1

� � � � � � � � � � � � � � � � � � � � � � � � � �

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ! � � " � " � � � � �

L. Johansson♣, L. Westerlund

# $ % & ' ( ) * + & , - * . / 0 1 2 & 3 4 ) 0 % 0 5 / 6 7 & 8 ' , . 9 & ) . 0 1 : & 3 4 ' ) * 3 ' % ; ) 5 * ) & & , * ) 5 6 7 * + * - * 0 ) 0 1 ; ) & , 5 /; ) 5 * ) & & , * ) 5 6 # $ % & ' 6 < = & > & )

? @ A B C D E BIn indoor swimming pool facilities the energy demand is large due to ventilation losses with the

exhaust air. Since water is evaporated from the pool surface the exhaust air has a high water contentand specific enthalpy. Because of the low temperature the heat from the evaporation is difficult torecover.In this paper the energy demand for the conventional ventilation technique in indoor swimming pools iscompared to two different heat recovery techniques, the mechanical heat pump and the open absorptionsystem. The mechanical heat pump is the most widely used technique in Sweden today. The openabsorption system is a new technique in this application. Calculations have been carried out on anhourly basis for the different techniques. Measurements from an absorption system pilot plant installedin an indoor swimming pool in the northern part of Sweden have been used in the calculations. Theresults show that with the mechanical heat pump, the electrical input increases with 63 MWh/year andwith the open absorption system 57 MWh/year. However, both a mechanical heat pump and an openabsorption system decreases the annual energy demand from 611 MWh to 528 and 484 MWhrespectively, which corresponds to a decrease of approximately 14 and 20% respectively. Changing theclimate in the facility has also been investigated. An increased temperature decreases the energydemand when using the conventional ventilation technique. However, when the mechanical heat pumpand the open absorption system is used the energy demand is increased when the temperature isincreased. Therefore increasing the temperature in the facility when using the conventional techniqueshould be considered the first measure to reduce the energy demand.

F G H I J K L MAbsorption system, Mechanical heat pump, Indoor swimming pool

♣ Corresponding author. Tel. +46 920 910 46, Fax. +46 920 910 47, Email: larsj@mt.luth.se

2

N O P Q R S T U V W X QA Area m2

cp Specific heat J/kg⋅Kh Specific enthalpy J/kg dry airM Mole mass kg/molemY Mass flow kg/sZ [ Rate of heat transfer W

P Pressure PaQ Energy demand Jr Evaporation heat J/kgT Temperature °C, Kx Humidity ratio kg H20/kg dry air

\ ] ^ ^ _ ` a b c d e `α Heat transfer coefficient W/m2·Kε Ratio of outdoor air mass flow rate to total air mass

flow rate-

φ Relative humidity %RHf g h i j k l m n i

a Dry airabs Absorbed water vaporair Moist airair heater Heater in the air circuitcirk Circulation airflowcomp Compressorcw Cold waterelectricity Electricity to compressor and generator

respectivelyevap Evaporation of water from pool surfacefacility Conditions in the buildingfrom Airflow from the facilitygen Generator/Boilerhp Conditions in heat pump evaporatorhw Hot waterl Liquidleak Leakage airflow to the facilitymixed Conditions of mixed airflowoutdoor Outdoor conditionspool Conditions in the poolrad Radiatorssolution Conditions of absorption solutionsurface Air closest to the pool surfaceto Airflow to the facilitytrans Transmission lossesv Vaporw Water

3

o p q r s t u v w x s y u r

An indoor swimming pool facility normally has a large energy demand and istherefore an interesting application for heat recovery systems.Water is evaporating from the pool surface and has to be removed from the facility inorder to avoid condensation on cold surfaces in the facility. The moisture is carriedaway by outdoor air, which lead to large energy demand for the facility. When waterevaporates, heat from the water in the pool is transferred to the air. The heat isdifficult to recover because of the low air temperature. In order to keep a good climateand quality of the air in the facility, a minimum outdoor airflow is supplied.According to Swedish regulations the minimum outdoor airflow for buildings is 0.35L/s and m2 floor area. By experience it has been shown that the outdoor airflow has tobe larger to retain good air quality. Rules of thumbs are often used that states aminimum outdoor airflow of 10 m3/h (2.8 L/s) per m2 pool area, during opening hours[1]. The climate in the facility, i.e. the temperature and relative humidity is dependenton the outdoor temperature. When the outdoor temperature decreases the relativehumidity is decreased in order to avoid condensation of moisture on cold surfacessuch as windows and outer walls.

There are different heat recovery systems on the market today that reduce theenergy demand for the facility. Mechanical heat pumps are most widely used,especially in the Scandinavian countries. In this paper the energy demand for themechanical heat pump and the conventional ventilation technique are compared.

An open absorption system also makes it possible to recover the latent heat in themoist air. Therefore the outdoor airflow can be reduced, which leads to reducedenergy demand for the facility. In the mid eighties a pilot plant of an open absorptionsystem were installed at an indoor swimming pool facility in the northern part ofSweden. To evaluate the technique, measurements were carried out during a couple ofyears. Results from the measurements have been published by Westerlund et. al [2].This paper will also compare the energy demand for the open absorption techniquewith the mechanical heat pump and conventional ventilation technique.

z { | } ~ � � ~ � � } ~ � � � � ~ � � � � � � } ~ � � � � ~ � � � �The conventional ventilation technique uses outdoor air to carry away the

evaporated water from the pool surface. Despite the outdoor air has a high relativehumidity, the humidity ratio is low compared to the air in the facility.

To decrease the energy demand the outdoor airflow is mixed with a circulationflow from the facility (see Fig. 1). The letters A-D stands for the different conditionsin the air circuit. The mixed airflow passes through an air heater, where heat issupplied to reach the required temperature of the air. The air heater consists of abundle of tubes enclosed in the air duct, with hot water from the district heatingsystem flowing inside the tubes and the air is heated by flowing on the outside of thetubes.

The outdoor airflow and circulation flow is regulated with dampers in the air ducts.Measuring the relative humidity in the exhaust air sets the outdoor airflow. If therelative humidity is too high, the outdoor airflow is increased and vice versa.

4

Fig. 1. Flow diagram of the conventional ventilation technique.

Fig 2. Pathways in a Mollier diagram for the conventional technique.

heaterair Q�

poolQ�

hwQ�

radQ�

transQ�

CA B

Airheater

Outdoor air

Facility

Pool

Exhaust air D

Tem

pera

ture

(°C

)

A

B

C

A: Outdoor airB: Mixing of outdoor airC: Air after air heaterD: Air from facility

D

Constant Enthalpy

Humidity ratio (kg/kg)

5

Measuring the temperature of the air from the facility sets the temperature byaltering the supplied heat in the air heater. Fig. 2 shows the pathways in a Mollierdiagram for the conventional ventilation technique.

Close to the pool surface the air is saturated. The difference in partial pressure ofthe water vapor in the air closest to the surface and in the facility leads to a masstransfer of water vapor from the pool to the air in the facility. The heat required forthe evaporation is taken from the pool water. Hence the evaporation pathway (C-D) isalmost horizontal in the Mollier diagram. The heater in the pool circuit consists ofbundle of tubes with hot water from the district heating system flowing inside thetubes and the pool water is heated by flowing on the outside of the tubes.

If the facility in total is taken into consideration heat is supplied to the pool, to theairflow, to the radiators and to the hot water production. The heat losses from thefacility consist of transmission losses, evacuation losses via the exhaust air and the hotwater from the showers to the sewer (see Fig. 1). The radiators are designed to coverthe transmission losses from the facility.

� � � � � � � � � � � � � � � � � � � �In many indoor swimming pools the air is dehumidified with a mechanical heat

pump (see Fig. 3).

Fig. 3. Mechanical heat pump dehumidifier installed in a facility.

compQ�

hwQ�

poolQ�

radQ�

transQ 

heaterair Q¡

F D

CBA

Evaporator

Airheater

Outdoor air

Condensate

Condenser

Valve Facility

Pool

Exhaust air

E

6

The mechanical heat pump consists of four parts, compressor, condenser,expansion valve and evaporator. The principle of the mechanical heat pump is that therefrigerant is evaporated in the evaporator. The heat of evaporation is taken from themoist air in the facility. The compressor increases the pressure and temperature of thevapor. In the condenser the refrigerant is condensed giving off heat. The pressure ofthe liquid is decreased through an expansion valve.When installed in the facility, moist air from the facility is passing through theevaporator where the air is cooled and water vapor is condensed. According to themanufacturer, approximately 20% of the total airflow are cooled in the evaporator and80% is by-passed [1].

The dehumidified airflow and the by-pass flow are mixed (F). The total airflow ismixed with outdoor air (B), before passing through the condenser, where heat issupplied to the air (see Fig. 3). The condenser consists of two parts. One part used topreheat the mixed airflow before entering the air heater and one, heat exchanger usedfor preheating of the water in the pool. The largest part of the condenser heat is usedto preheat the water in the pool. The mechanical heat pump is designed for theevaporation rate that occurs when the facility is closed [1]. During closed hours thereare no restrictions of outdoor airflow, hence all the air is circulated and no outdoor airis supplied.

During opening hours when the evaporation rate from the pool is larger, outdoorair is supplied. Fig. 4 shows the pathways in a Mollier diagram when the mechanicalheat pump is installed. It is running under on/off conditions, i.e. it is always operatingat maximum load since the efficiency is decreasing when it is operating at part load[3].

Fig. 4. Pathways in a Mollier diagram using the mechanical heat pump

The heat pump is automatically shut down when the evaporation rate is low andoperation is not necessary. The air heater in the air circuit is used to reach the set point

Humidity ratio (kg/kg)

AB

CD

E

F

A: Outdoor airB: Mixing with outdoor airC: Air after air heaterD: Air from facilityE: Air after evaporatorF: Mixing bypass and facility air

Constant enthalpy

Tem

pera

ture

(°C

)

7

values on the temperature. The heater in the pool circuit is used to reach the set pointvalue on the pool water.

The heat demand in the batteries is reduced due to the condenser heat. On the otherhand the electrical input is increasing since it is needed to operate the compressor.The other heat supplies consists of heat to the radiators and hot water production inthe same way as for the conventional ventilation technique.

¢ £ ¤ ¥ ¦ § ¨ © ª « ¬ ­ ® ¯ « ° ª ± ª ® ² ³When the open absorption system is used the airflow from the facility is passing

through the absorber where the water vapor is absorbed by an absorption solution.The humidity ratio is decreased and the temperature of the air is increased. Thediluted absorption solution is transported to the generator, or boiler, where heat issupplied. The absorption solution has a higher boiling point than water, hence thewater is evaporated and the absorption solution is regenerated. The concentratedsolution is brought back to the absorber. In the condenser the water vapor is cooledand the condensate is removed from the system. Figure 5 shows the open absorptionsystem installed in a facility. The open absorption system and different applicationsare described in Westerlund et al [2, 4] and Johansson et al. [5, 6].

Fig. 5. Open absorption system installed in a facility.

poolQ́

genQµ

radQ¶

transQ·

heaterair Q̧

E, F

CBA

Airheater Facility

Pool

Outdoor air

Condenser

Hot waterproduction

Exhaust air

Absorber

D

Condensate

8

The dried airflow (E, F) is mixed with the outdoor airflow before entering the airheater (B). The condenser heat is used for heating of the water in the pool. When thecondenser heat is less then required, an additional air heater is used. Since the heat ofevaporation is supplied to the airflow, it has to be cooled, to avoid an increased indoortemperature (see Fig. 6). If the absorption solution is supplied to the absorber at thesame temperature as the wet bulb temperature of the air, absorption will occur underadiabatic conditions and the enthalpy of the air will remain constant through theabsorber [7]. When the absorption solution is cooled, heat is transferred from theairflow during passage through the absorber. The heat transferred in the heatexchanger is used for preheating of hot water to the showers. Since there is adifference in time between high load of the evaporation rate (large swimming activity)and high load on the showers, heat storage has to be installed to store the heattransferred from the air. During closed hours the outdoor airflow is excluded and thetotal airflow is circulated since the absorption solution absorbs the evaporated water.

Fig. 6. Pathways in the Mollier-diagram using open absorption system.

¹ º » ¼ ½ ¾ ¿ À Á ÀTo compare the different techniques calculations were carried out for the facilitywhere the pilot plant absorption system was installed. Since measurements from acouple of years were available the comparison is made with some of the data fromthat facility. Similar calculations can be made for different indoor swimming pools.

The facility has six lanes of 25 meters. The total pool area is 258 m2. The totalventilation airflow in the facility is 8000 m3/h and the leakage of air was estimated to320 m3/h, which corresponds to 0.1 times the total air volume in the facility.

During opening hours, three different evaporation rates have been used. Theduration when these rates occur, have been established from statistical data of visitorsduring the day. Calculations of the total energy demand for the different techniqueshave been carried out hour by hour on annual basis.

Humidity ratio (kg/kg)

Tem

pera

ture

(°C

)

A

B

C D

E

F

A: Outdoor airB: Mixing with outdoor airC: Air after air heaterD: Air from facilityE: Absorption and coolingF: Adiabatic absorption

Constant Enthalpy

9

In prior measurements the outdoor temperature was measured every hour and usedin the calculations. The temperature is important since the climate in the facility isdependent on the outdoor temperature. The temperature in the facility was setconstant to 29 °C and the relative humidity is differing according to the change inoutdoor temperature. When the outdoor temperature was low the relative humidity inthe facility was decreased. The relative humidity in the facility was calculated usingEq. (1). The temperature in the pool was also set constant to 27Â Ã Ä Å Æ Ç È É Ê Ë Ì Ç Æ Í Î È Î Ç Æwas to supply the minimum outdoor airflow to the facility during opening hours.

Ï Ð Ñ Ò Ï Ï ÓÔ Õ Ö × Ø × Ñ ÙÛÚ⋅+= 25.055φ (%RH) (1)

The humidity ratio and the specific enthalpy of the dry air in the facility (point D inFig. 1-6) and the outdoor airflow (point A) were calculated using Eqs. (2). and (3).

( )ÜÝ Þ ÝÜßàâáááããä

−⋅= (kg water /kg dry air) (2)

where (Pv) is the partial pressure of the water vapor and (Ptot) is the total pressure. Inthe calculations the total pressure was set constant to 1 atm or 101325 Pa.

( )å æ çèéå æ çåéëêìíîêìï ⋅+⋅+⋅= ,, (kJ/kg dry air) (3)

The enthalpy of the air was calculated using 0°C as a reference temperature. Hencedry air at temperature 0°C has a zero value of the enthalpy.

Close to the water surface the air will be saturated with water at the sametemperature as the pool water ( ð ñ ò ó ô õ ö÷ ). Since the air condition in the facility was

known the evaporation rate was approximated to

( )ø ù ú û ü û ý þÿ � � ø ù ú �ù û ��� � � ü� � ù ����� −⋅

⋅=

,

α�(kg/s) (4)

Where (α) is the heat transfer coefficient (W/m2·K) [8]. Three different values onthe heat transfer coefficient have been used, to describe the activity during openinghours, see Table 1.

Table 1. Specific heat transfer coefficients for different evaporation rates.Evaporation

rateSpecific heat

transfercoefficient(W/m2⋅K)

Duration(Weekdays,

Opening hours8.00-21.00)

Duration(Saturdays,

Opening hours6.00-18.00)

Duration(SundaysClosedfacility)

Maximum 5.8 08.00-12.00,15.00-18.00

06.00-09.00,12.00-15.00

-

Medium 3.8 12.00-15.00,18.00-21.00

09.00-12.00,15.00-18.00

-

Minimum 1.8 00.00-08.00 00.00-06.00,18.00-24.00

00.00-24.00

10

When there is activity in the pool the surface area is increased due to watermovements, hence it is included in the different values. As mentioned earlier, theduration when the different evaporation rates occur, have been established fromstatistical data, see Table 1.

The different heat transfer coefficients have been established from measurementsin earlier work [2].

The heat release from the pool water due to evaporation was calculated using

( )� ���� � � �� � � ������� ⋅+⋅= ,

��(W) (5)

Since the leakage airflow to the facility and the total airflow were set constant thedry mass flow of air to the facility was estimated using

� � � �� � � ! �#"""%$$$ −= (kg dry air/s) (6)

The humidity ratio and the enthalpy of the air to the facility (point C in Fig.1-6)were established from a mass balance of water and an energy balance for the facility,Eq. (7). and (8).

& '( ) * +' , & - ' ' ./ ( * 01 * 2 3 / 3 & 41 . ' 5& '76686868:9999

−⋅−⋅= (kg water /kg dry air) (7)

; <= > ? @ A @ ;B; C > D EF G > H< I ; J < < CA F > K= > ? @ A @ ; B= C < L; <NM OOP

MPMPRQQQQQ

,−−⋅−⋅= (J /kg dry air) (8)

In Eq. (8) the heat transferred to the facility from the people, machines and lightingetc. is excluded. To make the energy balance complete, these heat supplies shouldalso be considered. For the mixed airflow (point B), the properties were calculatedusing mass and energy balances, Eq. (9). and (10).

S T U V W XU V W XT Y S Z T T WT Y S Z T T W[ V \ ] Z_^^`^``baaa

⋅+⋅= (kg water /kg dry air) (9)

c de f g he f g hd i c j d d gd i c j d d gk f l m jonnpnpprqqq

⋅+⋅= (J /kg dry air) (10)

Combining Eq. (9) and (5) makes it possible to calculate the outdoor airflow (pointA) as

s t u v w v x yx zs t u v w v x yz { x | z z } ~~~~

−−

=ε (-) (11)

where (ε) is the part of the total airflow to the facility

11

� �� � � � � � ��� ��

=ε (-) (12)

During opening hours the outdoor airflow was set to the minimum value (2580m3/h) even if a smaller airflow were needed to carry away the moisture. That leads tothat the evaporation rate from the pool increases since the outdoor air has a lowerhumidity ratio than the circulation air. The evaporation rate was recalculated andthrough iteration process using Eq. (4), (7) and (11), the conditions of the air suppliedto the facility were established.

When the conditions at the point of mixing and the conditions of the air to thefacility were known, the required heat to the air heater was calculated using Eq. (13).

( )� � � � �� �� �� � � � � � � � ������ −⋅= �� (W) (13)

The required heat to the pool was calculated using

( )� �� � � ����� � � �� � � �� � � �  � � � �� � � � ¡¡¢£¤¤¤ −⋅⋅++= ,,, ¥¥¥¥ (W) (14)

where the transmission losses from the pool was estimated to a fixed value of 8 kW[2].

The last term in Eq. (14) describes the preheating of supplied water to the pool.(Tcw) was set constant to 5°C.

Finally the transmission losses from the facility was calculated according to Eq.(15)

( )¦ § ¨ © ¦ ¦ ª« ¬ ­ ® ¯ ® ¨ °« ¬ ­ ® ¯ ® ¨ °¨ ª ¬ ± ²´³³µ¶· −⋅⋅= ∑,

¸(W) (15)

The (∑ ⋅ ¹º -value) was estimated to a fixed value of 0.44 W/ K. [9].

Calculations were carried out hour by hour on annual basis, assuming constantconditions during each hour. That assumption leads to that the annual energy demandfor the facility could be calculated using Eq. 16.

»¼¼¼¼¼½ »¾¼ ¿À Á À Â Ã Ä Å Â Å ÃÆ ÇÄ È ÉÊ Ë Ë ÁÆ À È Ã À ÄÈ Å Ä¿ À È Ä Ã Ë ÃÃ Ë Ã È Á ∆⋅

++++≈= ∑ ∑ ∑∑∑∫

8760

1

8760

1

8760

1

8760

1

8760

1

ÌÌÌÌÌ (16)

In Eq. (16) ( radQÍ ) is the heat demand due to transmission losses from the facility.

The radiators in the facility are designed to cover the transmission losses from thefacility and were included in the energy balance (Eq. 8). Therefore the transmissionlosses from the facility were covered by the supplied heat to the air heater in the aircircuit. However, during extreme low outdoor temperatures the radiator system mustbe used. ( hwQÎ ) is the heat demand for the hot water production and ( yelectricitQÏ ) is the

electrical input. With the conventional ventilation technique there is electricity inputto pumps and fans. In the calculations it is neglected since the same equipment areused in the other techniques. Therefore only the increased electricity input wascalculated for the mechanical heat pump and the open absorption system. When using

12

the mechanical heat pump the electricity input means supplied electricity to thecompressor and when using the open absorption system supplied electricity to thegenerator.

The required heat for the hot water to the showers was on average 17.8 kW duringone week. Measurements show that the annual heat demand for the hot waterproduction is 161 MWh/year.

Ð Ñ Ò Ñ Ó Ô Õ Ö × Ø Ù Õ × Ú Ö Ô × Û Ü Ý Þ ÜCalculations including the mechanical heat pump were carried out in a similar way

as the conventional ventilation technique. Data for the mechanical heat pump wereobtained from the manufacturer.

The electricity supply to the compressor is constant 8 kW, but only 6.25 kW isused as work on the refrigerant. The remainder consists of losses [1]. The condenserheat to the water cooled condenser is 16.7 kW and 2 kW to the air circuit, hence themechanical heat pump is giving off a total of 18.7 kW heat. That gives a COP of 2.3for the mechanical heat pump. The airflow through the evaporator was set to 2400m3/h.

The only difference in the calculations compared to the conventional ventilationtechnique was that the outdoor airflow was set to zero during closed hours. Thatimplies that the air circulated (point B), to the facility has the conditions that occurfrom mixing of the air through the evaporator and the air by-passed (point F).

Since the airflow through the evaporator and the heat transferred from the air to theworking medium was set constant, the enthalpy on the air leaving the evaporator wascalculated according to

ß àáß àâ á ã ä å ä æ çáéèêëë

,

ìì

−= (J/kg dry air) (17)

The cooling of the airflow leads to condensation of water vapor. The air after theevaporator (point E) is saturated (see Fig. 4). Using psychometric data, a function(T=f(h)) was established. The temperature on the air leaving the evaporator was thencalculated using

íîï ðñí ð òîï ðñòíîñí ð òîñó òñô ðõööööö

43

2

089457705103

0055349068292097275

⋅⋅+

⋅⋅ (°C) (18)

When the temperature and enthalpy of the air was known the humidity ratio wascalculated using Eq. (3).

The airflow through the evaporator and the by-pass flow are mixed and theconditions on the air circulated were calculated using energy and mass balances (pointF). The required heat to the air heater was calculated using Eq. (13) and reduced bythe condenser heat. The required heat to the pool was calculated using Eq. (14) andreduced with the part from the water-cooled condenser.

The required heat for the hot water production was set to 161 MWh/year and thetotal energy demand was calculated using Eq. (16).

13

÷ ø ù ø ú û ü ý þ ÿ � � � û � � � ý � � � � ü �The calculations when the open absorption system was installed were also made

with outdoor airflow set to zero during closed hours. All the air is passing through theabsorber hence the air circulated (point E, F), has the same conditions as the air at theabsorber outlet. The relative humidity of the air leaving the absorber is dependent onthe concentration of the absorption solution. Therefore the minimum outdoor airflowcan be used since the required condition can be achieved by altering the saltconcentration. During opening hours the outdoor airflow was therefore always set tothe minimum airflow, in order to optimize the heat recovery. During opening hours,the humidity ratio of the circulation air was calculated using Eq. (9) since thehumidity ratio in the outdoor air and to the facility were known.

The air is cooled through the absorber. To calculate the enthalpy of the air at theabsorber outlet the cooling demand must be known. The cooling demand wasestablished from measurements and the enthalpy after the absorber was establishedusing Eq. (19)

� � � �� � � � � � �� � � ���������

−= (J/kg dry air) (19)

Measurements showed that the heat demand for the hot water production ( � �� � ) inEq. (19) was varying roughly between 2 kW to 42 kW at minimum and maximumload [9]. The required heat was therefore set constant to three different values (2, 27and 42 kW) at the different evaporation rates, (see Fig 7). The measurements alsoshowed that the total heat supplied to the heat storage was 120 MWh/year. Since thetotal heat demand for the hot water production was 161 MWh/year during one year,an additional 41 MWh/year has to be supplied by an additional heater.

When the conditions of the circulation air were established the heat demand for theair heater and pool was calculated in the same way as for the other techniques.

The supplied heat to the generator is dependent on the evaporation rate accordingto Eq. (20).

( )� � ! " # � $� � ! " # � $%� � ! " # � $& ' �( ) $ *+,-,. ∆⋅⋅+⋅= ,

///(W) (20)

where (cp,solution ) and (∆Tsolution ) are properties for the absorption solution. Accordingto measurements the heating of absorption solution, solvent heat and transmissionlosses from the generator are less then 10% of the supplied heat [6]. Therefore a factorof 1.1 has been added to the first term in Eq. (20).

14

Fig. 7. Required heat for the hot water production during one typical week

0 1 2 3 4 5 6 7 4

Fig. 8 shows how the outdoor temperature is varying during a week from Monday00.00 to the next Sunday 24.00. That is 168 hours and most of the figures will showvariations during a week or during a day in order to show the fluctuations.

Fig. 8. Variation of the outdoor temperature during one week8The evaporation rate is varying during the week due to the activity in the pool and

climate in the facility. Fig. 9 shows the evaporation rate during one week of the year.The three different evaporation levels (max, mean, min) can be seen. The differentrates are approximately 47, 35 and 15 kg water/hour respectively.

60

50

40

30

20

10

0

Supp

lied

hea

t to

hot w

ater

(kW

)

160140120100806040200Time (hours)

Monday Tuesday Wednesday Thirsday Friday Saturday Sunday

-25

-20

-15

-10

-5

0

5

Out

door

Tem

pera

ture

(C)

160140120100806040200Time (hours)

15

Fig. 9. Evaporation rate for an average week.

The lowest value (about 15 kg/h) of the evaporation rate occurs during closedhours. During opening hours the maximum rate (about 47 kg/h) and mean rate (about35 kg/h) are shown. If the minimum outdoor airflow of 10 m3/h and m2 pool area wasnot to be taken into consideration, the mean evaporation rate would be lower. Sincethe outdoor air has a lower humidity ratio compared to the circulation airflow, theevaporation rate is increasing during opening hours.

Fig. 10 shows the outdoor air ratio during one day. The figure shows that theconventional ventilation technique requires most outdoor air. With the openabsorption system the outdoor air ratio is the lowest. Minimum outdoor airflow canalways be supplied because the condition of the air out from the absorber is dependenton the concentration on the absorption solution.

Fig. 10. Outdoor air ratio during one day for the different techniques.

70

60

50

40

30

20

10

0

Eva

pora

tion

rate

(kg/

h)

160140120100806040200Time (hours)

Monday Tuesday Wednesday Thursday Friday Saturday Sunday

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Out

door

air

rati

o

242220181614121086420Time (hours)

Conventional techniqueMechanical heat pumpAbsorption system

16

Fig. 11 shows the required heat to the air heater in the air circuit for the differenttechniques. The annual heat demand for the different techniques are 235, 217, 230MWh/year respectively. The heat demand is decreasing for the mechanical heat pumpand the open absorption system. When the airflow is cooled in the evaporator thecirculated airflow has a lower specific enthalpy than the air out from the facility.Since a minimum outdoor airflow has to be supplied, a larger heat supply in the airheater will be required during opening hours. When the open absorption system isused the airflow is cooled through the absorber. Therefore heat has to be supplied tothe air, to remain constant temperature in the facility. That explains why the reductionof the required heat is rather small compared to the conventional ventilationtechnique. Without the cooling the absorption process would occur with constantenthalpy of the air (see Fig. 6).

Fig. 11. Required heat to the heater in the air circuit.

Fig. 12. Comparison of supplied heat to the air heater for the conventional technique and the mechanical heat pump.

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Fig. 12 shows how the required heat to the air heater in the air circuit is varyingduring one day when the mechanical heat pump is in operation. It shows that therequired heat is increased with the mechanical heat pump during mean evaporationrate. During maximum evaporation the required heat is of the same magnitude, sinceonly about 2 kW of the condenser heat is supplied to the airflow. The remainder issupplied to the water in the pool. During closed hours the reduction of the requiredheat is larger due to the reduced outdoor airflow.

Fig.13 shows the same comparison for the open absorption system and theconventional ventilation technique. With the open absorption system the required heatin the air heater is less then for the conventional ventilation technique during closedhours. During opening hours the required heat to the air heater is larger then for theconventional ventilation technique, since the cooling demand in the absorber is larger,the required heat in the air heater becomes larger.

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Fig. 13. Comparison of supplied heat to the air heater for the conventional technique and the open absorption system.

Since the airflow has to be cooled when the open absorption system is in operation,the temperature in the facility will increase when closed during a long period of time.When there is no load on the hot water production the heat is used to charge the heatstorage. Therefore the open absorption system has to be shut down during the summerwhen the facility is closed for a period of 10 weeks. During this period the heatdemand to the air heater was approximately 20 MWh/year.

The temperature in the absorber outlet is varying since the absorption rate isvarying. Fig. 14 shows how the temperature is varying during one week. The lowesttemperature on the air is approximately 17°C. The temperature on the absorptionsolution is almost the same, which implies that the hot water can be heated toapproximately 10-15°C during those hours. The temperature of the cold water is about5°C. Hence heat storage of 3 m3 is needed.

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To avoid freezing on the surfaces in the evaporator when the mechanical heatpump is used, the temperature must exceed the freezing point. Fig. 15 shows how thetemperature after the evaporator is varying during one week. The lowest temperatureis about 14°C hence freezing won’ t occur.

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Fig. 15. Temperature of the air after the evaporator.

The temperature after the evaporator is varying due to different airflow through theevaporator. During closed hours a larger airflow is by passed which leads to a smallerairflow through the evaporator and therefore a slightly higher temperature. The heatdemand for the pool is dependent on the evaporation rate from the pool andtransmission losses from the pool. In the comparison the evaporation rate is the samefor the different techniques, hence the heat demand for the pool remains the same.The annual heat demand is 215 MWh/year for the conventional ventilation technique.Since the biggest part of the condenser heat is used to preheat the pool water, the heatinput to the heater in the pool circuit is decreased by 60% to 87 MWh/year when the

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mechanical heat pump is used. Also when the open absorption system is used, thecondenser heat decreases the heat demand for the heater in the pool circuit. It isreduced by 25% to 160 MWh/year, since the condenser heat is smaller for the openabsorption system (see Fig.16). During the summer when the open absorption systemwas turned off the heat demand to the pool was 29 MWh/year.

Fig. 16. Supplied heat to pool heater for the different techniques.

The mechanical heat pump and the open absorption system both require electricalinput to the compressor and generator (boiler) respectively. Fig. 17 shows theelectrical input for the different techniques. The annual electrical input for themechanical heat pump is 63 MWh/year and for the open absorption system 53MWh/year.

Fig. 17. Electrical input for the different techniques

Only the increase in electrical input has been calculated. Due to pressure dropthrough the absorber, more fan power will be required, hence the electrical input forthe fans increases. The same goes for pressure drop through the evaporator andcondenser when the mechanical heat pump is used. The increased electrical input dueto increased pressure drop has not been taken into consideration in the calculations.The total annual energy demand for the different techniques i.e. heat supply to air airheater, pool heater, electrical input and heat supply to the hot water production are611, 528, 484 MWh/year respectively (see Fig.18). The reduction of the annual

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energy demand is approximately 14% for the mechanical heat pump and about 20%for the open absorption system.

For the conventional ventilation technique and the mechanical heat pump, 161MWh/year out of the total energy demand consist of heat to the hot water production.For the absorption system it is reduced to 41 MWh/year.

Fig. 18. Total energy demand for the different techniques.

However, the total energy demand is dependent on the climate in the facility. If theair temperature is decreased and the relative humidity remains the same theevaporation from the pool will increase rapidly when using the conventionalventilation technique. That requires an increased outdoor airflow and therefore anincreased energy demand for the facility (see Fig. 19).

Fig. 19. Supplied heat versus temperature in the facility.

The increased evaporation from the pool is larger than the reduction of thetransmission losses from the facility due to the lower temperature. Calculationsshowed that the energy demand decreased when the mechanical heat pump or theabsorption system was used and the air temperature was lowered. Since part of the

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latent heat in the exhaust air is recovered with both systems the total energy demandbecomes lower, due to the reduced transmission losses from the facility. However,since the minimum outdoor airflow has to be supplied during opening hours thereduction of energy is small for different temperatures. The relative humidity in thefacility is set in order to avoid condensation of vapor on cold surfaces. When thetemperature is decreased the dew point is also decreased which leads to that a higherrelative humidity can be allowed in the facility. Approximately 5% higher relativehumidity can be allowed at the temperature of 25.5 °C. Using different climates in thecalculations gave an annual energy demands according to Fig. 19. For instance, at25.5 °C the energy demand was 644 MWh/year for the conventional ventilationtechnique and 527 MWh/year and 452 MWh/year for the mechanical heat pump andthe open absorption system respectively.

These results show the importance of keeping the required climate in the facility.For instance if the control system is recording a higher relative humidity than the setpoint value, the energy demand could be increased considerable. At lowertemperatures it seems that the heat recovery systems are more beneficial than at theclimate used in these calculations.

9 : ; < = > ? = = < @ A

Mechanical heat pumps are a well-established technique to reduce the energydemand for indoor swimming pools, especially in the Scandinavian countries. Thereare different manufacturers on the market today, which has led to that the technique iswell developed and optimized. The price of the mechanical heat pump of the size usedin these calculations, fully installed is about 500 kSEK. At the climate used thedecreased energy demand was 83 MWh/year. The price of heat in Sweden today isroughly 500 SEK/MWh. A non-discounted cash flow would then give a pay back timeof about 12 years.

The open absorption technique has not been commercialized in this application,hence it is difficult to compare the investment of the techniques. However,measurements and calculations have showed that the absorption system reduces theenergy demand more then the mechanical heat pump. At the same working conditionsas the mechanical heat pump the reduced energy demand was 127 MWh/year. If theinvestment cost for the open absorption system would be in the same magnitude as themechanical heat pump, the investment would be more beneficial. However, theinvestment for a fully installed absorption system is probably larger. To compare theinvestment for the two heat recovery systems, a more careful economic analysis has tobe made.

In the absorption system heat is supplied to the boiler. In the pilot plant electricitywas used but other heat sources are possible to use. For instance, methane and naturalgas etc, which make the technique flexible.

The water-cooled condenser in the mechanical heat pump was designed to give off16.7 kW heat. That is based on a water temperature of 25°C and a temperature raiseof 13 °C on the water. During operation these conditions might not be constant. In thecalculations this has not been taken into consideration.

The evaporation was approximated but the duration of the different evaporationrates was established from statistical data and is considered reliable. The heat transfercoefficients are not constant but have been established earlier and are generallyaccepted by manufacturers.

22

The minimum outdoor airflow is established from rules of thumbs and is muchlarger then the rules and regulations. Perhaps the minimum outdoor airflow could bedecreased and therefore reduce the energy demand for the facility, which leads to thatthe heat recovery systems will be more beneficial.

To make an adequate comparison the increased electricity input due to pressuredrop when using the mechanical heat pump and the open absorption system should betaken into consideration. Electrical input for fans and pumps has not been included inthe calculations.

If the open absorption system is used, cooling of the airflow is required preventinga temperature rise in the facility. Since there is no hot water demand when the facilityis closed heat storage is required. During the summer the system has to be shut downsince the facility is closed for a longer period of time, which has been taken intoconsideration in the calculations.

The amount of heat saved by using the heat recovery systems could be consideredlow. This is due to that transmission losses from the facility and the heat to the hotwater production stands for approximately 40-50 % of the total energy demand. Usingthe heat recovery techniques will not reduce the transmission losses from the facilityand the required heat to the hot water production is only decreased when using theopen absorption system. Therefore the heat saved is less then expected. The minimumoutdoor airflow during opening hours is rather high. That will also lead to that thesaving of energy by introducing heat recovery system is limited.

The largest source of errors in the calculations is the estimations of the differentevaporation rates. The evaporation is varying due to the temperature in the pool, thetemperature of the air, the relative humidity, water movements (activity in the pool).That is a lot of parameters and therefore the estimation of the evaporation rates isdifficult to make. However, since the three evaporation rates have been established inearlier work through measurements and are accepted by manufacturers theevaporation rates used here should be fairly accurate. The variation of the coolingdemand is another source of error. Since that was also established throughmeasurements they can be considered fairly accurate.

The equations used are based on well established psychometric theory and theaccuracy of the equations used is in the range of 2-3%. The assumptions made usingthis theory are mainly that moist air is considered a perfect gas and that the propertiesof the air, i.e. specific heat and density is constant in the temperature and pressurerange used in the calculations. The accuracy of the calculations should therefore be inthe range of +/- 10-15%.

Since the climate in the facility highly effects the energy demand, an increased airtemperature in the facility can save a lot of energy without investment of heatrecovery systems, i.e using the conventional technique. A general accepted fact todayis to use a temperature of the air that exceeds the temperature in the pool by 2°C.However, calculations showed when using the a mechanical heat pump or an openabsorption system a lower air temperature leads to a larger energy saving compared tothe conventional technique (see Fig. 19).

B C D E F G H I J K E F J

The annual energy demand is decreased considerable using both the mechanicalheat pump and the open absorption system. With the mechanical heat pump the

23

energy demand is reduced with 14% and with the open absorption system the energydemand is reduced by 20%

Even though the absorption system is able to reduce the energy demand more thanthe mechanical heat pump the latter technique have to be considered advantageoussince the technique is well established and is produced by a lot of differentmanufacturers. Calculations hour by hour is not necessary to calculate the annualenergy demand, but is a necessity to show the fluctuations of supplied heat during theyear.

The prizing of heat and electricity are of the same magnitude in Sweden today.Hence both alternatives reduce the total cost. If the electricity price would rise in thefuture it has to be considered that the heat demand is reduced but the electricitydemand is increased using both the mechanical heat pump and the open absorptionsystem. The energy demand using the conventional ventilation technique can bedecreased by increasing the temperature in the facility and should be considered as thefirst measure to reduce the energy demand.

L M N M O M P Q M R[1] Munters CO, S T U V W T X Y V Z [ Z \ ] T U V T ^ , Sweden, (In Swedish).[2] Westerlund L, Dahl J and Johansson L. _ ` a b c d b ` e f g h e i j b k ` e l g ` e c i c m ` a b a b g ` n b o g i n m c dp q r h e f r g ` a k s Energy, Vol 21, No. 7/8, pp. 731-737, 1996.[3] McQuiston F, Parker Jt u v w x y z { | } v z x y ~ w x y z { w z � w y � � � z � y x y � z y z { t � z w ~ � � y � w z � � v � y { z t John

Wiley & Sons Inc, Fourth Edition, Canada, 1994.[4] Westerlund L, Dahl J. � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � . Applied Energy, Vol. 48, pp.

33-49, 1994.[5] Johansson L, Westerlund L. � � � � � � � � � � � � � � � � � � �   � ¡ ¢ � � � � � £ � � ¤ ¥ � � ¦ � � ¥ � § � � ¦ � � � � § � § � � ¨ .

Applied Energy, Vol.67, pp. 231-244, 2000.[6] Westerlund L, Dahl J. © ª « ¬ ­ ® ¯ ° ± ª ² ³ ° ¬ ¯ ´ ¯ ² « µ ¶ · ¸ ª « ± ³ µ « ¬ ² ­ ¹ ¯ ² º » ´ ³ ¬ ­ ¹ ­ ® ° ± ­ ² ° ± ´ ª ³ ¹ ° ²ª ¹ ­ ¬ ² ¼ Applied Energy, Vol. 38, pp. 215-229, 1991.[7] Westerlund L, Dahl J, Hermansson R. ½ ¾ ¿ À ¿ Á  à ¿ Ä Ä À Å ¿ Á Ä Æ ¾ Å Ä Ç Ã È É ¿ À Ç Ê Á Ä Ê Æ À Ë ¾ ¿ Ì Ä Ê Å Í À Ç Ê ÁÍ Å Ê Î ¾ Ä Ä Ç Á ¿ Í ¿ Î Ï ¾ Â Ì ¾  ¿ Ì Ä Ê Å Ì ¾ Å Ð Applied Thermal Engineering, Vol. 18, pp. 1295-1308,

1998.[8] Pettersson F, Ñ Ò Ó Ô Õ Ö × Ø Ù Ú Ö Û × Ô Õ Ü Ý Þ Ú ß Ù Õ à á â ã Û á ä . Sweden, 1994 (in Swedish).[9] Westerlund L, Dahl J. å æ ç è é ê ë è ì ç ë ê í æ è î ì ï ð è ë ñ ç ê ï ì ò ó ì é è î ç ë ç î ô õ ö è ë æ ç î ÷ ê ï ð è ë ð ë êì ò í ø ð ö æ ù ð ó ó ð ë ô ì è è ø . Applied Energy, Vol. 49, pp. 275-300, 1994.