radiant products engineering guide

8/18/2019 Radiant Products Engineering Guide

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Please refer to the Price Engineer’s HVAC Handbook

for more information on Radiant Heating and Cooling

S E C T I O N H

Engineering GuideRadiant Products


2/43H-2All Metric dimensions ( ) are soft conversion. © Copyright Price Industries Limited 2011.Imperial dimensions are converted to metric and rounded to the nearest millimetre.

Radiant Products

Engineering Guide

Introduction To Radiant Heating and Cooling

Radiant heating and cooling systems offer anenergy efcient alternative to all-air systems.In most cases, the supply air volume ofthe air handling system is limited in sizeto satisfy only the ventilation and latentloads, with the radiant system making upthe balance of the heating and coolingloads. This comfortable method of heatingand cooling may save energy, space andbuilding maintenance costs. The followingpages offer an introduction to the products,systems and design methodology, as wellas the advantages and limitations of radiantheating and cooling.

Management of heat loads can generallybe classied into two different types: all-

air systems or hybrid systems. All-airsystems have been the most prominentin North America during the 20th centuryand have been in use since the advent ofair conditioning. These systems use air to

service both the ventilation requirement aswell as the building cooling load. In general,these systems have a central air handlingunit (or rooftop unit) that delivers enoughcool or warm air to satisfy the building load.Diffusers mounted in the zone deliver thisair in such a way as to promote comfort andevenly distribute the air. In many cases, theamount of air required to cool or warm thespace or the uctuations of loads makedesigning in accordance to these principlesdifcult. Draft is not uncommon, and someceiling diffusers have been known to “dump”at low capacities.

Hybrid systems have two components: anair-side ventilation system and a hydronic

(or water-side) radiant system. The air-sideis designed to meet all of the ventilationrequirements for the building, as well assatisfy all latent loads. It is a 100% outsideair system and because the primary function

of the supply air system is ventilation asopposed to cooling, it can be supplied athigher supply air temperatures than is typicalof overhead air distribution systems. Thewater-side is designed to meet the balanceof the sensible cooling and heating loads.These loads may be handled by water basedproducts, such as radiant panels, whichtransfer heat mainly by thermal radiation,and chilled sails, which transfer heat usinga combination of thermal radiation andnatural convection. Radiant panels havebeen used for sensible heating and coolingin North American buildings for over half acentury, and are a widely recognized andwell-established technology. Chilled sails

were originally developed in Europe inthe late 1990s, and are a relatively newtechnology in North America.


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© Copyright Price Industries Limited 2011. All Metric dimensions ( ) are soft conversion .Imperial dimensions are converted to metric and rounded to the nearest millimetre. H-3

Radiant ProductsEngineering Guide

Figure 1: Examples of radiant heating and cooling

Radiant heating and cooling systemsprovide an effective method for satisfyingthe heating and/or cooling loads of a spacewhile promoting a high level of occupantcomfort and energy efciency.

Hydronic systems have been successfullyused in several applications havingdramatically different characteristics.Some examples of areas where radiantsystems have been applied include:

• Green Buildings • Hospitals

• Burn Centers • Isolation Rooms

• Schools • Data Centers

• Ofce Buildings • Airports

• Cafeterias • Television Studios

• Theaters • Casinos

Benefits of Air-Water Systems

There are many benets to heating andcooling using radiant panels and chilledsails. Advantages of these water basedheating and cooling systems over othermechanical systems include:

• Energy and system efciency

• Reduced system horse power

• Indoor environmental quality

• Improved indoor air quality

• Increased thermal comfort

• Reduced mechanical footprint

• Lower maintenance costs

• Improved system hygiene

Radiant systems are a good choice where:

• Thermal comfort is a major designconsideration

• Areas have high sensible loads

• Areas require a high indoor air quality(100% outdoor air system)

• Energy conservation is desired

Energy Efficiency

The heat transfer capacity of water allows fora reduction in the energy used to transportan equivalent amount of heat as an all-airsystem (Stetiu, 1998). These reductions canbe found primarily through reduced fanenergy.

The higher chilled water supply (CHWS)temperatures used with active and passivebeam systems, typically around 58 °F[14.5 °C], provide many opportunities for areduction in energy use, including increasedwater-side economizer use. This increasedCHWS temperature also allows for morewater-side economizer hours than would bepossible with other systems where CHWStemperatures are typically ~45 °F [7 °C].

Concepts and Benefits

Indoor Air Quality

Depending on the application, under

maximum load, only ~15 to 40% of thecooling air ow in a typical space is outdoorair and is required by code to satisfy theventilation requirements. The balance ofthe supply air ow is recirculated air which,when not treated, can transport pollutantsthrough the building. Radiant systemstransfer heat directly to/from the zone andare often used with a 100% outdoor airsystem which exhausts polluted air directlyto the outside, reducing the opportunityfor VOCs and illness to travel between airdistribution zones.

Noise

Radiant systems do not usually havefan powered devices near the zone. This

typically results in lower zone noise levelsthan what is achieved with all-air systems. Insituations where passive beams are used inconjunction with a quiet air system, such asdisplacement ventilation, the opportunitiesfor noise reduction increase further.

Reduced Mechanical Footprint

The increased cooling capacity of wate

allows the transport system to be reducedin size. It is generally not unusual to be ableto replace ~60 ft² [6 m²] of air shaft with a6 in. [150 mm] water riser, increasing theamount of oor space available for use olease. Due to the simplicity of the systems(i.e. reduction in the number of moving partsand the elimination of zone lters, drainpans, condensate pumps, and mechanicacomponents), there tends to be less spacerequired in the interstitial space to supporthe HVAC system.

Lower Maintenance Costs

With no terminal unit or fan coil lters omotors to replace, a simple cleaning is althat is required in order to maintain the

product.



When To Use Radiant Systems

Hygienic System

With the elimination of the majority oflters and drain pans, there is a reducedrisk of mold or bacteria growth in the entiremechanical system.

Radiant systems such as radiant panelsand chilled sails are well-suited to someapplications and less so to others. As aresult, each application must be reviewedfor potential benets as well as the suitabilityof these types of systems. One considerationwhich can assist in the decision to employhydronic systems as opposed to an all-airsystem, is the air-side load fraction—orthe percentage of the total air supply thatmust be delivered to the zone to satisfycode and dehumidication requirements.Table 1 shows the load fraction for severalspaces. In the table the best applications forhydronic systems are those with the lowestair-side load fraction as they are the ones thatwill benet the most from the efcienciesof hydronic systems. Another factor whichshould be examined is the sensible heatratio or the percentage of the cooling loadthat is sensible as opposed to latent. Thelatent loads must be satised with an airsystem and offer some sensible cooling atthe same time because of the temperature ofdehumidied air. If the total sensible coolingload is signicantly higher than the capacityof the air supplied to satisfy the latent loads,a radiant system might be a good choice.

Commercial Office Buildings

In an ofce building hydronic heating andcooling systems provide several benets. Thelower supply air volume of the air handlingsystem provides signicant energy savings.In addition, the smaller infrastructurerequired to move this lower air ow allowsfor small plenum spaces, translating intoshorter oor-to-oor construction or higherceilings. The lower supply air volume andelimination of fans at or near the spaceoffers a signicant reduction in generatednoise. Often the lower air ow translatesto reheat requirements being reduced. Inthe case of 100% outside air systems, thelighting load captured in the return plenum

is exhausted from the building, lowering theoverall cooling load.

Schools

Schools are another application that canbenet greatly from radiant panels and chilledsails systems. Similar to ofce buildings, thebenets of a lower supply air volume to thespace are lower fan power, shorter plenumheight, reduced reheat requirement, andlower noise levels (often a critical designparameter of schools).

Hospital Patient Rooms

Hospitals are unique applications in thatthe supply air volume required by localcodes for each space is often greater thanthe requirement of the cooling and heatingload. In some cases the standard or coderequires these higher air-change rates forall-air systems only. In these cases thetotal air-change rate required is reduced ifsupplemental heating or cooling is used.This allows for a signicant reduction insystem air volume and yields energy savingsand other benets.

Furthermore, because these systems aregenerally constant air volume with thepotential to reduce the primary air-changerates, reheat and the cooling energy discardedas part of the reheat process is a signicantenergy savings opportunity. Depending onthe application, a 100% outside air systemmay be used. These systems utilize noreturn air and no mixing of return betweenpatient rooms, potentially lowering the riskof hospital associated infections.

Hotels / Dorms

Hotels, motels, dormitories, and similar typebuildings can also benet from hydronicsystems. Fan power savings often comefrom the elimination of fan coil units locatedin the occupied space. The energy savings

associated with these “local” fans is similarin magnitude to that of larger air handlingsystems. It also allows for the eliminationof the electrical service required for theinstallation of fan coil units as well as areduction in the maintenance of the drainand lter systems. The removal of thesefans from the occupied space also provideslower noise levels, which can be a signicantbenet in the sleep areas.

ApplicationTotal AirVolume (Typ.)

VentilationRequirement (Typ.)

Air-SideLoadFraction

Ofce 1 cfm/ft2 [5 L/s m2] 0.15 cfm/ft2 [0.75 L/s m2] 0.15

School 1.5 cfm/ft2 [7.5 L/s m2] 0.5 cfm/ft2 [2.5 L/s m2] 0.33

Lobby 2 cfm/ft2 [10 L/s m2] 1 cfm/ft2 [5 L/s m2] 0.5

Patient Room 6 ach 2 ach 0.33

Load-driven Lab 20 ach 6 ach 0.3

Table 1: Typical load fractions for several spaces in the United States

Limitations

There are several areas in a building wherehumidity can be difcult to control, suchas lobby areas and locations of egress.These areas may see a signicant shortterm humidity load if the entrances are notisolated in some way (revolving doors orvestibules). In these areas, a choice of acomplimentary technology such as fan coilunits or displacement ventilation is ideal.

Other applications may have high air ow/ ventilation requirements, such as an exhaustdriven lab. The majority of the benefitprovided by the hydronic system is linkedto the reduction in supply air ow. As such,these applications may not see sufcientbenet to justify the addition of the hydroniccirculation system, making them not likelyto be a good candidate for this technology.



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Products - Radiant Panels

Operation

Radiant panels mainly use thermal radiationto handle the heating or cooling loads of aspace. Thermal radiation heat exchange isbased on differences in surface temperaturesas discussed in Chapter 3—Introduction toHeat Transfer from the Price Engineer'sHandbook. Radiant panels add energy to orremove it from a room mainly using radiationwith surfaces in the room, but also directlyto occupants (Figure 2). To a lesser extent,the panels also heat or cool a room throughconvection of the room air as it is heated orcooled by the panel surface.

Because radiant panels can handle thesensible portion of a building load they mustbe paired with an air system for ventilationand latent load removal. In heating, forexample, heat from warm water is transferredto the panel surface via conduction. The heatpasses through the tubing, the mountingextrusion (the ‘n’), and the panel itself, tothe panel surface. At the surface, heat is bothradiated to other surfaces in the room andtransferred to room air via natural convection.

The heat transfer through a radiant panel caneasily be modeled with a thermal resistancecircuit, as in Figure 3. The resistance circuitrepresents the actual components of aradiant panel. The nodes represent varioustemperatures of the panel componentsurfaces, and the ‘resistors’ represent the heat

conduction through the panel componentsand to the surrounding room. The t w noderepresents the mean water temperature thattransfers through the copper tubing to theactual panel components. To achieve themaximum possible surface temperatureof the panel, T surf , the conduction from thepipe to the n to the panel surface mustbe maximized, or, inversely, the resistancemust be minimized. This can be achieved byusing materials that are highly conductivesuch as copper tubing and aluminum for then and panel. Even surface contact betweenthe water tubing and the ns decreasesresistance, along with thermal paste whichcan be applied between the n and the panelsurface to help spread heat evenly to the

panel surface (Figure 4)

RADIATION

Figure 2: Radiation pathways

Rconv,s

T air, ceiling

T panel, outer insulation

Rinsulation

AUST,ceiling

Rrad,s

Rconv,room

T air, room

Fin

Surface AUST, room

T surf, panel

Rrad,room

R fin

T fin, ave

R panel surface Copper tubing with

Towards slab

Towards room and occupants

AUST = Area-weighted temperature

of all indoor surfaces of walls, ceiling,

floor, windows, doors, etc.

Figure 3: Thermal resistance circuit diagram of a modular radiant panel

Without ThermalPaste With Thermal Paste

Figure 4: Surface temperature distribution of a radiant panel



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Products - Chilled Sails

Operation

Chilled sails provide a functional andunique alternative to conventional radiantpanels. Sails couple the radiant coolingeffects of standard radiant panels with aconvective component. In cooling mode,chilled sails create natural convection bycooling the surrounding air as it passes overthe surface facing the plenum. As the airfalls into the occupied zone, where warmair is pulled over the sail, the convectivecooling capacity of the sail is coupled withthe radiant capacity of the cool sail surface,resulting in a cooling capacity greater thanthat of standard radiant panels. In cooling,the approximate breakdown of heat mode

transfer of chilled sails is 30% by thermalradiation and 70% by natural convection.

A general air ow diagram of an exposedchilled sail in heating and cooling mode canbe seen in Figure 8. In certain applications,sails can also be used for heating. In heatingmode, the sails use radiation only to heatthe zone below. Because sails have noinsulation on their reverse side, heat isradiated not only towards the room, butalso towards the building structure. As theslab warms, it in turn helps heat the roomto a small extent by thermal radiation andnatural convection.

Like radiant panels, chilled sails can alsobe analyzed using a thermal resistance

circuit diagram, as seen in Figure 9. Theresistance circuit represents the actualcomponents of a chilled sail. The nodesrepresent various temperatures of the sailcomponent surfaces or the conditions of theroom, and the ‘resistors’ represent the heatconduction through the panel componentsor heat transfer between the sail and theroom. The mean water temperature, t ̅ w, noderepresents the mean water temperature thattransfers through the copper tubing to theactual sail. Most chilled sails are one singleextrusion, which means that the ‘n’ and‘sail’ are one solid piece of aluminum. Tomaximize heat transfer through the sail,or, conversely, to minimize resistance, amaterial with high thermal conductivity,

such as aluminum, is typically used.As seen in Figure 10, a chilled sail transfersheat to a room with a combination ofradiation and natural convection. Becausechilled sails have no insulation on theirreverse sides, heat is transferred from thecopper tubing/n to the slab and plenum.

The heat transfer from the sail to the roomhas three components: natural convectionwith the room air, thermal radiation with theroom surfaces, and thermal radiation fromthe top of the sail with either the suspendedceiling or the xed ceiling, depending on thedesign details.

Figure 8: Air ow pattern of an exposed chilled sail in cooling and heating mode

Cooling

Heating

Sail

Sail

Figure 9: Thermal resistance circuit diagram of a chilled sail

Figure 10: Typical chilled sail


Towards slab

Towards room and occupants

T air, room AUST, room

Rrad, room

Rrad, ceiling

Tair,ceiling

Rconv, ceiling

Rconv, room

R fin/sail

AUST, ceiling

Tsurface, fin/soil Sail/Fin

Copper tubing with

Top Bottom


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0

75

80

90

100

10 20 30 40 50 60 70

C a p a c i t y o f S a i l ,

%

Free Area as a Percentage of Sail Area

Figure 16: Typical piping of a sail with an odd number of passes

Figure 17: Typical even number of passes

Figure 14: Free area vs. active area of sail Figure 15: Clearance between chilled sail and slab

0

0 1 2 3 4 5 6

20

40

60

80

100

0 50 100 150

E f f e

c t i v e C a p a c i t y o f S a i l ,

%

Clearance Between Chilled Sail and Slab, in.

Clearance Between Chilled Sail and Slab, mm

Water Supply

Sail 1 Sail 2 Sail 3

Water Return

Flex HoseFlex Hose

Water Supply

Flex Hose

Products - Chilled Sails

The amount of free area vs. active area of sail will affect the performance of the sail system according to Figure 14. In all cases, theamount of space between the back of the sail and the structural slab will affect the level of circulation, and thereby the convective coolingcomponent. This capacity is affected according to Figure 15.


Connecting Chilled Sails

Depending on the width of the unit, the sailsmay have connection locations on oppositeends. Sails with an odd number of sectionswill have connections on opposite ends,and even number of sections will haveconnections on the same ends, as seen

in Figures 16 and 17 below. Flex hose isgenerally used to connect the water owbetween the units.



1. Determine the ventilation requirement

The ventilation requirement should be calculated to meet ventilatiown codes. For example, using ASHRAE Standard 62-2004 to determinethe minimum fresh air ow rate:

L1

where

Qoz = minimum fresh air ow rate, cfm [L/s]

R p = outdoor air ow rate per person, cfm/person [L/s(person)]

P z = zone population or maximum number of occupants in zone

Ra = outdoor air ow rate per unit area, cfm/ft2 [L/sm2]

A Z = zone oor area or net occupied area of the zone, ft2 [m2]

2. Determine required supply air dew-point temperature to remove the latent load

L2

where

q L = latent load, Btu/h [W]

Qs = supply air ow rate, cfm [L/s]

ΔW = difference in humidity ratio between the supply air and the room condition,

lbm,w /lbm,DA or gr/lbm,DA [kgw /kgDA or gw /kgDA]

Typically, the moisture content of the ventilation air will be sufciently low in the heating season to offset the internal gains.

3. Determine the occupied zone humidity ratio if there is excessive latent cooling

From equation L2:

L3

where W oz = humidity ratio of the room condition, lbm,w /lbm,DA or gr/lbm,DA [kgw /kgDA or gw /kgDA]

W SA = humidity ratio of the supply air, lbm,w /lbm,DA or gr/lbm,DA [kgw /kgDA or gw /kgDA]

If W oz is determined to be too low for comfort, humidication of the ventilation air should be considered.

4. Determine the supply air volume

The supply air volume is the maximum volume required by code for ventilation, and the volume required for controlling the latent load:

L4

where

Q L = air ow rate required for controlling the latent load, cfm [L/s]

5. Determine the heating capacity of the supply airIP L5

SI L5

where

qs,air = heating capacity of the supply air, Btu/h [W]

ρ = uid density, lbm/ft3 [kg/m3]

c p = specic heat at constant pressure Btu/hlb°F [kJ/(kgK)]

Qair = supply air ow rate, cfm [L/s]

Δt air = air temperature change (t return - t supply), °F [K]

Design Procedure – Heating



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6. Determine the heating required from the water side

L6

where

qs, hydronic = heating capacity of the water side, Btu/h [W]

qt = total sensible heating capacity, Btu/h [W]

7. Determine an appropriate temperature loss through the panels

Specify a panel surface temperature, then nd the related mean water temperature, t ̅ w.



Figure 18: Connection between mean water temperature and panel surface temperature or, t panel - t room = 0.74 (t ̅ w - t room)

0 10 120

0 10 20 30 40 50 60

0

10

20

30

40

50

60

70

80

90

0

5

35

45

50

t w - t r o o m [

R ]

t w - t r o o m [

K ]

20 30 40 50 60 70 80 90 100 110

10

15

20

25

30

40

t panel - t room [K]

t panel - t room [R]

8. Determine the heat transfer coefficients for the radiant panels

The natural convection coefcient is:

IP L7

SI L7

Where

hc,natural = natural convection coefcient, Btu/hft2°F [W/m2K]

t a = room temperature, °F [K]

t panel = panel temperature, °F [K]

Dh = hydraulic diameter, ft [m]

Dh = 4A panels / P panels L8

Where

A panels = surface area of active panels, ft2 [m2]

P panels = the pipe internal perimeter, ft [m]





The forced convection coefcient is:

IP L9

SI L9

Where

hc,forced = forced convection coefcient, Btu/hft2°F [W/m2K]

ach = air change rate, cfm/ft2 [m3 /hm2]

The total convection coefcient is:

L10

Where

hc,total = total convection coefcient, Btu/hft2°F [W/m2K]

9. Determine the specific capacity of the radiant panels

The convective heat transfer per square foot to the panel is determined:

L11

where

q ̋ c = convective heat ux or convective rate per cross sectional area, Btu/hft2 [W/m2]

qc = convective heat transfer rate, Btu/h [W]

A = surface area of the medium, ft2 [m2]

Assuming that the wall temperature is equal to the air temperature, the radiant heat exchange with the panel is determined:

IP L12

SI L12

where

q”r = radiant heat ux, Btu/hft2 [W/m2]

AUST = area-weighted temperature of all indoor surfaces of walls, ceiling,oor, windows, doors, etc. (excluding active panel surfaces), °F [°C]

The total heat transfer per unit of face area is

L13

where

q ̋ o = total heat ux, Btu/hft2 [W/m2]

10. Determine the area of panels required

L14

where

A panels = area of panels, ft2 [m2]


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Example 1 - Patient Room (IP)

In this case, the supply air is a mix of the return air and the ventilation air. This mixture of outdoor air (at the outdoor conditions, assumingsaturated air at 15 °F with a humidity ratio of 12.5 gr/lb) and return air (assuming that it is at the design conditions of 75 °F, 40% RH – 52.5gr/lb), will have more than enough capacity to handle the latent load. In applications where humidity is critical, further analysis may bedone to determine the requirement of humidication. For more information refer to Chapter 5—Introduction to Psychrometrics of thePrice Engineer's Handbook.

Determine the heating capacity of the supply air

Using equation L5:

Determine the heating required from the water side

Determine an appropriate temperature loss through the panels

Using a mean water temperature of:

Determine the heat transfer coefficients for the radiant panels

Using equation L7 and the relation for Dh from equation L8, the natural convection coefcient is determined:

Due to the conguration of the room, it can be assumed as a rst estimation that the panels will be arranged at the perimeter where theload is, and run the width of the exposure (10 ft). Assuming also a 2 ft width of panel:

Using equation L9, the forced convection coefcient is determined:

Using equation L10, the total convection coefcient is determined:



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Determine the specific capacity of the radiant panels

Using equation L11, the convective heat transfer per square foot to the panel is determined:

The outside air temperature has a signicant impact on the inside surface temperatures of exterior walls. The exterior wall temperatureis determined with an h value, convective heat transfer coefcient of a vertical wall, of 1.46 Btu/(hft2°F) and a U value, overall heat transfecoefcient, of 0.315 Btu/(hft2°F):

The average unheated surface temperature is:

t

Calculating the radiant heat exchange:

From equation L13, the total heat transfer per unit of face area is:

Determine the area of panels required

Using equation L14:




Therefore, the assumption of panel size (20 ft2) used to calculate the hydraulic diameter is appropriate.

The ow rate required to manage the load with a panel ΔT of 10 °F is:

For simplicity, a 2 ft × 10 ft Price RPL linear radiant panel is selected.

This panel with 0.41 gpm will have a pipe velocity of 0.55 fps, which corresponds to a Reynolds number of 1900, which is in the laminar

range. For a better selection, the ow rate is increased to 1.3 gpm, which corresponds with a Reynolds number of 6400, which is in theturbulent region. From the performance chart, this also increases the pressure drop from 0.31 ft to 3.7 ft, which will allow better owcontrol of the panel.

Recalculating the temperature loss in the panel as well as the capacity:

This increase in capacity will result in only requiring 15.7 ft2, though it is more practical to stay with the original size in order to maintainaesthetics (the panel will run the length of the perimeter) as well as a standard module size (24 in. wide). Panels can be designed tohave both active and inactive sections to maintain aesthetics.

When running the entire length of the room, the trim and series option will allow the panel to be trimmed on site if the room size variesslightly during construction.

PATIENT ROOM

Corridor

Panel



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Consider the patient room shown in the gure below. The patient room includes a television, monitoring equipment and overheadlighting. The temperature set-point is 24 °C with a minimum relative humidity of 40%. The room is 3 m wide, 6 m long, and has a 3m ceiling. There is one exterior wall and window. The supply air temperature in heating mode is reset to 35 °C and the heating watetemperature is 72 °C.

PATIENT ROOM

Corridor

3 m

6 m

1.75 m

2.25 m

Determine

The water ow rate and pressure drop for the heating panels required to handle the heating load, assuming -10 °C outdoor air temperature

Overnight in winter, the envelope loss is 1400 W and the internal gains at that time are limited to the patient load:

Design Considerations

Patient 50 W

Medical Staff/Visitors 0

Television 0

Medical Equipment 0

Overhead Lighting 0

Envelope -1400 W

Total -1350 W

Patient latent load 45 W

Determine the Ventilation Requirement

For this example, local code refers to ASHRAE Standard 170-2008 for the HVAC system. According to ASHRAE Standard 170-2008,patient rooms with auxiliary heating require 4 ach of supply air, of which two are outdoor air.

Determine the required supply air dew-point temperature to remove the latent load

From equation L2:

Using the ventilation rate:

Example 1 - Patient Room (SI)



In this case, the supply air is a mix of the return air and the ventilation air. This mixture of outdoor air (at the outdoor conditions, assumingsaturated air at -10 °C with a humidity ratio of 1.8 g/kg) and return air (assuming that it is at the design conditions of 24 °C, 40% RH – 7.4g/kg), will have more than enough capacity to handle the latent load. In applications where humidity is critical, further analysis may bedone to determine the requirement of humidication. For more information refer to Chapter 5—Introduction to Psychrometrics of thePrice Engineer's Handbook.

Determine the heating capacity of the supply air

Using equation L5:

Determine the heating required from the water side

Determine an appropriate temperature loss through the panels

Using a mean water temperature of:

Determine the heat transfer coefcients for the radiant panels

Using equation L.7 and the relation for Dh from equation L.8, the natural convection coefcient is determined:

Due to the conguration of the room, it can be assumed as a rst estimation that the panels will be arranged at the perimeter where theload is, and run the width of the exposure (3 m). Assuming also a 600 mm width of panel:






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Determine the specific capacity of the radiant panels

Using equation L11, the convective heat transfer per square foot to the panel is determined:

The outside air temperature has a signicant impact on the inside surface temperatures of exterior walls. The exterior wall temperatureis determined with an h value, convective heat transfer coefcient of a vertical wall, of 8.29 W/(m 2K) and a U value, overall heat transfecoefcient, of 0.055 W/(m2K):


Calculating the radiant heat exchange:



Using equation L14:




Therefore, the assumption of panel size (1.8 m2) used to calculate the hydraulic diameter is appropriate.

The ow rate required to manage the load with a panel ΔT of 5 K is:

For simplicity, a 600 mm × 3000 mm RPL linear radiant panel is selected.

This panel with 0.027 kg/s will have a pipe velocity of 0.24 m/s, which corresponds to a Reynolds number of 4300 with a pressure drop

of 1.2 kPa, which is a good selection.When running the entire length of the room, the trim and series option will allow the panel to be trimmed on site if the room size variesslightly during construction.

PATIENT ROOM

Corridor

Panel



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Design Procedure – Cooling

1. Determine the ventilation requirement

The ventilation requirement should be calculated to meet ventilation codes. For example, using ASHRAE Standard 62-2004 to determinethe minimum fresh air ow rate:

L1

2. Determine required supply air dew-point temperature to remove the latent load

L2

If the required humidity ratio is not practical, recalculate the supply air volume required with the desired humidity ratio.

3. Determine the supply air volume

The supply air volume is the maximum volume required by code for ventilation and the volume required for controlling the latent load

L4

4. Determine the sensible cooling capacity of the supply air

IP L5

SI L5

5. Determine the sensible cooling required from the water side

L6

6. Determine an appropriate temperature rise through the panels

A panel temperature correction is unnecessary because the temperature differential between the water and air is small in cooling modeFor panels and sails that are designed well, the surface temperature can be approximated to be the mean water temperature:

L15

where

t w = mean water temperature, °F [K]

t CHWS = chilled water supply temperature, °F [K]

t out = chilled water return temperature, °F [K]

7. Determine the heat transfer coefficients for the radiant panels

The natural convection coefcient is:

IP L16

SI L16



Design Procedure – Cooling

The forced convection coefcient is:

IP L9

SI L9

The total convection coefcient is:

L10

8. Determine the specific capacity of the radiant panels

The convective heat transfer per square foot to the panel is determined:

L11

Assuming that the wall temperature is equal to the air temperature, the radiant heat exchange with the panel is determined:

IP L12

SI L12

The total heat transfer per unit of face area is:

L13

9. Determine the area of panels required

L14



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Example 2 - Small Office (IP)

Consider a small ofce with a southern exposure. The space is designed for two occupants, a computer with LCD monitor, T8 orescenlighting, and has a temperature set-point of 75 °F. The room is 10 ft wide, 12 ft long, and 9 ft from oor to ceiling. The owner expressedinterest in using radiant panels.

12 ft

9 ft10 ft

SMALL OFFICE

Window

Space Considerations

One of the primary considerations when using a radiant heating and cooling system is humidity control. As previously discussed, it isimportant to consider both the ventilation requirements and the latent load when designing the air-side of the system.

The assumptions made for the example are as follows:

• Load/person is 250 Btu/h sensible and 155 Btu/h latent

• Lighting load in the space is 6.875 Btu/h/ft²

• Computer load is 300 Btu/h (CPU and LCD Monitor)

• Total skin load is 1450 Btu/h

• Specic heat and density of the air are 0.24 Btu/lb°F and 0.075 lb/ft³ respectively • Design conditions are 75 °F, with 50% relative humidity

• Design dew point is 55 °F


Occupants 2

Set-Point 75 °F

Floor Area 120 ft²

Exterior Wall 108 ft²

Volume 1080 ft³

qoz 800 Btu/h

ql 825 Btu/h

qex 1450 Btu/hqT 3075 Btu/h

Determine

a) The ventilation requirement.

b) The suitable supply air and supply water temperatures.

c) The total convective heat transfer coefcient for radiant panels.



Solution

a) Determine the ventilation requirement

The ventilation requirement should be calculated to meet ventilation codes. For example, using ASHRAE Standard 62-2004 to determinethe minimum fresh air ow rate for a typical ofce space:

b) Determine required supply air dew-point temperature to remove the latent load

From equation L2:


At the design conditions (75 °F, 50% RH), the humidity ratio is 65 gr/lb, requiring a difference in humidity ratio between the supply androom air of:

From the gure below, the dew point corresponding to the humidity ratio is 40 °F, which is too cool for standard equipment.

Evaluating the humidity ratio at several temperatures led to the selection of a dew point of 50 °F in order to use less expensive commonequipment while also minimizing the supply air volume required to control humidity.

Humidity Ratio

Dew Point lb/lb gr/lb

40 0.00543 38

45 0.0065 46

50 0.0075 53

55 0.0095 67

0.000

0.002

0.004

0.006

0.010

0.012

0.014

0.016

0.018

0.020

0.022

0.024

0.026

0.028

0.030

35 40

1 2 . 5

1 3 . 0

1 3 . 5 1 0 %

2 0 %

3 0 %

4 0 % 5 0 %

6 0 %

7 0 % 8

0 % 9 0 %

1 4 . 0

1 4 . 5

1 5 . 0

45 55 60 65 70 80 85 90 95 100 105 110 115 120

65

Dry Bulb Temperature, ºF

E n t h a l p y

- B t

u / l b

o f D

r y A i r

H u mi d

i t y R a t i o ,l b w / l b D A

70

75

80

85

1 5

2 0

2 5

3 0

3 5

4 0

4 5

5 0

6 0 6 5

Volume - ft3 /lb of Dry Air

Saturation Temperature, ºF

Relative Humidity

40

45

50

55

60

35

0.0075

50




25/43



The required air volume to satisfy the latent load is:

The supply air volume to the ofce is the maximum volume required by code for ventilation and the volume required for controllingthe latent load:

c) Determine the heat transfer coefficients for the radiant panels

For panels and sails that are designed well, the surface temperature can be approximated to be the mean water temperatureAssuming a chilled water supply temperature 2 °F above the dew point in order to minimize the potential for condensation and a

temperature rise of 4 °F through the panel leads to a mean water temperature of:

Using equation L16, the natural convection coefcient is determined:






Example 2 - Small Office (SI)

Consider a small ofce with a southern exposure. The space is designed for two occupants, a computer with LCD monitor, T8 orescentlighting, and has a design temperature set-point of 24 °C. The room is 3 m wide, 4 m long, and 3 m from oor to ceiling. The ownerexpressed interest in using radiant panels.

Space Considerations

One of the primary considerations when using a radiant heating and cooling system is humidity control. As previously discussed, it isimportant to consider both the ventilation requirements and the latent load when designing the air-side of the system.

The assumptions made for the example are as follows:

• Load/person is 65 W sensible and 55 W latent

• Lighting load in the space is 25 W/m²

• Computer load is 80 W (CPU and LCD Monitor)

• Total skin load is 425 W

• Specic heat and density of the air are 1.007 kJ/kgK and 1.3 kg/m³ respectively

• Design conditions are 24 °C, with 50% relative humidity

• Design dew point is 13 °C


Occupants 2

Set-Point 24 °C

Floor Area 12 m²

Exterior Wall 12 m²

Volume 36 m³

qoz 210 W

ql 300 W

qex

425 WqT 935 W

Determine

a) The ventilation requirement.

b) The suitable supply air and supply water temperatures.

c) The total convective heat transfer coefcient for radiant panels.

3 m

4 m

3 m

SMALL OFFICE

Window



27/43




Solution

a) Determine the ventilation requirement

The ventilation requirement should be calculated to meet ventilation codes. For example, using ASHRAE Standard 62-2004 to determinethe minimum fresh air ow rate for a typical ofce space:

b) Determine required supply air dew-point temperature to remove the latent load

From equation L2:


At the design conditions (24 °C, 50% RH), the humidity ratio is 9.5 g/kg of dry air, requiring a difference in humidity ratio between thesupply and room air of:

From the gure below the dew point corresponding to the humidity ratio is 5 °C, which is too cool for standard equipment.

Evaluating the humidity ratio at several temperatures led to the selection of a dewpoint of 10 °C in order to use less expensive equipmenwhile also minimizing the supply air volume required to control humidity.

Humidity Ratio

Dew Point g/kg

5 5.5

7.5 6.75

10 8

12.5 9.25

Dry-Bulb Temperature, ºC

E n t h a l

p y - k J

/ k g o f

D r y

A i r

30

25

20

15

10

5

0

50454035302520151050

0

4 5

4 0

3 5

3 0

2 5

2 0

1 5

5 0

5 5

6 0

6 5

7 0

7 5

8 5

9 0

9 5

1 0 0

1 0 5

1 0 5

1 1 0 1 1 5 1 2 0 1 2 5

Volume - m3 /kg of Dry Air

Saturation Temperature, ºC

Relative Humidity

H u mi d i t y R a t i o , g w / k g D A

30

10

5

15

20

0 . 7 8

0 . 8 0

0 . 8 2

0 . 8 4

0 . 8 6

0 . 9 0

0

. 9 2

0 . 9 4

0 . 9 6

1 0 %

2 0 %

3 0 %

4 0 %

5 0 %

6 0 %

7 0 % 8

0 % 9 0 %

25

8




The required air volume to satisfy the latent load is:

The supply air volume to the ofce is the maximum volume required by code for ventilation and the volume required for controllingthe latent load:

c) Determine the heat transfer coefficients for the radiant panels

For panels and sails that are designed well, the surface temperature can be approximated to be the mean water temperature. Assuminga chilled water supply temperature 1 K above the dew point in order to minimize the potential for condensation and a temperature riseof 2 K through the panel leads to a mean water temperature of:

Using equation L16, the natural convection coefcient is determined:





29/43



Performance

Radiant panels performance depends onseveral factors:

• The difference in surface temperaturesbetween the panel and the surroundingsurfaces

• The mean water temperature and thepanel thermal resistance

• View factor of the panel to the surfacesto be cooled/heated

• Water ow rate

• Emissivity and absorption of affectedsurfaces

The water ow rate in the coil affects

two performance factors. First, the heattransfer between the water and the panel isdependent on whether the ow is laminar(poor), transitional (inconsistent) orturbulent (good). Secondly, it also affectsthe mean water temperature.

The higher the ow rate, the closer thedischarge temperature will be to theinlet, thereby changing the average watertemperature imposed on the panel. Asthe separation between the mean watertemperature and the surrounding roomtemperature (ΔT) increases, so does thecapacity. In heating, the ΔT is limited bythermal comfort. In cooling, the ΔT is alsolimited by two factors, thermal comfort andcondensation prevention. Good practice

for panel selection in cooling avoidscondensation by limiting the enteringwater temperature to the room’s dewpoint + 2 °F [1 K]. The most commondesign condition for spaces in cooling is75 °F [24 °C] at 50% RH, producing a dewpoint of 55 °F [13 °C] and limiting enteringwater temperature to a minimum of 57 °F[14 °C].

Figure 19 shows the effect on the owrate, indicated by Reynolds number, onthe capacity of a typical radiant panel. Asindicated on the chart, increasing the owrate into the transitional range (Re > 2300,shown in blue on the graph) increases theoutput of the panel.

Product Selection

Figure 19: Radiant panel capacity vs. water ow

40

0 2000 4000 6000 8000 10000 12000

50

60

70

80

90

100

C a p a c i t y ,

%

Re

The water ow rate is largely dependent on thepressure drop and return water temperaturesacceptable to the designer. In most casesthe water ow rate should be selected tobe fully turbulent (Re > 4000) under designconditions. The difference between the meanwater temperature is dened as:

L17

and the room/surrounding surfacetemperatures are the primary driverof panel performance. The larger thisdifference is, the greater the radiant andconvective transfer rates are. As noted inequation L12, the radiant energy exchangebetween two surfaces is based on theabsolute temperature to the fourth power.Conversely, a lower temperature differencewill reduce the amount of potential energyexchange, and thereby capacity. As a result,it is desirable from a capacity standpointto select entry water temperatures as lowas possible in cooling, while maintaining itabove the dew point in the room to ensuresensible cooling only.

The location of radiant panels relative toloads in the space inuences their capacityand is greatly dependent on the view factoof the panel to the objects that are to beconditioned. When used in spaces withhigh solar gain, such as perimeter zonesthe capacity increases as the surroundingsurface temperature increases. As surfacetemperatures change throughout theday, panel capacity changes accordinglyFurthermore, as the distance between thepanel and the affected surface increases, theview factor diminishes, thus reducing direcradiant exchange between the two surfacesPanel placement is based on a combinationof surface temperature and distance to theoccupant in order to ensure an effectiveoperative temperature is achieved. Locatingpanels along glass perimeters without lowemissivity coatings may have a negativeeffect on energy use as some energy will belost to the outdoors through the glass.



Example 3 - Small Office Panel Selection (IP)


Occupants 2

Set-Point 75 °F

Floor Area 120 ft²


Volume 1080 ft³

qoz 800 Btu/h

ql 825 Btu/h

qex 1450 Btu/h

qT 3075 Btu/h

hc, total 0.823 Btu/hft°F

Qs 38 cfm

Ts 50 °F

tCHWS 57 °F

tpanel 59 °F

Determine

a) The area of panels required.

b) The area of panels required assuming 95 °F outdoor air temperature.

c) The ow rate for the panels from (b).

d) A practical layout and piping arrangement for the panels from (b).

12 ft

9 ft10 ft

SMALL OFFICE

Window

Consider the small ofce presented in the previous example.



31/43



Solution

a) Determine the sensible cooling capacity of the supply air

Using equation L5:

Determine the sensible cooling required from the water-side

Determine the specific capacity of the radiant panelsUsing equation L11, the convective heat transfer to the panel is determined:

Using equation L12 and assuming that the wall temperature is equal to the room air set-point temperature, the radiant heat exchangewith the panel is determined:



Using equation L14:

Using multiples of 4 ft2, which is a standard ceiling tile sized at 2 ft × 2 ft, the total area required is 76 ft2.





b) The area of panels required assuming 95 °F outdoor air temperature

The exterior wall temperature is determined with an h value, convective heat transfer coefcient, of 1.46 Btu/(hft2°F) and a U value, overallheat transfer coefcient, of 0.693 Btu/(hft2°F):


Recalculating the radiant heat exchange and total heat transfer from (a):


Using equation L14:



33/43




SMALL OFFICE

Light

Panel

PanelPanel

Panel Panel

SMALL OFFICE

Panel Panel

Panel

Panel

Light

c) The flow rate for the panels from (b)

d) A practical layout and piping arrangement for the panels from (b)

In order to t the panels from (b) in a lay-in ceiling, a 48 in. × 24 in. RPM modular panel is selected. Referring to the product datasheet, a ow rate of 1.02 (~1 gpm) has a water pressure drop of 0.17 ft.

Using these panels would require a quantity of:

If these panels are connected in series, the total loop pressure drop would be:



Example 3 - Small Office Panel Selection (SI)


Occupants 2

Set-Point 24 °C

Floor Area 12 m²


Volume 36 m³

qoz 210 W

ql 300W

qex 425 W

qT 935 W

hc, total 4.71 W/m2K

Qs 22.5 L/s

Ts 10 °C

tCHWS 14 °C

tpanel 15 °C

Determine

a) The area of panels required.

b) The area of panels required assuming 35 °C outdoor air temperature.

c) The ow rate for the panels from (b).

d) A practical layout and piping arrangement for the panels from (b).


3 m

4 m

3 m

SMALL OFFICE

Window



35/43



Solution

a) Determine the sensible cooling capacity of the supply air

Using equation L5:

Determine the sensible cooling required from the water-side

Determine the specific capacity of the radiant panelsUsing equation L11, the convective heat transfer to the panel is determined:

Using equation L12 and assuming that the wall temperature is equal to the room air set-point temperature, the radiant heat exchangewith the panel is determined:



Using equation L14:

Using multiples of 0.36 m2, which is a standard ceiling tile sized at 600 mm × 600 mm, the total area required is 6.48 m 2.





b)The area of panels required assuming 35 °C outdoor air temperature

The exterior wall temperature is determined with an h value, convective heat transfer coefcient, of 0.255 W/(m2K) and a U value, overallheat transfer coefcient, of 0.121 W/(m2K):


Recalculating the radiant heat exchange and total heat transfer from (a):


Using equation L14:



37/43




SMALL OFFICE

Light

Panel

PanelPanel

Panel Panel

SMALL OFFICE

Panel Panel

Panel

Panel

Light

c) The flow rate for the panels from (b)

d) A practical layout and piping arrangement for the panels from (b)

In order to t the panels from (b) in a lay-in ceiling, a 1200 mm x 600 mm RPM modular panel is selected. Referring to the product datasheet, a ow rate of 0.07 (~0.075 kg/s) has a water pressure drop of 0.69 kPa.

Using these panels would require a quantity of:

If these panels are connected in series, the total loop pressure drop would be:



Example 4 - Small Office Chilled Sail Selection (IP)



Occupants 2

Set-Point 75 °F

Floor Area 120 ft²


Volume 1080 ft³

qoz 800 Btu/h

ql 825 Btu/h

qex 1450 Btu/h

qT 3075 Btu/h

Cooling Capacity of Hydronic System 2049 Btu/h

tCHWS 57 °F

tpanel 59 °F

Determine

The required area and possible location of chilled sails.

Solution

The difference between the room air temperature and the mean panel temperature is:

Referring to the product data page, the specic capacity of the chilled sail is determined using this temperature difference:

12 ft

9 ft10 ft

SMALL OFFICE

Window



39/43



Example 4 - Small Office Chilled Sail Selection (IP)

Determine the area of sails required

Using equation L14:

Selecting a sail that is 10 ft long and 4.5 ft wide provides 45 ft 2 of sail area. From the performance table, this piped-in series will resulin a pressure drop of 2 ft.

24 in. × 96 in. Price CSA

(troom - t ̅ w), °F Capacity, Btu/h Water Flow Rate, gpm Head Loss, ft

14 635 0.35 0.356

16 740 0.41 0.488

18 848 0.47 0.642

20 959 0.53 0.816

Based on 4 °F water temperature drop

SMALL OFFICE

Sail



Example 4 - Small Office Chilled Sail Selection (SI)


3 m

4 m

3 m

SMALL OFFICE

Window


Occupants 2

Set-Point 24 °C

Floor Area 12 m²


Volume 36 m³

qoz 210 W

ql 300 W

qex 425 W

qT 935 W

Cooling Capacity of Hydronic System 557 W

tCHWS 14 °C

tpanel 15 °C

Determine

The required area and possible location of chilled sails.

Solution

The difference between the room air temperature and the mean panel temperature is:

Referring to the product data page, the specic capacity of the chilled sail is determined using this temperature difference:



41/43



Example 4 - Small Office Chilled Sail Selection (SI)

Determine the area of sails required

Using equation L14:

Selecting a sail that is 3 m long and 1.5 m wide provides 4.5 m2 of sail area. From the performance table, this piped-in series will resulin a pressure drop of 6 kPa.

600 mm × 2908 mm Price CSA

(t room - t ̅w), K Capacity, W Water Flow Rate, kg/h Head Loss, kPa

8 186 79 1.06

9 217 93 1.46

10 249 107 1.92

11 281 120 2.44

Based on 4 °C water temperature drop

SMALL OFFICE

Sail




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Society of Heating, Refrigeration and Air-Conditioning Engineers, Inc.

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Heating, Refrigerating and Air-Conditioning Engineers.

ASHRAE (2005). Standard 138-2005—Method of testing for rating ceiling panels for sensible heating and cooling .

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radiant products engineering guide

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