Thermodynamics of a 120 Gallon Mother Coral Reef Tank Build.

[fatman]

Thermodynamics of a 120 Gallon Mother Coral Reef Tank Build.

Introduction

In the past I have set up many salt water marine tanks for the display and propagation of live corals, in particular small polyp stoney (SPS) corals. These tanks have been set up with varying degrees of initial success. All of the tanks required modifications, some times the needed modifications were exstensive. Since setting up the majority of these tanks I have had the opportunity to return to school where I have obtained a Bachelors Degree in Civil Engineering. I have, due to this schooling, received instruction to some degree in the field of engineering Thermodynmics. I have taken two one semester coursse in Thermodynamics forr Engineers and a one semester course titled Energy and the Environment which dealt to a good degree in Thermodynamics, plus several semesters of Physics as well as obtaining a minor in Chemistry. Only the course in Thermodynamics for Engineers was intensive in Thermodynamics but it came no where near to engaging the full field of Thermodynamis but did provide me with enough basic education in the field to understand that I could use some of that knowledge to assist me in designing my next coral reef aquarium without the follow up necessity of quite so many modifications.

General Requirements of New Mother Colony Coral Reef Tank

● Use of standard 120 Gallon All Glass Aquarium with outside dimensions of 48 ½” x 24 ½” x 25 ½”.
● Use of sump of 40 gallon capacity with dimensions of 36 3/6” x 12 ¾” x 16 7/8”.
● Use of two HQI Metal Halide lights rated at 250 watts each.
● Use of two Power Compact Actinic lights rated at 96 watts each.
● Use of four LED “moon lights” rated at 5 watts each.
● Use of four circulation fans.

Thermodynamic Concerns of New Tank System

●Over heating of water.
Thermo gains of room where tank is to be housed.
●Size requirement of “chiller” system (water refrigeration) used for supplemental cooling.
●Water automatic top off requirements for maintaining proper water levels and therefore the proper salinity of water mixture.

Thermodynamic Principles Involved
The Evaporation of a Liquid
The average energy of the particles in a liquid is governed by the temperature. The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average.
Some of the more energetic particles on the surface of the liquid can be moving fast enough to escape from the attractive forces holding the liquid together. They evaporate.
The diagram shows a small region of a liquid near its surface.
I have not figured out how to post the picture yet.
Evaporation is an endothermic (heat-absorbing) process because molecules must be supplied with energy to overcome the intermolecular forces.
The fact that water is a low cost, convenient, and highly effective heat transfer medium, and that evaporation of a pound of water requires about 1000 btu, makes the thermo control of an aquariums tank fairly easier to predict and control.
General Assumptions
● Differences in specific heat capacities of differing materials contained within tank are negligible therefore the entire tanks contents will be calculated as water.
● No utilization of lamps light out put is converted by photosynthesis due to growth of corals or coralline algae. This is assumed both for ease of calculations as well as because the amount of growth of coral is judged over a period of months or even years and this articles calculations are concerned with a period of only hours.
● Silicone used in tank seam construction is minimal and is therefore provides negligible effects on the thermal flow out of tank.
● For reasons of ease of calculations the ¾ inch wide edges of the bottom of tank that sit on wood will be calculated as being exposed to room air rather than sitting on wood perimeter edges.
● Past tanks of this size with the exact same lighting size, style and set up had an evaporation rate of 3.5 gallons of water per day when tanks sumps were cooled by evaporative heating provided by two small 4 inch box fans run only during time lights were used.
● This tanks lighting fixture, as did the previous tanks contains two box fans.
● Room temperature will be maintained at 680 F true air conditioning/heating as is necessary through out the year.
● Temperature maintained in tank through combined use of evaporative fans, conductive heat losses and refrigeration (as needed) unit will be nominally 800 F.
Lighting Assumptions
● The HQI metal halides being used as powered by remote magnetic ballasts transfer all light rays as an equivalent thermal transfer watt produced for watt transmitted minus losses through radiant heat from reflectors (assigned a 10 % value) transferred away from aquarium by lighting hood cooling fans.
● Thermal output of halide lamp ballasts are remote and provide negligible effects to aquarium system.
● 70 % (Seventy percent) of remaining light output is transferred to aquarium systems water. The remaining light, and therefore the thermal energy is transfer to room containing aquarium. This is due to reflector design and placement of lighting seven inches above aquariums water level.
● It is assumed that all lighting thermal gain is to water and not aquariums corals, live rock or sandy substrate for ease of calculations.
● Halide lighting will run 10 hours per day, and power compact lights will run twelve hours per day, coming on one hour before halides and going off one hour after halides.

Data Used for Calculations
Specific Heat of Water: 1.00 Cal/g*K or 4180 J.kg*K
Specific Heat of Glass: 0.20 Cal/g*K or 840 J.kg*K
Evaporation of a pound of water requires about: 1000 btu
Sea Water Weight: 8.55 pounds
kglass: 1.0 k (w/m*k)
Glass Thickness: 0.5 inches or 0.0127 m
The BTU is a unit of heat energy defined as the amount of heat required to raise the temperature of one pound of water by one degree Fahrenheit. It is most often used in reef keeping as a way to gauge the power of chillers . One Btu is equal to about 251.996 calories, or 0.251996 kilocalories (the “calories” counted by dieters). One BTU equals approximately 778.169 foot pounds, 1.055 056 kilojoules or 0.293 071 watt hour.
The gallon is a unit of liquid volume. Its name derives from the Roman galeta, which originally meant a pail full. Gallons of various sizes have been used throughout history. In the United States, the liquid gallon is now defined as 231 cubic inches and holds 4 liquid quarts or 3.79 liters. A U.S. gallon of fresh water weighs about 8.33 pounds, and a gallon of salt water weighs about 8.55 pounds.
The joule is the SI unit of energy. It is defined to be the work done by a force of one newton acting to move an object through a distance of one meter. One joule is also the kinetic energy of a mass of two kilograms moving at a velocity of 1 m/s, or 107 ergs, 0.74 foot-pounds, 9.5 x 10-4 Btu, 0.24 (small) calories or 2.8 x 10-4 watt hour. The joule is named for the British physicist James Prescott Joule (1818-1889), who demonstrated that mechanical and thermal energy were equivalent in 1843.
The pound (lb) is a unit of weight that has evolved over the years. The unit now used in the United States is the avoirdupois pound. The avoirdupois pound is divided into 16 ounces
The meter is the metric and SI unit of distance. Originally, the meter was designed to be one ten-millionth of the distance between the Equator and the North Pole. Over the years it has become better defined, and is now defined in terms of the speed of light and the specific fraction of the length that light travels in a second in vacuum. One meter equals 1.09 yards, 3.28 feet or 39.4 inches. The unit is spelled meter in the U.S. and metre in Britain.
The watt is the SI unit of power. Power is the rate at which energy is used. One watt is equal to one joule of energy per second, 0.0013 horsepower (hp) or 0.738foot-pound per second (lbf/s). One watt is also the power produced by a current of one ampere flowing across an electric potential of one volt.
The watt hour is a unit of work or energy. It is the total energy delivered at a rate of one watt for one hour. A 100 W light bulb lit for one hour consumes 100 watt hours of energy. One watt hour equals 3.6 kilojoules (kJ), 3.4 Btu, 0.86 large Calories (kcal) or 2655 foot pounds.

Equations Used

TC = (TF - 32 ) * 5/9

Calculations

● Nominal Area (A) aquarium glass in contact with water: (2 x 47.5 x 24) + (47.5x 23.5) + (2 x 23.5 x 24) = 4524.25 sq inches = 2.91886 m^2
● Volume (V) of tank: (47.5 in x 23.5 in x 24 in) = 26790 in^3
● Gallons (G)contained within tank: (26790 in^3 / 231 in^3) = 115.97 gallons (kind of like false advertising I believe as it sells as a 120 gallon tank)
● Nominal Area (AS) aquarium glass in contact with water: (2 x 35.875 x 16) + ( 35.875 x 17.25) + (2 x 17.25 x 16) = 2318.84 sq inches = 1.49602 m^2
● Volume (VS)of tank: (35.785 in x 16 in x 17.25 in) = 9876.66 in^3
● Gallons (GS)contained within tank: (9876.66 in^3 / 231 in^3) = 42.76 gallons
● Thermal flow into tank from lights <all lights on> ((2 x 250 watts x 0.60)) + (2 x 96 watts x 0.60)) x 3.41 btu/watts = 1415.832 Btu/hr
● Thermal flow into tank from lights <actinics lights only> ((2 x 96 watts x 0.60) x 3.41 btu/watts) = 392.832 Btu/hr
● Btu removed by Evaporation of water from sump per hour lights are on: (3.5 gallon x 1000 btu/gallon)/12= 32.736 btu/hr
● Net initial btu/hr gain while halides and power compacts are on: (1415.832 Btu/hr)
● Net initial btu/hr gain while only power compacts are on: 392.832 Btu/hr

● Temperature increase per hour while all lights are on: (1415.832/(115.97 + 42.76)) = 8.92 degrees F

● Temperature increase per hour when only actinic lights are on lights: (392.832 / (115.97 + 42.76))= 2.475 degrees F

● Btu loss through thermal conduction through tank walls when all lights are on assuming refrigeration is not running:
Temp 0C = ((68 degrees F + 8.92 degrees F) - 32 degrees F) x (5/9) = 24.96 degrees C,
(68 degrees F -32 degrees F) x (5/9) = 20 degrees C
Pconduction = (1.0)(2.91886 m^2 + 1.49602 m^2)((24.96 degrees C - 20 degrees C) / 0.0127 m) = 1724.158 btu/hr

● Btu loss through thermal conduction through tank walls just actinic lights are on assuming refrigeration is not running:
Temp 0C = ((68 + 2.475 degrees F) - 32 degrees F) x (5/9) = 21.38 degrees C,
(68 - 32) x (5/9) = 20 degrees C
Pconduction = (1.0)(2.91886 m^2 + 1.49602 m2)((21.38 Degrees C - 20 degrees C) / 0.0127 m) = 479.705 btu/hr

● Final btu gain per hour with all lights on: (- 1724.158 btu/hr - 32.736 btu/hr) =
-1756.894 btu/hr

● Final btu gain per hour with just actinic lights on: (-479.705 btu/hr -32.736 btu/hr) = -512.441 btu/hr

● Expected temperature rise per hour that must be handled by refrigeration while all lights are on: -1756.894 btu/hr / ((115.97 gallon + 42.76 gallon)/( 8.55 lb/gallon))= -1.295 degrees F/hr

● Expected temperature rise per hour that must be handled by refrigeration while just actinic lights are on: -164.823 btu/hr / ((115.97 gallon+ 42.76 gallon)/( 8.55 lb/gallon))= -3.53 degrees F/hr

Subscript S signifies sump tank

CONCLUSION

All though this was an interesting project I had expected to find that I would need to fight a potential temperature gain, however, the final calculations show that I can actually expect a loss of temp by 1 to 3.5 degrees Fahrenheit if the room temperature is maintained at 68 degrees F. I am happy with the results even though a lot of minute details were left out of the calculations. For an example the calculations assume that all water evaporated is lost only when the lights and therefore the fans are running. This assuredly will not be the case so the tempratutres lost during the period of the lights running is skewed. There will a lot of plumbing lines on this tank system. Approximatley 35 feet of lightly pressurized PVC water pipe, mostly 1.5 too 2.0 inches in diameter,which carry a little over 9250 gallons of water per hour through them, which is the entire circulation system for the tank. Heat losses through those pipes are not calculayted. Three external pumps are used and no allowance is made for heat gain by the pumps as the cabinet is well ventilated when the lights are on. There is also a 100 watt Power Compact sump bulb, but as it is run on a reverse lighting cycle its heat output was left out of all calculations. Also there were things I would have liked to calculated but did not have adequate data to accomplish the calculations. One of these was to calculate what proportion of the evaporated water would come from the main display tanks air water interface and what proportion of evaporation would come from the sump tank. Also in the the tank their will be two Fractational skimmer/filters which draw large amounts of air through venturis which must surely provide some cooling as they have always provided a large boost to the dissolved oxygen levels in other tanks I have set up, which show they aerate the water very well which should equate as providing some cooling. There small in sump tank heat contrivb butions ewere left out of the calculations. Only two small 150 watt heaters will be provided in the sump tank, but they should only come on infrequently and mainly at night when no lights have been on for a period of hours not calculated, but which has proven minimal in previous tanks set up. Heat loss should be substantially less at night as there will be no fans running and there will not be as much diiference between the room temperature and the potential tank temperature from thermal inputs to drive increasing evaporation and conductive heat losses as during the lighting hours.
I am sure there are probably some mistakes within the paper as I am not a good proof reader or editor.

Fatman

[msbdiving]

Talk about doing research and planning, I'm impressed. Leave it to a Californian though, when reading your post I thought who is able to keep their house at 68? We use a moderate amount of the A/C during the summer to keep it at 80 if we want it cool. Then I saw you were in Alaska and the light bulb went on. It has been too many years since I took a physics or chemistry class that I will have to trust your calculations.

[fatman]

Not many sun tanned peple in Alaska.

[cgoodman381]

Wow, someone does their research! Great paper.

[cgoodman381]

double post

[Tru2nr]

so fatman i have come to the conclusion you are either an engineer or in the science field lol

[Frankie]

A tank build from Fatman.! Should be interesting!
Remember, Threads Are Pointless Without Pictures!

[fatman]

Originally Posted by Tru2nr

so fatman i have come to the conclusion you are either an engineer or in the science field lol

Civil Engineer, with emphasis in Environmental Engineering, a minor in Chemistry and Mathematics. Most of my work is in water treatment and waste water management, but I also work on fish hatchery projects and aquatic water projects as they relate to water discharges into natural water bodies IE effluent from industry and waste water treatment facilities. I have also worked on quite a few in situ remediation projects. I own and operate a reef tank lease company for which I maintain a SPS frag/grow out/feeding operation. I am small though being located in the state of Alaska, so I only have only two employees and one shop facility.

[BeanAnimal]

I hate to dredge this up from the depths of the past.... but in searching google for a published thermodynamics white paper , this thread came up near the top of google.

The post is full of errors. Misinformation spreads like wildfire on the internet.

From the original post:

I am sure there are probably some mistakes within the paper as I am not a good proof reader or editor.

● Btu removed by Evaporation of water from sump per hour lights are on: (3.5 gallon x 1000 btu/gallon)/12= 32.736 btu/hr

Doing the math above yields 291.66 BTU/hr but the answer is still wrong (have no idea where the 32.736 came from).

Evaporation was listed as ~1000 BTU per POUND. While this is correct, the OP neglected to calculate the POUNDS in a GALLON (about 8.5 for seawater) but used GALLONS in his calculations.

The calculation for 3.5 gallons evaporated over 12 hours should have read:
Btu removed by Evaporation of water from sump per hour lights are on: (3.5 gallon x 8100 btu/gallon)/12 = 2362 BTU/hr NOT 32.736 btu/hr

The OPs results are in error by a factor of about 72!

The OP also indicated 3.5 gallons of evaporation per day, but used 12 hours as the divisor. So is it 3.5 gallons ever 12 hours or 24 hours? Hrmmm...

Anyway the calculated thermal gains and losses per day are not accurate by a long shot. The thermal conductivity calculations are also way off. Only ~1700BTU/hr of heat loss in tank that is 12° F hotter than the room with 3.5 gallons of evaporation every day? The number should be closer to 5000 BTU/hr for the conduction, giving a total heat loss of around 7000 BTU/hr.

In other words, the entire opening post to this thread is full of errors and should not be used as a learning tool for thermodynamics and how they relate to your aquarium

[rbursek]

Bean Animal,
never sean you here, thought just a RC expert!!!! Nice to see you here!!!!

[reef dummy]

My aquarium is a 120g and has very similar lighting. My room temp. hovers close to 68-70 most of the time and I agree the calculations are wrong beananimal. I evaporate somewhere between 1.75 and 2g a day.

[mps9506]

Thanks for the corrections.

[zy112]

Haha... I'm an engineer too, mechanical. Yes, tons of thermo, heat transfer, thermo comp, ect classes. But I woulda just set the tank up and added a chiller if needed. Nice calcs though! The only thing I see is there should be heat transfer due to convection as well as conduction to get the sum of all DT.