration of pollution in the exhaustaircm= Mean concentration of pollution in theentire roomVentilation efficiency is a frequently usedparameter to evaluate a system's ability to

provide good atmospheric comfort.mdepends on the geometry of the room, locationof the valve and distribution of pollution in theroom.With ventilation by displacement, it is possibleto achieve ventilation efficiency exceeding100%, while the goal with ventilation by

diffusion is to achieve mof about 100%.

Air-change efficiencyAir-change efficiency is a measure of how fastthe air in a room is replaced. It depends on thedistribution of air in the room, i.e. location andsize of valves, room geometry, location of heatsources, etc.

a= n.100% where:

2.m

n= volume of fresh air supplied/room volume

m= average "age" of air in the room

With ventilation by displacement, it is possible

to achieve aof between 50% and 100%, while

full ventilation by diffusion yields aofapproximately 50%.

5.2 Ventilation by displacementVentilation by displacement is a ventilationform that has traditionally been applied for

ventilation of industrial premises. It can also beused to advantage for many types of so-calledcomfort ventilation. If properly dimensioned, it

can remove major heating effects and achievea high degree of ventilation efficiency. Beforethis principle is described in further detail, thefollowing terms should be defined:

Occupied zoneThe occupied zone is that part of the roomnormally occupied by people. For offices,schools, etc., it is normal to calculate with adistance of 0.5 m from an outer wall withwindow, 0.2 m from other walls and 0.1 m - 1.8m above floor level.

OCCUPIEDZONE

Near zoneThis denotes the zone near a low-impulsevalve where there is a floor draft. Velocity v =2.0 m/s is used as the draft limit for a low-impulse valve in a comfort system. The goal is

to achieve the smallest possible near zone fora low-impulse valve.

With ventilation by displacement, the supplymechanism is located at a low level, and air issupplied directly to the occupied zone at a lowvelocity. Convection flows from potential heatsources give the air an upward motion and thehot/polluted air is removed via exhaust valveslocated at the ceiling.


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Air is supplied to the occupied zone at a

temperature of 1 - 5C below the roomtemperature. It is important that the

temperature is not too low below the roomtemperature since this will yield a so-calledconvection draft from the cold surface.

Temperatures of more than about 2C belowthe room temperature place specialrequirements on the air-supply fittings.

Properly designed systems for ventilation bydisplacement yield air of very good quality.However, the principle has obvious restrictionssuch as:

* The supply valves are large and require a lot

of wall space* The supply valves are often covered* The near zone, where velocity exceeds0.2 m/s, becomes too large* The vertical temperature gradient becomes

too large; should not exceed 2-3C.

It should also be mentioned that several heatsources at various levels and with differenttemperatures complicate matters. Combinedwith movement in the room, this causesdisplacement of air from the upper layer to thelower zones. In practice, this will lead to"ventilation by diffusion."

Ventilation of auditoriumsA common method of supplying air toauditoriums is to supply air under the seats inthe belief that the air will float slowly upwards,thus achieving a displacement effect. This isnot correct however. The air behaves likewater and "flows" downwards where it collectsbefore it flows slowly upwards towards the rearoutlet. The supply air mechanisms could justas easily be installed at the front of theauditorium.Regardless of the location of supply airmechanisms, experience shows that is difficultto achieve any real displacement effect.Measurements and computer simulations for asolution as outlined above indicate aventilation efficiency of roughly 100%, whichentails complete agitation of the air.

Tests have also been carried out for traditionalventilation with (vortex) diffusers in the ceiling.This works satisfactorily, provided the outlet islocated towards the rear. Locating the outlet in

other spots may easily result in full short-circuiting.

Suggestions for sensible ventilation ofauditoriums:

VENTILATIONOFAUDITORIUMS

1. Ventilation by diffusion with air suppliedfrom the ceiling/rear by, for example, vortex

diffusers. Outlet at the rear.


2. Diffuse supply of air under or in front ofseats. Supply of air beneath the seats has notother significance except that it is oftenpractical to use the space below the seats as asupply chamber.


3. In addition to the above, the rear of theauditorium should be ventilated separately.This is particularly the case when it serves asan entrance area so that users are not met bya "wall" of hot, polluted air.


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Design of the supply air systems in ventilationby displacement

For displacement-type ventilation at premiseswith normal heat sources in the form ofpeople, light, etc., experience shows that thetemperature change between floor and ceilingis roughly as shown in the figure below:

t i

0.1

Hm

Roomheight

1.1m to

ttot

tCtg tutR

TEMPERATURECHANGE

BETWEENFLOORANDCEILING

tu= temperature at ceiling, Cti= supply air temperature, CQ = addition of heat in the room, WL = amount of air supplied, kg/scp= specific heat capacity of air, 1,000

Ws/kgC

In the following, the conditions when heat isadded in the form of pollution are reviewed.The temperature difference between floor andceiling is given by:

to= tg- ti, where:

tg= temperature 0.1 m above floor levelti= supply air temperature

Depending on the type of supply airmechanism, this yields:

to= K.ttot

K = 0.25 for supply air mechanisms with lowinduction, such a filter mat or perforated plateK = 0.50 for supply air mechanisms with highinduction.Furthermore:

ttot= tu- ti=

ExampleThe following data have been specified for anoffice:

Floor area 10 m2

Surplus heat 500 WHeight to ceiling 2.7 mSupply air mechanism K = 0.5The vertical temperature gradient must be less

than 3C per ma) Volume of supply air 5 m

3/h per m

2

b) Volume of supply air 10 m3/h per m

2

c) Volume of supply air 15 m3/h per m

2

Case a:

ttot= 500 = 30C

50 .1.2 . 1,000 3,600

to= 0.5.30 = 15C

This yields a vertical temperature gradient of

15/2.6 = 5.8C per m, which is too high.

Case b:

ttot= 15C

to= 0.5

.15 = 7.5C

This yields a vertical temperature gradient of

7.5/2.6 = 2.88C, which is satisfactory withrespect to the requirement.

Room temperature is often defined as thetemperature 1.2 m above floor level. If in this

case there is a supply air temperature of 15C,the temperature at the outlet becomes 30C,and 0.1 m above floor level, the temperature

becomes 15 x 7.5 = 22.5C. Furthermore:

tR= tg+ 1 . (tu- tg) 2,6

tR= 22.5 + 1 . (30 - 22.5) = 25.4C 2,6

This is usually an acceptable maximumtemperature.

pCL

Q

?ttot= tu- ti=


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Case cIf the supply air flow rate is increased to 15m

3/h, the temperature gradient becomes

1.9C/m and the room temperature 1.1 mabove floor level becomes 21.9C.These are excellent conditions.If the supply air temperature is increased to

17C, the temperature gradient remains thesame as before.

The room temperature rises to 24C, which isadequate in most cases.

In conclusion, it is possible to say that 15 m3/hper m

2yields excellent conditions.

Since outdoor temperatures during summer

only rise above 17C a few days of the yearmost places in Norway, artificial cooling is notnormally required.

Proposed ventilation for industrial premisesVentilation by displacement is well suited toindustrial premises with substantial pollutionand high heat loads.The measurement of dust concentrations inwelding shops with displacement-typeventilation indicates that the dust content of airat floor level is 1/3 of the dust content underthe ceiling.

y

COANDAEFFECT

5.3 VENTILATION BY DIFFUSIONVentilation by diffusion involves the injection ofone or several air jets outside the occupied

zone. The jet absorbs large amounts of indoorair and the velocity gradually decreases. Theoccupied zone lies in the return flow and thevelocity here is about 0.7 times the finalvelocity of the jet. In practice, there is often acombination of the aforementioned conditions.Definition of terms:

Coanda effectIf an supply valve is installed sufficiently near asurface, the air jet will cling to the surface. Airbetween the surface and the jet is carriedalong, and since new air cannot flow from

above, a slight under-pressure occurs at thesurface and the jet is subjected to an upwardmotion. For the jet to cling to the surface, Ymust not exceed 30 cm.

y

COANDAEFFECT

Air throw l0.2Air throw is defined as the distance from thediffuser to the point in the jet where centralvelocity has dropped to 0.2 m/s. Examples onhow large the air throw should be for typicalinjection valve locations are shown below:Rear injection l0.2= 0.7 L

Injection from ceiling l0.2= 0.5 L

L

H

VENTILATIONBYDIFFUSION


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Indoor air velocitySuitable air velocity in the occupied zonedepends on such conditions as activity,

cladding and air temperature. Empirical dataindicate that draft can be avoided in the

temperature range of 20 - 22C if the airvelocity is less than 0.18 m/s.

InductionA diffuser's ability to mix indoor air with the airjet from the diffuser.As an example, a high degree of inductionwhen supplying air at temperatures belowroom temperature is favourable in avoidingdraft problems.

Jet drop in case of obstructionThe change of direction of air jets under theceiling when obstructed by girders, lightfittings, etc. Lights must be installed at adistance from the ceiling to maintain a Coandaeffect.


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6. FANS Pyrox

The fan is an important part of any system fortransportation of air and gas.Some of the central points regarding fan

properties are reviewed below.

The most common wheel shapes are:1) Radial wheel2) Francis wheel3) Screw wheel4) Axial wheel/propeller wheel

>90

Forward-curvedblades

=90

Straight radial blades


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The characteristics of fans with backward-curved and forward-curved blades,respectively, are roughly as shown in figure6.3.

Pa

m/s3

Backward-curvedblades


(drumwheel)

FANSWITHBACKWARD-CURVEDBLADES

ANDFORWARD-CURVEDBLADES

Choke linesIn addition to the fan characteristics, so-calledthrottle regulation (choke) lines or operatinglines are included in the diagram. These areparabolic, but because a logarithmic axissystem is often used, the choke lines are

shown as straight lines. The lines arenumbered from 1 to 10 as shown in fig. 6.4The choke line number (l) can be calculated asfollows:

Pd= dynamic pressure at fan outletPt= the fan's total pressure increase


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Fan efficiencyThe efficiency is constant along a choke line.Diagrams often show efficiency instead of

choke lines, fig. 6.4.A fan's efficiency can also be shown as afunction of the choke line number, fig. 6.5.Fans with backward-curved blades have thehighest maximum efficiency, up to 85% andthe operating point should lie within the areabetween lines 1 and 2. Fans with forward-curved blades have a lower maximumefficiency, approximately 70%, and this isachieved in the area between lines 4 and 5,

Pa

m/s3

FANEFFICIENCY

12

34

5

10

(65%)

(75%)(80%)

(85%)(90%)

thus further to the right in the diagram.

Fan selectionFan selection is based on the calculated totalpressure increase required (sum of pressure

losses) and desired air flow rate per time unit.Normally, the desired operating point can beachieved with several fan sizes. If a small fanis selected, the operating point will lie far to theright of the diagram, fig. 6.4. The result will below efficiency. If a larger fan is selected, theoperating point will lie further to the left of thediagram and greater efficiency will beachieved. The initial cost of a larger fan istherefore often offset by lower operating costs.

Fig. 6.5 shows that fans with forward-curvedblades have higher efficiency at higher choke

line numbers.If because of space problems it is necessaryto choose a small fan (for example inaggregates) that functions at a high choke linenumber, a fan with forward-curved blades willoften be a correct choice.

m/s3

Backward-curvedblades


100

80

2 3 4 5

%


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7. ACOUSTICS Pyrox

7.1 BASIC CONCEPTSDefinitionsSound is the term for audible pressure

oscillation in an elastic medium. The soundpressure oscillates around the atmosphericpressure in a sinus-shaped sound movement,provided the tone is pure.

T

PTid

Pa

Soundpressure

Atmosphericpressure

The following values can be defined on thebasis of the above:P = the instantaneous value of the soundpressurePeff= sound pressure efficiency

Peff= Pa/2 - the value registered by soundmetersf = 1/T = frequency (Herz - Hz)c = sound velocity (m/s)= wavelength (m)

Whether a sound is audible to the human eardepends on its frequency and volume. Thehuman ear can hear sounds in the 20Hz -20,000 Hz frequency range.The lowest sound pressure the human ear canperceive lies at 2_10

5Pa at 1,000 Hz.

Sound that is perceived as undesired oruncomfortable is usually considered NOISE.

Measuring units for noiseSound power (N):The quantity of energy per time unit which isgenerated in the form of sound from a soundsource.

Sound intensity (l):The sound energy per time unit that flowsthrough a flat surface at a right angle to thesound's spreading direction.

Sound pressure (p):

The power the sound exerts on a surface at aright angle to the sound's spreading direction.

A logarithmic scale has been introduced toobtain more manageable figures. The

correlation between the numeric values on thisscale better express how the human earperceives sounds of various strength. Thesound power level and sound pressure levelare therefore defined as follows:

Sound power level (LW):

N = effect produced by sound sources (W)No= reference effect (W) (10-12

W)

Sound pressure level (LP):

p = sound pressure (N/m2= Pa)

po= reference pressure (Pa) (2*10-5

Pa)

Different manufacturers operate with differentreferences. It is therefore important for themanufacturer to specify the reference used inpresenting sound data.

Example (fan):The produced sound power from a fan isN = 10

-4W

LNref10-12

W= 10.log ( 10

-4) = 10

.log10

8= 80

dB 10

-12

LNref10-13

W= 10.log ( 10

-4) = 10

.log10

9= 90

dB 10

-13

These examples show that results may varydepending on the reference used.

))(log(10 dBNo

N?=

)log(20)log(102

2

Po

P

Po

P?=?=


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PyroxNoise criteria NR/dBANR criterion:This specifies the permissible sound pressurelevel as a function of the frequency.

dBA criterion:This furnishes the sound meter with an A filterto attenuate sound with frequencies of lessthan 1,000 Hz. There are also B and C filters.

An A filter is usually used in connection withventilation.It can often be of interest to determine therelationship between dBA and NR values.This can be calculated roughly as follows:dBA = NR + 5 (dB)

Addition and subtraction of dB valuesExample:

There are two sound sources of respectively:LW1= 40 dB and LW2= 38 dBThe resulting sound power level can bederived mathematically by logarithmic addition.However, it is more practical to use thediagram below for addition and subtraction ofdB values, respectively. As can be seen fromthe diagram, a sound level of LWR= 42 dBresults from the two sound sources.

ADDITIONDifference to be added to the highest dB valueDifference between dB values to be added

SUBTRACTIONValue to be subtracted from the highest dBvalue Difference between dB values to besubtracted

7.2 SOUND POWER AND SOUND PRESSURE LEVELThe connection between sound power leveland sound level

Sound power:This is independent of the location of themeasuring point and the characteristics of thesurroundings.

Sound pressure:This decreases with increasing distance to thesource. It depends on the room's acousticproperties.

Note: If the sound pressure level has beenspecified, additional information is required on

the acoustic properties of the room where themeasurements were taken. To avoiduncertainties with respect to the interpretationof sound data, it is therefore more appropriatethat the manufacturer specifies the soundpower level for his equipment.Sound measurements normally entailrecording the sound pressure level. It istherefore necessary to be aware of theconnection between sound power level andsound pressure level. The total sound picturein a room is affected by the location andopening cross section of the sound source, aswell as the room surfaces' ability to absorb orreflect sound.In addition, direction factor Q can be definedas a function of the size of the valve's crosssection, the measuring point's locationcompared to the opening cross section, thevalve's location in the room and, finally, thefrequency at which the sound is measured. Itis then possible to set up the followingcorrelation between sound pressure level LPand sound pressure level LW:

LP= LW+ 10.log ( Q + 4

.n)

4r2 R

The Q/4r2fraction represents the direct

sound from the sound source, while 4/Rrepresents the reflected sound.r = distance from sound source to measuringpointQ = directional factorR = room constant


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Pyrox

The room constant R is an expression of theroom's acoustic properties and can be definedas follows:

R = A [m2Sabine]

1_

A = the room's absorption (m2Sabine)

= mean absorption coefficient of the room

The following is normally understood byabsorption coefficient:

= la+ lt li

I= sound energy emittedIt= energy that penetrates the wallIa= energy that is absorbed by the wall

For an infinitely hard wall: = 0For an open window: = 1

This implies that the absorption coefficient willlie between 0 and 1.

The table below shows some typical meanvalues for , depending on type of room:

Type of room a

Sound studio, musicrooms

0.30-0.45

Housing, offices, hotelrooms, conferencerooms

0.10-0.15

Classrooms, hospitalrooms

0.05-0.10

Factories, indoorswimming pools

0.03-0.05

Value A has previously been defined as theroom's absorption. This value is derived fromthe product of surface F and absorption factor

_. This value is thus called the room'sabsorption and is actually measured in m

2. To

emphasise that an acoustic value is involved,m

2Sabine is used.

Near zone and reverberation fieldThe distance from a sound source to a pointwhere direct sound and reflected sound

equally contribute to the resulting soundpressure level is given by:

Near zone

At r


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Pyrox

Calculation exampleA suitable valve to supply air to a room is to beselected. You have the choice of a valve from

Pyrox or a valve from another manufacturercalled X. A comparison of the two with respectto sound is to be performed.

Sound level requirement for the room:NR30 (or close to 35 dBA)

According to manufacturer X's catalogue, thesound pressure levelis 30 dBA for theapplicable air flow rate. It is further stated thatthis value applies to the reverberation field of aroom corresponding to 10 m

2Sabine.

For Pyrox's valve, a sound power levelof 32

dBA is quoted for the corresponding air flowrate.

Unless you know the correlation between, andimportance of, sound pressure level andsound power level, it is easy to be misled intobelieving that manufacturer X's valve has alower sound level than the correspondingPyrox valve. As will be demonstrated, thecircumstances are a bit more complex.

The sound data specified for manufacturer X'svalve apply to a room with a clearly definedmeasuring point and clearly specified acousticproperties. Before the valves can becompared, the specified sound value must beconverted to a form where it is independent ofthe measuring point and acoustic properties ofthe room. It is therefore necessary to calculatethe sound power levelof the valve.

LW= LP- 10.log ( Q + 4)

4r2 R

The first fraction within the parenthesis equals0, since measurement was performed in thereverberation field. Furthermore, R_A = 10m

2Sabine since this is a relatively hard room,

so that can be set at approximately 0. Thiswill subsequently yield:

LW= 30 x 4 = 34 dBA

Accordingly, asound pressure levelof 30 dBAcorresponds to a sound power levelof 34 dBAfor manufacturer X's valve. The sound data

from both manufacturers are now available ina general form and it can be noted that in thiscase the Pyrox valve came out ahead.However, to determine whether the valvesmeet the specified sound requirement, it isnecessary to calculate sound pressure on thebasis of the sound power level for the actualroom where the valve will be located. In thiscase, the following data have been specified:

Room constant R: 38 m2Sabine

Direction factor Q: 2 (generally applies toceiling-mounted

valves)No. of valves: 2Measuring point: The soundrequirement

applies to 1 m belowone of the valves

The actual sound pressure level using a Troxvalve in the relevant room can now becalculated as follows:

LP= 32 + 10.log ( 2 + 4

.2) =

28dBA

4r2 38

This shows that the valve for the room inquestion ends up well below the 35 dBArequirement. Furthermore, it is important tonote that it is not possible to compare sounddata from various manufacturers unless themanufacturers operate with the samereferences.

As a general guideline in comparing thespecified sound data for Pyrox valves withvalves for which the sound pressure level isspecified at 10 m2Sabine, 4 dbA can besubtracted from the specified sound powerlevel to obtain the sound pressure level andthus obtain comparable values.


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Pyrox

7.3 Sound reductionSound reduction can basically be achieved intwo ways: by absorption and by reflection.

There are various stages of sound reductionin a ventilation plant, including:

* Sound reduction in straight ducts* Sound reduction in ducts with internal sound insulation* Sound reduction in bends and branch ducts* Sound reduction in connection with exhaustinto the chamber* Sound reduction through changes in cross-section* Sound reduction through end reflection. At

the point where a duct enters a room, part of the sound is reflected back to the duct. This actsas a sound reduction for the room.* Sound reduction through room absorption. The difference between LW- LPcan be considered as a sound reduction due to roomabsorption.* Sound reduction through sound absorbers. Ifthe natural attenuation in a system is notsufficient, sound absorbers must be added.

The degree of sound reduction as a result ofthe aforementioned stages can usually bederived from tables, diagrams and themanufacturer's data.The following will examine sound reduction inducts with internal insulation since this asimple method of reducing noise. Soundreduction can roughly be estimated as follows:

D = K.O

.

1.4[db/m]

A

K = correction factor, approx. 0.7 at 250 HzO = absorbing circumference (m)

A = free duct area_ = absorption coefficient

Q/A will vary with the shape of the valve'scross section, and sound reduction can beenhanced if the cross section is divided intoseveral ducts. This phenomenon serves as abasis for inserting so-called bafflers in sound

absorbers to enhance sound reduction.

7.4 Sound calculationsSound calculations to determine the requiredsound reduction in the main sound absorber

installed after a fan or, for example, after adamper. As a basis, use the room assumed tobe most exposed. This is usually the room withthe shortest main duct stretch from the fan tothe room. The following sound calculationsmust then be performed item-by-item asshown in the example below for an injectionsystem.

1. Use the room's permissible sound pressurelevel from the injection system as a basis.

2. First, check that the selected valves meet

the requirement of item 1. If this is in order,proceed to the next item.

3. The permissible sound pressure level fromthe duct system can now be determined bysubtracting the value of item 2 logarithmicallyfrom the value of item 1. As previously shown,a diagram is used for this purpose.

4. Add sound reduction due to roomabsorption to the value of item 3. (Standardaddition)

5. Add sound reduction due to end reflection tothe value of item 4. (Standard subtraction)

Having now proceeded from the room , pastthe valve and into the duct system, the valueof item 5 corresponds to the permissible soundpower level in the ducts right after the valve(since room absorption has been included).Continue with the following:

6. Sound reduction in the duct, bends,branches, etc. until the sound source isreached.The sum of the values in 5 and 6 (standardaddition) yields the permissible sound powerlevel just after the sound source.


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Pyrox

7. Find the sound source's sound power levelin the manufacturer's catalogue.

8. The difference between the values in item 7and item 6 yields the sound reduction requiredin the sound absorber.

A calculation as shown above must beperformed for each individual octave band.Sound is normally most critical at frequenciesof 250 Hz and 500 Hz. Accordingly, it usuallysuffices to perform calculations for thesefrequencies.

There is user-friendly software programavailable for these types of calculations,

which greatly facilitates the work.


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8. COMFORT CRITERIA Pyrox

8.1 TOTAL HEAT LOSS PER PERSONDEPENDING ON TYPE OF WORK/ACTIVITY

Activitylevel

Type of activity Total heatloss perperson,estimatedvalue in W

I Working in a sittingposition, such asreading and writing

100

II Light work standing,laboratory work,typing

150

III Moderate physicallabour

200

IV Heavy physicallabour

more than250

8.2 HEAT CONDUCTIVITY RESISTANCE OFCLOTHING

TYPE OF CLOTHING HEATCONDUCTIVITYRESISTANCE,R in m

2K/kW

Without clothing 0Light summer clothing 80Medium clothing 160Warm clothing 240

Note: The approximate value of heatconductivity resistance is about r = 160m

2K/kW = 1 clo.

8.2 COMFORTABLE INDOOR AIRTEMPERATURE RANGEThe following premises show how humans

experience comfortable indoor temperatures,depending on the outdoor temperature.With increasing outdoor temperatures,increasing room temperatures arerecommended.

CONDITIONS:* Activity levels 1 and 2* Medium to light clothing* Air temperature close to the surfacetemperature of the surrounding surface.

5. INTERNAL AIR TEMPERATURE

TOLERANCEAir-conditioning plants for ventilation andheating only should not have overall localdeviations in room temperature of more than2.0K in the horizontal measuring zone of theoccupied area.For air-conditioning plants with cooling, these

deviations must not exceed 1.5K, with theexception of impact from heat sources.


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Pyrox

4. ROOM AIR VELOCITYThe validity of this penetration curve is subjectto the specified conditions.

A shift in change of conditions may occur.Accordingly, the overall assumption of a roomair velocity of 0.2 m/s is often incorrect. Todetermine the limit value of air velocity, it isnecessary to be familiar with the room, theroom's functions and the activities of thepeople in the room.

THE CURVE APPLIES FOR:* Activity level 1* Medium clothing* Air temperature close to the surfacetemperature of the surrounding surfaces* With a 50W increase, the critical curve canbe increased by 0.04 m/s* With higher/lower heat conductivityresistance of clothing than 80 m

2K/kW, the

critical curve may be transposedcorrespondingly by 0.04 m/s.


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9. VENTILATION OF VARIOUS AREAS Pyrox

FOR VENTILATION OF VARIOUS AREAS,REFERENCE IS MADE TO THE 1987BUILDING GUIDELINES WHICH STATE:

47:52 PUBLIC BUILDINGS, PROFESSIONALBUILDINGSNecessary ventilation must be calculatedtaking the following factors into account:* human load* pollution from building materials, furnitureand equipment* pollution from special pollution sources

The volume of outdoor air that must be

supplied because of pollution is usually the

sum of the above items.

* Human load In rooms occupied by people for lengthyperiods, air must be supplied at a rate ofminimum 7 l/s per person

For rooms occupied by people on and off(e.g. schools, nursery schools), this volume ofair can be reduced according to the time thepremises are occupied.

In rooms with limited user time (assemblyhalls), air must be supplied at a rate ofminimum 4.5 l/s per person.

* Pollution from building materials, furnitureand equipmentBuilding materials, furniture and equipmentgive off odour and other irritating effects, andair must therefore be supplied at a rate ofminimum 0.7 l/s per m

2floor space. This value

has been established on a somewhatuncertain basis and assumes that acceptable"good" materials are used.This calculation basis also applies to corridors,auxiliary rooms and common areas. High roomtemperatures and unfortunate selection ofmaterials may increase the need for ventilationdue to increased discharge of waste gases tothe internal air.

A temperature of less than 22C isrecommended and has been used as a basisfor the calculations.

Experience dictates that emission of wastegas and particles from building materials maynecessitate considerably more air to achievean acceptable dilution. In the first year ofoperation, the air-conditioning plant shouldtherefore run 24 hours per day at full capacity.The same applies after extensive renovation.To avoid a high level of pollution at the start ofthe day's work, the air-conditioning plantshould also run outside of working hours (ifnecessary at reduced capacity or switched onearly enough for the pollution to be sufficientlydiluted before work begins in the premises).

* Pollution from special pollution sourcesExtraction by hoods and/or encasing shouldbe used in areas with special, local pollutionsources to limit dispersion in the room. If it isnot possible to sufficiently limit dispersion, theair flow rate required to dilute the pollutionmust be calculated.

Normally, the air flow rate will depend on thenumber of people. For rooms where thenumber of people cannot be determined in anyother way, the values in table 47:521 can beused as a guideline.To obtain a proper indoor climate, the outdoorair flow rates must be limited in each case onthe basis of pollution in a), b) and c). Table47:522 shows examples of air flow rates in a)and b) for various premises. In addition tomechanical ventilation, there should bewindows that can be opened.The law on protection against the harmfuleffect of tobacco contains regulations that maybe of importance to the design of the air-conditioning plant.

As an example, air supply to separate,enclosed offices with air transfer to an exhaustarrangement in the corridor will not beacceptable in buildings where smoking ispermitted in individual, enclosed offices. Inaddition, pollution from tobacco smoke will initself require greatly increased air flow rates,see table 47:522.


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REFERENCE IS ALSO MADE TO THEFOLLOWING GUIDELINES IN THEWORKING ENVIRONMENT ACT ISSUED BY

THE DIRECTORATE OF LABOURINSPECTION:

* Administrative standards for pollution in a working environment

* Climate and ventilation at indoor workplaces

table 47:521

PREMISES No. of people per 100 m

2floor space

Assembly halls 150Gymnasiums 30Educational facilities 50Offices 10Sales premises 30Restaurant premises 100

No. of peopleThe number of people (number of people perm

2of floor space in the room) is determined

on the basis of the assumed function of theroom. If the number of people cannot bedetermined in any other way, the values in thetable can be used as a guideline.

Examples of air flow ratesNotes:

Air flow rates are specified in litres per secondper person for column a), in litres per secondper m

2floor space for column b), and in litres

per second per shower/WC/urinal for vents. (1l/s 3.6 m

3/h).

The figures in column b are based on the useof building materials, equipment and furnitureof acceptable "good" quality with respect toodour and other irritating effects.

Air for showers and washrooms can be takenfrom adjacent rooms if this air is of relativelyhigh quality.

table 47:522PREMISES External air flow rates due to

pollution fromVents

a)people

l/(sp)

b)buildingandfurniture

l/(sm2)

c)specialpollutionsources

l/(s equip)

Offices andpremises forlight andaverage work

7.0 0.7 CopyingWeldingSoldering

Sales premises 7.0 0.7

Educationalpremises andactivity rooms innurseries

5.5 0.7 ChemistryroomsWelding/Soldering

Gymnasiumfacilities

7.0 0.7

Assembly halls 4.5 0.7

Hotel rooms 7.0 0.7

Restaurants 7.0 0.7 Kitchen

Patients'rooms/residentrooms athospital andhealthinstitutions

7.0 0.7

Showers 17

WC, urinals 17

In rooms where smoking is common, theexternal air flow rate should normally be atleast 20 l/s per person.

47:43 HOUSINGThe above evaluations on pollution frombuilding materials, equipment and furniturealso apply to housing.


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47:531 GENERALThere are no requirements with specific

figures for outdoor air supply in connectionwith housing. Ventilation is based on a certainventing, natural or mechanical, from roomswith greater air pollution or humidity loads,such as kitchens, bathrooms and WCs.

An outdoor air flow rate corresponding to theoverall venting is supplied to the housing unitvia a separate supply air unit, through outdoorair valves, windows that can be opened,building irregularities, etc.

Total venting from the kitchen, bathroom, WCand laundry room according to table 47:53 will

normally secure an outdoor air flow ratecorresponding roughly to 0.5 air changes perhour. This is considered the minimumrequirement for housing ventilation. See alsounder 47:533.

To always ensure that indoor air of adequatequality, ventilation should be maintained evenwhen the house or rooms are not in use.

For laundry rooms with openings to thekitchen, the ventilation requirements willnormally be met if the kitchen has ventilationas specified for kitchens.

When the basement is partly above ground,the ventilation for basements can be met byvents in several outer walls so that thebasement is provided with a draft.

In housing exposed to radon, it is important forthe ventilation plant to cause as little underpressure as possible. Balanced ventilationshould therefore be used for these houses.See "NBI Building Details, sheet A 520.706Radon" Construction measures/details. ?

Radon is crucial in housing and particularly forthe floor at ground level. In accordance withthe law on municipal health services, thehealth authorities must procure an overview ofhealth-related problems. This also includesradon occurrences, including a duty to makesure radon occurrences are surveyed inconnection with new buildings.

The World Health Organization (WHO), hasrecommended 200 Bq/m

3as the maximum

limit for radon concentration in housing.Concentrations above this limit providereasons for simple measures such as asealing layer against the ground.Concentrations of 800 Bq/m

3should not occur.

If this is the case, measures such asventilation of the ground in addition to thesealing layer must be implemented. TheNational Institute of Radiation Hygiene haspublished various information with guidelineson building surveys, measurement in existinghousing and on health hazards and preventivemeasures.

47.532 SUPPLY OF EXTERNAL AIRThe supply of outdoor air to individual roomsshould be adapted to the rooms' outdoor airrequirement and arranged in such a mannerthat it does not create uncomfortable draftproblems.

In buildings with mechanical ventilation,special precautions must be taken to avoidunder pressure in the rooms. Fireplaces andother open hearths that require extensivesmoke suction capacity call for air supply of150-300 m3per hour (42-84 l/s), whichcorresponds to overall external air vents of atleast 300 cm

2, preferably distributed

throughout the various rooms of the house.

47:533 EXHAUST ARRANGEMENTSTable 47.53 gives the recommended ductdiameter, air flow rate, etc. for rooms where anexhaust arrangement is required. For smallerhouses where the sum of the specifiedexhaust air flow rate yields unduly highbackground ventilation, it should be possible toreduce the exhaust air flow rate, provided itcan be increased according to need, e.g.during cooking, taking a batch, etc.

The kitchen should have a suction hood overthe cooker to keep cooking smell fromspreading throughout the building.


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Table 27.53ROOM Natural

ventilation

Duct crosssection in cm2

Mechanicalventilation Air

volume rate inl/s

Air supply, valves,diffusers, ducts

Living room,bedroom

- - Window that can be opened and/or adjustable valve with a valve with a freeaperture of 100 cm

2in the outer wall

Kitchen 150 17 As for the living room, or with diffuser above,in or beneath the door from adjacent roomswith an aperture of 100 cm

2

Bathroom with/withoutWC 150 17 Diffuser above, in or beneath the door froman adjacent room, with an aperture of 100cm

2

Separate shower or WC 100 11 Diffuser above, in or beneath the door froman adjacent room, with an aperture of 100cm

2

Laundry room forindividual apartments

150 22 150 cm2diffuser/valve from adjacent room

150 cm2adjustable valve in outer

wall/window that can be opened and adjustedCommon laundry roomassumed used by:a) one family per dayb) several families perday but not

simultaneouslyc) several familiessimultaneously? per family

30Notrecommende

dNotrecommended

3355

44

Duct for supply of external air must have across section that is 80% of the duct sectionrequired for ducts in natural ventilation c)

Basement with windowsthat cannot be opened

3 cm2pr m

2

floor space0.3 l/s perm

2floor

space

Valve cross section of 3 cm2per m

2floor

space

Lift shaft 50 cm2

perm

2of shaft

area

0.5 l/s perm

2of shaft

area

Cross section for external air supply:50 cm

2per m

2of shaft area

Garbage chute andrelated refuse roomStorage rooms/attics

Exhaust and supply must be designed to avoid under pressure

Adequate ventilation


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47:54 SMALL HOUSES

47:541 GeneralSee guideline for 47.531

Reference is made to "NBI Building Details"sheet A 552.304 Ventilation of small houses.Systems and components.

The regulation that mechanical ventilation canbe carried out by a separate fan and duct,freely routed from each room, entails that theexhaust duct above the roof can be eliminatedand replaced by an electrically driven exhaustfan located in the outer wall. This regulationapplies first and foremost to detached houses.For semi-detached houses, and small housesin groups or rows, etc. exhaust through the

outer wall can cause annoying smells inneighbouring apartments. As such, this type ofsolution is particularly ill-suited to horizontallydivided houses.

47:55 GARAGESTo achieve efficient airing, air supply andexhaust must be distributed as equally aspossible throughout the premises. Ventingfacilities should be found at both floor andceiling level.

Air transferred from other premises can beused to supply air to garages, provided it is ofadequate quality.

With respect to ventilation of service stations,reference is made to Directorate of LabourInspection regulations.

AS A MINIMUM, GARAGES SHOULD BEVENTILATED AS FOLLOWS:

1. UP TO 50 m2FLOOR SPACE

With natural ventilation: outdoor air intake andoutlet, each with an area of minimum 2% ofthe floor space.

2. MORE THAN 50 m2FLOOR SPACE

* Garages in rows for 1 row of cars, asspecified in item 1.Other garages should have mechanicalventilation with the following capacity:* Garages for long-term parking of private carsand with persons present less than 1 hour/day:q = 2 x M l/s, but minimum 0.85 l/s per m

2of

floor space.

* Other garages:q = 4 x M l/s, but minimum 1.7 l/s per m

2of

floor spaceq = air changes in l/sM = exhaust in l/s emitted from engines in thegarage

For private cars, M is calculated according tothe following formula:M = (20+0.1s1)n1+0.1n2s2)1/3.6 l/s wheren1= number of parking spaces in thegarage/garage section.n2= number of other parking spaces withentrance and exit through the garage section.s1= total average driving stretch in meters forthe entrance and exit of a car in thegarage/garage section.s2= average driving stretch in the garagesection for the entrance and exit of a carwhose entrance and/or exit takes placethrough the garage section.For trucks and buses in garages where thereis no need to count on traffic queues, n1and n2are specified as 1/25 of the area of the garagein question.x) A garage may have outer walls and thesemay be arranged in such a fashion that thegarage is sufficiently aired without specialventilation.


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10 ELECTRICITY Pyrox

1.10.1 Ohm's lawU = R

.I

P = U.I

U = voltage in V (volts)R = resistance W (ohm resistance)I = current in A (Amperes)P = power in W (Watts)

EXAMPLE:How to control the amperage of anelectric battery, 1 phaseelectric heating power, 3 kWvoltage 230 V

l = P .l = 3,000 = 13.04 A U 230

For three-phase multiply by a factor of 3.

EXAMPLE:How to determine the amperage of a fanmotor:

P = U.

I.

cos

COS = Loss factor for a motor; the value isalways less than 1.P = 0.25 kWU = 400 V

COS= 0.7What is the amperage?

I = 250 = 0.89A 400

.0.7

EXAMPLE:

What is the power?I = 10AU = 400 V400 x 1.73 x 10 = 6920 W

1.10.1 Control of heating equipmentOn/off thermostatsThe simplest and cheapest method of

maintaining an even temperature in a room tobe heated, is to use a thermostat that sensesroom temperature and switches the hot air fanon or off. See figure

TP

C

25

24

1

0

ON/OFFTHERMOSTATS

An important concept when discussingthermostat control is the dead band. A

thermostat that switches on at 24C andswitches off at 25C has a dead band of 1C.

A small dead band is often desired since itrepresents stable room temperature. Seefigure

TP

C

25

24,5

1

0

DEADBAND

An excessively large dead band can causetemperature fluctuations. Several types ofthermostats are available. Capillary tubethermostats, which exploit the principle thatliquid expands as temperature rises, areaccurate and inexpensive. Electronicthermostats involve slightly higher costs buthave two advantages. They offer theopportunity to set/adjust the dead band with ahigh degree of accuracy. The dead band mustnot be too small when heat is controlled bycontactors. The contactors will switch on andoff at short intervals, leading to wear and

noise.

IUP ??= 3


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10. ELECTRICITY PyroxMulti-stage thermostatsFor hot air systems with higher power values, switching full power on/off is seldom desired. Thecommon practice is to divide the power into two or several equal parts and switching on as muchpower as required. The simplest way to control a hot air fan is to install a multi-stage thermostatwith three contactors. The thermostat functions in the following manner:

When the temperature drops below the set temperature, contactor no. 1 switches on 1/3 power.When the temperature drops further, contactor no. 2 switches on another 1/3. If the temperaturecontinues to drop, contactor no. 3 switches on the remaining power. This yields a smaller networkload during control of higher output and a more even temperature from the hot air system. Theprinciple is shown in figure

TP

P

P

P

C

25

24

23

22

13

2

1

0

1

0

1

0

MULTI-STAGETHERMOSTATS

Program controlFor higher effects/output it is possible to use program connection. The duct coil elements are thendivided into a number of groups. If group 1 is of 1 kW, then group 2 is of 2 kW, group 3 of 3 kWand group 4 of 4 kW. These four power groups yield 7-stage power control. Power can be

adjusted according to 1kW stages from 1 to 7 kW.This principle is illustrated in figure

2

2 2

2 2

4

4 4 4 4

3 5 6 71

1

1

1

1

Stage

P

PROGRAMCONTROL

A program controller can be used for control of stationary hot air fans. With this type ofequipment, it is possible to control the power to each apparatus in the system, with up to 12heating stages.Night set-back of room temperature is a standard feature. Program controllers can also bedelivered with functions for control of damper motors in mixing dampers to control outdoor andsurrounding air, respectively.


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PyroxInfinitely variable temperature controlTemperature controllers of the Pulser type or TTC are often used as an alternative to powercontrol with on/off or 3-stage thermostats. Pulser 220 and TTC use a Triac in the control circuit.Triac is a semi-conductor that can switch power on/off to the electric heating elements without thedisturbing noise which would otherwise occur with contactor-type control. A connection and

disconnection cycle takes one minute or less. Output is increased by increasing the "on" time andreducing the "off" time correspondingly.The power controller is also equipped with a temperature controller and potentiometer. Thetemperature sensor can be of the built-in or external type. Triac controllers cannot be used tocontrol contactors and can therefore not be used for control of control AVR hot air appliances andLPX/LG air gates. On the other hand, they are well suited for control of ceiling-mounted heatersand duct batteries.

1 min

1 min

TEMPERATURECONTROLLERS

The principle is illustrated in figure


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11. BUILDINGS`POWER REQUIREMENT FOR HEATING Pyrox

General

The total power requirement for heating a building is determined by three factors:

HEATLOSSINA BUILDING

INFILTRATION

VENTILATION

TRANSMISSION

Transmission heat loss is caused by the stream of heat from inside the building going through thebuilding's structure. The greater the difference between indoor/outdoor temperature, the greaterthis heat loss will be. Another determining factor is the building's thermodynamic properties.Generally, efforts are made to reduce this kind of heat loss by insulating the building.

Air infiltration heat loss is due to cold outdoor air seeping in through leaks in the structure of thebuilding. The extent of this type of heat loss is difficult to calculate precisely. Normally empiricaldata is used to calculate this type of heat loss for various types of building.

Finally, heat is required to heat ventilation air, either via direct application of heat to the heatingcoil in a mechanically balanced system, or indirectly to the inside air in a building with naturalventilation.

Heat loss due to transmission

Heat loss will occur through various components of the building such as exterior walls, windows,floors and ceilings. This heat loss can be expressed by the following equation:

Un= total heat transfer coefficient for surface no. n, [W/m2

x K]An= area of surface no. n, [m

2]T = difference in temperature at the building component, [K]

[ ]WTAU nn ??=


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Pyrox

If we are familiar with the construction of the building, the U factor for a construction consisting ofseveral layers can be calculated using the following equation:

[ ]UR R R R R

W m Ki y n

=+ + + + +

?1

1 2

2

...... /

U-FACTOR FOR A CONSTRUCTION

OUT

ROOM

Ri= internal heat transfer resistivity between the air and the building's constructionRy= external heat transfer resistivity between the air and the building's construction

R1..Rn = heat transfer resistivity for each individual material layer

The following values can be used for R iand Ryin the calculations:

Ri= 0.13 m2K/W

Ry= 0.04 m2K/W

Rnequals the heat transfer resistivity for each individual material used in the building's

construction, and is calculated on the basis of Rn= sn/nwhere s stands for the thickness of the


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layer and is the material's thermal conductivity. The values for most construction materialscan be found in Norsk Standard 3031 (NS 3031) as well as in technical manuals.

When we are required to calculate the transmission heat loss for a building in practice, we mustbegin with the building regulations' U factorrequirements. However, in older buildings, and insome exceptional cases, it may be necessary to calculate the U factor.

The building regulations' requirements are shown in the following table:

Component U factor in W/m2

x K

Facades: Applies at tinside>18C Exterior wall 0.30

Window 2.40 Door, entrance 2.00

Ceiling/roof: 0.20

Floor:

Above outside air 0.20

Above non-heated room 0.30

On ground 0.30

When the difference in temperature at the component is less than 5C, transmissionheat loss willbe negligible. This is usually the case at inside walls. The difference in temperature at a

component adjacent to outside air is calculated as T = tr- t0, where:tr= room temperaturet0= dimensioning outside temperature

The room temperature is usually set at 20C for rooms for normal living/working areas. Inswimming baths, operating theatres and treatment rooms, 2530C is often chosen. For special-purpose rooms, it is important to discuss what the dimensioning temperature should be.

The dimensioning outside temperature is the lowest outside temperature measured over a 3-dayperiod in the course of a 30-year period. Dimensioning outsidetemperatures for various places inNorway can be found in NS 3031 or in the Byggforsks Manual No. 33, VVS-tekniske klimadata forNorge.

When calculating the heating requirement of a room adjacent to an unheatedroom, thetemperature in the unheated room must be estimated, or calculated by setting up a heatingbalance for adjacent rooms.


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PyroxHeat loss due to air infiltration

Outside air will infiltrate the building through leaks in the building's construction. Thisair then hasto be heated to room temperature. Calculating this type of heat loss is difficult because it willdepend on many factors such as how draught-proof the building is, wind pressure and airtemperature.

The volume of infiltrated air is calculated using the air change rate n, which expresses the ratiobetween the volume of outside air L in m3/h and the volume of the room V in m

3. In a new

building, n can be calculated as 0.2, while in older buildings air change rates in the range of n =0.30.8 have been measured.In new buildings in particularly exposed locations, an air change rate higher than n = 0.2 shouldsometimes be used. NS 3031 stipulates the recommended values in such cases.

Heat loss through air infiltration is calculated using the following equation:

( ) [ ] = ? ? ? n c V t t W p r 0

n = air change rate, [h-1]V = volume of room, [m3]cp= the air's heat capacity = 0.335 Wh/m

3Ktr= room temperaturet0= dimensioning outside temperature

Heat loss due to ventilation

Calculations of heat loss due to ventilation depend on what kind of ventilation is in use in thebuilding:

Natural ventilation (upthrust ventilation):This kind of ventilation system is based on thermal currents and the volume of air is determinedby the difference in weight between outside air and inside air. For this reason, the volume of airwill vary, and it is therefore difficult to calculate the heat loss precisely.

The regulations state that n = 0.4 h-1

must be calculated without any addition for air infiltration.This type of heat loss is calculated according to the same equation as for heat loss due to airinfiltration. The heat loss calculated is added to the heat loss due to transmission for eachindividual room.


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PyroxMechanical exhaust ventilation:This type of ventilation system sucks in the outside air through vents and the used air is extractedthrough a system of ducting and a fan. The volume of air depends on the size of the fan. Theoutside air which is drawn in must be heated to room temperature and the heat requirement iscalculated using the following equation:

HEATLOSSDUETOMECHANICALEXHAUSTVENTILATION

MECHANICAL

EXHAUST

SUPPLY

AIR

HEATER

HEATER

HEATER

INFILTRATION

TRANSMISSION

L = exhaust air flow rate in m3/h.The heat loss calculated is added to the heat loss due to transmission and air infiltration for eachindividual room, since there are heating elements in each individual room which mustcompensate for the heat loss.

Balanced ventilation:In balanced ventilation systems, the fresh air is drawn in centrally and heated in a separateheating coil, sometimes after a process of heat exchange with the exhaust air. The necessarypower supply to the heating coil is calculated using the following equation:

( ) ( ) [ ] = ? ? ? L t t Wi0 335 10,

ti= the air temperature after the heating coil, usually equal to room temperature= heat return efficiency


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HEAT LOSSDUETO BALANCEDVENTILATION

HEATER

HEATER

HEATER

INFILTRATION

TRANSMISSION

In this kind of system, there is no addition in each individual room due to the ventilation system.The heat requirement for each room is therefore equal to the sum of the transmission heat lossand the air infiltration heat loss.


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PyroxAddition to/deduction from the heat loss:It is not usual to make any deduction in the calculated power to allow for lighting and other heatsources. This is partly because the building needs to be heated outside normal hours of use, andpartly because it is desirable to incorporate a certain security margin.NS 3032 states the recommended values for power requirements for heating and ventilation in a

whole building for various types of building:

Type of building Max. simultaneous power requirement [W/m2] Low Medium High

Office building, new 35 40 50

Office building, existing 40 50 60

Nursing home, new 60 70 80

Nursing home, existing 70 80 90

Hotels, new 60 70 80

Hotels, existing 70 80 90

Bathrooms, swimming pools, new 100 140 180

Bathrooms, swimming pools, existing 120 160 200

Detached houses, new, + 20 cm insul. 55 70 85

Detached houses, existing 70 90 110

Blocks of flats, new 40 45 50

Blocks of flats, existing 45 50 60

Schools, new, with gymnasium 40 50 60

Schools, existing, with gymnasium 45 60 70

Example

You have been asked to calculate the necessary power requirement to heat a room in a detachedhouse. The house is in Oslo which has a dimensioning outside temperature of -20C. The desiredroom temperature is 20C. The room is below an unheated attic (-20C) and the floor is above abasement (0C). The house plans provide the following details:

Floor area: 44 m2

Ceiling area: 44 m2

Exterior walls: 29 m2 (less the window area)

Window area: 5 m2

Volume of room: 100 m3

The house has a mechanical exhaust ventilation system. The total exhaust air flow rate is 150m3/h. The exhaust air flow rate from the room in question is 50 m

3/h. It is assumed that the

building's construction complies with the statutory requirements in terms of U factor. This givesthe following data:

Exterior wall: 0.30 W/m2KCeiling: 0.20 "Floor: 0.30 "Window: 2.40 "


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This data generates the following calculations:

Transmission heat loss:

Exterior walls: = 0.3 x 29 x 40 = 348 WCeiling: = 0.2 x 44 x 40 = 352 WFloor: = 0.3 x 44 x 20 = 264 WWindows: = 2.4 x 5 x 40 = 480 W

Total transmission heat loss: 1,444 W

Air infiltration heat loss:

Calculated with n = 0.2 air changes

Air infiltration heat loss: = 0.2 x 100 x 0.335 x 40 = 268 W

Ventilation heat loss:

Exhaust air flow rate 50 m3/h

Ventilation heat loss: = 50 x 0.335 x 40 = 670 W

Total power requirement for heating the room is 1,444 + 268 + 670 = 2,382 W.

This gives approx. 54 W per m2floor area.


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12. BUILDINGS' ENERGY REQUIREMENTS Pyrox

General

When calculating the annual energy requirements of a building for heating rooms and heatingventilation air, we must take the following into account:

_ transmission and air infiltration heat loss_ ventilation heat loss

In the overall calculation we also need to take the building's heat gain into consideration, i.e.:

_ heat from the sun_ heat from electric lighting_ body heat_ heat from machinery and equipment_ heat from fans

This section demonstrates how to calculate the annual energy requirements of a building,including how to take the above-mentioned factors into account:

Definition of central concepts in energy calculations

To be able to carry out an energy calculation, we need to know something about the climaticconditions, and how they vary during the course of the year at the site of thebuilding. Threeconcepts are important to this process:

_ temperature duration curves_ length of the heating season

_ number of degree days

The temperature duration curve is plotted by counting the number of days with a meantemperature below a certain temperature. A temperature duration graph can be generated byplotting consecutive values of the temperature and number of days. If this graph is plotted as amean graph covering ten years, it will give an approximately correct picture of the climate for theplace in question. An example of this kind of graph is given below:


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PyroxTemperature duration graph:

TEMPERATUREDURATION GRAPH

TEMPERATUREDURATION

GRAPH

DAYS

The heating season means the part of the year when the building needs to be heated. The lengthof the heating season is usually calculated as lasting until the meanoutside temperature is 10C.If the temperature duration graph for the location is known, the length of the heating season canbe worked out. The diagram above shows that the length of the heating season will be approx.240 days. The fact that it is generally unnecessary to heat the building outside this period, despitethe outside temperature being lower than room temperature, is due to internal supply of heat andsupplementary heat from the sun.

Degree days means the area which is delimited by the duration graph and an isotherm. The trueterm is degrees multiplied by days, but is usually abbreviated to degree days.

The number of degree days if the desired room temperature is 20C is shown by theareadelimited by the duration curve and the isotherm 10C plus the rectangle marked between theisotherms 10 and 20C as shown in the figure above. For example, if the desired roomtemperature is 17C, the number of degree days would be the area below the isotherm 10C plusa rectangle delimited by the isotherms 10 and 17C.The number of degree days for different places can be found in tables of climatic data or may beobtained on request from the Meteorological Institute.


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PyroxCalculations of annual energy requirement for heating and ventilation

If the power requirement and the above figures are known, the annual energy requirement can becalculated as follows:

( ) ( ) ( ) [ ]Q

t tG

t tG

t tG Q kWh

T

R o

I

R o

V

i o

v s t=

? ? +

? ? +

? ? ?

T= transmission heat loss (kW)I= air infiltration heat loss (kW)V= ventilation heat loss (kW)G = number of degree days (C x days)

= operative time for heating system, usually 24 hourstR = dimensioning room temperatureto= dimensioning outside temperaturetI= inflowing temperature for the ventilation system, usually the same as tRv= operative time for the ventilation system, in the case of natural ventilation equal to

s= takes into account non-operation of ventilation system at weekends. s= 5/7 in the case of 5 days operation.Qt= heat from the sun, body heat, heat from equipment and other internal sources of heat.

If there is insufficient data to calculate Qt, it may be assumed that this quantity reduces thecalculated energy consumption by approx. 2025%.

NS 3032 states the recommended values for the energy requirement in various types of building:

Type of building Energy requirement (kWh/m2floor area)

Low Medium High

Office building, new 90 110 130

Office building, existing 120 150 180

Nursing home, new 190 220 250

Nursing home, existing 220 270 300

Hotels, new 210 240 270

Hotels, existing 240 280 320

Bathrooms, swimming pools, new 300 500 700

Bathrooms, swimming pools, existing 500 700 900

Detached houses, new, + 20 cm insul. 100 130 170

Detached houses, existing 150 190 240

Blocks of flats, new 100 120 150

Blocks of flats, existing 150 180 220

Schools, new, with gymnasium 100 120 140

Schools, existing, with gymnasium 100 140 180


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13.RADIANT HEAT Pyrox

Heat transfer

Transfer of heat always occurs from a higher temperature to a lower temperature.There are fivebasic methods by which this transfer can take place:

_ conduction_ convection_ radiation_ mass transfer_ phase change

Conduction occurs when heat spreads during constant lowering of temperature from a hotter partof a body to a colder part of the same body. Heat can also be conducted from one body toanother if there is close contact between the two.

Convection occurs when heat from a liquid or a gas is conducted to a solid body (or vice versa).

Radiation occurs when heat is transferred from one body to another without heating the mediumpresent between the two bodies, e.g. air. For instance, the sun heats the earth without heatingthe atmosphere between the two.

Mass transfer occurs when the heat content of a fluid alters if it runs through surroundings with adifferent temperature than the fluid itself.

Phase change occurs during evaporation and condensation. In the case of evaporation a certainamount of heat must be applied for phase change to occur, while in the case of condensation acertain amount of heat is released.

In practice, the heating elements used to heat rooms will transfer heat using a combination ofconvection and radiation, depending on the surface temperature of the heating element.

In water-filled radiator systems convection will account for 7080% of the heat output, while therest of the heating occurs through radiation. In radiant heaters, the ratio will be reversed, i.e.radiation will account for more than 80% of the heat transfer.

General

In actual fact, radiant heat consists of electromagnetic rays like radio waves and visible light rays.Heat radiation may or may not occur, depending on the wavelength of the electromagnetic rays.Part of the electromagnetic spectrum is represented below, showing where temperature radiationwill take place:


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PyroxThe electromagnetic spectrum:

UV radiation Visiblelight

Temperature radiation(infrared heat)

Microwaves Radio waves

1 nm 1 m 1 mm 1 m

Radiant heating elements are divided into groups according to how high the elements' surfacetemperature is. One group includes short-wave radiation with temperatures of approx. 750C, andlong-wave radiant heaters with a surface temperature of approx. 100C are used in rooms withlow ceilings. In the middle is a group with temperaturesof approx. 300C, used for ceiling heightsof 2.5 to 20 m.

Heating with radiant heat can have many advantages over other types of heating. The radiatedheat passes unimpeded through the air and does not emit heat until it meets surfaces such asfloors and walls. These surfaces in turn emit convection heat to the air in the room, therebyheating the entire living/working area. As such, it is possible to avoid heat collecting under theceiling. Some of the radiated heat will also be absorbed directly by the people in the room.

This effect makes it possible to tolerate an air temperature 23C lower than with convectionheating, while experiencing the same degree of comfort.

These factors mean that correctly dimensioned radiant heating is energy-saving.

Definitions

In connection with planning of radiant heating and the use of catalogue data, it is important todefine what is meant by a number of concepts and quantities. We shall therefore offer definitionsof the following:

_ heat radiation_ intensity of radiation_ radiation temperature contribution/asymmetry of radiation_ radiation temperature_ air temperature_ operative temperature_ thermal gradient

Heat radiation from a radiant heating panel increases to the power of 4 of the absolutetemperature. This means that the radiant heating panels can be manufactured in very small sizes,while still being capable of emitting considerable heat. The radiated heat emitted from a hot

surface can be calculated using the following equation:

( ) [ ]P A T T W s s o= ? ? 4 4_

A = surface area of the radiant heating panel, m2

Ts= absolute temperature of the radiant heating panel, KTo= absolute temperature of the surr

HeatVentTutorial - Pyrox

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