Siphon-Heat Exchange-Electric Generation

[Catach]

Hello All
I am new to this great forum. I would like some thoughts on my idea which is:

I am designing a passive solar greenhouse for aquaponics. I am concerned about excess heat in the summer (I am in western NY Sate). A requirement is air exchange for a healthy interior environment. I am also trying to keep the greenhouse as protected from outdoor pests and airborne plant disease (closed controlled environment).

Ok, that said, I have a small steam - feeder creek from hill runoff - so limited flow.

I would like to create a siphon by placing a 2" pvc flexpipe upsteam then go upgrade to the greenhouse, pass through a exchange unit (possibly a truck radiator), blow air through the radiator to provide cooling from the 50-60 degree F water, continue flow back to the creek completing the siphon action and placing a turgo or pelton style micro-hydro generator at the flexpipe outlet to the creek.

That is the concept.

Distance from pipe inlet to radiator @800 ft
Elevation lift from pipe inlet to radiator @15 ft
Temp differential in greenhouse @35 degrees - hotter air temp to cooler water temp
Elevation drop from radiator to pipe outlet/generator @45 ft
Distance from radiator to creek @ 250 ft
Pipe ID 2" full flow only using siphon action

My first concern is can I maintain a siphon over that distance with a radiator interrupting the peak of the siphon. Would cavitation occur? If not, then first hurdle crossed.

Second, is there enough temperature differential and is a radiator a practical inexpensive way to go for a cooling exchange unit?

Third, with a siphon, I would assume the only working head that I can calculate for micro-hydro generation is the difference between the original pipe inlet and the outlet or 30 ft, not the drop from the radiator to the outlet or 45 ft? Would the first 15 ft (or more) of head be used just to maintain the siphon?

Any input would be greatly appreciated.

 

[tyroman]

Welcome Catach,

Seems to me the first step in design would be to calculate the cooling load and the generator load in order to know the required water flow rate.

You seem to have concluded that the pressure (head) drop in 2" flex pipe over 1250 feet plus the drop across the radiator and the drop across the generator will be less than 30 feet of head at the required flow rate... Is this true?

If so, what is the flow rate you have calculated?

 

[Catach]

Hi
I haven't done the calculations. I was hoping to get some help from those of you that are well versed in hydraulics. I have limited the pipe dia to 2" as a good sized truck radiator inlet and outlet is 2" (at least I haven't found any larger yet) as well as the most economical flex pipe is 2".

So, max flow rate of 2" flex pipe, no pressure-just gravity, I am finding @55 gpm.
I am under the impression that using 4" flex pipe and limiting it to a "bottleneck" of the radiator at 2" wouldn't work for a siphon. Let me know if I am incorrect here.

Here again it is gut feeling that I would be losing head with the restriction of the radiator and the friction loss in that length of pipe. I am guessing that I may only end up with 20-25 ft of head (wishful thinking?)

As far as the generator, it would be at the outlet (45 ft below radiator). I was hoping to get some input here as well, as to the remaining gpm and hea, so that I could calculate the potential wattage. If there isn't enough to warrant the expense of a pelton wheel, than I will just use the siphon for cooling. That is if a siphon can be maintained with the radiator restriction?

 

[Catach]

Does anyone know if a siphon can be maintained with the information I have given?

 

[tyroman]

Hi,

Some observations on your plan;

1. Have you considered a solar powered fan such as;
http://shop.solardirect.com/product_info.php?products_id=805
in lieu of the micro-hydro generator? Solar might be the best option since cooling demand will coincide with sunlight and the head/flow rate available to a generator is questionable in your case, see;
http://www.altestore.com/howto/HydroPower/Micro-Hydro-Power-Systems-Overview/a27/
"A simple equation estimates output power for a system with 53% efficiency, which is representative of most micro hydro systems:

Net Head* (feet) x Flow (US gpm) / 10: Output (Watts)"

* this will be the head remaining after friction losses in all the piping (including the radiator).

2. Cooling load will be a function of the efficiency of the radiator, the volume of the greenhouse, surface area and thermal properties of its walls and roof, etc. So more info is needed to calculate your cooling needs (and thus your flow rate needs).

3. Not to be a "wet blanket", but from a practical standpoint, have you considered the maintenance issues involved in using your water source; "a small steam - feeder creek from hill runoff - so limited flow"?
a) If the runoff is intermittant, you will lose your siphon's prime every time the source dries up.
b) Being runoff, the supply is likely not to be very free of sediment, or other things which may require periodic back-flushing of the radiator tubes.

4. Cavitation damage is very unlikely at your flow rates... however, any opening to atmospheric pressure in the piping system will break the siphon, see;
http://www.engineeringtoolbox.com/cavitation-d_407.html

 

[Catach]

Thank you for the response
My main objective was a never ending flow of cool water, with no outside energy expended, that went through the greenhouse for cooling. Afterthought was, hmmm, I may even be able to get some electricity, however small, as a side benefit.

If cavitation isn't an issue, then I can maintain the siphon, if no air leaks are present. So, I have cool water in a warm environment. Isn't there an equation to calculate the potential cooling (btu's ?) if the water temp, gpm, temperature differential, and rate of air flow through through the radiator are known? The space it cools shouldn't be a factor, should it?

As far as the stream/creek it never totally dries up and trash catchment and screening should alleviate most potential problems.

Thanks again

 

[tyroman]

OK, let's take it apart into two pieces, the hydraulics and the thermodynamics...

Hydraulics;
Please obtain this useful reference "Cameron Hydraulic Data" which

can be found at http://www.cameronbook.com/
or downloaded at http://download-zzz.com/?mod=server&id=qwertyweaxaecaw&name=cameron_hydraulic_datafull.

Your system will have 30 feet of head available... (vertical distance between source and discharge).

In order to flow, the total flowing pressure drop in the system must be less than 30 feet of head. As flow increases, pressure drop increases.

Just a straight 1250 foot run of pipe (no screens, filters, bends or radiators) will allow you a loss of 2.4 feet of head per 100 feet (2.4x12.5=30).

I don't know the hydraulic characteristics of "flex pipe", but in 1250 ft. of (old*) 2" ID tubing your max flow rate (not allowing for any other losses) is about 30 gpm. (See Cameron table for friction losses in tubing... 2 inch type L tubing has 2.45 ft. head loss per 100 ft. at 30 gpm.)

* old tubing is assumed since the internal surface roughness increases with age and; alas, everything is eventually "old".

So, excluding the drop across the inlet filter, radiator and other components, your max flow rate through the radiator will be less than 30 gpm.

Thermodynamics;
The water temp is 50-60 degrees F and the air temp in the greenhouse is (85-95 degrees F?).

If only convection moves the hot air across the cool radiator you will achieve "some" cooling depending on the location of the radiator within the space. I do not have any experience with such a design, perhaps another forum member could help with this... I do feel confident that the efficiency of this design would be very low.

If you opt to use a fan (solar powered or otherwise) to move the air across the radiator, the heat exchange would be significantly improved.

I realize that these comments are very general and qualitative... however; without more specific information, a more quantitative reply is not possible.

Good luck and let us know how your plans work out.

 

[Catach]

If I understand you correctly, the 1250 ft of 2" pipe friction loss is equivalent to 30 ft head loss, which is the same as the total starting head of 30 ft, leaving me zero head at exit. When the loss is calculated in for screens, bends, radiator I will not be able to maintain the siphon as the total flowing pressure drop will now be greater than 30 ft?

I am planning to either put a fan on the radiator and/or have an exhaust fan that will create a negative pressure on the greenhouse space with the only inlet of air to the structure is through the radiator.

Thanks for the detailed input and the links that you have provided

 

[tyroman]

Guess I could have been more clear in my explanation... let me try again.

Adding components to the system will not STOP the flow (unless its a valve)... they will REDUCE it because they increase friction. (Think of what happens to flow when you start to get a kink in your garden hose.)

(Note: Caps are for emphasis... not a shout)

I think the sentence which may have confused you is as follows.
Quoting myself:
"In order to flow, the total flowing pressure drop in the system must be less than 30 feet of head."
Re-wording the above:
The system will flow at a rate such that the total flowing pressure drop in the system is equal to 30 feet of head.

Let me know it this does not clear up the question.

 

[gmax137]

Hi
I have limited the pipe dia to 2" as a good sized truck radiator inlet and outlet is 2" (at least I haven't found any larger yet) as well as the most economical flex pipe is 2".

There is no reason to think that the piping size needs to match the radiator inlet nozzle. Increasing the pipe size will dramatically reduce the friction losses. These losses go as the fifth power of diameter - so using a 3 inch pipe (ID = 3.068 inch) vs. 2 inch pipe (ID = 2.067) reduces the head loss by

$$(\frac{2.067}{3.068})^5 = 0.14$$

or a factor of seven. (Note these diameters are for normal schedule 40 pipe, since I don't know what the sizes are for 'flex pipe' - and if 'flex pipe' is that corrugated stuff, you should re-think the plan, as I'm sure the losses through that will be much higher than for common steel pipe.

There will be some increased shock loss at the reducers at the radiator inlet & outlet but considering the length of pipe you have you can determine if the larger pipe size is 'worth it.'

But back to the beginning, I think you need to find out what the stream flow rate is first, (and make sure it is all 'yours' to divert). Then you can size the pipe; and then you can see if it is capable of providing the cooling you need. For the hydraulics, get Cameron (as suggested above) or the Crane 410 book. These have all of the charts & equations needed.

Regarding 'cavitation' what you need to do is calculate the pressure at the high point in your pipeline. If it is above the vapor pressure at the flowing temperature you will be OK. A more realistic concern will be leaky joints - if air can get in it could break the siphon.

Finally, the turbine idea seems weak to me (but I haven't done any calcs) other than this: using your 30 gpm at 30 feet, the water horsepower is

$$\frac{30 * 30}{3960} = 0.22 hp$$

Not much to work with, and the actual power produced will be lower due to the turbine efficiency and all of the other losses in the system.

Good luck

 

[Catach]

This is how I understand it and please correct me where needed.

Total length of pipe 1,050 ft, let’s just round off to 1,000 ft, (800 ft up and 200 ft down). If I go uphill 15 ft and then drop 45 ft, I have a gross head of 30 ft.

According to engineeringtoolbox.com/pvc-pipes-friction-loss-d_802.html , a 2” pvc pipe has 0.07 loss/100 ft of pipe at 5 GPM flow and a velocity of 0.5 ft/sec

With a flow of 5 gpm * 7.48 gal/cu ft = 37.4 cu ft/min = 0.62 cu ft/sec

With a friction loss of 0.07/100 ft and a total length of 1,000 ft = 0.7 ft head loss

Total head of 30 ft – 0.07 = 29.3 ft net head

With an efficiency factor of 60%, for electric generation:

Hp = (0.62*29.3*0.6)/8.8 = 1.245 (8.8 is a correction factor given in a book I have on water power)
kW = (37.4*29.3*0.6)/708 = 0.929 (708 is a constant factor given in a book I have on water power)

No radiator loss calculated. (I have no idea how to figure that one in).

Let’s say we lose 9.7 ft head for the radiator ? giving a total head loss of 10 ft. The net head is now 20 ft.

Hp = (0.62*20*0.6)/8.8 = 0.850
kW = (37.4*20*0.6)/708 = 0.634

So, back to one of my questions, can I maintain a siphon?
If so, how can I figure my cooling potential?

Do I use a formula like one of these? Or Both? Or something different? I assume I need to calculate the water flow as well as the air flow combined?
Q= 1.08*CFM*(T2-T1) or Q=500*gpm*(T2-T1).
Q= BTU/h
CFM= air flow - Cubic ft/min
gpm = gal/min
T1 and T2 air the 2 temperatures

From the initial post, Temp differential in greenhouse @35 degrees. Let’s use a 1,000 cfm fan
Q = 1.08*1,000*35 = 37,800 BTU/hr (water flow)
Q = 500*6*35 = 105,000 BTU/hr (air flow)
Hmmm, big discrepancy, so the formulas must need to work in conjunction somehow?

 

[tyroman]

Good!

You have used one of the tools provided... the likelyhood of sucess with your plan has increased significantly. :-)

HYDRO

Now, please realize that the PVC pipe tables you found are for rigid, smooth walled pipe... not flexpipe. PVC is apparently even smoother inside than copper tubing! That, or Cameron's tables are more conservative than engineeringtoolbox.com's (Cameron does not offer data on PVC).

Flexpipe is probably going to be your best choice for various reasons (terrain, economy, DIY ease of installation etc.) but it is not smooth inside and will therefore present more resistance to water flow. Can you get data on flexpipe through the vendor/manufacturer?

By the way, I noticed an error since my last post... I think I added 200 ft to 1,050 as a 20% preliminary design cushion - and then promptly forgot that I had done so...:-(

I haven't gone through your generator calcs (can you say where you found the equations?) But, your solutions seem to give quite high values for power compared to what I calculate (seven points of efficiency aside)...
see; http://www.altestore.com/howto/HydroPower/Micro-Hydro-Power-Systems-Overview/a27/
"A simple equation estimates output power for a system with 53% efficiency, which is representative of most micro hydro systems:

Net Head* (feet) x Flow (US gpm) / 10 = Output (Watts)"

Solving for your new case, I get:
29.3 (feet) x 5 (GPM) / 10 = 14.65 (Watts)

Also, see gmax137's calculation above where gets only 0.22hp @ six times the flow rate.

My gut feel is that solar is your best bet.

You said "No radiator loss calculated. (I have no idea how to figure that one in)." That is a difficult thing to calculate (even when you know the size and number of tubes involved) so the manufacturer determines the performance of their equipment experimentally and publishes the data in tabular form. They do the same for thermal performance. When you decide what you want to consider using, be sure that such data is available and that it covers the operating range you need. Selection of equipment is an iterative process... get ALL the data for everything and finalize a design which will work before you buy anything.

THERMO

From your post:
"Q= 1.08*CFM*(T2-T1) or Q=500*gpm*(T2-T1)"
Not sure where these equations came from but I suspect that they are to be used differently than you have... The CFM formula must apply to air where the two temps are the air before and after the heat exchange has occurred and the gpm formula likewise with water temp before and after.

Theoretically, if you could get 100% efficiency out of your radiator/fan system you could raise X GPM of 60 degree water up to the temperature of the air (95 degrees) as the water passes from the cold inlet side of the radiator to the hot outlet side (this is impossible in practice since the rate of heat transfer from the hot medium to the cold medium goes to zero as the two temperatures get closer).

One BTU is the heat energy required to raise the temperature of one lb of water one degree F. One BTU is equivalent to 778.0 ft-lb of energy.

Water weighs 8.34 lb per gallon @ 60 degrees F, so X GPM of water is equal to 8.34X pounds per minute. Let's say that your creek flows at least 10 GPM 24-7-365 and that you are able to design your system such that all of this flow is achieved. Then the "theoretical" BTUs taken out of the greenhouse air would be calculated as follows:

(10 GPM)(8.34 lb per gallon) = 83.4 pounds per minute

(83.4 pounds per minute)(35 degrees F) = 2919 BTU per minute

(2919 BTU per minute)(60 minutes) = 175,141 BTU per hour

BUT REMEMBER... this is theoretical and assumes 100% efficiency in the exchanger! Think about a car radiator... it does not lower the water temp to the air temp nor does the air temp rise to the water temp. The fan/radiator system needs to be designed, its efficiency determined and applied to this theoretical calculation. Also, the effect on the bulk air temperature in the greenhouse can not be determined unless we know the volume of air involved and the dynamics of solar heat input through the walls and ceiling. Knowing only the volume of air in the greenhouse, we could calculate how long it would take to lower its temp from, say, 95 degrees F to 85 degrees F... after the sun goes down... but that is a trivial solution. I'm sure you want to know what to expect from sunrise to sunset...

GENERAL

You ask "So, back to one of my questions, can I maintain a siphon?". Yes. Have you read my last post? Please do so and ask again if its still not clear.

Do you have a response to gmax137's great contribution? He raises some valid issues... Water rights could be a significant stumbling block, as could flow constraints... these issues are for you to address, Catach. gmax137 was quite correct in saying that they are THE FIRST priority. To determine the stream capacity, I would suggest a 5 gallon paint bucket and a stop watch. Measure flow at several different times under different weather conditions and come up with the lowest rate you can find, then design to (80% ?) of that rate... no more. I'm sure that after the first time you get the system primed and the siphon started you will not want it to ever "out-run" the supply.

 

[Catach]

Thanks for all the info
In response to gmax137, Yes, I have consulted with our County’s water and soil dept. As long as I get permission from the 2 landowners below me I can dam the stream. It is irrelevant in my case, because I will not be interrupting the flow at all and any dam that I would put up, would be a small containment area, not one that due to failure would cause a major flooding issue.

The flexpipe that I am looking at is not the same as the corrugated flexible pipe. It is smooth and according to one manufacturer – Flexpipe Systems – in Canada, the Flexpipe - Hazen-Williams Coefficient is 150.

The flow rates that I measured, over a 3 month period, during the driest time of the year, gave me an average flow of 8.92 gpm. The lowest flow, after 14 days of no precipitation, was 0.35 gpm. A small containment area should alleviate the flow dropping below 2 gpm and possibly maintaining the 5 gpm that I have used in the above calculations.

For 1,000 ft of flexpipe using the Hazen-Williams Coefficient of 150, flow: 5 gpm pipe dia: 2”
Head loss (psi) 0.3
Head loss (ft water) 0.664
Velocity: 0.51

“I haven't gone through your generator calcs (can you say where you found the equations?)”, tyroman. The calc for net horsepower are given by Robin Saunders, mechanical engineer, from a section of a book on water power, called “Energy Primer” published by Portola Institute 1974. The calc for kW was from a book “Harnessing Water Power for Home Energy” by Dermot McGuigan, Garden Way Publishing, 1978. The calcs for cooling were posted by “Alex” on solarpowerforum.net/forumVB/archive/index.php/t-1589.html

I realize that my electric generation possibilities are extremely low, if not totally impractical. Which I have said, was basically an afterthought – wishful thinking – out the door with that one!

Main objective: potential cooling
Main requirement: maintain siphon

The statement by tyroman, "In order to flow, the total flowing pressure drop in the system must be less than 30 feet of head." I understand this to mean, that if the total head loss (flowing pressure drop) is greater than 30, the flow will stop.

So, let’s go with a lower flow (from my above measurements) of 2 gpm.
For 1,000 ft of flexpipe using the Hazen-Williams Coefficient of 150, flow: 2 gpm pipe dia: 2”
Head loss (psi) 0.1
Head loss (ft water) 0.122
Velocity: 0.2

For 1,000 ft of flexpipe using the Hazen-Williams Coefficient of 150, flow: 2 gpm pipe dia: 10”
Head loss (psi) 0.0
Head loss (ft water) 0.0
Velocity: 0.01

So, should I go with the 10” pipe so that I have no head loss? I don’t think so. I don’t think there is enough volume of water to fill the pipe completely; therefore a siphon can’t be maintained.

I don’t know how to find the minimum flow required, to completely fill a pipe of a given dia., to even think about maintaining a siphon

The cross section of a 2” pipe is equal to 3.14 sq in, 1 foot of said pipe contains: (12” * 3.14 sq in) = 37.68 cu in = (37.68 cu in * 7.48 gal/cu ft water) = 0.16 gal. At a velocity of 0.2 ft/sec = 1.96 gpm with 12 complete exchanges of water through the 1 ft of 2” pipe/min

I don’t know how to use this information!

To begin calculations on the cooling/heating I need to find the largest possible size of pipe I can use to maintain the siphon. gmax137 suggests going with a larger pipe. How large is too large with a 2 gpm flow over 1,000 ft, to still maintain a siphon? Of course, I would like to go with the largest pipe size possible.

 

[tyroman]

Guess I could have been more clear in my explanation... let me try again.

Adding components to the system will not STOP the flow (unless its a valve)... they will REDUCE it because they increase friction. (Think of what happens to flow when you start to get a kink in your garden hose.)

(Note: Caps are for emphasis... not a shout)

I think the sentence which may have confused you is as follows.
Quoting myself:
"In order to flow, the total flowing pressure drop in the system must be less than 30 feet of head."
Re-wording the above:
The system will flow at a rate such that the total flowing pressure drop in the system is equal to 30 feet of head.

Let me know it this does not clear up the question.

Did you miss this post?

 

[Catach]

The statement by tyroman, "In order to flow, the total flowing pressure drop in the system must be less than 30 feet of head." I understand this to mean, that if the total head loss (flowing pressure drop) is greater than 30, the flow will stop.

Yes, I did read it and responded. So I guess I do not understand it.
Because:

For 1,000 ft of flexpipe using the Hazen-Williams Coefficient of 150, flow: 2 gpm pipe dia: 10”
Head loss (psi) 0.0
Head loss (ft water) 0.0
Velocity: 0.01

Is less than 30.
It is 0.0
As I understand your statement the flow will not stop, however, I don't think there is enough water to fill the pipe.

So, I am extremely thick in the skull. Please clarify.
Thank you for your patience with me

 

[tyroman]

Catach,

In the context where it originally appeared;

"In order to flow, the total flowing pressure drop in the system must be less than 30 feet of head."

was intended as a generality to say;
In order to flow {at any given flow rate}, the total flowing pressure drop in {any given pipe size} system must be less than 30 feet of head.

for example in a given case;
In order to flow {at 30 GPM}, the total flowing pressure drop in the {2" ID tubing} system must be less than 30 feet of head.

So that (in the 2" ID tubing case) the total flowing pressure drop at 30 GPM would be greater than 30 feet of head (adding a radiator, strainer and a generator), therefore flow rate will be less than 30 GPM.

I hope this clears up the way the pipeline will work... the system will find its own flow rate depending upon the pressure head available and the total flowing pressure drop it experiences.

Quoting myself again;
"The system will flow at a rate such that the total flowing pressure drop in the system is equal to 30 feet of head."

***************************************************

Now, moving to the real world, in light of the new fact that the water source is very limited, you cannot allow the system to find its own flow rate... you must design the system around a flow rate which the stream (plus some "small containment area") can supply.

A 10" line for example might be able to flow at hundreds of GPM... IF! it is supplied by a RIVER! In your case however, it would quickly drain the containment area - air would enter the siphon - the siphon prime would drain out at the discharge end of the line - and the flow would stop; necessitating the cumbersome task of re-priming the line.

If you design for 2 GPM and then for 4 days the stream flow drops to 0.5 GPM your "small containment area" will have to contain over 8,000 gallons of water.

1.5 GPM x 60 min x 24 hrs x 4 days = 8,640 gal

Would a dam failure in this design "cause a major flooding issue"? Guess that would depend on what / who is downstream of the dam.

Perhaps more importantly; will 2 GPM of 60 degree F water have a noticeable effect on the air temp in your greenhouse which is X feet long by Y feet wide and Z feet high? And which is made from a material whose thermal and transmission properties are "who knows"?

I'm being facetious... Seriously, a little more information?

 

[tyroman]

Catach,

While you absorb my last post, let me offer an outline of things to be done and the order in which they might be accomplished;

Feasibility

1) Decide a reasonable design flow rate based on water supply and operating variables.

This will involve your measurements of stream flow and a realistic size for your containment pond. Also consider how closely the system can be monitored. If 24/7 monitoring of the reservoir level is possible, it would be feasible for you to use a higher design flow rate than might otherwise be realistic (for an un-manned system), on the assumption that you would close valves on the inlet and outlet ends of the line before the reservoir ever empties (thus preserving the siphon prime). Unfortunately, I would expect that low flow in the stream will coincide with high cooling needs in the greenhouse; no rain = no clouds = high solar heat gain.

2) Having decided on a design flow rate, calculate the expected cooling capacity of a radiator / fan system in the greenhouse for 60 degree F water at that flow rate.

This will require preliminary selection of the radiator and fan equipment and a preliminary design for the greenhouse itself. Make some reasonable assumptions on efficiency if needed (this is the "Feasibility" stage). The heat-load variables to be determined in greenhouse design will include its volume, N-S orientation, your Latitude, the surface area and thermal properties of wall and ceiling materials, etc. An important design decision will be how to install the fan / radiator system. Will the fan blow ambient (hot inside) air across the radiator or will it draw cooler outside air through the radiator. The first case (hot air versus cold water) will be more efficent but the second case will provide for air exchange.

3) Come up with a "ball park" estimate (with design cushion ie; assume 1,250 ft. pipe length) for all materials and installation costs and compare this with your budget (and the cost of any alternative cooling methods).

***If the Feasibility stage is acceptable, the next step is;

Detail Design

4) Select pipe size, strainer, valves, radiator, fittings etc. which will result in a total flowing pressure drop of 30 feet of head at the previously determined design flow rate. The preliminary (Feasibility stage) selection of the radiator and fan equipment and the preliminary design for the greenhouse itself are subject to change during the Detail Design stage. A running check on cost should be done. I would suggest sizing everything to allow flow at 110% of design and then plan to use the final outlet valve as a manual flow control valve to "de-tune" the system back to no more than the design rate.

Does this outline seem do-able to you?

 

[Catach]

Catach,
In order to flow {at any given flow rate}, the total flowing pressure drop in {any given pipe size} system must be less than 30 feet of head.

In picturesque terminology; A 55 gallon barrel of water can only flow so fast through a pipe the size of a straw because of the resistance of the straw. It would flow faster if you just tipped the barrel over and let it flow through the 2 ft diameter open top, because there is less resistance?

Are you speaking of Head loss (ft of water) or Head loss (psi)

For 1,000 ft of flexpipe with a Hazen-Williams Coefficient of 150:
Flow - GPM: 60.0
Pipe Dia: 2”
Head loss (ft water): 66.228 (loss Greater than 30 – but OK because it is not pressure loss?)
Head loss (psi): 28.7 (loss less than 30 – OK until more resistance is thrown in that puts it over 30?)
Velocity: 6.13

When you speak of the containment requirements, I know exactly how much water I can impound in my situation, as I have surveyed the land, drawn contour maps, and calculated potential gallons that could be contained. That is not an unknown factor that I am trying to calculate.

If I go with my worst case scenario, as stated above, with only 0.35 gpm, I do not need a containment area and with the following data I am below 30 Head loss, whether it is ft of water loss or psi loss

For 1,000 ft of flexpipe with a Hazen-Williams Coefficient of 150:
Flow - GPM: 0.35
Pipe Dia: 0.75”
Head loss (ft water): 0.57
Head loss (psi): 0.2
Velocity: 0.25

However, I still don’t know if there is sufficient water to fill the pipe completely. I am still trying to find out the largest possible size of pipe for a given flow rate that fills the cross-sectional area of the pipe in order to maintain the siphon.

Again with picturesque terminology; a dripping kitchen sink faucet will not fill the drain to its maximum capacity therefore the sink never fills up. (The resistance of the drain isn’t sufficient to allow it.) If I have only 0.35 gpm flow, I imagine that it would fill the cross-sectional area of a straw but not a 20” diameter pipe. I don’t know how to figure the largest possible size of pipe for a given flow rate that fills the cross-sectional area of the pipe in order to maintain the siphon; therefore I can’t proceed with any calculations.

 

[tyroman]

You said:
"I know exactly how much water I can impound in my situation,"

Good, take that number and together with the stream flow rate which you feel you can comfortably count on, calculate your design flow rate. Until you have done this calculation, it is useless to go further with the design.

You said:
"However, I still don’t know if there is sufficient water to fill the pipe completely. I am still trying to find out the largest possible size of pipe for a given flow rate that fills the cross-sectional area of the pipe in order to maintain the siphon."

I detect a possible misunderstanding on your part about how a siphon is started and why it continues to work after it is started. In your project, the siphon will be "primed" (filled with water) manually.

Consider the following steps;

1. With a closed valve on the inlet and a closed valve on the outlet, take the fill cap off the radiator (the high point in the siphon) and pour water into the radiator. The water will flow downhill from the radiator in two directions to the fill pipe (toward the inlet and toward the outlet). Air in each of the legs of the pipeline will be displaced by the water and exit the system through the radiator fill pipe.

2. When all the air has been displaced and no more water can be poured into the fill pipe, replace the radiator cap. You have just "primed" the siphon.

3. Now open both the inlet and outlet valves. Because the outlet is 30 feet lower than the inlet, the siphon will begin flowing from the inlet, through the radiator and to the outlet. It will flow at a rate which depends upon the flowing friction it encounters (higher friction = lower flow rate). It will maintain flow at that rate as long as the inlet end is submerged in water. If air can enter the inlet end (ie; the stream and impound are dry) that air will displace the water all the way to the outlet. The prime is gone and the siphon stops.

Does the explanation above clear up anything for you?

 

[Catach]

There is no reason to think that the piping size needs to match the radiator inlet nozzle. Increasing the pipe size will dramatically reduce the friction losses.

I am still trying to find the largest dia. pipe for a given flow rate. How big can I go with the worst case scenario of 0.35 gpm?

There must be a calculation to come up with this answer.

I understand all about starting a siphon but to maintain it, I need to keep the cross-sectional area full at all times. So, as suggested by gmax137 I am trying to find out the largest possible size with my given flow rate.

Your friendly thick sculled question asker.

 

[gmax137]

Are you speaking of Head loss (ft of water) or Head loss (psi)



For 1,000 ft of flexpipe with a Hazen-Williams Coefficient of 150:
Flow - GPM: 0.35
Pipe Dia: 0.75”
Head loss (ft water): 0.57
Head loss (psi): 0.2
Velocity: 0.25

you also seem to be confused by feet and psi head loss. These are just two units for the same thing. The relationship between the two depends on the density of the fluid:

$$head\ in\ feet = \frac{144}{density}\ * head\ in\ psi$$

where density is in pounds per cubic foot, and 144 is the number of square inches in a square foot. I didn't check your Hazen-Williams calculated value, but if the result is 0.57 feet, that would be (for cold water)

$$0.57\ *\ \frac{62.4}{144} \ =\ 0.25\ psi$$

Now, you calculated the available head as 30 feet, because the elevation difference between inlet & outlet is... 30 feet. So the simplest thing to do is forget about psi and just work in feet.

if you do convert back & forth from feet to psi, just remember that atmospheric pressure (14.7 psi) is about 34 feet, so roughly speaking the feet is twice the psi. If your answers aren't like that you made a mistake.

 

[gmax137]

I am still trying to find the largest dia. pipe for a given flow rate. How big can I go with the worst case scenario of 0.35 gpm?

There must be a calculation to come up with this answer.

I understand all about starting a siphon but to maintain it, I need to keep the cross-sectional area full at all times. So, as suggested by gmax137 I am trying to find out the largest possible size with my given flow rate.

Your friendly thick sculled question asker.

If you prime the pipeline as described above, and keep the inlet end submerged, the diameter is immaterial to maintaining the siphon. What might be a problem with the larger line would be permitting so much flow that the inlet end sucks your reservoir dry - and then pulls air in, breaking the siphon. That's why the first consideration is the stream's flow rate, because you siphon must be sized to not exceed that flow rate.

And now that you are talking about flow rates as low as 2 gpm, the friction loss through a 2-inch line or a 3-inch line are both very small so I don't think you need to be concerned that the 2 inch line is too small. In other words, if the friction loss is 0.6 feet or 0.1 feet doesn't matter much since it far exceeds the available 30 feet.

 

[Catach]

So, I should not be trying to find the largest possible pipe dia. I should be trying to find the smallest possible pipe dia. to maintain my given flow rate and still not exceed the 30 ft head loss?

 

[gmax137]

So, I should not be trying to find the largest possible pipe dia. I should be trying to find the smallest possible pipe dia. to maintain my given flow rate and still not exceed the 30 ft head loss?

Yes, that's what I'm thinking now. I know it's not what I was saying upthread, but back then my mental picture was a big brook with a little pipe. But if you are really going to divert the entire flow into the siphon then you need to ensure the siphon doesn't suck the stream dry (because that would break the siphon and you'd lose your flow). You should put valves on each end of the line (as discussed above for priming) and you will also use the downstream end valve to control flow. If you end up throttling the valve, it means your pipe is too big, and bigger pipe generally costs more.

Of course there's other thoughts - you may want to size the line for max flow conditions (so that you can take advantage of them when they occur); larger pipe is more rugged (nobody will crush a 2-inch line by stepping on it); etc.

 

[tyroman]

gmax137,

Thanks for your assistance in getting the concepts across.

Catach,

This is very close to correct:

So, I should not be trying to find the largest possible pipe dia. I should be trying to find the smallest possible pipe dia. to maintain my given flow rate and still not exceed the 30 ft head loss?

I would make the following editorial changes;

"So, I should not be trying to find the largest possible pipe dia. I should be trying to find the smallest <cheapest> possible pipe dia. <which, together with all other components in the system (radiator, etc.) will> maintain <(allow) no more than> my given <design> flow rate and still not exceed the 30 ft head loss?"

You have two "givens" in your situation... they are the design flow rate and the availible pressure head. The first (design flow rate) is determined by the stream / containment volume and is for you to decide (see suggestions in my earlier posts). The second is 30 feet of head.

Using these two values, you will select a pipe size and other components which together will allow no more than the design flow rate with 30 feet of head loss.

I will not proceed here with an example of the way to size the pipe, because I would have to "guess" at what your design flow rate is. I would be happy to do this if you tell us what design flow rate you have decided upon.

As I have said before, the next logical step after deciding on a design flow rate is to check the cooling performance which might be expected from 60 degree F water at that flow rate (see my earlier posts). If the cooling benefit is negligible, the siphon concept must be reappraised. This reappraisal might include designing a larger reservoir (increasing the design flow rate) or building a smaller greenhouse (lowering the thermal load) or adjusting some other design variable.

 

[Catach]

Thanks for the previous inputs.
I am in an information gathering mode.
I am having trouble finding specs on radiator head loss. After tons of searching the best I could find was:

Following is a typical Winston Cup engine at 100 GPM:
Lower radiator hose = 1.5 PSI
Block and cylinder head - each (at 50 GPM) = 8.5
Outlet manifolding = 1.25
Top radiator hose = 2.25
Radiator = 1.5
Total = 15.00 PSI
From: stewartcomponents.com/tech_tips/Tech_Tips_6.htm

If I take out the engine, manifold and hoses, there is a psi loss of 1.5 for the radiator at 100 GPM. I am not sure how to adjust this. Being that this is a racing engine, possibly with a smaller radiator, but also very efficient design, I would double the psi loss and make it 3 psi.

So, at 100 GPM with a loss of 3.0 psi, how does this translate to a 0.35 GPM or a 2 GPM flow in additional head loss, over my current calculated head loss without the radiator?

 

[tyroman]

Catach,

Stewart Components' figure of 100 GPM for an automotive radiator's design flow rate is a tip-off that such equipment will probably not perform effectively in your application... the diameter of the tubes inside the radiator will be so large compared to your flow rate (2 GPM is 1/50th of their design) that flow will be laminar and not reach turbulent velocities.

see;
http://stewartcomponents.com/tech_tips/Tech_Tips_5.htm

"Coolant must flow through a radiator tube at a velocity adequate to create turbulence... The turbulence allows the water in the center of the tube to be forced against the outside of the tube, which allows for better thermal transfer between the coolant and the tube surface."

You may have to go to a custom designed/built radiator or a specialty (non automotive) exchanger. I've no idea where to direct you on such equipment, but would expect the process or HVAC industry will be a place to start.

In an automotive radiator, your head loss at 2 GPM is negligible (for the reason described above). In a custom radiator which would be thermally efficient, head loss may be significant.

***

So, have you decided to use 2 GPM as your design flow rate?

 

[tyroman]

Catach,

See;
http://www.heatexchangersonline.com/airtowater.htm

These are exchangers designed to use hot water on the tube-side to heat ambient air on the fin-side; however, the reverse will work, just not clear how well. Call 1.866.543.2839 to find out.

One of their models (HTL12 x 12) is shown in the performance chart to have a design water flow rate of 6 gpm and an air flow rate of from 500 to 1,016 CFM.

At 6 gpm water and 1016 CFM air there will be a pressure drop (pd) of 0.49psi in the flowing water and a pressure drop of 0.195 inches of water column (pd in.wc) in the flowing air. The chart shows that hot water at 160 degrees F can deliver 49,200 BTU/hr of heat to 1016 CFM of air in this equipment. Let's say they have assumed entrance air temp in the range of 80 degrees F, that gives them a delta T of 80 degrees F between the air and water. In your application, that delta T will be 35 degrees F (BUT in the opposite direction).

If, (and this is a BIG IF) the following assumptions can be made;

1) The equipment can be used as efficiently to cool air as to heat it. This is probably not a valid assumption unless the air to be cooled has zero humidity. With humid air, some of the cooling capacity of the stream water will be used in condensing water vapor out of the air... not in cooling the air.
2) The heating (or cooling) BTU/hr performance is linear within the range of delta T's between 80 degrees F and 35 degrees F. This is probably not true... Google "LMTD".
3) The equipment will cause turbulent flow at one third (2 gpm) of its design flow rate (6 gpm).

Then, (using their exchanger and a solar-powered fan at 1016 CFM) your design flow rate of 2 gpm at 35 degrees delta T will yield (1/3rd of 35/80ths = .1458 or) 14.6% of 49,200 BTU/hr of cooling. This would be about 7,175 BTU/hr of cooling... about 1/2 ton, not very much.

Call their 866 number to see if the equipment might work.