Tarm Solo Plus 40 Data

[jebatty]

This might be interesting to some. The following chart shows data from operation of my Tarm yesterday.

Data: As with all data, it's only as good as the inputs. All water temperature readings are taken with probe milk frothing thermometers zip tied to the pipes and then wrapped with insulation. Before using the thermometers, I placed all of them on the same heat source and then adjusted them so that they all read the same, my way of calibrating them. I don't know absolutely how accurate they are. A Rutland probe thermometer is in the flue.

The 8 gpm for the boiler loop is taken from the Taco 009 pump curve chart and is assumed, as total head is estimated to be 5-10 feet. The 7.5 gpm on the system loop is calculated based on measured flow with a flowmeter reading 5.8 gpm, which has about a 2 psi pressure drop, and then operating the circulator without the flowmeter which achieves about a 4.6 foot decrease in head. Maximum gpm's for a 009 is about 8.

System: Tarm Solo Plus 40 with 50% antifreeze. Plate HX 5 x 12 x 30 plate; boiler:1.25 copper to 1.00 copper. System heat load is 100% to pressurized storage, 1000 gal.; 1.00 copper to 3/4 pipe.

Test: The boiler was cold started. An aquastat on the boiler return line after the Termovar is set to ON at 155F. This turns on both the boiler and system circulators at the same time. Since the aquastat is fed by gravity flow, when it turns ON, the top of boiler water temperature actually is greater than 155F.

Temperature readings were taken every 10 minutes during an ON or OFF cycle.

Example: At 12:31 the aqustat turned on both circs. This first ON cycle was 76 minutes long before the boiler cycled off. Boiler water input to the HX was 170 and HX return to boiler was 130. The return to boiler after the Termovar was 148, which meant that the Termovar was mixing some boiler output directly back to boiler return. Boiler side BTU's were BOILER INPUT TO HX - BOILER RETURN (deltaT) X 8 gpm x 500 x 0.85. The 0.85 factor relates to 50% antifreeze solution, which has about 85% of the heat capacity of water at 160F.

On the system side, tank input to the HX was 130 and HX output to the tank was 145. System side BTU's were HX OUTPUT TO TANK - TANK INPUT TO HX X 7.5 gpm x 500.

Note: in this first entry BTU's available on the boiler side were 70,400 but that BTU's extracted by the HX were 56,250, for an 80% transfer efficiency. Note also that as system input temperature rises, HX heat transfer efficiency decreases.

Note: that as system input temp rises, the boiler starts to cycle through idle periods. With the given gpm's and HX specs, the system side cannot keep up with boiler output as system temp rises. More gpm's and/or a larger HX would reduce cycling.

Note: if water rather than antifreeze were used in the boiler loop, boiler BTU output to HX would increase by about 18%, and HX output to system also would rise adjusted by the heat transfer efficiency.

Other: the data is what it is. I don't think I made any errors. Obviously, to the extent that temp readings are inaccurate, or gpm's are inaccurate, the BTU calculations will not be accurate.

Have fun for this if it makes any sense to you. Ask questions if you wish.

 

[Nofossil]

Not sure that I have all of this, but BTUs don't just disappear. Heat coming out of the boiler has to equal heat into the loads. If the only load is storage, then the heat gained by storage has to equal the heat produced by the boiler. There's a TINY amount lost from the plumbing between the two, but it's insignificant unless you have a long run of buried pipe.

Heat exchangers are 100% efficient in the sense that the BTUs into side A exactly equal the BTUs out of side B. HX performance is more about reducing the delta T between side A supply and side B return.

 

[jebatty]

Your comments really piqued my interest, as there appeared to be a logic, yet my gut tells me something in what you said is not right. I spent some time trying to find info on plate hx efficiency, of which there is plenty, but the math is beyond me. Then I found this website dealing with use of plate hx's with solar systems, and came across this statement:

Flat plate heat exchanger sizing guidelines are based on an effectiveness of 0.70-0.80 that provides the optimum balance between system efficiency and unit cost. http://www.sunearthinc.com/sunplate.htm

At this time I am not able to critically analyze this statement but will take it at face value. Assuming its correctness, and assuming this is the correct calculation, if I divide hx system output BTU'S by boiler output BTU'S, performance shown by my chart is on target, ranging between about 0.60 on the low side and 0.92 on the high side.

I also applied the GEA flat plate selection software on this chart example: Boiler I=193, Return=160; System I=145, O=170; Boiler flow=8 gpm; System flow=7.5gpm; Boiler 50%=antifreeze, System=water. Boiler BTU=105600; System BTU=93750.

First calculation, leaving Load blank and leaving Boiler input temp blank.
Side A leave temp=160
Side B enter temp=145
Side B leave temp=170
Recommended hx=5 x 12 x 40; Load=91,825 BTU/hr

Second calculation, leaving Load blank and leaving Side B flow rate blank.
Side A enter temp=193
Side A leave temp=160
Side B enter temp=145
Side B leave temp=170
Recommended hx=5 x 12 x 40; Load=115,180 BTU/hr

My hx is 5 x 12 x 30. It seems that performance in this example also is on track with performance indicated by the selection software, if "fudged" by my hx which is smaller than recommended.

Further discussion is warranted, but it seems to me that my chart provides real world info which may be useful in helping boiler users understand their system performance when using a plate hx.

The most uncertain variable in my calculations, besides simple accuracy of the thermometers, is my gpm assumptions. I think I'm pretty close on these, but, if in the example above system gpm's=8 rather 7.5, then system BTU's=100,000; or if gpm's=7, then system BTU's=87,500.

The Tarm 40 is rated at 140,000 BTU for water. I think only one data point on my chart achieved this output: Boiler output=121,600 BTU with antifreeze at 198F; with water output=121,600/.865= 140,578 BTU (antifreeze heat capacity at 200F=0.865).

For those using the chart only to help understand real world boiler output, if you divide the boiler BTU column by 0.85, then you will have BTU output using water at assumed 8 gpm.

 

[Nofossil]

This is a great example of why ordinary language is inadequate for mathematical problems. 'Effectiveness' and 'efficiency' sound like they might be synonyms. In this case, they are not.

Basic thermodynamics: Energy cannot be created or destroyed. Every bit of chemical energy in the log you burn goes somewhere - either up the flue as heat or as chemical energy, or out through the sides of the boiler, or into the water. The heat energy that leaves the boiler has to go somewhere - that is, is has to be transferred to some other object. The challenge is to figure out how much energy is leaving the boiler and where it's going.

The heat energy is going into raising the temperature of other stuff. Some of that might be the air around the pipes, but mostly it will be the water in your storage. It can't be lost in the heat exchanger, because it can't be lost at all. It can't be absorbed by the heat exchanger because that would result in the temperature of the heat exchanger rising continuously which clearly doesn't happen.

It might be helpful to think about the 'black box' approach. Draw a conceptual black box around and object, subsystem, or system. The energy in the black box can only change by the amount entering or leaving. For instance, if you draw a black box around your boiler, energy can come into the box via wood fuel (chemical). It leaves the box via three paths: up the flue (thermal and chemical), into the surrounding air (thermal) and via circulated water (thermal). In a steady-state situation where the boiler temp remains constant, the energy in has to add up exactly to the energy out. In reality, there are other small scale nuances such as electricity and inlet air, but hopefully this makes sense.

If you draw a black box around the heat exchanger in a steady-state situation (heat exchanger temp is constant) the same rule applies: the energy in MUST equal the energy out.

Let me take a shot at the difference between efficiency and effectiveness. Efficiency usually describes the percentage of energy that's transformed from one form to another in a usable way.

For instance, a gasifier under ideal conditions transforms about 90% of the chemical energy in a piece of wood into heat energy in the water at the boiler outlet. An outdoor boiler system might deliver 60% of the chemical energy in a piece of wood as heat into the remote building. In each case, energy is being 'lost' only in the sense that it's leaving the system in an undesirable way. The amount of energy balances perfectly.

Effectiveness means something slightly different. It usually describes the percentage of the desired work that was actually accomplished. If you wanted to transfer 50,000 BTU of heat energy but you were only able to transfer 40,000 then your effectiveness is 80%. The other 10,000 BTU is not lost, it just never got transferred.

Here's an illustration. Let's assume that we have an outdoor boiler that produces 100,000 BTU per hour at the boiler outlet. It's used to heat a remote pressurized storage tank. After three hours, the tank temperature has increased by 240,000 BTU. The other 60,000 BTU has been lost to the ground surrounding the buried pipes. The system efficiency is 80%. You'll have to burn enough wood to generate 100,000 btu for every 80,000 BTU that your house actually needs.

Next example: Indoor 100,000 BTU boiler with unpressurized storage and a heat exchanger. No piping loss, but the heat exchanger is undersized such that it can only transfer 80,000 BTU/hr at average storage temperatures. What happens in this case is that the boiler temperature will rise - it's producing 100,000 BTU/hr, but only 80,000 BTU/hr is being transferred to storage. Very soon, the boiler will idle until its average output drops to 80,000 BTU/hr. The 20,000 BTUs are not lost - they stayed in the boiler side of the system.

Don't know if this helps, but if not the failure is in my powers of explanation, not in the laws of thermodynamics ;-)

 

[jebatty]

I think we are in precise agreement on the theory. My data chart was intended to provide real world "effectiveness" of a plate hx in with my Tarm under the circumstances identified. The theory is very useful to understand, and actual BTU's delivered is useful in providing the needed heat energy.

 

[Nofossil]

I think we are in precise agreement on the theory. My data chart was intended to provide real world "effectiveness" of a plate hx in with my Tarm under the circumstances identified. The theory is very useful to understand, and actual BTU's delivered is useful in providing the needed heat energy.

I was pretty sure that you had the theory firmly in hand. However, there are a lot of people who believe that heat is lost in a heat exchanger. Thanks for letting me use your thread as a platform for a soapbox.

To make up for any trauma that I may have unintentionally inflicted, here's a spreadsheet that I use to study what's happening in my system. This is a 24 hour snapshot, 8:00am to 8:00am.

On this particular day it was cold and we were away until 8:30 in the evening. That meant that I couldn't start a fire until then. You'll see that I started another fire right about 8:00 the next morning.

Apropos to this thread, I calculate for the period how much heat is produced during the burn (366,166 BTU), how much heat is used by the loads (434190 BTU) and the difference in the amount of heat in storage (-101858 BTU). These should sum up to zero: Heat used + storage delta - heat produced = 0. That never works out exactly (surprise!) and on this day I used 33,834 more than I can account for. This is almost certainly heat from storage used to preheat DHW. I have no way of measuring that heat load, so it shows up as an error.

Sitting behind the calculations is a model that takes into account the baseboard output as a function of water temperature. There are some crude approximations - I only have three temp sensors in my storage and only one in my DHW tank.

 

[Jackpine Savage]

Thanks for taking the time to put together that data guys. That will be very helpful to compare to when I finally get mine on-line.