∫∫∫ Fountain Cooling

Updated from earlier post – thank you Emma for your helpful feedback! If you have questions, you may always email us at info {at} carbonlighthouse.com. We’ll do our best to answer as soon as we can!

Roof top cooling towers are a critical aspect of energy efficiency in buildings. They can provide a way to remove heat that is “free” and environmentally-charming (environmentally-friendly is getting to be a boring phrase).

For the following exercise, imagine a 10-story office building. The cleverness of cooling towers starts with a loop of water that runs from the cooling tower into the building and back up again. Once inside the building, the cold water then runs over hot air or hot refrigerant liquids. This interaction heats up the cold water and cools off the air or refrigerant liquid. The warmed water is then pumped up to the rooftop cooling tower, sprayed over a series of metal fins, and blown by gigantic fans that cause the hottest water particles to evaporate. What is left behind is cool water that is cycled back into the building.

This turns out to be an extremely efficient way to provide cooling, because a Big Old Chunk of energy intensive work is happening by evaporation. N.b.: Big Old Chunk is not a standard unit of measurement. And because the atmosphere, not electricity or gas, is responsible for evaporation, the utility cannot charge you for it.

In some buildings, however, the roof is not large enough to hold a cooling tower that can meet the building’s entire cooling needs. What if, though, a nearby body of water could be used for cooling?

Recently, an office complex approached Carbon Lighthouse because they wanted to determine if they could use the fountain in front of their high-rises as an additional cooling tower. After all, the fountain was already built and already constantly circulating water. Could it be connected to the building to carry away some excess heat? If so, how much?

Other buildings have used large bodies of water as cooling towers. The City of Toronto undertook a large project to this end. We have heard unsubstantiated reports that the Bellagio hotel in Las Vegas uses its famous fountains for water-cooling, as well. (If they do not, they probably should.)

What follows is a discussion of how we approached and solved this fun energy problem. The magic of science and engineering happens in the equations, so this will be a quantitative overview. If you ever wanted to be a magician, keep reading.

Before we begin the legerdemain, let’s review one piece of nomenclature.

A refrigeration ton (RT) is a standardized unit of cooling. It comes from days of yore when letting a block of ice melt cooled buildings. In melting, the ice absorbed a large amount of heat from the environment. An RT is the amount of heat that a block of ice weighing a ton “removed” from the building via cooling over a 24-hour period. It is equivalent to 12,000 BTU/hour. Cooling towers are rated in refrigeration tons. So the answer we are looking for regarding the fountain will also be in RT.

Okay, onto the good stuff. When analyzing heat flows in a situation like this there are two important things to consider: 1) heat being released into the earth through the fountain’s base and walls and 2) evaporation from the fountain surface. We will take these one at a time.

Heat radiating from the fountain into the earth

The situation we were faced with involved redirecting 80°F condenser heated water moving at 400 gallons per minute into a 15,000 gallon fountain. Given the high flow rate, within an hour of pumping the water the fountain will heat up to close to 80 degrees and stay there as long as the hot water keeps flowing. Heat will radiate through the concrete walls and into the earth, but not nearly fast enough to cool the water. (Imagine filling your bathtub with cold water. Once it is full, you switch the tap to hot water but also open the drain. Water flows in and out at the same rate, but over time the water becomes very hot and remains that way).

To determine the steady state heat flow from the fountain into the earth, we will use the (first magical) formula:

q = U · A · (T2 – T1) where:

  • q is the heat released (or removed)
  • U is the overall heat transfer value and
  • A is the area of the fountain touching the earth (base and sides)
  • T2 is the temperature of the earth, 57°F
  • T1 is the steady state temperature of the water, 80°F

This equation tells us how quickly heat radiates outward across the area, A, from the fountain boundaries and into the earth.

The formula can be derived in a number of ways, but the easiest way to think about it is qualitatively: heat transfer should depend on the temperature difference between two materials, the area of the material transferring the heat, and some aspects of the material itself (e.g. aluminum will transfer heat very differently than, say, a tree). To make the units work out we measure ‘U,’ the intrinsic heat transferring property of any given material, in such a way that the units work out.

Another aspect of this magical equation that a thermodynamically informed reader will notice is that there are three types of heat transfer – conductive, convective, and radiative – yet this equation seems to represent all of them. Furthermore, radiative heat transfer is related to the difference in temperature raised to the fourth, not linearly, and this equation does not capture that. That is true, but this equation is still an amazingly good approximation for a reasonable temperature range.

Back to the fountain: U is determined from the thermal conductivity of the concrete and its thickness, and happens to be equal to ~0.2. The relevant area, A, where heat exchange is taking place is determined from the as-built drawings and is about 2100 ft2. The change in temperature is (80-57)=23°F. Multiplying it all together we get q is about 10,000 BTU/hour. About 0.8 tons of cooling. This sounds hefty, but unfortunately it is not a whole lot (about enough to cool a small apartment). Let’s see if heat change through evaporation is better.

Heat evaporating off the fountain’s surface

Evaporation is the type of cooling used in cooling towers, but cooling towers have giant fans in them to speed up airflow and ridged fins to maximize surface area. A fountain does not have these advantages, though the higher it shoots upwards the better (ours only rose a few feet). Nevertheless, it should have some cooling impact. But how much?

ASHRAE tells us that the rate of evaporation from a fountain occurs at:

W = (Pw – Pa) · (0.089 + 0.0782V) · / Y   where:

  • W is the rate of heat leaving the fountain
  • Pw is the vapor pressure of the water at the water’s temperature
  • Pa is the pressure of the vapor in the air at the air temperature
  • V is the velocity of the air over the water, and
  • Y is the latent heat of vaporization of water

Pressures are in units of kPa, and Y is in kJ/kg, giving a W in SI units of kg / (s · m2).

For water, Y = 2260 kJ / kg.

We also need the wind speed, V. Fortunately, we can use satellite data from a Typical Meteorological Year (TMY) which tells us wind speed, temperature, and relative humidity for our chosen location. The average summer wind velocity is 4 meters/second (9 miles per hour).

Pw for water at 80 degrees Fahrenheit (27°C) is 3.6 kPa. An analysis of TMY data tells us that the average afternoon, summer temperature at the fountain location is 77°F (25°C). At 100% humidity this would mean Pa = 3.2 kPa. But to get the pressure of the water vapor in the air, we need to multiply by the relative humidity, which we analyze TMY data to determine is 60% in the summer. So 3.2 · 0.60  = 1.9 kPa = Pa.

So plugging everything in we get an evaporative rate, W, of: 0.000302 kg / (s · m2).

The surface area of the fountain was about 1400 ft2 or 130 m2. Multiplying by W, we obtain a rate of .0393 kg/s of evaporation. We know each kg of water that evaporates carries 2260 kJ with it, giving us 89 kJ/s or 89 kW. Over an hour this would mean the evaporation of 89 kWh of energy. We convert to BTU (1 kWh = 3412 BTU) to obtain 303,000 BTU/hour or 25 tons of cooling. Now that’s exciting!

So though cooling through the sides and bottom of the fountain (like a geothermal heat pump) only provides 1 ton of cooling, evaporation provides 25 tons! Currently, each building has about 200 tons worth of cooling towers on their roofs, so 25 tons would have an impact.

Whether it would be worth it to redirect the pipes and remove load from the cooling towers, however, requires pondering some other considerations. Imagine, for instance, walking out of your building on a hot summer day to sit by the cool fountain mist, only to discover the water is almost as hot as a Jacuzzi. Many property owners find such an outcome suboptimal. Furthermore, drilling through concrete is expensive.

Individual circumstances of buildings often limit what energy solutions are feasible and fountain cooling is no exception. Nevertheless, using fountains as cooling towers remains an exciting opportunity to reduce carbon emissions, cut energy usage, and save money, all using existing infrastructure.

9 Responses to ∫∫∫ Fountain Cooling

  1. Great post and fun question. If the fountain is actively spraying water in the air, would the velocity of the water be added to the wind speed? Also, if drops of water are in the air, would that increase the surface area of the fountain?

    Reply
    • Good question: yes! The velocity of the water shooting into the air should be added to the wind speed (vector addition since the wind velocity is horizontal, whereas the water velocity is vertical), but that would only happen if we were including the effects of the spraying fountain at all. As it is currently approximated, we’ve pretended the entire fountain was a still pool with a replacement rate of 400 gpm. Adding in the effects of the water column would increase evaporation and hence cooling capacity, but given that we do not know the thickness, height, or speed of the streams, this seemed difficult to model at all accurately. We’ve opted to be conservative instead.

      Reply
  2. Kolenia says:

    The content matter here is magnificent, appreciate your efforts. You should keep it up forever! Good Luck.

    Reply
  3. This evaporative cooling technique brings up the interesting idea of dual-use applications. In this case, the aesthetic value of the fountain could be added as a wanted side benefit to the heat transfer purpose.

    It’s important to remember that this cooling approach requires an understanding of humidity. This is similar to the reason why Houston heat feels much worse than Phoenix heat. At the same temperature, it’s harder for our bodies to reject heat in an environment with higher humidity. Likewise, the efficiency of this cooler will depend on the relative dryness of the local climate.

    Reply
  4. For your conduction of heat into the soil: 1) Why did you omit the thickness of the concrete as I thought you need to divide that into your equation? 2) Why did you not take the soils thermal conductivity?

    Reply
    • Hi Bryant,

      Good questions.

      1) The thickness of the concrete does matter, and we included that in the U value for the concrete. Had the concrete been thicker, we would have used a lower U value. Had it been thinner, we would’ve needed a higher U value.

      2) For the soil conductivity, we’ve made a simplifying assumption here: we assumed the temperature of the soil touching the fountain base was constant. In real life, the soil immediately touching the concrete would slowly heat up, and we would then need to dump heat from that soil into soil farther away. We’ve ignored this effect and essentially assumed infinite conductivity in the soil. In the case of the fountain cooling, the dominant cooling mechanism is not conductivity through the base but evaporation from the surface, so this assumption is not super important in the overall calculation. If you are concerned about this effect, use a U value of .18 instead of .2 for the concrete to be a little more conservative.

      Best,
      Brenden

      Reply
  5. What would be the effect of say making this a waterfall and blowing the air through the falling water. Say if you had the water pumped to the second floor and use a duct fan to draw cool air from the fall like a swamp cooler?

    Reply

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