Increase in Evaporation Caused by Running Spa Jets swhim.com

Increase in Evaporation Caused by Running Spa Jets swhim.com Nomenclature A pipe cross-section area, m D water inlet diameter of the venturi tube nozzle, mm diameter of small end of the throat of the venturi tube nozzle, mm D 1 D 2 diameter of large end of the throat of the venturi tube nozzle, mm d air inlet diameter of the venturi tube, mm E rate of evaporation from spa, kg h -1 m -2 L pipe inside diameter, m venturi tee inside diameter, m L vt ln natural logarithm ф relative humidity, % 100-1 p atm standard atmospheric pressure, 10132 Pa p ws saturation pressure of water vapor in air at given temperature, Pa partial pressure of water vapor in air at given temperature, Pa p w ρ H2O density of water in pipe, kg m -3 ρ r density of moist air at room temperature and humidity, kg m -3 ρ w density of moist air saturated at surface water temperature, kg m -3 Q v air flow entering venturi tube air inlet, L s -1 Q w water flow entering venturi tube water inlet, L s -1 R universal gas constant, 8314.41 J kg -1 K -1 R da gas constant for dry air, J kg -1 K -1 R w gas constant for water vapor, J kg -1 K -1 Re Reynolds number T temperature, K (unless specified C) μ dynamic viscosity of water in pipe, N s m -2 v specific volume of dry air, m 3 kg -1 V velocity of water in pipe, m s -1 W r specific humidity of moist air at room temperature and humidity, kg moisture kg -1 air specific humidity of moist air saturated at water surface temperature, kg moisture kg -1 air W w Subscripts atm atmospheric r at ambient room temperature and relative humidity w at water temperature and saturation s saturated at given temperature. 1

Assumptions 1. 'Typical' spa is assumed to have the following conservative measures: a) Area = 6 m 2, b) Average Water temperature = 38 C c) Jets = 4 d) Pumps = 1 e) Jet Pump Timer = 15 min. max. f) Plumbing = 2 in. g) Tap Water temperature = 20 C 2. Spa Location = California - Indoor (outdoor spas experience significantly greater evaporation due to wind). 3. The average temperature of tap water equals the average temperature of the climate. 4. Standard atmospheric pressure = 101325 Pa 5. Average Air humidity = 60% 6. Average Air Temperature = 25 C 7. Ambient airflow over water of unjetted spa is the same as unoccupied jetted spa. 8. Unjetted spa water surface is undisturbed (i.e., indoor with no wind). 9. Jets run for a total average of 30 min/day while spa is unoccupied (i.e., as few as 3 people leaving the spa unoccupied with 10 min. remaining on the timer). 10. The spa is open to the public an average of 30 days month -1. 11. No blower is used to create bubbles. I. Introduction Spas are major energy consumers, and, by virtue, they represent a significant opportunity for energy conservation. The vast majority of energy consumed by spas is to counteract the effects of evaporation. Evaporation occurs when liquid molecules near the surface experience collisions that increase their energy above that needed to overcome the surface binding energy. When those higher energy molecules break away from the water into vapor, the remaining water molecules move slightly slower. Thus when water evaporates, the spa drops in temperature. As a result, the spa needs to be reheated to the intended temperature. Evaporation from indoor spas impose a greater concern by increasing the load on HVAC systems, shortening HVAC and spa equipment life-cycles, as well as the usable life of building materials. The evaporation also increases the risk of exposing users to chemical and microbial hazards. 1 1 See generally, World Health Organization (2006) Guidelines for safe recreational water environments. Volume 2, Swimming pools and similar environments. http://www.who.int/water_sanitation_health/bathing/srwe2full.pdf. Last accessed May 1, 2015. 2

There is no known prior study of the effects of spa jets on evaporation. II. Baseline Evaporation The baseline evaporation is the amount of water evaporating when the spa jets are off. Many empirical studies since the late 19 th century model evaporation from pools. Recently, computer Computational Fluid Dynamics software has been coupled with these empirical studies to determine the best formula for swimming pool evaporation. 2 One such study compared the most commonly cited evaporation formulas and concluded M.M. Shah's model for evaporation from pools and spas to be most reliable. 3 Shah's formula has also been validated by the American Society of Heating, Refrigerating and Air-conditioning Engineers (ASHRAE). 4 While Shah has further developed his well known formula for evaporation from unoccupied pools to more precisely account for ambient air flow, because we assume ambient airflow to be constant between our unjetted and jetted spa comparisons, we will use Shah's simpler non-airflow formula: 5 E = 35ρ r (ρ r ρ w ) 1/3 (W w W r ) (1) The specific density of moist air (ρ), the mass of moist air per volume of moist air, is the inverse of specific volume (v) of moist air, the volume of moist air per mass of moist air: 6 ρ = (1/v) (2) The specific volume (v) of moist air is: 7 v = RT / 28.9645(p atm p w ) (3) The pressure of water vapor (p w ) relates to the assumed relative humidity (ф) as: 8 2 See generally, Li, Z., Heiselberg, P. (2005): CFC Simulations for Water Evaporation and Airflow Movement in Swimming Baths. Instituttet for Bygningsteknik: Aalborg Universitet. http://vbn.aau.dk/files/19914391/cfd_simulations_for_water_evaporation_and_airflow_movement_in_swimming_baths. Last accessed May 1, 2015. 3 Id. at p. 25. 4 See generally, Shah, M.M., (2012): Calculation of Evaporation from Indoor Swimming Pools: Further Development of Formulas, ASHRAE Transactions 2012, Volume 118, Part 2. http://mmshah.org/publications/pool %20Evap%202012%20ASHRAE%20Trans.pdf. Last accessed May 1, 2015. 5 Id. Equation (3). 6 National Aeronautics and Space Association. https://www.grc.nasa.gov/www/k-12/airplane/specvol.html. Last accessed May 1, 2015. 7 ASHRAE Fundamentals Handbook 2001, Chapter 6, Equation (11). http://systemssolution.net/cadtechno/0%20sample/specs%20&%20details/books%20mechanical/hvac/ashrae %20HVAC%202001%20Fundamentals%20Handbook.pdf. Last accessed May 1, 2015. 8 Id. Equation (13). 3

ф = p w / p ws (4) The saturation vapor pressure (p ws ) at temperature (T) is: 9 Where 3 ln p ws = C i + (C 4 lnt ) (5) i = 1 C -1 = 5.800 220 6 E+03 C 0 = 1.391 499 3 E+00 C 1 = 4.864 023 9 E 02 C 2 = 4.176 476 8 E 05 C 3 = 1.445 209 3 E 08 C 4 = 6.545 967 3 E+00 And the humidity ratio (W) is: 10 W = 0.62198 p w / (p atm p w ) (6) Thus, using Equation (5), the saturation pressure of moist air saturated at the assumed ambient room temperature of 25 C (p rs ) and the air saturated at the assumed water temperature of 38 C (p ws ) are: p rs = 3169.216447 Pa p ws = 6631.472431 Pa Using Equation (4), the water vapor pressure at room temperature and humidity (p r ) and the saturated water vapor pressure at water temperature, respectively, are p r = (0.60)3169.216447 = 1901.529882 Pa p w = (1)6631.472431 = 6631.472431 Pa Using these pressure values into Equation (6), the moist air humidity ratio at room temperature and humidity (W r ) and the saturated water vapor humidity ratio at water temperature (W w ), respectively, are: W r = 0.00987605 kg kg -1 W w = 0.04355782 kg kg -1 Using the pressure values in Equation (3), the moist air specific volume at room temperature and humidity (v r ) and the saturated water vapor specific volume at water temperature (v w ), respectively, are: v r = 101325(25+273.15) / 28.9645(101325 1901.529882) 9 Id. Equation (6). 10 Id. Equation (22). 4

v w = 101325(38+273.15) / 28.9645(101325 6631.472431) Using Equation (2), the respective specific volumes can be converted to specific densities: ρ r = 1.16168586 ρ w = 1.06019334 Using these specific densities and humidity ratios in Equation (1), the typical spa's evaporation under the assumed conditions, while jets are off, is: E = 35(1.16168586)(1.16168586 1.06019334) 1/3 ( 0.04355782 0.00987605) E = 0.54803 kg h -1 m -2 III. Evaporation Caused by Running Jets Aeration is the bubbling of water (like jetting a spa). Aeration promotes evaporation in two ways: within the bubbles and at the surface where the bubbles burst. The greater the airflow rate, the greater the increase in evaporation due to aeration. 11 Spas typically create jetted bubbles by using the venturi effect. Here, we conservatively assume the more energy efficient venturi effect is the only method used to create bubbles. Venturi tubes use converging and diverging tube angles to change water pressure and velocity (as shown in Figure 2). Venturi jets have an air inlet which pulls air into the jet mixing the air and water to produce jetted bubbles (as shown in Figure 1). When the water inlet angles reduce the water pressure below atmospheric pressure, a pressure vacuum pulls air into the tube. The air and water join in the diverging portion of the tube where velocity is converted back to higher pressure air/water mixture which emerges into the spa as bubbled water. Fig. 1 Venturi effect: mixing air with water to form jetted bubbles. 11 See generally, Helfer, F., Lemckert, C., Zhang, H.. (2012): Influence of bubble plumes on evaporation from non-stratified waters. Journal of Hydrology, 438-439, 84-96. http://www98.griffith.edu.au/dspace/bitstream/handle/10072/46754/78054_1.pdf?sequence=1. Last accessed May 1, 2015. 5

A venturi tube's air/water volume mixing depends on the Reynolds number, the relative sizes of the air and water inlets, and the throat diameter. 12 For Reynolds numbers between 1x10 5 and 5x10 5, the typical venturi air/water flow ratios range between 0.1 and 0.5 with the most efficient designs approaching 0.6. 13 Fig. 2 Specifications of a typical public spa venturi tee. Airflow (Q v ) in venturi tees is: 14 Q v / Q w = X 1 (Re x 10-4 ) x 2 (d/d2 ) x 3 X4 (Re x 10-4 ) X5 (d/d2) (8) The D 1 / D 2 ratio for a typical public spa venturi is near 0.5 (where D 1 = 9 and D = 17 based on the typical public spa venturi tee Figure 2 15 ). Where D 1 / D 2 = 0.50, as in the typical venturi tee, the constants in Equation (8) are 16 X 1 = 0.481 X 2 = 1.242 X 3 = 0.506 12 See generally, Baylar, A., Ozkan, F., and Unsal, M., (2010): Effect of Hole Diameter of Venturi Tube on Air Injection Rate, KSCE Journal of Civil Engineering, 14(4), pp. 489-492. https://www.researchgate.net/publication/225538525_effect_of_air_inlet_hole_diameter_of_venturi_tube_on_air_injection_rate. Last accessed May 1, 2015. 13 Id. at Figures 3-6. 14 Id. at Equation (5). 15 Lasco Fittings Inc. Venturi Tee part # 473210 with inserted nozzle Part # 11125. 16 Id. at Table 1. 6

X 4 = 0.833 X 5 = 0.162 The Reynolds Number (Re) of the water flowing through the typical spa venturi tee is found using; 17 Re = ρ H2O VD/μ (9) The density of the water (ρ H2O ) is: 18 Where 5 7 ρ H2O = ( F i T i ) / ( F i T i 6 ) (10) i = 0 i = 6 F 0 = 0.240 336 020 1 E+04 F 1 = 0.140 758 895 E+01 F 2 = 0.106 828 765 7 E+00 F 3 = 0.291 449 235 1 E 03 F 4 = 0.373 497 936 E 06 F 5 = 0.212 037 87 E 09 F 6 = 0.342 444 272 8 E+01 F 7 = 0.161 978 5 E 01 Using Equation (10), the density of the water (ρ H2O ) at the assumed 38 C is: ρ H2O = 992.97626 kg m -3 The dynamic viscosity (μ) is: 19 μ = 0.6612(T 229) -1.562 Thus, the dynamic viscosity of water in the pipe (μ) at our assumed 311.15 K (38 C ) spa water temperature is: μ = 0.6612(311.15 229) -1.562 = 0.000675641 N s m -2 A typical public spa pump's maximum speed is 3450 RPM. 20 While spa jets typically run at the highest speeds, we assume here the pump runs at the lower 3110 RPM factory preset speed. 21 At 17 Bergman, T. L., Incropera, F. P., (2011): Fundamentals of Heat and Mass Transfer, 7 th Edition, Wiley, Hoboken, New Jersey, Equation (8.1). http://www.ualberta.ca/~seyedsha/fundamentals%20of%20heat%20and%20mass %20Transfer-Incropera.pdf. Last accessed May 1, 2015. 18 Hyland, R.W., Wexler, A., (1983): Formulations for the Thermodynamic Properties of the Saturated Phases of H 2 O from 273.15 to 473.15 K, ASHRAE Transactions, 89, Part IIA, 550-519, Equation (5). 19 Gawin, D., Jajorana, C.E., and Schrefler, B.A. (1999): Numerical analysis of hygro-thermal behavior and damage of concrete at high temperature. Mechanics of Cohesive-frictional Materials, 4:37-74. Equation (51). 20 Id., Performance Curves link. 21 Id. 7

this speed, the pump in a typical installation is taken to have a flow rate of about 20 m 3 h -1. 22 Thus, Q w = (20 m 3 h -1 )(1/3600 h s -1 ) = 0.005556 m 3 s -1 The typical pipe cross-section area (A) is: 23 A = πr 2 = π(d/2) 2 = 3.1415(48mm(1m/1000mm)/2) 2 = 0.0018095 m 2 The water velocity (V) in the pipe before reaching the venturi tee is: 24 V = Q w / A V = Q w / A = 0.005556 m 3 s -1 / 0.0018095 m 2 = 3.07045 m s -1 The pipe diameter (D) is: D = (17mm)((1m/1000mm) = 0.017 m Using the derived values ρ H2O = 992.97626 kg/m 3 V = 3.07045 m s -1 D = 0.017 m μ = 0.000675641 N s m -2 in Equation (9), the Reynolds Number (Re) of the water flowing through the typical spa venturi tee is: Thus, using Re = 76,714 D 1 = 9 mm D 2 = 17 mm d = 21 mm in equation (8) Re = (992.97626)(3.07045)(0.017) / 0.000675641 = 76,714 Q v / Q w = 0.481(76,714x10-4 ) 1.242 (13/17) 0.506 (0.833) (76,714x10-4 ) (0.162) (13/17) = 0.3241 (this 0.3241 air to water flow ratio (Q v / Q w ) of a typical spa's venturi tee where the outlet to inlet ratio (D 1 / D 2 ) is near Baylar's tabled 0.50 and the Reynolds number is near Baylar's 75,000 22 Id. 23 ASHRAE Fundamentals Handbook 2001, Chapter 2, Section 2.8 on Flow Analysis. http://systemssolution.net/cadtechno/0%20sample/specs%20&%20details/books%20mechanical/hvac/ashrae %20HVAC%202001%20Fundamentals%20Handbook.pdf. Last accessed May 1, 2015. 24 Id. 8

agrees well with Bayar's findings. 25 ) Thus, Q w = (20 m 3 h -1 )(1000 L m -3 )(1/3600 h s -1 ) = 5.556 L s -1 Thus, the typical spa pump airflow is: Q v = (0.3241)(5.556 L s -1 ) = 1.8 L s -1 Helfer measured water evaporation increases, on an absolute basis relative to baseline evaporation, by 200% 26 when bubbling 0.19 L s -1 into a tank with a surface measuring 0.90 m by 1.80 m (the air diffuser side) containing 22 C water. 27 Helfer further found that the evaporation caused by bubbling increases as the water temperature increases 28 (this reflects the increased saturation vapor density caused by higher temperatures as well as the increased convection of water molecules due to the higher temperatures). Here, though our assumed spa water temperature of 38 C is 16 C higher than the 22 C tested by Helfer, and though we know the evaporation is greater at the higher temperature, we conservatively assume the evaporation increase is the same as that measured by Helfer at the lower temperature. More significantly, Helfer calculated the evaporation due to bubble saturation when bubbling airflow at 0.19 L s -1 in water at 22 C to be 0.12 L/day. 29 Given Helfer's tank (the side with the air diffuser) measures 0.90 m by 1.80 m, 0.12 L/day water loss is: 0.12 L/day = 0.12 L/day(0.001 m 3 /L) = 0.00012 m 3 /day From Helfer's tank (the air diffuser side) dimensions we can calculated the water loss to be: 0.00012 m 3 /day / (0.90 m)(1.80 m) = 0.000074074 m/day = 0.074074 mm/day Helfer found the total increase in evaporation when bubbling airflow at 0.19 L s -1 in water at 22 C to be 0.80 mm/day. 30 Thus, the ratio of evaporation due to bubble saturation to the total increase in evaporation is: (0.074074 mm/day) / (0.80 mm/day) = 0.0925925 Therefore, less than 10% of the evaporation increase caused by bubbling is due to evaporation into the bubbles. Over 90% of the evaporation increase is due to the physical effects of bubbles 25 See generally, Baylar, A., Ozkan, F., and Unsal, M., (2010): Effect of Hole Diameter of Venturi Tube on Air Injection Rate, KSCE Journal of Civil Engineering, 14(4), pp. 489-492. https://www.researchgate.net/publication/225538525_effect_of_air_inlet_hole_diameter_of_venturi_tube_on_air_injection_rate. Last accessed May 1, 2015. 26 Id., Figure 4(b). 27 Helfer, Figure 1. 28 Id., Figure 3. 29 Id., Figure 10. 30 Id., Figure 4(a). 9

bursting at the surface. Bubbles cause small capillary waves at the surface which increase the surface area of the water, and evaporation is directly proportional to surface area. Bubbles also cause surface winds which increase evaporation; these Helfer measured to be 0.29 m s -1 at a height of 2 cm from the water surface when 0.19 L s -1 is bubbled in 22 C water. 31 Helfer's experimental apparatus had a single diffuser which bubbled vertically from the bottom of the tank. 32 The typical spa, on the other hand, has jets pushing bubbles horizontally before they emerge at the surface. This difference has two significant effects: (1) While the bubbles break the surface the same in both setups, the horizontal force of the jets, however, create a form of wind in the relative motion between the water and the air. (2) The bubbles in a horizontally jetted spa break the entire surface area of the spa (rather than just above the vertically rising bubble diffuser 33 ). As mentioned, Helfer measured a greater net evaporation due to bubbling when the water temperature was increased from 20 C to 22 C. 34 These differences cause substantially greater evaporation due to bubbling in the spa at 38 C than that measured by Helfer. As mentioned,helfer bubbled a tank with a surface area of 1.62 m 2 surface area with a single airflow diffuser which caused surface winds over only half of the tank's surface area (the winds that account for over 90% of the increase in evaporation due to bubbling). 35 But even if the typical spa's bubbles emerged vertically from the bottom (having less effect on the surface than horizontal jets do), because the assumed typical spa has 6 m 2 surface area and 4 jets, each jet would bubble a surface area of 1.5 m 2. Thus, because more of the surface area is perturbed by spa jets than Helfer's diffuser affects, the relative increase due to bubbling in a spa will again be higher on this account than that measured by Helfer. But for simplicity, and to be conservative, we extrapolate Helfer's measurements for evaporation due to bubbling at Helfer's various tested airflows by using the typical spa jet's airflow. The 1.8 L s -1 airflow calculated for the typical pump, on a per jet basis, is: Q v = (1.8 L s -1 pump -1 )(1 pump/4 jets) = 0.45 L s -1 jet -1 Figure 3 charts the extrapolated 436 % increase in evaporation (relative to baseline) at the typical spa's calculated per jet airflow is 0.45 L s -1 jet -1. 31 Id., Figure 8. 32 Id., Figure 1. 33 Id., Figure 7. 34 Id., Figure 4. 35 Id., Figure 7. 10

Fig. 3 Extrapolated evaporation increase for typical four-jet spa with airflow at 0.45 L s -1 is 436%. Given the typical spa's absolute increase from baseline evaporation is 0.54803 kg h -1 m -2 (found in Section II), and the typical spa's extrapolated evaporation increase is 436 % (found in Section III), the total increase in evaporation, from baseline, is: (0.54803 kg h -1 m -2 )(436/100) = 2.3894 kg h -1 m -2 For our assumed 6 m 2 spa, that is (2.3894 kg h -1 m -2 )(6 m 2 ) = 14.3364 kg h -1 For our assumed 30 min/day of jetting while spa is unoccupied, the amount of water that evaporates is: (14.3364 kg h -1 )(0.5 h day -1 ) = 7.1682 kg day -1 IV. Conclusion There is a more than five-fold increase in evaporation in typical public spa when the jets run. For such a spa, that amounts to over 7 kg day -1. 11