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Most all of the information on spa insulation is incorrect when comparing a thermal sealed spa to a fully foamed spa. It is nothing more than sales nonsense.  The concept for the thermal sealed spa came about sometime in the early 90's. It evolved out of several spa designs. The full foam sales people are incorrect when they group a thermal closed design in with the other "non full foam" designs. As soon as you see open vents, louvers, or big holes in the bottom, you know it isn't a thermally efficient spa.
 
The thermal closed cabinet design has four basic beneficial features that make it a superior design for insulation:  energy efficiency, better filtration , quiet operation, and  the most freeze protection available.  
One of the basic concepts of this design is to stop wasting energy from the spa pump motors that is typically lost in most spas. When any pump runs, it puts out heat exhausted and radiated from the motors. There are three types of heat generated, radiant from the heated induction winding wires inside the motor that heat the frame,  forced convection heat from the motor's fan,  and frictional heat generated by moving parts and the water going through the water pipes. As one engineer put it: "All energy ultimately is dissipated as heat". When nearly all of the motor heat is available to the spa water, the spa becomes more efficient, because the spa's electric water heater is not used nearly as much.  

The other part of this concept is warm air thermal barrier created by the warm air in the cabinet.
This is designed to stop the loss of heat from the sides of the spa vessel (the seating area). This is accomplished by using the warm air from the spa motors as well.

The third beneficial concept of this is actually a by product of the energy conservation of this system. There is a much better filtration achieved by moving more water through the filter or filters per day at higher circulation rates. Because the heat from the motors is used to eliminate a large quantity of the water heater energy, we can run the motors and have more filtering with cleaner water. The resistive electric heater that uses more power than any other component in the spa is the most expensive component to operate. If we could totally eliminate that heater, then the spa would be near perfect in design.  In Haven Spas we come very close to that "perfection" with the DAIT
 
 

Heated air creates "infinite" conduction "R Value" insulation*

Here is a basic law of thermodynamics:

If one form of matter, solid, liquid, or gas, is in contact with another form of matter, the heat from one to the other is transferred by going in the direction of warmer matter to the cooler matter. Heat transfers from hot to cold is a simpler way of putting it.

For instance, if the water in the spa vessel is 102 degrees, and the shell is 100 degrees with no extra insulation, the heat is being transferred from the water through to the shell and then to the outside environment. The degree of insulation in this case is less than R1 and the heat loss is immense. If the air on the outside is 70 degrees then the 100 degree shell will transfer its heat out to the outside environment at a predictable rate.

Here is another law of energy transfer. It takes a lot less energy to warm "one gallon" of air than one gallon of water. I looked up the formula for this and it is shown below. The difference is tremendous, more than 3,100 times more energy required.

Since air weighs in at .075 pounds per cubic foot while water weighs in at 63 pounds per cubic foot. ( a box of air vs. a box of water), water weighs 840 times more than air. A simple demonstration of this is to put a bubble of air under the water and watch how fast it moves to the top. Fill a box with water and lift it. "Fill" a box of air and lift it. This means that it takes about 3100 times as much energy to heat the same amount by volume of water from 40 degrees F. to 102 degrees F as it does to heat the same volume of air.

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Here is the formula Since water is the basis of these calculations, the mass factor for water is 1 (one). The source of this information comes from http://www.watlow.com a manufacturer of heater equipment.

Equation 1

Heat Required To Raise The Temperature of A Material

Q1 = Heat required to raise temperature

W = Pounds of material

CP = Specific heat of material (Btu/lb-ºF)

T = Temperature rise of material (TFinal - TInitial) ºF

Water:

There are 3412 BTU's per kilowatt hour of electricity. So how many kilowatts does it take to heat one cubic foot of water from 70 degrees to 102 degrees; a 32 degree differential?

63 (the mass of water) X 1 ( base factor ) X (102 - 70) / 3412 = .591 KWH

It takes .591 kilowatts to raise one cubic foot of water 30 degrees. ( For reference that is equivalent to about 6 each, 100 watt, light bulbs burning for one hour.)

Air:

.075 (the mass of water at 70 degrees F.) X .240 ( base factor ) X (102 - 70) / 3412 = .0001688 KWH  (I don't know of a light bulb that uses 0.16 watt hours.  It is a very small amount of energy.)

So it takes .591 / .0001688 = 3143 times more energy to heat the same volume of water. Or water is 3143 time harder to heat than air.  This is something that is primary to understanding how a thermally closed spa insulates.

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So air is a lot less dense than water and costs a lot less (minuscule energy) to heat the same volume of air than water.

Picture  a container  split it into two halves by a thin membrane of near zero R value plastic  with infinite insulation all around. This is like taking a jar and dividing it in half, putting air in one side and warm water in the other side. As soon as the air reaches the water temperature, the heat transfer stops from the water to the air.

In other words; when the air reaches the water temperature, or is warmed to the water temperature, the heat transfer out of the water is stopped dead in its tracks.

This is the big "secret" of a thermal closed cabinet with the motors inside the closed cabinet. You could have zero insulation on the outer wall of the spa, and put all the insulation on the outer cabinet, but it has several drawbacks that I'll go into later.

As the motors run, they heat the air chamber inside the spa contained between the shell and the outer insulated wall in a thermal sealed cabinet designed spa. When the air inside the chamber becomes warm as the water, there is an " R infinite" insulation created. All of the heat loss on the outside of the shell is stopped completely as soon as the air is the same temperature as the water. Because it costs so much less (3143 times less) to heat the air, we are using a very economical heat retainer, warm air. As the pumps run inside the cabinet, ( doing a great job of filtering ) the warm air is created, when the pumps stop, the air cools down, but never as cool as the outside air.

So the common arrangement of this design is a closed cabinet with about 2 to three inches of foam on the shell and one inch foam boards in the outer wall of the spa, then the wood or plastic skirt. There are no venting holes purposely placed in the cabinet.

The trick is to not have too much foam on the shell and to not have too much foam on the outer skirt.

This design can become too heat retentive by not allowing the spa to cool between filter cycles. Because we are literally putting all the heat possible into the spa water, we must not overdo it . It just so happens that the only time the jet pumps are on high at the peak of thermal efficiency, the cover is off, losing the most heat out the top. You would not want to run the jets on high while the cover was on, this never happens under normal use. Why would anyone run the jets on high with the cover on, unless it was for a thermal test?

The biggest concern with this design is that it  does not become too heat retentive in the summer. It is recommended to not run the filtering during the heat of the day, but to filter at night and early morning. This gives us the optimum energy efficiency (for any season), and clean water with out the use of ridiculous 7 to 3 GPM tiny circ pumps.

As the air in the cabinet, heated by the pump, gets above the temperature of the water, the motor heat will start to transfer into the water. It is drawn in by the air controls, the exposed equipment, exposed pipes, and through the wall of the spa and through the air pump. Now even though the air is less dense, about 840 times less dense, we are taking the heat from a confined machine, an electric motor of about 0.23 cubic feet, the heat is dissipated into a volume of 16 cubic feet or more. So it is 16/0.23 ratio for heat dissipation. 69 time the volume. As the temperature rises to, say, 10 degrees above the water temperature, the heat is transferred much more quickly.

Another rule about heat transfer, is the greater the difference between two masses in temperature the faster the heat is transferred. The barrier of heat resistance between the two masses has a predictable effect on the heat transfer.

For instance, which has more heat loss, a spa shell with  2 of inch foam with 102 degree water inside and 0 degrees F outside the foam, or the same with 102 inside and 70 degrees outside?

I think the answer is obvious. The higher the temperature differential, the faster the heat transfer. If you never have a great temperature difference, then the heat loss is minimized.

Inside a thermal pane or thermal sealed cabinet, the residual heat from the water actually helps to insulate itself. The water heat in the vessel will begin to warm the air in the chamber when the pumps are turned off. This keeps the temp inside the spa above 70 degrees, even at sub zero weather outside.
 
 

Equation 4

Heat Losses By Conduction Through Materials

Where:

Q4 = Conduction heat losses

K = Thermal Conductivity (Btu · in/ft2 · ºF · hour)

A = Heat transfer surface area (ft2)

X = Thickness of material (inches)

T = Temperature difference across material (T2 - T1)ºF

t = Conduction time (hours)

In a full foam spa with say 6 inches of foam at the top, and four inches in front near the door, 12 inches at the bottom, the overall insulation value is less than a total of three inches of foam plus the wood in the cabinet in a thermal sealed spa. So full foam is actually less insulation than 3 inches of foam, dead air, and wood in a thermally closed cabinet.

8 inches of polyurethane foam is about R 48. Two inches ( or six inches or fifty inches....so on ) of foam with warm air outside the vessel and inside the cabinet is about R-100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 and so on. As far as the water vessel is concerned the conduction heat loss is gone from the sides.

Which is better, six inches of foam, or two with warm air on the outside?

In the equation above, the delta "T" (T2 - T1) degrees F. is zero for a zero difference from the inside to the outside. If you multiply by "ZERO", the whole heat loss is "ZERO".

So, as long as the cabinet temperature is as warm or warmer than water, we have an extremely good insulation scenario. If you filter for eight hours per day, the spa has the equivalent of "infinite" conduction insulation for about eight hours per day. In the winter, if it is super cold outside, the more you filter, the less it costs to operate filtration. In the summer the spa use is normally less, so the filtering can be reduced as necessary. ( I call this the "Gestalt" or overall integration of all the parts of this system. Each of the engineering principles or parts in this system integrate with each other part extremely well.)
 
 

The heat loss when the pump is off in a thermal closed spa.

Here is another area where the laws of thermal dynamics work in the favor of the thermally closed cabinet. The temperature differential in the above equation, showing heat loss through materials, works in the favor of the thermal pane or closed cabinet design.
The water at 102 and the wall of the spa with average of 3 inches of polyurethane foam has an R value of about 18.  But the temperature differential between the filter times ( when the pump is off ) is less than 15 degrees average, the extremes high temperature being 115 F degrees, and lowest of 70 F degrees.

The heat loss is strongly affected by the "DELTA T" in the equation. If it is an average of 15 degrees, then we are not dealing with more than double that in a full foam spa. In other words the 3 inches of foam in a thermal sealed spa is equal to 6 inches in a full foam (R 36) with cool air on the outside. The average temperature in most places is about 68 degrees, or 34 degrees cooler than the water.

So, for six to eight hours per day, we have infinite insulation, and the rest of the time we have an equivalent of R 36 or more by using an actual R - 24.

Here is the formula for a thermal closed spa, using the KWH ( kilowatt hours) equation 4 above;

Where:

K = .228 for polyurethane foam.

K = .18 for air ( second equation)

A = 1 square foot

X = 3 inches on the foam. ( average of 10 inches Air space).

Delta T = 15 degrees average.

t = 16 hours with the pump off during the day.

.228 X 1 x 15 x 16 / 3412 X 3 = .005345838: kilowatt hour loss in 16 hours per square foot of surface. The rest of the time there is no loss at all.

Now, we have a heat loss formula we can convert to any size of spa. 70 Square feet would be a good size spa. This is the inside square footage of water contact surface. 70 x .0053458 = .374 Kilowatt hours out the sides of the spa in 16 hours. Not much, considering that an average KWH is about 12 Cents per KWH across the country. That is 4.5 cents per day , 1.35 per month, loss from the sides. (Now the top is another story.)

What we are trying to do more than anything with a thermal closed spa is have really clean water and have a low electric bill.

Here's another scenario. In a typical fully foamed spa, the pipes, heater, pump housings, all have square footage of heat loss as long as the temperature of the air inside the cabinet is cooler than the spa water. There is no foam insulation on the equipment or pipes in the equipment area, and cool air is allowed into the cabinet. This is nearly all the time in winter, and most of the time in summer. Plus the spas on all the sides will never have zero heat loss, because the temperature outside the spa foam is always cooler than the spa water, almost all the time. Even when it is 90 degrees outside, the spa is still losing heat if the water is 91 or above.  In a full foam spa the only time there is 100% insulation is when the water is the same temperature as the outside.  Since nobody runs the temperature up and down, there is almost always heat loss.

Since the average thickness of the insulation is about 6 inches on an average full foam spa, top bottom and sides. There may be 12 inches on a diagonal from the seat to the outside, and two inches on the wall between the equipment and the outside three inches in the upper shell. That averages at R 36. The companies often brag about 30 inches of foam, but that is measured diagonally in the largest area of the spa.  The type of foam is also part of the equation.  An open cell foam with low density is a poor insulator.  Many of the spa companies, including one of the largest, uses this type of foam.  It's R-value is less than R-2 per inch in some foams.

When the "tiny circ" 24 hour pump is running in the cabinet ,which is cooler than the spa water, there is heat loss in all those surface areas. It is a "cooler" representing about 20% or more of the spa's total cubic feet of volume. I call it the "cooler". When I was a child they had these things called "coolers" in houses. It was a open backed cabinet in the kitchen. On the back was the outside world. It kept things cooler as long as the outside was cool. In winter the vegetables were moved to the cellar, because they would freeze in the cooler, just like spa equipment does. When the water inside the equipment freezes, it expands and breaks the equipment, one of the most expensive repairs there is.

When you figure mathematically the total non insulated, water containing parts, the average "Delta T" and the time involved, the equipment in the front of the spa is similar to a radiator, constantly loosing heat.  While the spa's heater is on the air in the equipment area is also cooling the water at the same time.  This is similar to placing your home heater and sitting it on the open non heated porch.

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The "air" as insulation:

According to the data from the heater manufacturer, air has a heat transfer factor of .18, while polyurethane insulation has a heat transfer factor of .223. The lower number is better.

So, if we figure in the above equation, the air insulation in the spa between the outer skirt and the foam on the wall of the shell, we come up with this formula:

Average "thickness" of the air is about 10 inches in a large thermal closed spa, six in an average spa.

Figuring

K = .228 for polyurethane foam.

K = .18 for air ( second equation)

A = 1 square foot

X = 3 inches on the foam. ( average of 10 inches Air space).

Delta T = 15 degrees average.

t = 16 hours with the pump off during the day.

The Second Equation for heat transfer through air.

.18 ( heat loss factor for air) X 1 x 15 x 16 / 3412 X 6 = .0021101 (kilowatt loss in 16 hours per square foot of surface.)

In 16 hours that is a total loss of .0337616 kilowatt hours or $.00405 per day 12.15cents per month. With an average temp of 15 from the foam on the spa to the foam on the wall. We are now down to 72 degrees at the inside of the outer wall.

From the outer wall to the outside is a delta T of three degrees average. The insulation R value of the outer wall is R-6 + R 4 for the wood. R-10.
 
 
 
 

One of the most important issues in heat loss is that heat from a spa vessel will be lost faster out the top of the vessel than from the sides or bottom. Heat rises by convection, so the top needs the most insulation. This is why we always tell people to get a thermal blanket. It is one of the smartest things we sell. It should be on the spa, winter, spring, summer, or fall, because it also captures the ozone, making the spa a great contact chamber for the ozone. http://www.spaspecialist.com/cover.html I will be publishing the data on heat loss on an open spa in one of the next articles.

There are a couple of notes here.

This illustration is very conservative. The pumps will come on at some time between the filtering, especially in winter. The controls we use are set up to operate on a plus and minus 1 degree range. The heater runs until the water temp is one degree above the setting and it will come back on at one degree below the setting. Since the sensor is also inside the water vessel, it minimizes the number of pump starts  to heat between the filter cycles. When the pump comes on, the air inside the cabinet is warmed and the insulation becomes infinite again until the air drops below the water temperature, then the delta T in the formula starts working again. The point is, the temperature is never at 80 degrees for any long period of time, and it is normally above 80 degrees, making the "delta T" not much and the heat loss a lot less. Because the pump comes on to run when the spa calls for heat in the dead of winter, the warm air is created quickly to stop heat loss. Remember, it is a lot cheaper to heat air than water.

By breaking it up into a sort of average temperature difference, it gives us something to calculate. Even with this conservative calculation, we are talking about less than $1.50 per month heat loss out the sides of the spa. Now do you understand why I say full foam is wasting heat?  If you do not understand now read the next articles. Most of the heat loss in any spa is out the top!!!!

I get this question from time to time.   "Isn't there some heat loss by the fact that the air on the outside of the spa is causing a heat transfer to the outside world and what about the convection inside the cabinet?  Here are some of the factors that make this heat loss non linear.   If you place a sensor near the spa vessel, average 102, the air is nearly the same temperature as the water.  If you place a sensor in the middle of the thermal cabinet it is cooler, average 95 degrees.  If you place a temperature sensor on the outer wall it is normally about 85 degrees.  You may have noticed that I used the lowest average temperature in the cabinet as an average in the above formula.  This is to account for the fact that there is convection inside the cabinet.  In a closed box the warm air will rise and the cool will fall, as is commonly known.  Since this is a totatally unpredictable part of this, the above foumula accounts for the lowest average temperature in the cabinet.   The outer wall area has measured as high as 115 degrees and the lowest 68 degrees in a large thermally closed spa.  

In the DAIT insulated spa the foam on the shell is less and the foam on the outer wall is much thicker.  This helps to keep the temperature inside the cabinet much warmer and makes the "Delta T" much less.

One of the things that most engineers seem to forget in this formula is the effect of warming the air and slowing the heat loss on the outside of the cabinet.

No foam on the shell; why not?

There are some spas with no foam at all on the shell, and all the foam on the outer wall. This is not a completely thought out design.

Whenever I see this, I know the spa is not fully insulated, because the heat from the vessel leaves into the dead air chamber at a bit higher rate. If the temp becomes even 80 degrees in the formula above with no insulation, (a high loss factor), the spa will use more electricity. As the spa's air chamber cools, so does the water. This will turn on the pump sooner, and shorten the life of the motor, and cause the air to warm again, but as soon as the pump is off again it cools too quickly. Like I always tell people it is best to "dress in layers".  You must not put on too much foam, because the spa will not "heat transfer" correctly.

In the event of a power outage or equipment failure in the dead cold of winter, the lack of foam on the shell becomes even more critical. The down time is shortened considerably.

I often describe it this way to people who have no engineering understanding, but do understand cold: When I was young, we used to go to "Gramma's house" in Kansas. When we went to bed we slept up in the "sleeping loft". It was unheated, as in many farm houses built in the late 1800's or early 1900's.

In order to keep from "freezing" and to get some warmth in the bed, rocks or a hot water bottle was used at the foot of the bed. Instead of a "raw" hot water bottle or heated rock, they were wrapped in a blanket. The blanket kept the rock from burning my feet, and gave duration to the heat while at the same time allowing heat transfer. If the rock was just put in the bed, it would have gotten too hot under the blankets, but the heat would also leave the rock and I would have to go downstairs and sleep on "Grandpa's" old bent wood rocker, next to the old wood stove. (It was actually not meant to sleep on .)

A spa without foam on the wall lacks duration in the event of a shut off in the cold of winter.

The other reason why not having foam on a shell is not so good:

The smaller plumbing going to the jets needs to be held from shaking around. It is really very simple. If the tubing were allowed to shake, it wastes pump energy, as well as helping to shorten the life of the tubing. The tubing works better, and the water flows smoother and quieter if it is imbedded in foam.
  

*In engineering the term "infinite R-Value" does not exist.  In this demostration we use the term "infinite" as an illustrative word, because it is as close of an exact word in terms of the actual effects of this system.  The heat loss out side the cabinet is zero, making the effective R-value immeasureable and beyond the limits of all numbers associated with the ability to stop heat transfer.  

Next article. Heat loss in the equipment and plumbing areas of Fully Foamed spas.