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convective flow
- Thread starter Dogma72
- Start date
A heat transfer text should have that sort of information. You might try:
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FAQ731-376
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FAQ731-376
- Thread starter
- #3
hey, thank you very much!
When I was clicking the link, i thought for sure i was going to some place where i could purchase the appropriate text - i was super excited when i saw that i could just download the thing!
Awesome help!
I'll be back though if i need some clarification!
Wes
When I was clicking the link, i thought for sure i was going to some place where i could purchase the appropriate text - i was super excited when i saw that i could just download the thing!
Awesome help!
I'll be back though if i need some clarification!
Wes
- Thread starter
- #4
Basically what i want is something that i can just plug and chug into.
I want to essentially have water/ethylene glycol mixture 30 ft x 10ft x .25 inch solar panel. I want to have a 25 or 30 ft high vertical pipe connected to it, where it would flow upwards into a heat exchanger and then through a fin fan heat exchanger such that all the solar energy that went into the fluid would be removed... then fluid would then fall into a pelton turbine. I want to play around with different sizes so that i can design a system around the pressure / flow requirements for the turbine.
If anybody has a simple way that i can approximate this and do some rough calculations without forcing me to break out my Ti-89 - that would be awesome. In the meantime, I will continue to go through this text and make my brain hurt by going through all the calculus (blech!)
Thanks again guys!
wes
I want to essentially have water/ethylene glycol mixture 30 ft x 10ft x .25 inch solar panel. I want to have a 25 or 30 ft high vertical pipe connected to it, where it would flow upwards into a heat exchanger and then through a fin fan heat exchanger such that all the solar energy that went into the fluid would be removed... then fluid would then fall into a pelton turbine. I want to play around with different sizes so that i can design a system around the pressure / flow requirements for the turbine.
If anybody has a simple way that i can approximate this and do some rough calculations without forcing me to break out my Ti-89 - that would be awesome. In the meantime, I will continue to go through this text and make my brain hurt by going through all the calculus (blech!)
Thanks again guys!
wes
- Thread starter
- #6
the other thing that i'm noticing while reading through this is that the focus is primarily on heat transfer. In my applicatioon, i really do not care about that. What i care about is the fluid flow, in gallons per minute, based upon an experimentally determined rate of heat input. i.e. if i have X number of BTU's/sec which result in a temperature rise of Y degrees and a difference in height of Z feet, how many gallons per minute will be flowing? How much difference in head will there be between the hot leg and cold leg?
I want basically buy a stock water turbine, and they are going to ask me how much head to you have and how much flow? I'd like to be able to calculate that.
Thanks again!
Wes
I want basically buy a stock water turbine, and they are going to ask me how much head to you have and how much flow? I'd like to be able to calculate that.
Thanks again!
Wes
- Thread starter
- #7
I don't believe there are any commercial solar panels out there that are designed to do what i want to do.
Basically the issue with solar panels is that they are only 10 to 15 percent efficient. Whereas a water turbine is upwards of 70 percent and as long as i can minimize heat loss through insulation on the hot leg, I'm thinking i can get considerably more usable power out of a convective flow system.
Basically the issue with solar panels is that they are only 10 to 15 percent efficient. Whereas a water turbine is upwards of 70 percent and as long as i can minimize heat loss through insulation on the hot leg, I'm thinking i can get considerably more usable power out of a convective flow system.
I think the system would be very inefficient, although I don't know that I could explain why.
As a starting point for your analysis, forget the heat exchange part. Assume one vertical leg of your loop is hot, one side is cold. Look up the densities at each temperture, use those temperatures to calculate a differential pressure from one side to the other. Use that pressure with a Moody chart or similar tube/pipe flow equations to calculate flow rate.
I think the problem is that for the amount of heat you put into the fluid, you get very little expansion, as compared to air, so it's hard to extract much energy back out of it. You're mainly just heating and cooling a fluid.
As a starting point for your analysis, forget the heat exchange part. Assume one vertical leg of your loop is hot, one side is cold. Look up the densities at each temperture, use those temperatures to calculate a differential pressure from one side to the other. Use that pressure with a Moody chart or similar tube/pipe flow equations to calculate flow rate.
I think the problem is that for the amount of heat you put into the fluid, you get very little expansion, as compared to air, so it's hard to extract much energy back out of it. You're mainly just heating and cooling a fluid.
- Thread starter
- #9
So,
How would the difference in height affect the differential pressure?
density is usually in kg/m^3 or lb/ft^3
pressure is psi or metric equivalent
so, if the cold leg is something like 980 kg/m^3 and is at 25ft in the air and the hot leg is 900 kg/m^3 and is at 0 ft relative elevation...
so, you'd have 80 kg/m^3 of "relative differential density"
would you multiply that by 7.62 meters to get 609.6 kg/m^2
... or is that going down completely the wrong path?
I know there is precedent for this - US nuclear subs have an emergency cooling system that operates in this fashion and i know that the thermal capacity they are designed for is something around 7 percent of the thermal output of the reactor - which is pretty substantial.
So, i guess, where do i go from here?
Wes
How would the difference in height affect the differential pressure?
density is usually in kg/m^3 or lb/ft^3
pressure is psi or metric equivalent
so, if the cold leg is something like 980 kg/m^3 and is at 25ft in the air and the hot leg is 900 kg/m^3 and is at 0 ft relative elevation...
so, you'd have 80 kg/m^3 of "relative differential density"
would you multiply that by 7.62 meters to get 609.6 kg/m^2
... or is that going down completely the wrong path?
I know there is precedent for this - US nuclear subs have an emergency cooling system that operates in this fashion and i know that the thermal capacity they are designed for is something around 7 percent of the thermal output of the reactor - which is pretty substantial.
So, i guess, where do i go from here?
Wes
And what is the physical meaning of kg/m2?
They type of flow you are talking about is simply driven by the difference in density. The volume of water in the hot leg weighs less than the volume of water in the cold leg.
The density of water can be written as a function of temperature.
The rate of temperature change is a function of your solar collector's ability to make water hot. Assume that you can size the heat rejection to match.
This gives you everything you need to know to calculate the driving force and power input.
Opposing the driving force is pipe friction and the pressure drops through the collector, heat rejector and turbine. All of those pressure drops can be written a functions of flow.
Equate and solve for flow.
They type of flow you are talking about is simply driven by the difference in density. The volume of water in the hot leg weighs less than the volume of water in the cold leg.
The density of water can be written as a function of temperature.
The rate of temperature change is a function of your solar collector's ability to make water hot. Assume that you can size the heat rejection to match.
This gives you everything you need to know to calculate the driving force and power input.
Opposing the driving force is pipe friction and the pressure drops through the collector, heat rejector and turbine. All of those pressure drops can be written a functions of flow.
Equate and solve for flow.
- Thread starter
- #11
perfect.
Thanks a bunch. BTW, I'm assuming that there is a general design guideline for sizing pipes vs. volumetric flow rate. In my scenario, I want to minimize head loss so, I would want to oversize everything to the extent practical from a cost standpoint as to minimize non-useful energy loss from the system (i.e. i want as much energy as possible available to spin my turbine.)
Wes
Thanks a bunch. BTW, I'm assuming that there is a general design guideline for sizing pipes vs. volumetric flow rate. In my scenario, I want to minimize head loss so, I would want to oversize everything to the extent practical from a cost standpoint as to minimize non-useful energy loss from the system (i.e. i want as much energy as possible available to spin my turbine.)
Wes
If you live near a university with an engineering school, you should go pay them a visit. A brief chat with a professor would be a good exercise. You need a bit of an education about density flows.
The point that you don't seem to have is that the energy you are removing from this system with a turbine is some of the energy required to get the fluid lifted back up 25 feet. Hot water doesn't rise on it's own, it has to have colder, denser water to force it up. You need to sit face-to-face with someone who can show you the equations involved in determining whether your proposed system could produce excess energy.
Another point that you may not have grasped is that in order to force the hot water up 25 feet, you will need to have cooler water flowing down. The heat transfer from your system may require long holding time, or a very large tank in order to realize a temperature differential.
Most of the help available on this site is for a very specific question about some aspect of a system, not how to design the whole system. If you think your idea is a money-maker you will need to hire an engineer, or visit the university.
The point that you don't seem to have is that the energy you are removing from this system with a turbine is some of the energy required to get the fluid lifted back up 25 feet. Hot water doesn't rise on it's own, it has to have colder, denser water to force it up. You need to sit face-to-face with someone who can show you the equations involved in determining whether your proposed system could produce excess energy.
Another point that you may not have grasped is that in order to force the hot water up 25 feet, you will need to have cooler water flowing down. The heat transfer from your system may require long holding time, or a very large tank in order to realize a temperature differential.
Most of the help available on this site is for a very specific question about some aspect of a system, not how to design the whole system. If you think your idea is a money-maker you will need to hire an engineer, or visit the university.
The ONLY way to make anything halfway useful, if I understand what the problem is, is to use the solar energy to boil water into steam, and have the steam turn a motor or whatever.
Even at that, it's questionable whether that's any more efficient than a photovoltaic solar panel that converts to electricity to drive motors or pumps.
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FAQ731-376
Even at that, it's questionable whether that's any more efficient than a photovoltaic solar panel that converts to electricity to drive motors or pumps.
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FAQ731-376
The bottom line is that anything that simply converts solar energy into heat that must be extracted elsewhere is a losing proposition.
Even in the apparent example of convection currents in water, almost all the energy is simply shuffling the water around, and much of that is lost in friction and the inevitable countercurrents. Very little of it can be extracted for mechanical advantage.
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FAQ731-376
Even in the apparent example of convection currents in water, almost all the energy is simply shuffling the water around, and much of that is lost in friction and the inevitable countercurrents. Very little of it can be extracted for mechanical advantage.
TTFN
FAQ731-376
- Thread starter
- #15
I tend to disagree. You can remove 7% of a nuclear reactors power output through natural circulation and that is without an obscene difference in height. Granted you are not using mechanical energy to spin a turbine. But, that is more an issue of designing the system to minimize head loss and managing the heights of the hot leg / cold leg to produce enough of a pressure differential to spin the thing. The question is - how high would those legs need to be if i needed 11 psi to spin the turbine?
Insofar as heat removal - i don't see that as an issue. I already have a use for the thermal energy that would be put into the system via the sun. The excess would get dumped to another thermal load or if its truly excess i would just dump it to the atmosphere via a fin fan HX, thus creating my falling column of water to push the hot water up.
Insofar as heat removal - i don't see that as an issue. I already have a use for the thermal energy that would be put into the system via the sun. The excess would get dumped to another thermal load or if its truly excess i would just dump it to the atmosphere via a fin fan HX, thus creating my falling column of water to push the hot water up.
Nuclear reactors generally boil water to turn turbines. Their waste heat carrier is usually steam whose heat is removed by condensation.
However, it sounds like you now talking about heat pipes. Heat pipes are used to remove heat from lots of systems, but that's the extent of it; heat pipes are not used to generate mechanical energy.
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FAQ731-376
However, it sounds like you now talking about heat pipes. Heat pipes are used to remove heat from lots of systems, but that's the extent of it; heat pipes are not used to generate mechanical energy.
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FAQ731-376
- Thread starter
- #18
well
in that case
i'm sorry that i asked the question in the first place.
although, i really don't see where the big hang up is - i'd really rather have that explained to me rather telling me that i'm living in some kind of lala land. That is completely unhelpful.
cold water falls, pushes hot water up. Circulation is dependant upon temperature difference. Head is determined by temperature difference (therefore difference in density) and relative difference in elevation.
insofar as heat removal from a nuclear reactor. The primary system stays single phase. The secondary system undergoes a phase change. So, the point of how the heat is removed is moot isn't it?
Why can't a heat pipe system using natural circulation be used to spin a water turbine? I'm not talking about a huge turbine to power a quarter of a city, i'm talking about a water turbine that could provide a percentage of electricity for a single home. These turbines can operate with as little as 6 ft of head. That's not alot - a little over 2.6 psi.
let me restate or state if i haven't made myself clear before. I'm all for living in the real world, but, don't tell me that an idea is dillusional without backing it up with some kind of calculation or condition that would make the construction costs and difficulty so ridiculous as to render it impractical beyond any possible gains that it could offer (i.e. 500 ft tower, to power a turbine that would only produce 5 percent of a home's need).
On a second thought - what about having this convective loop to 2 phase instead of single phase. I don't think thermal energy input is a problem - it's just a question of sq. footage which can be built for cheap.
in that case
i'm sorry that i asked the question in the first place.
although, i really don't see where the big hang up is - i'd really rather have that explained to me rather telling me that i'm living in some kind of lala land. That is completely unhelpful.
cold water falls, pushes hot water up. Circulation is dependant upon temperature difference. Head is determined by temperature difference (therefore difference in density) and relative difference in elevation.
insofar as heat removal from a nuclear reactor. The primary system stays single phase. The secondary system undergoes a phase change. So, the point of how the heat is removed is moot isn't it?
Why can't a heat pipe system using natural circulation be used to spin a water turbine? I'm not talking about a huge turbine to power a quarter of a city, i'm talking about a water turbine that could provide a percentage of electricity for a single home. These turbines can operate with as little as 6 ft of head. That's not alot - a little over 2.6 psi.
let me restate or state if i haven't made myself clear before. I'm all for living in the real world, but, don't tell me that an idea is dillusional without backing it up with some kind of calculation or condition that would make the construction costs and difficulty so ridiculous as to render it impractical beyond any possible gains that it could offer (i.e. 500 ft tower, to power a turbine that would only produce 5 percent of a home's need).
On a second thought - what about having this convective loop to 2 phase instead of single phase. I don't think thermal energy input is a problem - it's just a question of sq. footage which can be built for cheap.
Dogma72, you sound like a bright person who is just on the edge of haivng enough knowledge to analyze this system. I really would encourage you to find a university professor to sit down with you and discuss your proposed system. I would actually encourage you to enroll in a couple of engineering classes like thermodynamics, fluid dynamics, and heat transfer and start dreaming up all kinds of things that are possible but not currently done.
For someone on this site to go through all of the calculations and back and forth posting of discussion to develop or analyze your system is not a realistic expectation. If you can break your questions down to very specific ones (posted separately), then you might get more help.
For someone on this site to go through all of the calculations and back and forth posting of discussion to develop or analyze your system is not a realistic expectation. If you can break your questions down to very specific ones (posted separately), then you might get more help.
Dogma72
Just one clarification:
In pressurized water nuclear reactors, a secondary side cycle is set-up where the steam removed from the steam generator drives a turbine-driven pump which feeds cold water to the steam generator. This, along with the design of the generator and piping, provides the heat differential needed to move water through the primary side. Without the secondary side loop there would not be natural circulation.
BTW, my credentials are that I'm a former power plant system engineer for secondary side systems, previous engineering inspector and currently working for a federal agency involved with nuclear power and over 25 years of experience in this area.
Patricia Lougheed
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.
Just one clarification:
In pressurized water nuclear reactors, a secondary side cycle is set-up where the steam removed from the steam generator drives a turbine-driven pump which feeds cold water to the steam generator. This, along with the design of the generator and piping, provides the heat differential needed to move water through the primary side. Without the secondary side loop there would not be natural circulation.
BTW, my credentials are that I'm a former power plant system engineer for secondary side systems, previous engineering inspector and currently working for a federal agency involved with nuclear power and over 25 years of experience in this area.
Patricia Lougheed
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.
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