Estimating head loss from pipe slope?

[yahastu]

I am trying to calculate head loss for a sloped pipe.

I found this calculation here, which seems to be what I want..or at least a start:
https://www.pumpsandsystems.com/pumps/april-2015-calculating-head-loss-pipeline

My confusion/skepticism arises from the fact that the equations in the above link do not seem to take into account the orientation (ie, slope) of the pipe at all. It seems like slope should be very important in this calculation...if the pipe is horizontal then it's orthogonal to the force of gravity, causing a large amount of friction loss, but if the pipe is vertical then it aligns with gravity and friction of the pipe becomes almost irrelevant (ignoring other forms of attraction).

Can anyone provide insight as to:
1) why isn't slope included in the above equations? is it because these equations are only valid for horizontal pipes?
2) how can I properly take into account slope in my head loss calculations?

Thanks!

 

[russ_watters]

...if the pipe is horizontal then it's orthogonal to the force of gravity, causing a large amount of friction loss, but if the pipe is vertical then it aligns with gravity and friction of the pipe becomes almost irrelevant (ignoring other forms of attraction).

It isn't clear to me why you think that is true -- it isn't. Gravity has nothing to do with fluid/pipe friction.

 

[yahastu]

It isn't clear to me why you think that is true -- it isn't. Gravity has nothing to do with fluid/pipe friction.

If the pipe is oriented vertically (parallel to gravity), then a droplet of water in the center of the pipe never comes in contact with the pipe walls, so the roughness on the pipe could not possibly slow the flow of water. If the pipe is horizontal, then the force of gravity is pushing the water molecules against the surface normal, creating friction.

 

[russ_watters]

If the pipe is oriented vertically (parallel to gravity), then a droplet of water in the center of the pipe never comes in contact with the pipe walls, so the roughness on the pipe could not possibly slow the flow of water. If the pipe is horizontal, then the force of gravity is pushing the water molecules against the surface normal, creating friction.

You are correct of course; if a drop of water free-falls in a vertical pipe, it doesn't touch the sides, so there is no friction. But the Darcy equation is for friction in a full pipe. Also, it isn't the same as dry friction, so it does not rely on contact force/pressure to create the friction. That's why it isn't a function of pipe pressure.

 

[jrmichler]

1) why isn't slope included in the above equations? is it because these equations are only valid for horizontal pipes?
2) how can I properly take into account slope in my head loss calculations?

The head loss equation takes into account only friction losses in a horizontal straight pipe without bends or diameter changes. You need to calculate additional friction loss due to diameter changes, entrance loss, bends, fittings, and static head. Good search terms are Moody chart and friction loss pipe fittings. Those additional losses are added to the friction loss for the straight pipe. Friction is friction, and is dependent only on velocity, pipe properties, and fluid properties. It does not depend on horizontal or vertical pipe. The equations do assume that the pipe is running full.

Static head is the difference in liquid level between the two ends. If you are pushing the liquid uphill (destination is higher than beginning), the difference is added to the friction loss. If downhill, the difference is subtracted from the friction loss.

Friction loss is normally calculated in feet of head. Then the static head, in feet, is directly added to the friction loss.

 

[Tom.G]

Rule-of-thumb: Head Loss = 0.45 psi for each foot of vertical distance.

or make it easier;

ROUGH rule-of-thumb: Half a pound per foot of rise.

or to make it really simple;

Head Loss(psi)= rise(ft)/2

 

[Baluncore]

Can anyone provide insight as to:
1) why isn't slope included in the above equations? is it because these equations are only valid for horizontal pipes?
2) how can I properly take into account slope in my head loss calculations?

A “head loss” is really a pressure drop due to friction in the horizontal pipe. Consider a horizontal pipe with a steady flow. The flow must be the same everywhere along the pipe. As energy is steadily lost due to friction of flow in the pipe, the pressure must fall steadily along
the pipe. Energy_lost = pressure_drop * flow_volume.

If the pipeline was sloped gently downwards, so the hydrostatic pressure increase counters exactly the frictional pressure drop, then there will be no pressure change along the pipeline. If the flow is then slowly turned off at the bottom, the pressure there will rise to the hydrostatic head of water in the pipe. That may surprise the operator.

 

[russ_watters]

Rule-of-thumb: Head Loss = 0.45 psi for each foot of vertical distance.

or make it easier;

ROUGH rule-of-thumb: Half a pound per foot of rise.

or to make it really simple;

Head Loss(psi)= rise(ft)/2

I wouldn't call that a "loss". It's a pressure gradient that exists even when the fluid isn't moving. And for a closed system you can generally safely ignore it.

[edit] And in an open system, it is still an equal trade.

 

[yahastu]

...The equations do assume that the pipe is running full.

You say that the equations assume the pipe is running full which I presume means that the results cannot be trusted if the flow rate is small relative to pipe diameter...but I don't see a minimum flow rate listed for this equation. What, then, does it mean to be running "full"? How can I know if the equations are valid or not, for my flow rate?

According to the equation, the larger the pipe diameter is relative to the flow rate, the lower the head loss will be. Below is a calculation I did on my proposed pipeline location, using the head loss equation:

Flow(gpm)​	126.60000​
Pipe Length(f)​	2,689.60000​
Fluid density(lb/f^3)​	62.40000​
Pipe Diameter(inch)​	10.00000​
Fluid viscocity​	1.10000​
Pipe Absolute Roughness (in)​	0.00006​
Re (Reynold's number)​	36,382.35404​
f (Darcy friction factor)​	0.02235​
HeadLoss(f)​	29.96657​
Head(f)​	444.32520​
Effective Head(f)​	414.35863​
​	​

Based on this, it looks like I would only have about 30 ft of head loss using a 10" diameter pipe, which sounds great. However, at this flow rate, the cross sectional area of water available at the inlet is only about half the cross sectional area of the pipe inlet. Does this mean that the pipe is "not full" and therefore, the calculation is invalid? And in the case of a not full pipe, should I assume that the equation over or under-estimates the actual head loss?

I am confused now, wondering if, for a given flow, I would get less head loss by using an oversized pipe (so that water is just running along a channel in the bottom), or if I should intentionally use a smaller pipe and try to evacuate air bubbles so that the pipe is "full". It seems contradictory to the nature of the equation to do that, which seems to suggest larger diameter is always better.

If keeping the pipe full is critical to reducing head loss, then it seems a very delicate balance, because if the pipe is sized just right to be "full" at one position, then slightly further down the pipeline it's going to be too small and cause massive friction loss...and the only way to avoid that would be to have the diameter down the entire length of the pipeline tuned exactly to 1 particular flow rate. Surely it can't be that difficult?

On the other hand, if it's ok to have an over-sized pipe, then should I also insert air breathing holes periodically?

If the flow is then slowly turned off at the bottom, the pressure there will rise to the hydrostatic head of water in the pipe. That may surprise the operator.

To make sure I am understanding you...in the limit as I restrict the flow in the bottom of the pipe towards zero, the head loss asymptotically goes to zero, right?

 

[anorlunda]

You say that the equations assume the pipe is running full which I presume means that the results cannot be trusted if the flow rate is small relative to pipe diameter.

No. Flow rate is not the issue. A full water pipe has only water in it, and no air. The picture shows paritally filled pipes.

 

[Baluncore]

To make sure I am understanding you...in the limit as I restrict the flow in the bottom of the pipe towards zero, the head loss asymptotically goes to zero, right?

The head loss is a pressure drop due to flow, so yes.

 

[russ_watters]

You say that the equations assume the pipe is running full which I presume means that the results cannot be trusted if the flow rate is small relative to pipe diameter...but I don't see a minimum flow rate listed for this equation. What, then, does it mean to be running "full"? How can I know if the equations are valid or not, for my flow rate?

Full as in no air, not full as in maximum flow. A pipe that is filled with water can have any flow, including zero, so there is no minimum. All that means is that for any given flow rate, as you increase the pipe size, the velocity drops in [inverse] proportion to the cross sectional area.

According to the equation, the larger the pipe diameter is relative to the flow rate, the lower the head loss will be.

Right; larger pipe means slower flow, means less [viscous] friction.

However, at this flow rate, the cross sectional area of water available at the inlet is only about half the cross sectional area of the pipe inlet.

Does this mean that the pipe is "not full" and therefore, the calculation is invalid? And in the case of a not full pipe, should I assume that the equation over or under-estimates the actual head loss?

I don't understand; why would that be true? It sounds like you have a mismatch/miscalculation in velocity vs pipe size. Or, you have a system in mind that you haven't described to us, with constraints you haven't told us. It is possible that all of this analysis is invalid for your situation.

So I think at this point the question has changed from "how does this concept work?" to "how does my system work?" In order to answer that, you'll need to describe your system to us in detail.

 

[anorlunda]

If the pipe is oriented vertically (parallel to gravity), then a droplet of water in the center of the pipe never comes in contact with the pipe walls, so the roughness on the pipe could not possibly slow the flow of water. If the pipe is horizontal, then the force of gravity is pushing the water molecules against the surface normal, creating friction.

Maybe this is the source of your confusion. Resistance to flow in a pipe is not just a matter of friction with the pipe walls. The fluid is viscous. That alone is a form of friction in fluid flow that is not associated with the container walls.

 

[gmax137]

Just to be clear, the way these problems are done is to take the bernoulli equation at, say the upstream and downstream ends of the pipe: the upstream sum of pressure, velocity, and elevation heads equals the downstream sum.

But -- real pipe flow includes friction not accounted for in Bernoulli. So, you calculate the "head loss" term (as shown in your linked paper). And then subtract that from the head at the downstream end. This modified Bernoulli equation will then account for the energy losses in the pipe flow. There are other "head loss" terms to account for bends in the pipe, varying cross section area, valves, and other fittings which may be present.

 

[yahastu]

Full as in no air, not full as in maximum flow. A pipe that is filled with water can have any flow, including zero, so there is no minimum. All that means is that for any given flow rate, as you increase the pipe size, the velocity drops in [inverse] proportion to the cross sectional area. Right; larger pipe means slower flow, means less [viscous] friction.

Ah, ok...I see now.

So I think at this point the question has changed from "how does this concept work?" to "how does my system work?" In order to answer that, you'll need to describe your system to us in detail.

Basically, my goal is to come up with a design for the pipeline that delivers the most hydro power possible for a given stream (or at least, to understand how to estimate power as a function of pipe design, so that I can do cost/benefit analysis between different pipe designs and overall system production).

I understand that power = flow * pressure, and since I cannot increase the flow (it is limited by the stream), my focus has been on getting the maximum pressure (ie, head) at the outlet while preserving the maximum flow provided by the stream input. Basically, I've been thinking about how to design a pipe that will accelerate the water particles as much as possible without reducing their flow.

Last I checked, the cross sectional area of the stream where the inlet would be is about 45 inches^2, and the flow rate about 1 ft/sec...so I estimate 126 gpm. The total elevation drop from the top of the stream to the bottom is 444 ft, and the length of pipe would need to be 2689 ft. My calculations so far suggest, that for the flow I measured, a 10" diameter pipe would only have about 30 ft of head loss, which would be great if it's true.

Of course, a 10" diameter pipe has a cross sectional area of 78 inch^2, which is about twice that of the stream, so if I were to just funnel the entire stream into the pipe, the pipe would only be about half full. The water would just run down the bottom of the pipe for the length of the pipeline with the top half of the pipe being air. From your comments so far, I now understand that the head loss equation is only intended to be used for pipes that are free of air...but it's not clear to me if this is bad from a power production perspective or now. If I have a very large pipe relative to flow, then my pipe would basically function as an open-air trough/slide for the water (and I could drill holes in the top of the pipe to prevent air bubbles from forming any blockages). How would such a pipe compare to a pipe that was fully filled with water?

On the other hand, if I were to use a pipe fully filled with water, then I would need to first close off the outlet, let the pipe fill completely with water, then open an outlet valve such that the flow through the pipe is less than or equal to the flow of the stream (so that the pipe would remain filled).

 

[russ_watters]

Basically, my goal is to come up with a design for the pipeline that delivers the most hydro power possible for a given stream (or at least, to understand how to estimate power as a function of pipe design, so that I can do cost/benefit analysis between different pipe designs and overall system production). [snip]

Last I checked, the cross sectional area of the stream where the inlet would be is about 45 inches^2, and the flow rate about 1 ft/sec...so I estimate 126 gpm. The total elevation drop from the top of the stream to the bottom is 444 ft, and the length of pipe would need to be 2689 ft. My calculations so far suggest, that for the flow I measured, a 10" diameter pipe would only have about 30 ft of head loss, which would be great if it's true.

Ok, that helps a lot (ambitious project!) and the answer seems reasonable, not including any elbows or fittings you might have.

I understand that power = flow * pressure, and since I cannot increase the flow (it is limited by the stream), my focus has been on getting the maximum pressure (ie, head) at the outlet while preserving the maximum flow provided by the stream input. Basically, I've been thinking about how to design a pipe that will accelerate the water particles as much as possible without reducing their flow.

You're more or less on the right track, but pressure and flow velocity are interchangeable per Bernoulli's equation; as you increase the velocity, the pressure drops. What the Darcy equation tells you is that you should keep the velocity as low as possible for as long as possible to reduce the pressure loss. Then just before you use the water to generate power, accelerate it up to whatever inlet conditions your turbine requires.

Of course, a 10" diameter pipe has a cross sectional area of 78 inch^2, which is about twice that of the stream, so if I were to just funnel the entire stream into the pipe, the pipe would only be about half full.

That's not how you'd want it to work. The water can flow freely into the pipe at the top, but you'd regulate the output at the bottom so the pipe stays entirely full of water. You want the inlet pipe to be entirely below the surface of the stream.

 

[jrmichler]

Ah, ok...I think I see now. If, that is, this diagram is correct:

You only have the full 444 feet of static head if the pipe is running full AND you are measuring to the tailwater level. Your head at the turbine is the static head minus pipe friction loss.

if I were to use a pipe fully filled with water, then I would need to first close off the outlet, let the pipe fill completely with water, then open an outlet valve such that the flow through the pipe is less than or equal to the flow of the stream (so that the pipe would remain filled).

Yes, it is your job to operate the turbine so as to get the full head. The pipe can be any diameter. Larger is better. If your turbine flow is less than stream flow, it will run full. If your turbine flow is greater than stream flow, the pipe will run partially full and you will have much less than 444 feet of heat at the turbine. It's all about regulating the flow through the turbine. The diameter of the pipe only affects the friction loss in the pipe.

Also, you need a stilling pond at the pipe entrance so that sticks and stones coming downstream have a place to settle out. And the pipe entrance needs a strainer to stop everything too large for the turbine.

 

[anorlunda]

Maximum head at the turbine is achieved with the biggest diameter pipe, smoothest pipe surface, no diameter changes, and the fewest turns, bends, and elbows. But that can also be maximum cost. So you don't want max head, you want optimum-head-per-dollar.

I'm skeptical that you will be able to calculate the optimum in advance. There are too many parameters that you would need to measure by experiment.

From what I've seen of other micro hydro projects (there are dozens of videos on Youtube), the biggest efficiency gains are not in the penstock (that's what we call the pipe for a hydro plant), but in the turbine. You'll likely want a Pelton wheel turbine and the design of the water scroll around the turbine, and the design and angle of the injection nozzles. There are many design parameters for Pelton wheels. One series of videos I viewed showed the owner making 5-10 iterative improvements in his project's efficiency. It was very instructive.

I think the best thing for you is to get in touch with other micro hydro owners, and get their advice. We know physics here, but as far as I know, none of us have done a micro hydro power project. But others, like me, might be envious of you. It sounds like a marvelous project.

Here's an example video.

 

[yahastu]

Maximum head at the turbine is achieved with the biggest diameter pipe, smoothest pipe surface, no diameter changes, and the fewest turns, bends, and elbows. But that can also be maximum cost. So you don't want max head, you want optimum-head-per-dollar...

I think the best thing for you is to get in touch with other micro hydro owners, and get their advice. We know physics here, but as far as I know, none of us have done a micro hydro power project. But others, like me, might be envious of you. It sounds like a marvelous project.

My wife and I have just bought this undeveloped land. After buying the property, we realized that there are no power lines on the main road...and our neighbor who owns property next to the road wasn't willing to give us 5 ft of easement onto a tiny corner of their 200+ acre property that we needed for the power line clearance region. Neighbor suggested that if we wanted power we should use solar, so I decided that is what I would do. In addition to building our own house, it's my responsibility to get it powered 100% by renewable power on site...including all heating during winter, all cooling during summer, and charging of EVs.

For the past year, I have been designing my system to be primarily solar powered, because that is where I have the most experience (I installed 5 kw of solar panels on the roof of my old house myself in a grid tied system, and we also designed and built an off grid survival trailer that runs off 5 kw of solar power on the roof, and has an EV charger for an electric truck to tow the trailer).

I've honestly been bashing my head against the wall trying to figure out how to make this off grid thing work with solar power. The problem is that where I live, it can sometimes be up to 2 weeks in the dead of winter with basically zero solar production, so extended battery backup is necessary...and there just aren't economical battery solutions for that duration of time.Maybe someday flow batteries will make it possible, but it seems that the technology is just not there yet.

My first plan was simply to massively over-provision the solar system...eg, put up 100x as much solar power as I expect we need in the steady state, so that if production goes down to 1% in the winter, we still have enough power to keep things going. But that is very expensive and wasteful. If only there was some way to store large volumes of energy in an economical way that doesn't degrade...

In search of a low-cost scalable battery, my next thought after flow batteries was hydrogen. I have a supply of water on the property, and you can buy large scale electroolysis machines to store it in hydrogen, so increasing the battery is as simple as getting more giant propane tanks. Then use fuel cell to convert it back into energy. After being initially excited by this idea, I realized it was not for me when I saw the manufacturer's expected service life of a $20,000, 5 kw hydrolizer..less than 1 year of continuous operation! Way too complicated of a system and difficult to maintain I think.

Then I finally had the big brain idea to think of hydro, an idea I initially rejected because the stream flows are quite weak. I didn't know everything on the land, but I got access to a lidar scans from the state from which I am able to assess all the topography using Quick Terrain Reader. By analyzing the lidar scans, I've identified three different streams spanning several thousand feet and having up to about 400 ft of head. My plan is to install Turgo Turbines in each one at the point where I have the greatest power potential. Based on my calculations on the flow of a stream is at it's weakest, I may still be able to get about 5 kw from it...and the water can probably get up to 100 times this flow. The water stored in the pipe (about 10,000 gallons) will serve as a battery, and possibly a dam at the top if I want more.

The only downside of hydro is that the maximum power potential is limited by the site. Maybe it's more than enough, but there is a possibility that other family members may want to build their own homesteads and I would need to provide power to them as well...so I wanted a system that could be infinitely scaled up.

My conclusion is that the only way to do this is by sourcing my energy from solar, and then utilizing the water as a scalable low cost battery. There is a very large river at the bottom of my property that my streams feed into, so my plan is to use the river as a source for pumped storage, making a large water battery at the top of the hill. It makes more sense to locate the solar panels near to the storage pump (aka, water battery charger) because the majority of the time the solar energy will be surplus.Then of course, running it down through pipeline to turbine at the bottom when it needs to be reclaimed.

From what I've seen of other micro hydro projects (there are dozens of videos on Youtube), the biggest efficiency gains are not in the penstock (that's what we call the pipe for a hydro plant), but in the turbine. You'll likely want a Pelton wheel turbine and the design of the water scroll around the turbine, and the design and angle of the injection nozzles. There are many design parameters for Pelton wheels. One series of videos I viewed showed the owner making 5-10 iterative improvements in his project's efficiency. It was very instructive.

Thanks, I'll have to check those out. I have read that Turgo turbines are slightly preferred over Pelton. I'm looking at Rickly systems, they have Turgo turbines ranging from 5kw up to 30 kw...probably more efficient than most DIY designs, I would hope?

https://ricklyhydrosystems.com/turgopower-turbine-system-5-kw/

Ideally, I want my turbine to output high voltage DC power, but so far it seems they mostly output AC power...I may want to transform it, since it will be sent several thousand feet up the hill in buried cables. I'm thinking 1 AWG at least.

At the homesite, I will plug the HV DC into a MPPT charge controller, that will keep a relatively capacity battery -- either Lithium ion or lead-acid, I don't think makes a huge difference, as it's only for power stabilization. The idea is that the real battery will not be at the homesite but at the generation facility, which will be automatically controlled remotely by the homesite. When the local battery voltage drops below a threshold, it triggers the hydro plant to increase flow, and vice versa when battery is charged it will restrict flow at the hydro plant. Should be scalable system if multiple homes connect to it.

My biggest concern with the system is potential for freezing during winter, because my only source of long term battery is water based, and I need the flow to be low to preserve it. I could paint the pipes black and that would help, but maybe not enough, especially if I need to stop the flow. I think the only solution might be to bury the entire pipeline underneath the frost line.

 

[anorlunda]

Wow. Now I'm even more envious.

Yes, ice in the winter could be a major problem. You should verify by observation that all the streams continue flowing during the coldest weather. The playlist here discusses combating ice in micro hydro if I remember right.

400 feet of head is huge. Many landowners make do with 4 feet for hydro projects.

Beware of overpressure at the turbine end. Be especially aware of what will happen if you try to turn off the flow too fast. With 2689 feet of penstock the inertia is very big and rapid shut off will destroy all your equipment. You may even need a surge pipe with a blow-out diaphragm to protect against overpressure.

Are you serious about the dimensions? 10" diameter and 2689 feet, 444 feet vertical, 126 gpm, 1kw? Double check your numbers. What is the total mass of water in the pipe? What is the velocity of the water?

Be aware that PF rules forbid us discussing dangerous topics. Depending on the numbers, safety might be a major problem for you.

 

[morrobay]

The head loss equation takes into account only friction losses in a horizontal straight pipe without bends or diameter changes. You need to calculate additional friction loss due to diameter changes, entrance loss, bends, fittings, and static head. Good search terms are Moody chart and friction loss pipe fittings. Those additional losses are added to the friction loss for the straight pipe. Friction is friction, and is dependent only on velocity, pipe properties, and fluid properties. It does not depend on horizontal or vertical pipe. The equations do assume that the pipe is running full.

Static head is the difference in liquid level between the two ends. If you are pushing the liquid uphill (destination is higher than beginning), the difference is added to the friction loss. If downhill, the difference is subtracted from the friction loss.

Friction loss is normally calculated in feet of head. Then the static head, in feet, is directly added to the friction loss.

Related question: Is the static head the only driving force for flow , in diagram would the horizontal pipe and sloped pipe discharge water at same rate?

 

[russ_watters]

Related question: Is the static head the only driving force for flow , in diagram would the horizontal pipe and sloped pipe discharge water at same rate?

In that diagram, the static head forcing water through the horizontal pile is close to zero, so the flow rate of an uncontrolled flow would be much lower.

...unless this is some sort of trick question.

 

[256bits]

Related question: Is the static head the only driving force for flow , in diagram would the horizontal pipe and sloped pipe discharge water at same rate?

Adding two more cases for you to think about.
Case 3 is a vertical 444 foot pipe extension to Case 1
Case 4 is a vertical pipe.
Neglecting pipe friction ( ie length of pipe ) and the elbow loss, which case, 3 or 4, or 2 should have the greater flow?
Or are they all the same by Bernoulli?

 

[russ_watters]

Case 4 is a vertical pipe.
Neglecting pipe friction ( ie length of pipe ) and the elbow loss, which case, 3 or 4, or 2 should have the greater flow?
Or are they all the same by Bernoulli?

Per Bernoulli, they are all the same.

 

[yahastu]

Adding two more cases for you to think about.
Case 3 is a vertical 444 foot pipe extension to Case 1
Case 4 is a vertical pipe.
Neglecting pipe friction ( ie length of pipe ) and the elbow loss, which case, 3 or 4, or 2 should have the greater flow?
Or are they all the same by Bernoulli?

View attachment 275751

Case 3 converts gravitational potential energy into pressure by stacking up a long tower of water molecules that weigh down the water column. Case 4 converts gravitational potential energy into kinetic energy by allowing the water to accelerate through open air. In theory both methods could work just as well but the friction losses will be different.

In case 4 the friction losses are due to the drag force, in case 3 the friction losses are due to the pipe walls. Both are proportional to velocity squared, but the expense of creating a horizontal pipe to setup the drop is too expensive.

If you want a pipe that follows surface of land, then you could either use the pressure method with a water filled pipe flowing as slow as possible or use the kinetic energy method by trying to get the water moving as fast as possible. however the friction in pipe would slow it down a lot and kinetic energy would not be as efficient as the pressure method. This is actually how i thought pipelines worked at first and the reason why i was initially thinking that slope should be so important in my initial post

 

[gmax137]

444 feet vertical, 126 gpm, 1kw? Double check your numbers.

Yes check your numbers, I think you are off by a factor of ten.

 

[russ_watters]

Yes check your numbers, I think you are off by a factor of ten.

Where are you guys seeing 1kW? I see 5 kW in post #19. That seems reasonable based on 50% extraction efficiency.

 

[gmax137]

Where are you guys seeing 1kW?

For the given conditions (126 gpm and 444 feet) I get ~10.5 kW. As @russ_watters alludes, the this would be a maximum theoretical value with real-world considerably less.

 

[256bits]

Per Bernoulli, they are all the same.

Thanks for answering, but it was mainly for @morrobay to think about in response to his question.

 

[256bits]

Case 4 converts gravitational potential energy into kinetic energy by allowing the water to accelerate through open air

Case 4 adds a vertical pipe. You are referencing case 1.
Since they are all gravity fed, the energy of the water is the same at the end of the 444 drop in elevation, barring, as I said, any friction loss in the pipes., which in the real world would have to be taken into account.
Your penstock could be of either .
A free falling column of water would be difficult to control onto the turbine buckets.

 

[yahastu]

Beware of overpressure at the turbine end. Be especially aware of what will happen if you try to turn off the flow too fast. With 2689 feet of penstock the inertia is very big and rapid shut off will destroy all your equipment. You may even need a surge pipe with a blow-out diaphragm to protect against overpressure.

This is a good point, but I'm a bit confused about how a surge tank would be made. If I'm understanding correctly, the surge tank would need to be located at the bottom of the pipeline, and would need to be higher than the head, in order to prevent water from coming out during normal operation. In other words, if I have 444 feet of head, then my surge pipe would need to be greater than 444 feet tall. Would that even work? It seems like the 444 ft of head pressure in the surge tank would prevent it from rapidly absorbing pressure spikes. I must be missing something .

Perhaps a simpler option in this case would simply be to install a pressure relief valve?

Are you serious about the dimensions? 10" diameter and 2689 feet, 444 feet vertical, 126 gpm, 1kw? Double check your numbers. What is the total mass of water in the pipe? What is the velocity of the water?

Not sure where you got 1 kw from. Some of my earlier calculations were a little bit off, but I think my current numbers below are good. I'm seeing about 10 kw power available, of which the turbine will probably capture about 80%.

One thing I notice is that the water PSI is now 176 which exceeds the 150 PSI limit for PVC (and that's without taking into account pressure surges). Therefore, I would either need to use a steel pipe, or shorten the pipeline to stay safely below 150 PSI. During actualy construction, I would need to be careful to measure the PSI and see how close reality is to theory.

Resource​	​
Head(f)​	444.32520​
Stream velocity(ft/sec)​	1.00000​
Stream cross section(in^2)​	45.00000​
Flow(gpm)​	140.25971​
​	​
Pipeline​	​
Pipe Length(f)​	2,689.60000​
Pipe Diameter(inch)​	10.00000​
Pipe cross section(inch^2)​	78.53982​
Pipe Absolute Roughness (in)​	0.00006​
Pipe volume (gal)​	43,894.17240​
​	​
Constants​	​
Fluid density(lb/f^3)​	62.40000​
Fluid viscocity​	1.10000​
​	​
Intermediate calculations​	​
Re​	40,307.88717​
F (Darcy friction factor)​	0.02183​
​	​
Results​	​
Head Loss(f)​	35.91983​
Effective Head(f)​	408.40537​
Water PSI​	176.83952​
Water velocity through pipe (ft/sec)​	0.57262​
Total pipe travel time (min)​	78.28398​
Max power (kw)​	10.79508​
Gallons per kwh​	779.57561​
Rain barrels per kwh​	15.59151​

 

[gmax137]

Just to be pedantic, you should know that in general pipe inside diameter is not the same as the nominal pipe size. Pipe comes in "schedules" which are different wall thicknesses. 10-inch pipe has an outer diameter of 10.750 inches. Normal or standard is schedule 40; for 10-inch sch 40 pipe the ID is 10.02 inch, transverse area is 78.86 in^2. So in this case you can get away with using 10 inch ID in your calculations.

I have never seen PVC in anything but schedule 40. You can look at the pipe manufacturer's website to see what they supply; they usually have useful tables of "Pipe Data" showing the dimensions for the various schedules in each nominal size. There is some ASME specification that defines the dimensions.

 

[anorlunda]

If you have 176 psi working pressure, the burst pressure should be on the order of 800 psi. Then pressure relief valves and/or burst diaphragms should try to limit the pressure to 400 psi.

That much 10 inch steel pipe, I couldn't find the price, but at least several hundred thousand dollars.

 

[anorlunda]

This paper has a lot of info, formulas, and data relevant to your project.

Micro Hydro Penstock Design
https://ir.library.oregonstate.edu/downloads/x633f2363

 

[gmax137]

I see 10-inch sch 40 PVC online at about $13 per foot, so there's $35,000. Plus 150 pipe couplings, glue, supports, etc... The turbine, the controls, wiring back up to the house...

our neighbor who owns property next to the road wasn't willing to give us 5 ft of easement onto a tiny corner of their 200+ acre property that we needed for the power line clearance region.

Maybe you can resume negotiations with the neighbor. Grid power at about 20 cents a kW-hr is a bargain.