Energy analysis for a snow-free surface A technical analysis of the benefits of insulation under the heating pipes Ying Song Spring 2018

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FACULTY OF ENGINEERING AND SUSTAINABLE DEVELOPMENT

Department of Building, Energy and Environmental Engineering

Energy analysis for a snow-free surface

A technical analysis of the benefits of insulation under the heating pipes

Ying Song

Spring 2018

Student thesis, Bachelor degree,15 HE Energy Systems

Master Programme in Energy Systems

Supervisor: Arman Ameen Assistant supervisor: Roland Forsberg

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Preface

This project is the final work of my bachelor’s degree in Energy system programme at Gävle University.

The project was done with significant help from supervisor Roland Forsberg from SWECO AB. In addition, Stefan Jonsson at Gävle Energi has assisted with the theoretical background and data on which the project is based on. Jan Akander at the University of Gävle has assisted with technical help of the project. I am obliged to you all for your assistance during the project process.

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Abstract

Snow-free surfaces is needed for parking place, platform, and playground and even in city center square. With energy prices rising, energy saving is becoming a hot topic. Meanwhile environmental problems are becoming more and more serious, thus, the ways to saving energy is becoming an eye-catcher. So burring heating pipes

underground has been a popular way to get ice-free surfaces. Using heating pipes for melting snow is much more efficient and more benefit for the environment comparing with using other methods.

In this project, an energy analysis of a football pitch with an area of 5000 m2 is carried out under a series of conditions between insulated and uninsulated construction. All calculations are done with the so-called finite element method (FEM), in the COMSOL. COMSOL is used for simulating and calculating the energy use with outdoor

temperatures of -5 ºC and -10 ºC. Top layer materials concrete, grass and stone are also discussed. The ability of XPS and EPS insulation material is compared and noted. The models are divided into two parts, one is with snowfall and the other is without

snowfall.

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Nomenclature

k = thermal conductivity of the material [W m ∗ K⁄ ] v_air = vapor content [kg

m3]

ṁ_evap = rate of mass transfer [ kg s.m2]

αconv = the convective heat transfer coefficient [ W m2_.K]

ρ_air = the density of air [kg m3]

c_{p air} = the specific heat capacity of air [ J kg.K] M_v = the molar mass of vapor [ kg

kmol] R = the universal gas constant [ J

kmol.K] d = interior diameter [m]

q = fluid volumetric flow rate [l s⁄ ] dt = temperature difference [K] R = heat resistant [m2_{∗ K/W]}

Q_evap = energy required to evaporated water [J] Mevap = mass [kg]

Re = Reynolds number

μ = dynamic Viscosity [Pa*s] ν = kinetic viscosity [m2/s] ti = interior temperature [K] t₀ = exterior temperature [K] r_i = interior radius [m] ro = exterior radius [m]

E_evap = latent heat of vaporization [J] M_snow = mass of snow [kg]

h = heat power [W] ∆Pf = pressure drop [Pa] 𝐿 = latent heat [J] 𝑧 = depth [m]

ℎ𝑚 = snowpack water equivalent in meters [m]

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𝜆_𝑓 = latent heat of fusion [𝑀𝐽 𝑘𝑔⁄ ] 𝛽 = steam transition coefficient [𝑚 𝑠⁄ ] 𝑣_𝑠𝑎𝑡 = vapor content [𝑘𝑔 𝑚_⁄ 3_]

𝑉 = volume [𝑚3]

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1. Introduction

1.1 Background

With the development of the society, district heating is becoming more and more common. District heating is supplied by a central plant which can use advanced methods to run many different fuels, so district heating is a smart, environmentally sustainable way to heat homes, schools and other premises. It is also better than every building having its own boiler.

The roads can be more dangerous during wintertime especially when they are slippery. The number of road accident decrease dramatically by winter road maintenance. There are different methods to remove ice and snow from the roads. Salting and sanding are the most common ways to accomplish winter road maintenance.

District heating system is another way which can be used to make the roads safer during the winter. In district heating systems, pipes contain warm water and warm up the road surface and melt the ice or snow. This system is not very common and is used only in small scale. In this system different kind of energy sources can be used.

District heating has undergone steady growth over the years and now accounts for more than half of all heating for homes and other premises in Sweden. With the global climate change the world is currently facing, the application of district heating will be furthermore expanded. Sweden has long tradition for using district heating for public heating in urban areas. According to the data from Swedish association of district heating, it is said that about 60% of private and commercial houses were heated by district heating in 2015 (Fjärrvärme, 2015). In this thesis, the district heating for public places is simulated and optimized.

1.2 Objective and scope

This project is an investigation and simulation of ice free ground by using the common heating pipes buried underground. Simulation is done by COMSOL 5.3.

In the simulation, effects of the insulation under the underground pipes are studied. Also, different type of surface materials to show the effects of the type of materials on the project efficiency.

The goal of this project is to examine an ice-free football pitch. The pitch which is simulated is located in Gävle city. The aim is to study and analyze the energy use for the football pitch during winter time and to investigate the effect of changing the pitch profiles, like surface materials, outdoor temperature and insulation materials. The different type of ground conditions was discussed at the same time.

Therefore, energy usage calculation is done in two ground conditions, two outdoor temperatures, and three kind of ground surfaces. The two ground conditions are ground with snow and when snow was falling. Two kinds of outdoor temperatures are 5ºC, -10ºC. Three types of ground are stone, grass and concrete. Two types of insulation materials are also discussed.

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1.3 Limitation

This project focuses on the heat transfer and some other parameters between the pipes and the soil surface. There are some important factors which should be considered and calculated for having this system, but they are not reflected in this study.

During the water flow process some energy is lost. This energy lost effects on the total energy usage in the system is not considered in this project.

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2. Theoretical background

Underground district heating can be used to melt the ice and snow on the road and public squares. This system can be worked by different kind of energy sources. For example, fossil fuels or electricity can be used as an energy source, which are

expensive, environmental harmful, limited and non- renewable. In recent years, some cheaper and environmental friendly energy have been introduced, such as biofuels, geothermal energy or solar energy. Changing the heat source to environmentally sustainable sources may reduce carbon dioxide emissions and emissions of Sulphur, Nitrous Oxide and particles, which can lead to a reduction of the pollution in the area (Fjärrvärme, 2015).

This kind of district heating is common and even have some successful implementation around the world. This system can keep the roads free from ice.

There are some alternatives to reduce the risk of accident when the roads are slippery. The traditional ways are based on removing the ice and snow of the road surface mechanically.

Additionally, using sand to increase friction and sprinkling salt to reduce freezing point and to make snow melt faster. Though these solutions can make the roads safer, they cost a lot and are not environmentally sustainable.

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2.1 The Impact

2.1.1 Social impact

During the winter time, icy and slippery roads are the most dangerous situation which threatens the road safety and the comfort of driving. In Sweden, 74% of the total accidents in 2005-2006 happened when the roads were in slippery conditions, mostly when it was snowing or had snowed.

Figure 2.1 illustrates the relation between the total number of accidents and the number of accidents when the road was slippery. According to the Figure below, 60% of all accidents in January 2005 and 90% of all accidents during January 2006 occurred in slippery conditions (Andersson, 2010).

Figure 2.1: The relationship between the total number of accident and the number of accidents that occurred during the slippery conditions (Andersson, 2010).

Since the road conditions are very critical and cause many and even severe traffic accidents, then accurate winter maintains is a crucial factor to guarantee the normal mobility on the icy roads.

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Figure 2.2: Traffic hold-up caused by heavy snow fall. Even maintenance vehicles are stopped (Eugster, 2007).

In addition, carports, ramps or car access to buildings or a garage need to be clear from snow and ice, but this might be delayed due to maintenance vehicles clearing the roads. Public areas need to be cleaned up in case of accidents. Icy airports or runway may cause delayed flights and passengers inconvenience (Eugster, 2007).

2.1.2 Economic impact

A traffic situation in winter leads to reduced speeds, traffic jams and therefore to loss of time. Every jam hour costs about 10 Euros per vehicle. Every improvement in the status Quo has a positive effect on the roadworthiness, the traffic processes and on our society as a whole (Klotz & Balke, 2004).

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Table 2.1: The costs of road accident in different countries as a percentage of GNP. Values are in national currencies, amounts are in millions (Abbasi, 2014).

2.1.3 Environmental impact

One of the most common ways for removing icy surface from roads is using the mechanical snow cleaners or snowplow with a combination of using sand and salt spilling on the road. It is estimated that the yearly salt usage cost about 450 million Euro in Germany (Eugster, 2007). And in Sweden, during 2007-2008, 184000 tons of salt is used for winter road maintenance (Andersson, 2010).

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Figure 2.3: The effect of salt on the leaf and shoot (Eugster, 2007).

The water which carries this kind of salt is corruptive. This ability can corrode the roads made with bitumen and concrete.

Finally, the salty water can flow into drainage and cause rusty pipes, and even mix with the groundwater or water surface and increase the salt concentration in the drinking water (Nordin, 2011)

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Figure 2.5: The effect of salt on drainage pipes (Zhihu, 2016).

2.2 Previous study

There are several successfully snow melting projects that are operating around the world where they are using district heating as the main heat source. The oldest recorded of these melting systems was built in 1948 in Klamath Falls, Oregon by the Oregon Highway Department. Geothermal water was used in that system to heat up the surface (Lund, 1999).

In this oldest system, the grid consisted of 3/4-inch diameters iron pipes embedded three inches below the concrete pavement. The grid system was connected to a geothermal well and got heat transferred by a heat exchanger to a 50-50 ethylene glycol-water solution and circulated at 50 gpm (gallons per minute). This result in the temperature in the grid drop approximately 30 to 35 oF with the supply temperature varying from 100 to 130 o_{F. This system supplied a maximum of 3.5 x 105 Btu/h (British thermal unit per} hour) at the original artesian flow of 20 gpm and 9.0 x 105 Btu/h at the pumped rate of 50 gpm. The latter energy rate could provide a relative snow free pavement at an outside temperature of -10 °F and a snowfall up to three inches per hour, at a heat requirement of 41 Btu/ft2/h (British thermal unit per hour and square feet). Due to a temperature drop in the well from 143 to 98 °F, the well was rehabilitated in 1992 (Thurston, Culver, & Lund, 1995).

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1.25- to 2.5-inch insulated black iron pipe, which in turn was connected to the downhole heat exchanger. The header pipe has brass manifolds placed at about 40-foot intervals in concrete boxes, to allow for four supply and return PEX pipes to be attached.

The heated bridge deck and pavement covers 22 000 ft2 and is designed for a heat output of 50 Btu/ft2/h. The DHE (district heating enterprises) supplies 100 °F glycol mixture to the grid with a temperature drop of 24 °F, estimated to increase to 30 °F once the ground and concrete temperatures reach equilibrium. This is supposed to keep the deck clear during heavy snowfall down to -10 °F. The renovated deicing system appears to be operating effectively, based on substantial snowfalls in January and February 1999.

Figure 2.6: Detail of the loop system for the Klamath Falls Project (Lund, 1999).

One of the well-known in this subject is at Sapporo in Japan, it has worked from 1966 until now. The main aim of this project is to prevent the ice formation on the road by using the water from Jozankei Spa (Sato & Sekioka, 1979).

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Figure 2.7: Laying polytubene tubes at a parking area in Jozankei, Sapporo (Sato & Sekioka, 1979).

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2.3 Ice formation

There are three important parameters which cause the ice making. The first one is temperature and the others are humidity and pressure. Since the ice is the solid form of water, no ice is made in the dry cold weather.

2.4 Soil properties

The heat flow in soil can be calculated by the sum of heat conduction and convection, by equation

𝑞 = −𝑘𝑠 𝜕𝑇

𝜕𝑧+ 𝐶𝑤𝑇𝑞𝑤 + 𝐿𝑣𝑞𝑣 [1]

Where v means vapor and w means liquid water and s means soil q = heat flux [𝑊 𝑚⁄ 2] k = thermal conductivity[𝑊 𝑚 ∗ 𝐾⁄ ] T = soil temperature[𝐾] C = heat capacity [𝐽 𝑘𝑔 ∗ 𝐾⁄ ] L = latent heat [𝐽] z = depth [𝑚]

During the heavy snow day and when the snow is melting, heat flow by convection can be calculate by (𝐶_𝑤𝑇𝑞_𝑤). The rest of the convection term can be calculated by 𝐿_𝑣𝑞_𝑣 which is the latent heat flow by water vaporization.

2.4.1 Heat conduction

In conduction, the energy is transmitted due to internal vibrations of molecules, without a net displacement of the molecules themselves.

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Figure 2.9: Heat transfer by convection: place a metal into an open flame (BBC, 2014).

Heat conduction depends on some factor: temperature gradient, cross-section of the material, length of the travel path and physical material properties.

Since the main part of the heat transfer is done by conduction, the thermal conductivity of soil is very important. The higher thermal conductivity, the higher heat transfer. The thermal conductivity is depended on the soil aggregates, material and atomic structure. Thus, different layers of soil have different thermal conductivity. The thermal

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experiments. The thermal conductivity of a soil layer is improved by adding aggregation with different thermal conductivity.

In this project, stone and concrete is added on the soil surface. The thermal conductivity of concrete is 1.7 W/(m*K), soil is 2 W/(m*K) and stone is 1 W/(m*K) (ToolBox, 2003).

2.4.2 A pipe with transverse heat flow

In this project, heat for melting the ice is from the pipes buried in the soil, hot water flows through the pipe and release heat. According to the equation below, conductive heat loss through the wall of the pipe can be expressed as:

𝑞 =2𝜋𝑘(𝑡𝑖−𝑡𝑜) 𝑙𝑛(𝑡𝑜−𝑡𝑖) [𝑊/𝑚] [2] Where 𝑡_𝑖 = interior temperature (K) 𝑡0 = exterior temperature (K) 𝑟𝑖 = interior radius (m) 𝑟𝑜 = exterior radius (m)

q = heat transferred per unit time per unit length of cylinder or pipe (W/m) k = thermal conductivity of the material (W/m*K)

2.5 Insulation properties

The current insulation board for insulation system are: EPS insulation board and XPS insulation board.

2.5.1 System performance comparison

1) Thermal insulation properties:

The same thickness of EPS and XPS insulation performance is gradually increased. EPS is 0.033 W/(m*K) and XPS is 0.036 W/(m*K). Therefore, to achieve the same effect of insulation, the thickness of XPS sheet is thicker than the EPS sheet, and the price of pure sheet XPS board is much more expensive than EPS board. If considered the full process and building height, the price of XPS per square meter will be more expensive than EPS. The current EPS or XPS thin plastering systems fully meet the requirements of the heat conservation effect.

2) Strength:

The strength pointed out here should be tensile strength. The bulk density and tensile strength of polystyrene board have an absolute relationship. The tensile strength of general EPS bulk density of 18 kg/m3 is 110 ~ 120 kPa, while that of 20 kg/m3 is about 140 kPa. XPS’s bulk is normally from 25 kg/m3_{~ 45 kg/m}3_{, the intensity from 150 kPa} ~ 700 kPa.

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14 3) Weatherability:

Weatherability refers to the ability of the insulation system to adapt to weather change, refers to whether the insulation effect has changed due to a series of issues related to quality under different climatic conditions and the overall stability of the system. Weatherability is a very important indicator of the insulation system.

Since EPS board has a higher ability in water absorption than XPS, the weatherability of EPS is worse than XPS. However, the ductility of the EPS board is better than that of the XPS board, which can overcome certain defects.

In addition, the resistance to wind pressure is also an important factor in weatherability because the more cavities the system was, the worse the wind resistance is. And both XPS board and EPS board can reach the requirements.

4) Breathable:

In terms of the material itself, the breathability of EPS is much better than XPS. XPS has almost no breathability, it is easy to make water vapor condense on both sides of the board with large indoor and outdoor temperature difference.

5) Bond strength:

For the thin plastering system, this indicator will directly affect the use of sheet EPS board’s strength is low, shear strength is also low, and sheet damage may not be in the bonding surface, but the direct destruction of the middle plate, so XPS has better strength performance. Tensile strength 0.1 MPa equivalent to say that the tensile strength of each square meter is 10 tons of force, which is very large.

However, there is a big problem with the XPS: the interface finish is high, and it is hard to stick if it is not interlaced with a very high polymer emulsion, which result in

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15 6) The surface smoothness

XPS is difficult to ensure the smoothness of the outer wall. The EPS board is much better due to its relatively soft property, so it performs better to ensure the flatness of the wall than the XPS board.

2.5.2 Advantages and disadvantages of two systems

EPS insulation system has the following advantages:

1) Mature technology. It has been used in Europe and the United States for nearly three decades. The technology has formed a system, adhesive layer, insulation layer and finishes can be used in combination.

2) Thermal insulation effect is good.

3) Due to the use of expanded polystyrene insulation materials, the price is not very expensive but rather affordable. It is also user-friendly.

4) It does not have complex construction technology. Disadvantages of EPS system:

1) Due to the property of the plate itself, its strength is not high, the load-bearing capacity is low, need to be strengthened when used as external tile.

2) EPS needs to be placed for some time before use in order to go through a period of maturity. If the aging time is not enough, the quality of the plate cannot be guaranteed, the plate would shrink after construction and result in system cracking.

The advantages of XPS system are:

1) XPS board has a dense surface and closed-cell structure of the inner layer. Its thermal conductivity is much lower than EPS with the same thickness, so it has better thermal insulation performance than EPS. For the same building façade, its thickness can be less than other types of insulation materials.

2) Due to the closed-cell structure of the inner layers, it has good moisture resistance. In wet conditions it can still maintain good thermal insulation properties and it is suitable for cold storage and other special requirements of the insulation construction. It can also be used for exterior cladding materials for the construction of brick or stone.

Disadvantages of XPS system:

1) XPS board has higher strength, but more brittle. It is also not easy to bend, the stress will be concentrated when stress is put on the board, which makes it easy to be damaged and crack.

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16 3) The price is higher than the EPS system.

4) It is just commonly used in recent three or four years.

5) Its structure of poor scalability, temperature and humidity may result in the material deformation and starting to drum off from the insulation layer.

6) Poor adhesive, bonding strength is not enough.

2.6 Ice melting

2.6.1 Heating

In a homogenous layer with constant surface temperature, heat flow is expressed as: 𝑞 = 𝜆𝑇1−𝑇2 𝑑 [𝑊/𝑚 2_] _[3] Where 𝑇_𝐼 = interior temperature [K] 𝑇₂ = exterior temperature [K] 𝑑 = thickness [m]

𝜆 = thermal conductivity of the material [W/m*K] q = heat transferred per unit time [W/m2]

At stationary conditions, the heat flow is constant throughout the design and distribution of temperatures that follow a straight line. For the temperature in a given point (x), it can be expressed as follow:

𝑇(𝑥) = 𝑇₁−𝑥

𝑑(𝑇1− 𝑇2) [𝐾] [4]

Where

x = the distance from the surface with higher temperature and into the structure. T1 = interior temperature [K]

T2 = exterior temperature [K] d = thickness [m]

When the construction consists of several layers, the heat flow is assumed the same through different layers. On the other hand, the temperature difference in the different layers changes according to the following equation:

𝑞 =𝜆1 𝑑1(𝑇1− 𝑇2) = 𝜆2 𝑑2(𝑇2− 𝑇3) [𝑊/𝑚 2_] _[5] Where 𝑇𝐼, 𝑇2, 𝑇3 = temperature [K] 𝑑 = thickness [m]

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However, the surface temperature should be known for each surface, which can be difficult to read in a multilayered design. The expression can be reformed with total heat flow based on the surface temperature of each side of the entire structure according to: 𝑞 =(𝑇1−𝑇3) 𝑑1 𝜆1+ 𝑑2 𝜆2 [𝑊/𝑚2_] _[6] Where 𝑇_𝐼, 𝑇₂, 𝑇₃ = temperature (K) 𝑑 = thickness [m]

𝜆 = thermal conductivity of the material [W/m*K] q = heat transferred per unit time [𝑊 𝑚⁄ 2]

𝑑1 𝜆1 = 𝑅1, 𝑑2 𝜆2 = 𝑅2 [ 𝑚2∗𝐾 𝑊 ] [7] Where R = heat resistance [𝑚2_{∗ 𝐾/𝑊]} 𝑑 = thickness[m]

𝜆 = thermal conductivity of the material [W/m*K]

The total heat flow in the construction can be expressed as: 𝑞 =Δ𝑡

Σ𝑅 [𝑊/𝑚

2_] _[8] Where

q = heat transferred per unit time [𝑊 𝑚⁄ 2] R = heat resistance [𝑚2_𝐾/𝑊]

𝛥t = temperature difference [K]

2.6.2 Evaporation and melting

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The energy required for melting snow as well as vaporization of water is shown in Table 2.2:

Table 2.2: Properties for melting snow into water, and evaporated water into vapor. Specific melting

heat

Ice - Water 334 000 𝐽 𝑘𝑔⁄

Evaporation heat Water -Vapor 2 500 000 𝐽 𝑘𝑔⁄

Specific heat capacity Water 4 184 𝐽 𝑘𝑔. ℃⁄ Specific heat capacity Ice (snow) 2 100 𝐽 𝑘𝑔⁄ . ℃

The rate of evaporation is dependent on the temperature of the wet surface and the temperature and humidity of surrounding air. In essence, the vapor content of surrounding air can be expressed as 𝑣_𝑎𝑖𝑟[𝑘𝑔

𝑚3], where the unit tells how much actual

vapor content is per cubic meter of air. The vapor content at a wet surface is commonly assumed to be saturated, thus called 𝑣_𝑠𝑎𝑡[𝑘𝑔

𝑚3], where the relative humidity is set to be

100%. Thus, the rate of mass transfer 𝑚̇_{𝑒𝑣𝑎𝑝}[ 𝑘𝑔

𝑠∗𝑚2] can be expressed as:

𝑚̇𝑒𝑣𝑎𝑝 = 𝛽(𝑣_𝑠𝑎𝑡− 𝑣_𝑎𝑖𝑟) [𝑘𝑔/𝑠 ∗ 𝑚2_] _[9] Where

𝛽 = 𝛼𝑐𝑜𝑛𝑣

𝜌𝑎𝑖𝑟𝑐𝑝𝑎𝑖𝑟 [10]

With

𝛼𝑐𝑜𝑛𝑣 = the convective heat transfer coefficient [ 𝑊 𝑚2_𝐾]

𝜌𝑎𝑖𝑟 = the density of air [ 𝑘𝑔 𝑚3]

𝑐𝑝 𝑎𝑖𝑟 = the specific heat capacity of air [ 𝐽 𝑘𝑔∗𝐾] Table 2.3: Properties of air.

Convective heat transfer coefficient 𝛼_{𝑐𝑜𝑛𝑣} 20 𝑊 𝑚⁄ 2. 𝐾 Density of air 𝜌_𝑎𝑖𝑟 1.2 𝑘𝑔 𝑚_⁄ 3 Specific heat capacity 𝑐𝑝 𝑎𝑖𝑟 1010 𝐽 𝑘𝑔. 𝐾⁄

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19 𝑣_𝑠𝑎𝑡 = 𝑝_𝑠𝑎𝑡∗ 𝑀𝑣

𝑅∗(273.15+𝑡) [𝑘𝑔/𝑚

3_] _[11]

Where the saturation partial pressure of vapor is calculated according to the equations below: 𝑝_𝑠𝑎𝑡 = 𝑎 (𝑏 + 𝜃 100) 𝑛 [12] If 0 ≤ 𝜃 ≤ 30℃ then 𝑎 = 288.68 𝑃𝑎 ℃𝑛 , 𝑏 = 1.098 , 𝑛 = 8.02 Else −20 ≤ 𝜃 < 0℃ then 𝑎 = 4.686 𝑃𝑎 ℃𝑛 , 𝑏 = 1.486 , 𝑛 = 12.3

𝑀_𝑣 = the molar mass of vapor = 18.02 [ 𝑘𝑔 𝑘𝑚𝑜𝑙] 𝑅 = the universal gas constant = 8314.3 [ 𝐽

𝑘𝑚𝑜𝑙.𝐾]

The rate of mass transfer (𝑚̇𝑒𝑣𝑎𝑝) indicates the rate of vapor flow that the air can receive, but evaporation requires enthalpy. The latent heat of vaporization is 𝐸_{𝑒𝑣𝑎𝑝}. The energy required to evaporated water, 𝑄𝑒𝑣𝑎𝑝 [𝐽] for the mass 𝑀𝑒𝑣𝑎𝑝 [𝑘𝑔] will be:

𝑄_{𝑒𝑣𝑎𝑝} = 𝑀_{𝑒𝑣𝑎𝑝}∗ 𝐸_{𝑒𝑣𝑎𝑝} [13]

The rate of latent heat of evaporation is the function of the difference in vapor content at the boundary layer next to the wet surface. This is expressed as:

𝑞̇_{𝑒𝑣𝑎𝑝} = 𝐸_{𝑒𝑣𝑎𝑝}𝛽(𝑣_𝑠𝑎𝑡− 𝑣_𝑎𝑖𝑟)[𝑊/𝑚2_] _[14]

𝑞̇𝑒𝑣𝑎𝑝 can be treated as a heat sink at the wet surface which in turn is a function of the temperature of the same surface.

The amount of energy needed to melt 𝑀𝑠𝑛𝑜𝑤 [𝑘𝑔] is: 𝑄_{𝑠𝑛𝑜𝑤} = 𝑀_{𝑠𝑛𝑜𝑤}𝐸_{𝑖𝑐𝑒−𝑤𝑎𝑡𝑒𝑟} [15]

Where the enthalpy for phase change from ice to water is 334 000 𝐽 𝑘𝑔⁄ . The wetness layer can be set as:

𝑄_{𝑠𝑛𝑜𝑤} = 𝜌𝑉𝑐_𝑝Δ𝑇 [16]

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2.7 Pressure drop in pipes

Having determined the required flow and return water temperatures of underfloor heating system, the calculation of the water flow rates and pressure drops for each loop leading from the manifold is needed.

These must be calculated as early as possible during the design process in order to be provided to the heat pump manufacturer tasked with checking the supplied data with the design flow rates of the heat pump.

Hopefully, the flow rates and pressure drops supplied to the manufacturer will be within the limits of the heat pump. However, if the pressure drops and flow rates exceed the limits of the heat pump, then additional pumps, piping, controls, and wiring is required. In this situation, the heat pump manufacturer will be able to advise what is required. When fluid flows through a pipe there will be a pressure drop that occurs as a result of resistance to flow. There may also be a pressure gain/loss due a change in elevation between the start and end of the pipe. This overall pressure difference across the pipe is related to a number of factors:

• Friction between the fluid and the wall of the pipe • Friction between adjacent layers of the fluid itself

• Friction loss as the fluid passes through any pipe fittings, bends, valves, or components

• Pressure loss due to a change in elevation of the fluid (if the pipe is not horizontal)

• Pressure gain due to any fluid head that is added by a pump

Hagen-Poiseuille Pressure drop calculation for laminar flow: Δ𝑃_𝑓= 128𝜇𝐿𝑞

𝜋𝑑4 [𝑃𝑎] [17]

Where

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21

Darcy-Weisbach Pressure calculation for turbulent flow: Δ𝑃_𝑓= 8𝜆𝜌𝑞2

𝜋2_𝑑5 [𝑃𝑎] [18]

Where

𝑑 = interior diameter [𝑚] 𝑞 = volumetric flow rate [𝑚3_{⁄ ]}_𝑠 𝜌 = density [𝑘𝑔 𝑚⁄ 3]

Colebrook-White

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22

Volumetric flow rate in a heating system is calculated from:

𝑞 = ℎ

𝑐𝑝∙𝜌∙𝑑𝑇[

𝑚3

𝑠 ] [26]

Where

𝑞 = volumetric flow rate [𝑚3_{⁄ ]}_𝑠 𝜌 = density [𝑘𝑔 𝑚_⁄ 3_]

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23

3. Modeling

The project was designed and analyzed by using COMSOL 5.3. After that the model was analyzed dynamically over time. This model is concerned about the ability to store energy in soil, the impact of outdoor temperature, surface insulation, and how the energy is affected by different insulation under the pipes.

The model is built up of 0.05 m surface insulation, 0.3 m sand, 0.1 m filler, and 0.25 m drainage. The 20 mm diameters pipe is set in the middle of the filler, which has a center distance of 200 mm between two parallel pipes. The model is insulated when 0.25 m drainage is replaced by 0.1 m XPS insulation material and 0.15 m drainage, or 0.1 m EPS insulation material and 0.15 m drainage.

3.1 Finite element methods

The COMSOL 5.3 is a program that was used to analyze the problem based on a so-called "finite element method" or FEM. This method is based on using numerical procedures to solve partial differential equations. Mathematically, the method is based on finding approximate solutions to differential equations and integrals. The accuracy of the approximate solution can be adjusted by refining the elements of the program. The model is divided into smaller simple squares or geometries (Figure 3.1), so-called "finite elements". Therefore, heat conduction equations are set between all elements and solutions are given for the given time. Thermal conduction in a construction is

calculated by allowing the program to solve the equations between all the elements (Kraft, 2011).

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24

3.2 Project modeling

3.2.1 ‘Dry ‘model- without snowfall on the surface

In order to calculate the approximate heat conduction in the structure, a model was created in COMSOL. The model was first constructed as a loop whose average temperature was due to the flow temperature and return temperature. The pipes were placed with 200 mm spacing between the supply and return pipes. Then, the different materials properties in the different layers were obtained and inserted in COMSOL. Values of these can be found in Appendix.

When materials got their material properties, several conditions were also created for the model to simulate. The surface of the model under the statutory conditions holds a dimensioning outdoor temperature, which for Gävle is -5 °C and -10°C. In ‘dry’ model, there is no precipitation on the surface, thus heat is calculated for keeping the surface warm.

3.2.2 ‘Wet’ model- with snowfall on the surface

This model is defined as a ’wet’ model. In this model, moisture is taken in solid form. Now it takes energy to melt a certain volume of snow into water and then evaporate this moisture in addition to the energy required to heat surface. This condition will be more related to the reality and suitable for the football ground, when the heating system under the sports facilities intends to keep the ground surface immediately above 0 °C and snow and ice-free.

The energy in the heater pipe, which is called Qs, is regulated differently depending on the model which is simulated. According to Figure 3.2, energy in heater pipe is set to be controlled from the snowfall where Qs (as regulates the effect) depends on the volume of snow. For the dry model, Qs is defined as 0.

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25 Figure 3.2: Parameters in COMSOL.

To get the moisture in the model and its influence the precipitation is divided into two parts, where the first part of the precipitate is evaporation and the other part is melting. The moisture "cools" the surface when energy is used to evaporate it. In this case the vaporized proportion and melting are all taken into account, which is the so-called" wet "model.

1) Melting

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26

The snowmelt process requires absorption of energy Q of an amount equal to the sum of the energy absorption associated with the three snowmelt phases:

𝑄 = 𝑄_𝑐𝑐+ 𝑄_𝑚2+ 𝑄_𝑚3 [27]

For warming phase, energy is added to raise the temperature to 0 °C. The required energy represents the “cold content” (𝑄_𝑐𝑐) of the snowpack. This energy total, in J*m-2 shown below:

𝑄_𝑐𝑐 = −𝐶_𝑖𝜌_𝑤ℎ_𝑚(𝑇_𝑠− 𝑇_𝑚) [28]

For ripening phase, net energy (𝑄_𝑚2) in J*m-2 is required, which is shown below:

𝑄_𝑚2 = 𝜃_𝑟𝑒𝑡ℎ_𝑠𝜌_𝑚𝜆_𝑓 [29]

Once the snowpack is ripe, further energy input results in meltwater that percolates downward to become water output. The energy (𝑄_𝑚3) in J*m-2 required to melt the snow remaining at the end of the ripening phase is:

𝑄𝑚3 = (ℎ𝑚− ℎ𝑤 𝑟𝑒𝑡)𝜌𝑚𝜆𝑓 [30] Where

ℎ_{𝑤 𝑟𝑒𝑡} = 𝜃_𝑟𝑒𝑡ℎ_𝑠 [31]

is the liquid-water retaining capacity of the snowpack and ℎ𝑚 is the liquid water equivalent of the snowpack.

𝐶_𝑖 = heat capacity of ice, 2 100 𝐽 𝑘𝑔 ∙ 𝐾⁄ 𝑇_𝑠 = average temperature of snow [K] 𝑇_𝑚 = melting point of ice [K]

𝜌𝑤 = density of water,1 000 𝑘𝑔 𝑚⁄ 3

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27 2) Evaporation

From equation previously mentioned, the rate of vapor flow that the air can receive is related to the difference between the vapor content at a water surface and the vapor content of surrounding air. Thus, the mass of evaporation (𝑀_{𝑒𝑣𝑎𝑝}) is defined as: 𝑀_{𝑒𝑣𝑎𝑝} = 𝛽(𝑣_𝑠𝑎𝑡− 𝑣_𝑎𝑖𝑟)∆𝑡 [𝑘𝑔/𝑚2_] _[32]

where

∆t is the time, having the unit second (s)

Insert the data in Table 2.3, equation can be write as: 𝑀_{𝑒𝑣𝑎𝑝} = 0.06(𝑣_𝑠𝑎𝑡− 𝑣_𝑎𝑖𝑟) [𝑘𝑔 𝑚_⁄ 2_]

This means the amount of water that is vaporized is 0.06 kg per square meter of surface area at one vapor density of 1 𝑔 𝑚⁄ 3.

Since the latent heat of vaporization is 𝐸𝑒𝑣𝑎𝑝 = 2 500 000 𝐽 𝑘𝑔⁄ the heat required to evaporate water becomes:

𝑞̇𝑒𝑣𝑎𝑝=41.2(𝑣𝑠𝑎𝑡−𝑣𝑎𝑖𝑟)[

𝑊 𝑚2]

The equation shows that 41.2 W per m2_{of surface area at one vapor difference.} The saturated vapor content 𝑣_𝑠𝑎𝑡 depend on whether the temperature is above 0 ° C or not. To calculate the heat flow, 𝑣_𝑠𝑎𝑡 can be calculated by the equation [11]

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28

Then the vapor content of surrounding air is set, which can be found from Figure 3.3. By detecting the vapor content in the air as well as the saturation vapor content at the surface, it is also possible to model how much energy is required to evaporate the proportion vaporized according to equation previously mentioned. This energy is written as a negative value which creates a "cooling" effect of the surface of the model. Add heat for melting and heat for evaporation together to get a value 𝑄𝑠, then insert the value −𝑄_𝑠 into the constant of COMSOL which is shown below.

Figure 3.4: Energy for melting snow is inserted as a negative heat source in COMSOL.

3.3 Pressure drop in pipe

To calculate the pressure loss in a pipe it is necessary to compute a pressure drop, usually in fluid head, for each of the items that cause a change in pressure. However, to calculate the friction loss in a pipe for example, it is necessary to calculate the friction factor to use in the Darcy-Weisbach equation [18] which determines the overall friction loss.

The friction factor itself is dependent on internal pipe diameter, the internal pipe roughness and the Reynold's number which is in turn is calculated from the fluid viscosity, fluid density, fluid velocity and the internal pipe diameter.

There are therefore several sub-calculations that must take place to calculate the overall friction loss. Working backwards we must know the fluid density and viscosity

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29

number and use this to calculate the friction factor using the Colebrook-White equation. Finally plug in the friction factor into the Darcy-Weisbach equation to calculate the friction loss in the pipe.

In this part, pressure drop is calculated in five parts. The football pitch is 100 m length and 50 m wide. The main pipe, and it has three zones of pipes, and each zone is separated with two columns, and each column has several branches with pipe spacing 200 mm as show in below figure.

Figure 3.5: Piping for the football pitch.

In the following result, section 1 presents the main pipe from the pump to the nearest zone; section 2 presents the main pipe from section 1 to the side zone; section 3 presents the main pipe from section 2 to the last zone; section 4 present one of the column in one zone; and section 5 presents one branch in one column. The pipe properties and length are shown below in Table 3.1.

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30 Table 3.1: Pipe properties and length of each section.

Pipe type Length (𝒎) Pipe roughness (𝒎𝒎) Diameter (𝒎𝒎) Interior diameter (𝒎𝒎) Section 1 PEH 33 0.030 200 180 Section 2 PEH 66 0.030 200 180 Section 3 PEH 66 0.030 200 180 Section 4 PEX 100 0.007 90 65.4 Section 5 PEX 33 0.007 20 14.4

3.3.1 Outdoor temperature -5 °𝐶

Heat capacity of water 𝑐_𝑝 = 4 184 𝐽 𝑘𝑔 ∙ ℃⁄ , density of water 𝜌 = 1 000 𝑘𝑔 𝑚_⁄ 3_, supply temperature 𝑇_𝑠 = 30 °𝐶, return temperature 𝑇_𝑟 = 25 °𝐶, so the temperature difference is 5 °𝐶. Inserting the values into equation [26], volumetric flow rate for each section is defined with unit (𝑚3⁄ ). 𝑠

Kinetic viscosity of water at 30 °𝐶, ν=0.8005*10-6_m2_{/s, and pipes properties is shown} above in Table 3.1.

Inserting the variables above into equation [19-25], Reynold's number and related coefficient are obtained, and pressure drop in pipe per meter is calculated by using the Darcy-Weisbach equation [18].

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3.3.2 Outdoor temperature -10 °𝐶

Heat capacity of water 𝑐_𝑝 = 4 184 𝐽 𝑘𝑔 ∙ ℃⁄ , density of water 𝜌 = 1 000 𝑘𝑔 𝑚_⁄ 3_, supply temperature 𝑇𝑠 = 45 °𝐶, return temperature 𝑇𝑟 = 40 °𝐶, so the temperature difference is 5 °𝐶. Insert the values into equation [26], volumetric flow rate for each section is defined with unit (𝑚3_{⁄ ).}_𝑠

Kinetic viscosity of water at 45 °𝐶, ν=0.603*10-6_m2_{/s, and pipes properties is shown} above in Table 3.1.

Inserting the variables above into equation [19-25], Reynold's number and related coefficient are obtained, and pressure drop in pipe per meter is calculated by using the Darcy-Weisbach equation [18].

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4. Result

This chapter presents the power requirements for each model, energy use and the heat transports occurring in the field constructions with and without 100 mm insulation. In addition, the pressure drop in pipes are calculated.

Power demand and energy use describe how much energy in Watt is used to heat the ground surface under the two conditions without snowfall and with snowfall. In

addition, the effect on the ground heating system varies depends on different positions. Putting two types of insulation materials under the pipes and a case with no insulation is also reported.

Figure 4.1: Difference in heat between section without insulation (left pillar) and with insulation (right pillar).

4.1 Dry model- without snowfall

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4.1.1 Outdoor temperature -5 ºC

According to the calculation carried out in COMSOL for the dry model with outdoor temperature -5 ºC, it turns out that the energy use of the insulated condition is lower than the effect of the uninsulated condition.

From the data calculated by COMSOL which is shown below in Table 4.1, when the construction has a concrete top layer, 159 W per square meter power is needed for the uninsulated design, 154.7 W per square meter power is needed for the XPS insulated design, and 154.5 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

Table 4.1 indicates that the uninsulated design has 21.3 kW difference compared with the XPS insulated design and it has 22.2 kW difference compared with the EPS

insulated design. Besides, the XPS constructed design has more energy use than that in the EPS constructed design, which has the difference 0.9 kW.

Table 4.1: Data showing the difference between insulated and non-insulated construction with outdoor temperature -5 ºC and concrete top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 158.957 154.696 154.525 Area (𝒎𝟐₎ _{5 000} _{5 000} _{5 000} Heat loss (𝒌𝑾) 794.786 773.481 772.627

From the data calculated by COMSOL which is shown below in Table 4.2, when the construction has a stone top layer, 146.7 W per square meter power is needed for the uninsulated design, 142.4 W per square meter power is needed for the XPS insulated design, and 142.3 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

Table 4.2 indicates that the uninsulated design has 21.3 kW difference compared with the XPS insulated design and it has 22.1 kW difference compared with the EPS

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35

Table 4.2: Data showing the difference between insulated and non-insulated construction with outdoor temperature -5 ºC and stone top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 146.700 142.449 142.279 Area (𝒎𝟐) 5 000 5 000 5 000 Heat loss (𝒌𝑾) 733.499 712.244 711.397

From the data calculated by COMSOL which is shown below in Table 4.3, when the construction has a grass top layer, 161.9 W per square meter power is needed for the uninsulated design, 157.6 W per square meter power is needed for the XPS insulated design, and 157.4 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

Table 4.3 indicates that the uninsulated design has 21.3 kW difference compared with the XPS insulated design and has 22.1 kW difference compared with the EPS insulated design. Besides, the XPS constructed design has more energy use than that in the EPS constructed design, which has the difference 0.8 kW.

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36

Diagram 4.1: The difference in total heat loss in the tubes with and without insulation for a 5000 m2 _surface.

The diagram above indicates that uninsulated construction has more energy use than the insulated construction. The insulation material EPS has better insulating ability than XPS. The surface with stone top layer has lower heat loss than concrete and grass, and grass top layer surface has highest energy loss.

The differences in temperatures in the ground a few meters down are also apparent. It depends on the effect of the insulated and uninsulated construction, and different top layer.

Diagram 4.2: The difference in heat loss downward in the tubes with and without insulation for a 5000 m2 surface.

733.5 712.2 711.4 809.3 788.1 787.2 794.8 773.5 772.6 660 680 700 720 740 760 780 800 820

Without insulation XPS EPS

H

e

at L

o

ss (kW)

Total Heat Loss

Stone Grass Concrete

41292 20467 19553 40727 20174 19273 40836 20231 19327 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 Without insulation XPS EPS Heat Loss (W)

Heat Downward in Y-Direction

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37

As shown in Diagram 4.2 the differences are apparent. The heat loss for uninsulated section scale on the Y axis is approximately 2 times greater than the insulated sections. So, the insulation materials have obvious insulating abilities to prevent the heat going downwards.

Diagram 4.3: The difference in heat loss upward in the tubes with and without insulation for a 5000 m2 surface.

As shown in Diagram 4.3, there is slight difference between insulated and uninsulated constructions. So, the insulation materials have weak effect on the heat loss upwards.

4.1.2 Outdoor temperature-10 ºC

According to the calculation carried out in COMSOL for the dry model with outdoor temperature -10 ºC, it turns out that the energy use of the insulated condition is lower than the effect of the uninsulated condition.

From the data calculated by COMSOL which is shown below in Table 4.4, when the construction has a concrete top layer, 250 W per square meter power is needed for the uninsulated design, 243.2 W per square meter power is needed for the XPS insulated design, and 243 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

692207 691778 691844 768616 767872 767944 753950 753250 753300 640000 660000 680000 700000 720000 740000 760000 780000 Without insulation XPS EPS Heat Loss (W)

Heat Upward in Y-Direction

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Table 4.4: Data showing the difference between insulated and non-insulated construction with outdoor temperature -10 ºC and concrete top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 250.029 243.210 242.951 Area (𝒎𝟐₎ _{5 000} _{5 000} _{5 000} Heat loss (𝒌𝑾) 1 250.145 1 216.052 1 214.753

From the data calculated by COMSOL which is shown below in table 4.5, when the construction has a stone top layer, 230.8 W per square meter power is needed for the uninsulated design, 224 W per square meter power is needed for the XPS insulated design, and 223.7 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

Table 4.5 indicates that the uninsulated design has 34 kW difference compared with the XPS insulated design and has 35.4 kW difference compared with the EPS insulated design. Besides, the XPS constructed design has more energy use than that in the EPS constructed design, which has the difference 1.4 kW.

Table 4.5: Data showing the difference between insulated and non-insulated construction with outdoor temperature -10 ºC and stone top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 230.773 223.970 223.698 Area (𝒎𝟐) 5 000 5 000 5 000 Heat loss (𝒌𝑾) 1 153.863 1 119.848 1 118.489

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Table 4.6: Data showing the difference between insulated and non-insulated construction with outdoor temperature -10 ºC and grass top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 254.607 247.792 247.526 Area (𝒎𝟐) 5 000 5 000 5 000 Heat loss (𝒌𝑾) 1 273.036 1 238.959 1 237.629

Diagram 4.4: The difference in heat loss in the tubes with and without insulation for a 5000 m2 surface.

The diagram above indicates that uninsulated construction has more energy use than the insulated construction. The insulation material EPS has better insulating ability than XPS. The surface with stone top layer has lower heat loss than concrete and grass, and grass top layer surface has highest heat loss.

The differences in temperatures in the ground a few meters down are also apparent. It depends on the effect of the insulated and uninsulated construction, and different top layers. 1153.9 1119.8 1118.5 1273.1 1238.9 1237.6 1250.1 1216.1 1214.8 1000 1050 1100 1150 1200 1250 1300

H

e

at L

o

ss (kW)

Total Heat Loss

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40

Diagram 4.5: The difference in heat loss downward in the tubes with and without insulation for a 5000 m2 _surface.

As shown in Diagram 4.5, the differences are apparent. The heat loss for uninsulated section scale on the Y axis is approximately 2 times greater than the insulated sections. The insulation materials have obvious insulating abilities to prevent the heat going downwards.

Diagram 4.6: The difference in heat loss upward in the tubes with and without insulation for a 5000 m2 _surface.

66214 32823 31357 65326 32364 30918 65495 32452 31002.5 0 10000 20000 30000 40000 50000 60000 70000 Without insulation XPS EPS Heat Loss (W)

Heat Downward in Y-Direction

Concrete Grass Stone

1087649 1087025 1087132 1207709 1206595 1206711.025 1184650 1183600 1183750 1020000 1060000 1100000 1140000 1180000 1220000 Without insulation XPS EPS Heat Loss (W)

Heat Upward in Y-Direction

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41

4.2 Wet model

In order to make the football pitch ice and snow free, calculation on the heat required for melting snow is needed. This means the heating system provide heat not only for keeping the surface temperature above 0 ºC, but also melting the snowfall accumulated on the surface. Since it has a phase change, a larger amount of energy is required in this section than that in the ‘dry’ section.

4.2.1 Outdoor temperature -5 ºC

The precipitation is set to be 200 mm snowfall when outdoor temperature is -5 ºC. According to the calculations performed in COMSOL for the wet model show that the effect for the insulated surface is lower than the effect of the uninsulated surface. From the data calculated by COMSOL which is shown below in Table 4.7, when the construction has a stone top layer, 391.2 W per square meter power is needed for the uninsulated design, 387.3 W per square meter power is needed for the XPS insulated design, and 387.2 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

Table 4.7: Data showing the difference between insulated and non-insulated construction with outdoor temperature -5 ºC and stone top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 391.207 387.299 387.158 Area (𝒎𝟐₎ _{5 000} _{5 000} _{5 000} Heat loss (𝒌𝑾) 1 956.036 1 936.495 1 935.791

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42

From the data calculated by COMSOL which is shown below in Figure 4.9, when the construction has a concrete top layer, 415 W per square meter power is needed for the uninsulated design, 411 W per square meter power is needed for the XPS insulated design, and 410.9 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

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43

Diagram 4.7: The difference in heat loss in the tubes with and without insulation for a 5000 m2 surface.

The diagram above indicates that uninsulated construction has more energy use than the insulated construction. The insulation material EPS has better insulating ability than XPS. The surface with stone top layer has lower heat loss than concrete and grass, and grass top layer surface has the highest energy loss.

The differences in temperatures in the ground a few meters down are also apparent. It depends on the effect of the insulated and uninsulated construction, and different top layer. 1956.1 1936.5 1935.8 2102.2 2082.6 2082.1 2074.9 2055.2 2054.6 1850 1900 1950 2000 2050 2100 2150

H

e

at L

o

ss (kW)

Total Heat Loss

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44

Diagram 4.8: The difference in heat loss downward in the tubes with and without insulation for 5000 m2 surface.

As shown in Diagram 4.8, the differences are apparent. The heat loss for uninsulated section scale on the Y axis is approximately 2 times greater than the insulated sections. So, the insulation materials have obvious insulating abilities to prevent the heat going downwards.

Heat Downward in Y-Direction

1923850 1920750 1920750 2071150 2067450 2067500 2043550 2039950 2040000 1800000 1850000 1900000 1950000 2000000 2050000 2100000 Without insulation XPS EPS Heat Loss (W)

Heat Upward in Y-Direction

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45

As shown in Diagram 4.9, there is slightly differences between insulated and

uninsulated construction. So, the insulation materials have weak effect on the heat loss upwards.

4.2.2 Outdoor temperature-10 ºC

The precipitation is set to be 100 mm snowfall when outdoor temperature is -10 ºC. According to the calculations performed in COMSOL for the wet model show that the effect for the insulated surface is lower than the effect of the uninsulated surface. From the data calculated by COMSOL which is shown below in Table 4.10, when the construction has a stone top layer, 347.5 W per square meter power is needed for the uninsulated design, 340.8 W per square meter power is needed for the XPS insulated design, and 340.6 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

Table 4.10: Data showing the difference between insulated and non-insulated construction with outdoor temperature -10 ºC and stone top layer in dry model calculations. Without insulation With XPS insulation With EPS insulation Heat flux (𝑾 𝒎_⁄ 𝟐₎ 347.493 340.814 340.561 Area (𝒎𝟐₎ _{5 000} _{5 000} _{5 000} Heat loss (𝒌𝑾) 1 737.465 1 704.070 1 702.804

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From the data calculated by COMSOL which is shown below in Table 4.12, when the construction has a concrete top layer, 371.9 W per square meter power is needed for the uninsulated design, 365.2 W per square meter power is needed for the XPS insulated design, and 364.9 W per square meter power is needed for the EPS insulated design, which can keep the surface temperature above 0 ºC.

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47

Diagram 4.10: The difference in heat loss in the tubes with and without insulation for a 5000 m2 _surface.

The diagram above indicates that uninsulated construction has more energy use than the insulated construction. The insulation material EPS has better insulating ability than XPS. The surface with stone top layer has lower heat loss than concrete and grass, and grass top layer surface has highest energy loss.

The differences in temperatures in the ground a few meters down are also apparent. It depends on the effect of the insulated and uninsulated construction, and different top layer. 1737.5 1704.1 1702.8 1887.8 1854.4 1853.2 1859.4 1825.9 1824.7 1600 1650 1700 1750 1800 1850 1900 1950

H

e

at L

o

ss (kW)

Total Heat Loss

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Diagram 4.11: The difference in heat loss downward in the tubes with and without insulation for a 5000 m2 _surface.

As shown in Diagram 4.11, the differences are apparent. The heat loss for uninsulated section scale on the Y axis is approximately 2 times greater than the insulated sections. So, the insulation materials have obvious insulating abilities to prevent the heat going downwards.

Heat Downward in Y-Direction

1675600 1673500 1673600 1827100 1824400 1824500 1798400 1795850 1795900 1550000 1600000 1650000 1700000 1750000 1800000 1850000 Without insulation XPS EPS Heat Loss(W)

Heat Upward in Y-Direction

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49

4.3 Pressure drop calculation

In this part, flow rate and pressure drop in each part of the system are shown in the tables as well as the characteristics of the pump.

The football pitch is 100 m length and 50 m wide. The main pipe is connected to three zones of pipes, and each zone is separated with two columns, and each column has several branches with pipe spacing 200 mm.

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50

4.3.1 Dry model

4.3.1.1 Outdoor temperature -5 ºC

Table 4.13: Energy use for heating pipe.

Without insulation

XPS insulation EPS insulation

Stone (W) 733 498.9 712 244.3 711 397.2

Grass (W) 809 343.5 788 046.0 787 217.0

Concrete(W) 794 785.5 773 480.5 772 627.0

Table 4.14: Volumetric flow rate.

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51 Table 4.15: Pressure drop in pipe in Pa/m.

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52 Table 4.16: Pressure drop in pipe in Pa.

Without insulation XPS insulation EPS insulation Stone (𝑷𝒂) 1 2 772 1 2 623.17 1 2 617.23 2 2 599.08 2 2 460.48 2 2 455.2 3 724.68 3 687.06 3 685.08 4 40 297 4 38 189 4 38 107 5 1 483.35 5 1 411.41 5 1 407.78

Total 47 876.11 Total 45 371.12 Total 45 272.29 Grass (𝑷𝒂) 1 3 335.97 1 3 172.62 1 3 166.02 2 3 121.14 2 2 970 2 2 964.06 3 867.9 3 826.32 3 824.34 4 48 246 4 45 947 4 45 859 5 1 754.61 5 1 676.07 5 1 673.43

Total 57 325.62 Total 54 592.01 Total 54 486.85 Concrete (𝑷𝒂) 1 3 223.77 1 3063.06 1 3 056.79 2 3 017.52 2 2868.36 2 2 862.42 3 839.52 3 798.6 3 796.62 4 46 669 4 44 405 4 44 315 5 1 700.82 5 1 623.6 5 1 620.63

Total 55 450.63 Total 52 758.62 Total 52 651.46 Specific pressure drops in section 4 is the largest of all, and the ground with grass top layer has the largest pressure drop than the other two type of top layer, and the

uninsulated construction has larger pressure drop than the insulated ones. EPS insulated construction has a better ability to prevent pressure drop.

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53

Table 4.17: Operating pressure and flow rate for pump with -5 ºC outdoor temperature.

Without insualtion

XPS insualtion EPS insulation

Stone 35.0621 𝑙 𝑠⁄ ; 0.5 𝑏𝑎𝑟 34.0461 𝑙 𝑠⁄ ; 0.5 𝑏𝑎𝑟 34.0056 𝑙 𝑠⁄ ; 0.5 𝑏𝑎𝑟 Grass 38.6876 𝑙 𝑠⁄ ; 0.6 𝑏𝑎𝑟 37.6695 𝑙 𝑠⁄ ; 0.6 𝑏𝑎𝑟 37.6299 𝑙 𝑠⁄ ; 0.6 𝑏𝑎𝑟 Concrete 37.9917 𝑙 𝑠⁄ ; 0.6 𝑏𝑎𝑟 36.9733 𝑙 𝑠⁄ ; 0.6 𝑏𝑎𝑟 36.9325 𝑙 𝑠⁄ ; 0.6 𝑏𝑎𝑟 Table 4.17 indicates that a pump with 39 l/s flow rate and 0.6 bar pressure

characteristics is suitable for the system to provide enough heat for the football pitch under any conditions.

4.3.1.2 Outdoor temperature -10 ºC

Table 4.18: Energy use for heating pipe.

Stone (W) 1 153 862.95 1 119 847.60 1 118 489.24

Grass (W) 1 273 035.71 1 238 958.52 1 237 629.02

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54 Table 4.19: Volumetric flow rate.

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55 Table 4.20: Pressure drop in pipe in Pa/m.

Energy analysis for a snow-free surface A technical analysis of the benefits of insulation under the heating pipes Ying Song Spring 2018

FACULTY OF ENGINEERING AND SUSTAINABLE DEVELOPMENT