Use. Worth noting. Key figures. installation. lindab atrium plana

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1 Contact, Overview, Index Guideline heating and cooling Plexus Professor / Professor Plus Premum / Premax / Solus Architect Polaris I & S Plafond Podium Celo Cabinett Capella Carat Fasadium Atrium / Loggia Regula Drypac Lighting TEKNOsim 00 Lindab Ventilation All forms of reproduction without written permission are forbidden is the registered trademark of Lindab AB Lindab's products, systems, product and product group designations are protected by intellectual property rights (IPR)

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3 Use Lindab s heating and cooling panels are mounted in the ceiling and primarily provide heating by radiation The radiation share for the panels exceeds 0% to 0%, compared to approx % for conventional fin coil products Since the air velocities can be kept low, the result is a draught-free environment Radiant heating can also be used with high ceiling heights since the radiant heating, despite the high placement, heats the underlying surfaces directly without any losses to the air The lower temperature gradient and the increased radiation temperature provided by the panel results in an energy-efficient heating alternative compared to other heating systems Installation is installed either suspended or recessed into a suspended ceiling Worth noting A high radiation quotient results in low air velocities and provides excellent results, even when high installation heights apply The low weight ensures quick and effortless installation Lindabs radiant panels are tested according to EN-0/ EN-0 and are CE-marked Key figures Length: Width: Height: Capacity: 00, 00, 00, 00, 000 mm 00, 00, 00 mm 0 mm Cooling: W Heating: W Calculation setup Room temp: C/ C, Water temp: - C/- C

4 The Lindab is a simple yet powerfull radiant panel, with an uncomlicated design that ensures that the does not draw attention on the building and interior design The water pipes are made of copper Nevertheless, water should be oxygen-free to prevent corrosion Heating, H When warm water passes through the copper pipe, heat is transferred to the aluminium plate, with very little temperature loss thanks to the unique welding technique The panel is warmed and it then radiates the heat into the room The thermal radiation travels through the air without any loss of energy on its way to the floor, walls and room objects In this way, you avoid heating a large air mass that, when warm, sticks to the ceiling Instead, the heat goes where it is needed the most It is mainly the floor, walls, furniture and fittings in the room that are heated The temperature of the room surfaces becomes higher than that of the room air and thus transfers its heat to the air By heating primarily the room surfaces instead of the air, you can save a lot of energy A more detailed description of how ceiling heating works is available in Lindab s Ceiling Heating Guide Cooling, C When cold water passes through the panel, the heat of the aluminium plate, obsorbed from the hot room air, is transferred to the cold water, with very little temperature loss The panel partly chills the warm room air on its cold surfaces and, partly absorbs heat from the room via low-temperature radiation In this way, the room is chilled via both radiation (approx 0 %) and convection The absorption of low-temperature radiation means that the surfaces of the room, and above all the floor, walls, furniture and fittings have a lower temperature than if then cooling was only convective This means that storage of cooling energy is greater Heating+Cooling, H+C By adding an extra set of copper pipes, a single Atrium Plana panel can be used for both heating and cooling Data Variants Width: The panels are available in two different widths for either cooling or heating (width 0) and (width 0) The panels with combined heating and cooling are available in widths (width 0) and (width 0) Lengths: Available in lengths 0 0 m Water connection: Available with vertical or horizontal DN connection Surface treatment: The panels are powder-coated Colour The product is available as standard, in signal white RAL 00 or in pure white RAL 0, gloss value 0 Other RAL colours on request Accessories Delivered separately Control: Refer to the chapter Regula Suspension options: Hangers are included as part of the standard package Alternatively the panel can be delivered with threaded rods and attachment hooks If the panel is to be installed on the ceiling, securing plates are available as an accessory

5 Dimensioning of heating panels There are a number of things to consider when dimensioning heating panels To achieve the best energy efficiency and as small a temperature gradient as possible and to utilise the high radiation quotient, you should take the following points into consideration: The length and width of the ceiling surface should be greater than the height of the room and the room s furnishings should not consist of high walls or storage shelves If this is the case, there is a risk that the thermal radiation is absorbed before it reaches the floor The capacity requirement in the lower part of the room should not be less than 0% of the total capacity requirement Otherwise, the convective heat will not be utilised and the temperature gradient will rise There should not be any large uncoated horizontal metal surfaces under the heating strips, as these will reflect the thermal radiation The floor should be insulated or positioned above the ground (Otherwise, there is a risk that the floor will not be warmed sufficiently) If some of these dimensioning recommendations are not fulfilled, the heating panels should be supplemented with ceiling fans that blow the warm air down from the ceiling to the occupied zone

6 Dimensioning Heating capacity water Follow the instructions below, to calculate the heating capacity P W provided by the heating panel Calculate temperature difference Δt rw Product length L minus 0 m, to optain the active length L act Find the specific heating capacity P Lt, relative to Δt rw, in diagram Multiply the specific heating capacity with the active length L act Example What is the heating capacity of a 0 m suspended panel H-0? The room s winter temperature is assumed to be t r = ºC The heating water temperature in/out is 0/ºC Δt rw = (t wi + t wo )/ - t r = (0 C + C) / - C = K L act = 0 m - 0 m = m Read off from diagram P LT = 0 W/m = 0 W/m x, m = W Width Heating Capacity Specific heating capacity P Lt [W/m] Multiplication Factor H+C 0 0 H+C NB! Please use the multiplication factor to calculate the capacity when using other than H-0 Definitions: P a = Cooling capacity air [W] = Cooling capacity water [W] P tot = Cooling capacity total [W] q ma = Air mass flow rate [kg/s] q a = Primary air flow rate [l/s] c pa = Specific heat capacity air [,00 kj/kg K] t r = Room air temperature [ C] t wi = Water inlet temperature [ C] t wo = Water outlet temperature [ C] t ra = Temp diff, room air and primary air temp [K] t rw = Temp diff, room air and mean water temp [K] t w = Temp diff water circuit [K] e tw = Capacity correction for temperature e qw = Capacity correction for water flow P Lt = Specific cooling capacity [W/(m K)] Width 0 suspended Width 0 Direct to ceiling (EN 0) Temperature difference Δt 0 rw [K] Diagram, specific heating capacity per active length at nominal flow, nom = 00 in relation to Temperature difference Δt rw

7 Dimensioning Cooling capacity water Follow the instructions below, to calculate the cooling capacity P W provided by the cooling panel Calculate temperature difference Δt rw Product length L minus 0 m, to optain the active length L act Find the specific heating capacity P Lt, relative to Δt rw, in diagram For suspended installation, more than 0 mm from the ceiling, the capacity increases by %, so multiply by Multiply the specific heating capacity with the active length L act Example What is the cooling capacity of a 0 m 0 suspended installation? The room s summer temperature is assumed to be t r = º C The cooling water temperature in/out of the is /º C Answer: Temperature difference: t rw = t r (t wi + t wo )/ t rw = - (+) / = K Active length: L act = 0 m - 0 m = m Read off, from diagram : P Lt = W/m Multiply the specific cooling capacity by the factor for suspended installation > 0 mm: => P Lt = W/m x = W/m The cooling capacity is: = W/m x m = W NB! The capacity diagram applies at a nominal flow in one strip nom = 00 l/s Follow the steps in example, to obtain the right capacity at other flows Width Cooling Capacity Multiplication factor H+C 00 0 H+C NB! Please use the multiplication factor to calculate the capacity when using other than C-0 Specific cooling capacity P Lt [W/m] Width Temperature difference t rw [K] Diagram installed direct to the ceiling, specific cooling capacity P Lt per active length at the nominal flow in one strip nom = 00 l/s

8 Dimensioning Capacity correction for water flow e qw Follow the steps below: Calculate the water flow with the current capacity Calculate the water flow in one strip Read off the capacity correction for waterflow e qw from diagram Multiply the capacity by the capacity correction e qw Repeat steps through with the new capacity Example, Heating: H-0 gives = W (from example ) Temperature difference was: t w = 0 C - C = K To calculate the water flow rate, use formula: = / ( c pw x Δt w ) = W / (00 Ws/(kg K) x K) = 00 l/s Read off the capacity correction e qw from diagram The value is e qw = 0 Calculate the new capacity: = W x 0 = W Use the new capacity to calculate the water flow rate: = W / (00 Ws/(kg K) x K) = 00 l/s The new capacity correction e qw will then be 0 and the new capacity is calculated to be: = W x 0 = W Example, Cooling : C-0 gives = W (from example ) Temperature difference was: t w = C - C = K To calculate the water flow rate, use formula : = / (c pw x t w ) = W / (00 Ws/(kg K) x K) = 00 l/s Read off the capacity correction e qw from diagram The value is 00 Calculate the new capacity: = W x 00 = W Use the new capacity to calculate the new water flow rate: = W / (00 Ws/(kg K) x K) = 00 l/s Read off the capacity correction e qw from diagram again The value is 0 Calculate the new capacity: = W x 0 = 0 W Note! Your calculations will be more accurate the more times you keep repeating step -! Capacity correction e qw,, Heating Cooling 0 0, 0 0, 0 0, 0 0, 0 0, ,0 00 0, ,0 0, ,0 0,0 00 0, ,0 00 0,0 0, ,0 Diagram Capacity correction e qw as a function of waterflow Waterflow [l/s]

9 Pressure drop in water circuit, width 0 Water flow rate [l/s] 0 0, 0, ,0 00 0,0 00 0,0 00 0,0 00 0,0 00 0,0 00 0,0 0 0, m, m, m, m 0,0 m 00 0,0 00 0,0 0, 0 0, , 0,0, 0,, Diagram 0, pressure drop at 0 C For pressure drops at temperatures other than 0 C, the pressure drop is multiplied by the pressure drop factor (see diagram ) Example : 0 m provides a capacity from: = W at t w = K t wio = 0 x (t wo + t wi ) t wio = 0 x (0 C + C) = C = / (c pw t w ) = W/(00 Ws/(kg K) K) = 00 I/s The pressure drop p w in the water circuit is read off at kpa from diagram Read off the pressure drop factor at t wio = º C from diagram The value is 0 Calculate the new pressure drop: p w = x 0 = kpa Multiplication factor Width Heating Cooling H+C 0 H+C NB! Please use the multiplication factor to calculate the pressure drop when using other than H-0 or C-0 Pressure Drop p w [kpa] Definitions: = Water flow rate [l/s] = Cooling/heating capacity water [W] c pw = Specific heat capacity water [00 Ws/(kg K)] Δt w = Temperature difference water circuit [K] t wio = Mean water temperature [K]

10 Pressure drop factor Pressure drop factor Diagram Temperature adjusted pressure drop factor Mean water temperature t wio

11 Weight and water volume Width 0 Width 0 Width 0 H+C Width 0 H+C Weight, [kg/m] Water content, [l/m] Copper pipes, quality EN - CU-DHP Pressure class PN Table, weight and water volume Width, height and length, mm Connection option (Øx) 0 Connection option (Øx) X 0 0 Ø 0 X 0 m mm m mm m mm m mm 0 m mm, L - mm, X 0 Ø 0,, L - mm Figure is manufactured as standard in width 0 and 0 is manufactured as standard in lengths 0 0 m Tolerances for deflection, 0 = mm, 0 = mm

12 Width, height and length, mm H+C 0 0 H+C 0 0 X 0 Ø H+C 0 0 X 0 Ø L - mm H+C 0 Cooling Cooling Heating Heating 0 L - mm Figure H+C is manufactured as standard in width 0 and 0 H+C is manufactured as standard in lengths 0 0 m Tolerances for deflection, 0 = mm, 0 = mm Cooling Heating Minimum permitted flow min (l/s) Max working pressure (bar) Max test pressure (bar) Table, recommended minimal water flow rates min NB! If the recommended minimal water flow rates min are maintained, the air becomes agitated No separate bleeding is then required for As normal, the bleeding of the coolant system takes place at peak points, in the equipment bay

13 Control Lindab offers control equipment that is very simple to use To avoid heating and cooling being activated at the same time, the systems are controlled sequentially (Regula Combi) For the technical data, refer to a separate brochure, Regula Programme text Panels from Lindab Qty Product: C 0 0 m Plus features: Colour, RAL xxxx Accessories: Regula Combi 0 Regula Secura 0 Control valve, cooling 0 Actuator, cooling 0 Designations Product: Width, cm: C, H, H+C 0 ( mm), 0 ( mm), 0 ( mm only H+C) Connection dim Water, mm: Coupling options:, Length, m: 0 0 m Order code Product C 0 m Type: C, H, H+C Width: 0, 0, 0 cm Water connection: mm Connection type: Length: 0 0 m Example: C 0 m