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# Martian Radiator Design

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Member Posts: 2
Hello,
I am a mechanical engineering student working on designing a radiative hot water heater for a small room designed to withstand Martian temperatures. I am working on the earth analog system and am looking for guidance on a few design parameters. The room is small (8mx4mx9m), and is underground and well insulated (0.1524m of xps insulation). Analysis has shown that heat transfer into concrete and dirt is negligible. The system will have access to 95C hot water. The heating requirements for the system is to cyclically heat the system from a low steady state to a high steady state (Delta of 11C). Using first order approximations, to heat this room in 2 hours, it would require around 600W. This number is currently a liberal estimate due to waiting on final calculations of interior thermal mass and other heat loss elements. I would expect this to at a max require 200 more watts. Assuming 800 watts required over 2 hours, using q=mc(T1-T0), the water mass flow rate required is 1.72 L/m with a water temperature delta of 10C. This number doesn't account for inefficiencies in the radiator, so I am assuming a safety factor of 1.5 on this calculation. The calculated radiative surface area required is .504 m^2, assuming an overall heat transfer coefficient of 45.

My main questions are:
Is using a 10C delta in the water temperature reasonable?
Is 1.72 L/m a reasonable flowrate for a hot water radiator?
Is .504 m^2 a reasonable surface area required?
Is an overall heat transfer coefficient of 45 reasonable, or is there a resource to find a value or range of values for this application?

• Member Posts: 9,840
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This sound like you're trying to get us to do your exam or homework for you. the transfer i to concrete and the earth is not negligible, that is only something that would be done on a question in a class.
• Member Posts: 2
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I'm not, I am asking if my numbers make sense, I don't have much experience in room heating design. The transfer to concrete and earth is negligible, I did the nodal analysis of the system to prove it. The minimal heat transfer pass the insulation is accounted for in conservative design choices made in the calculations
I have found useful and relevant information on this topic to be somewhat difficult to find. I have done the work, I am asking if these numbers are within reason for people who have more experience designing these systems than I do. I won't be able to experimentally confirm any of this for 3-4 months. This is an actual system that is actually being built.
• Member Posts: 3,356
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Hi @Airlaz , A lot of the people who visit this place don’t work or think in metric units. I’m imagining that you might get more responses if you presented your questions in a way that didn’t need converting to easily understand.
Yours, Larry
• Member Posts: 23
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Delta T is typical and should be fine.
Flow rate for such a small heat load and area seems okay as well.
200w is not much heat so it would seem the radiative surface is adequate.
I am not sure about the transfer coefficent. 95C is high enough temp that it should work fine as a radiator.
There should be some convection as well.

I assume all surface including base are insulated the same if the heat load is only 600w?
Tom Gocze
• Member Posts: 2
edited May 2023
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I'm a bit confused by the plot. It appears everything is starting at about 275K... and the Node 0 (which I assume is the surface of the insulation exposed to the interior of the room) quickly heats to something like 296K. That's a 21K deltaT in a short period of time - whereas your original statement is looking for an ultimate 11K dT. Second comment is that the heating water dT will vary as a function of room air and room surface temperature (i.e. assuming circulator flow is constant). As the room heats up, there is less driving force to pull heat from the heating water through radiative and/or convective heat transfer. Anyway, the problem at hand is to isolate the modes of heat transfer and describe them using appropriate heat-transfer equations and bundle that all up in an arithmetic or numerical solution. Additionally, I think your comment about negligible heat loss through the concrete and dirt does not make sense. In fact, the thermal mass (think specific heat) of the concrete will play a large role in the dynamics of heating this room - as you must bring up the temperature by some amount to achieve thermal equilibrium in the room. Same for the insulation. Perhaps you can assume the dirt is at a constant temperature - just like on earth... (i.e. the dirt underneath the floor of my unfinished basement is always about 55F.) Conduction is involved in heating up the concrete and insulation solids. It will have a temperature gradient that needs to be figured out... What is the thermal conductivity? Are there interface resistances (between discrete layers) to include in the analysis?
• Member Posts: 110
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John Gray, an author and PhD, Wrote a book tiled: "Men are from Mars, Women are from Venus". Never read it 'cuz I'm not from Mars nor do I plan to visit. As for other men and women, I'll let them decide.
Where on Mars can anyone play golf, go hiking, fishing, camping, ride a bike or motorcycle, take a Sunday drive to the mountains or ocean, visit a museum or even have a barbecue with family and friends? I dunno!
I like earth best. It might be, "The Devil I know is better than the Devil I don't", But It's comfortable. I can take a shower, go out in the backyard with a hot cup of coffee and watch the birds and squirrels, smell the fresh air and relax without a spacesuit. Then go out and fix a broken steam boiler.

Dave
• Member Posts: 23
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He states the room is insulated with 6" of XPS/styrofoam. I assume that means all six faces. Soil over that also has some insulative value.
Tom Gocze
• Member Posts: 55
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Where's the node for the room air?  I assume that would be the most important variable of all.
• Member Posts: 178
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Airlaz said:

I'm not, I am asking if my numbers make sense, I don't have much experience in room heating design. The transfer to concrete and earth is negligible, I did the nodal analysis of the system to prove it. The minimal heat transfer pass the insulation is accounted for in conservative design choices made in the calculations
I have found useful and relevant information on this topic to be somewhat difficult to find. I have done the work, I am asking if these numbers are within reason for people who have more experience designing these systems than I do. I won't be able to experimentally confirm any of this for 3-4 months. This is an actual system that is actually being built.

As others have noted, the transfer into the earth and concrete is not negligible, and I don't know how you built your model for heat loss that concluded that. Given that you present a chart of temperatures vs time, I think you are confusing a lack of change in temperature with a lack of heat transfer. For example, you will see an ice bath remain fixed at the freezing point while tons of heat is going into the bath, and melting the ice in the process. Only once the ice is all melted would you see a change in temperature, yet vast amounts of heat have been transferred.

What is the temperature of the dirt surrounding your room? Your graphs seems to show something like 275K, which is ~35°F, which seems way too warm for Martian soil. Maybe something more like 200K (-100°F) would make sense? That value may well remain a constant, but it will absorb heat at a very non-zero rate, and you need to figure that value into whatever heat you plan to add to the room. Given the insulation stated, what is your steady-state heat loss that you need to maintain the equilibrium at your low temp? Same question at your high temp. They will be different, and you need take those values into consideration with regards to how much heat you need to add when you want to raise the temperature of the room.

Your "small" room is about 26.25'x29.5'x13'. That is the size of many small houses in the US. Your design radiator is about 5.4 sq-ft. which seems very small in comparison to steam radiators that folks on this forum are familiar with. I would expect to see at least 100 sqft of steam radiation for a 775 sqft house in New England.

In my own home, the calculated steady-state heat loss for my 21'x40'x8' basement (3 sides in contact with the earth using 2" XPS and heated environment above) is 13000 BTU/hr (3812W) with interior temperature of 68°F and outdoor design temp of 1°F. I think Martian soil is a lot colder than what we have around here, so that number would be much larger (like many multiples of 3800W) if I put it on Mars. You have better insulation, but your surface area in contact with the heat sink is 75% larger, so I'm thinking you need an order of magnitude more than 800W just to maintain a reasonable interior temperature.

To answer the question you asked: 10C ∆T is a reasonable design parameter. 95C water temperature is fine. For 800W of heating the radiator also seems like it could be fine, but I don't think you've done the right calculations to understand what the steady-state heat loss will be, and thus that 800W figure and the size of the required radiator are probably way short of the mark.

If you're actually going to build it, and use these parameters for heating it, you'd better plan to dress for a very cold environment.
• Member Posts: 8,153
edited May 2023
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Let me call My Uncle Martin. He might have some first hand experience.

Seriously, you will need to know what the temperature below the surface of Mars is. Only then can you know what the heating load inside the conditioned space is going to be.

If you are building this structure below the surface of the earth and the ground temperature is 40°F. (4.44° C) and you want to maintain a 70°F (21°C) in the conditioned space, then your load calculation must be based on a 30°F ∆T (about 16.7°C ∆T). depending on how deep you go underground, the ground temperature will change. Also the Latitude on the Earth will have different temperature ranges. I would think that the Martian subterranean temperatures are much lower than what you might find on Earth.

Once you know the inside and outside temperatures you can then calculate the U factors of the exterior walls, plug them into a load calculation form and do the math. It all seem quite simple to me.

Edward Young Retired

After you make that expensive repair and you still have the same problem, What will you check next?

• Member Posts: 8,153
edited May 2023
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Since @Airlaz is not a professional HVAC system designer, I understand that what we all take for granted as normal operating procedure, may not be as easy to grasp by someone who is planning to visit Mars. A quick heal loss course might be in order. This is the slide in my Hydronics class that introduces the need for a proper load calculation.

(This is in Fahrenheit and imperial measurements like 12" = 1 ft.)
I then go on to explain that by measuring the exterior walls that are exposed to the outdoors must be measured and the R value of that assembly must be known. (in @Airlaz 's case that will be easy since they are building with only one wall floor and ceiling material) Then the outdoor temperature must be known. In the trade we call it the design temperature. (For @Airlaz that will be the ground temperature and it will stay mostly constant). Then we must account for infiltration. (On Mars I don't think there will much outside air infiltration). Get the total square footage of the entire envelope (walls, floors, ceilings, doors, windows etc.) and multiply by the U value then multiply that by the ∆T. ...There is your load calculation.

That is the crash course and then I sell the students on taking the Manual J Class we offered. It’s not a comprehensive discourse on the subject but I was teaching Hydronics, Not Manual J

I hope this helps @Airlaz with the project. If you build this experiment properly, It might be a great place to stay during a nuclear attack.

Or we could all just Vote smarter.

Mr. Ed

Edward Young Retired

After you make that expensive repair and you still have the same problem, What will you check next?

• Member Posts: 23
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He states the room is insulated with 6" of XPS/styrofoam. I assume that means all six faces. Soil over that also has some insulative value.
Assuming no infiltration and no soil insulation, the heat loss is roughly about 4,000 btus/hr.
Tom Gocze
• Member Posts: 23
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I figured a Delta T of 100F.
Tom Gocze
• Member Posts: 3
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You say: Analysis has shown that heat transfer into concrete and dirt is negligible.
I suggest 0.1524m of xps insulation, (6 inches for most of us), is negligible for the extreme temperature difference between a 20°C interior and -80°C exterior.

https://www.aeronomie.be/en/encyclopedia/mars-climate-important-temperature-difference-between-day-and-night ... +27°C during the day, -133°C during the night ... 100°C delta T assumed soil is average temperature, ( which is probably incorrect as ... https://www.nasa.gov/mission_pages/msl/multimedia/pia16081.html ... indicates soil temperature variation is more extreme. I misplaced the link for insulation value of Martian soil, but recall it as about 6°C per kilometer, so pretty much negligible. Also, our radiators are misnamed, as most of the energy transferred is by conduction. Also we conduct the heat to air, I suspect there will be little nitrogen in your room, and it may not be at 1 Bar.