liquid friction equation

Since water has a very low viscosity this approach will not be very effective. It is better to use other ways to transfer mechanical energy as thermal to water. Use same principle as in the hydrodynamic brakes or use a pump and hydraulic orifices to heat water due to the pressure drop and high shear stresses in the orifice flow.

nick name -- You're right on this. I have a feeling our OP got a very low power result from his calculation and that is what got him to wondering about it. He probably ought to go take a look at Testla turbines. Conceptually they look appealing; but when you get deep into that technology you find the impracticality of manufacturing the required stacks of thin plates at uniform close clearances. Best he take a look at the way water brake dynamometers are designed because they do essentially what he wants to do.

Actually as you suggested using the wind turbine to drive a small centrifugal pump recirculating water in a tank and controling the flow with a globe valve is probably a good way to start.

There is a caution in this that I can see and it has to do with the pump losing needed net positve suction head (NPSH) it requires as the water heats up. The resultant drop in load from a steam bubble in the pump suction could cause a runaway of the wind turbine in high winds unless there is an additional overspeed safety device in the system. I'm pretty sure that a bit of internet searching will disclose commentaries by people who have had direct experience with such setups and can offer better advice than I can.

Ed Weldon

Looks good if you are using, for example, the change in water temp to measure wind turbine output, as an experiment.

Also in principal, a wind turbine could be used to generate heat in a fluid and that fluid could be used to heat something else, such as water or air. In certain scenarios, the efficiency could be quite high. A VAWT could have a shaft directly into the top of a very well insulated water tank, and you could heat the water very efficiently -- as long as you can keep the temperature in the usable range, and it does not sit around in the tank for a long time, eventually cooling.

A water brake for measuring wind turbine output could be made along these lines, too.

hi thanks for your reply

i had looked into centrifigul pumps but didnt pursue it any further as i couldnt find anyway to work out the heat produced

i have since found this website http://www.engineeringtoolbox.com/pumps-temperature-increase-d_313.html

my goal is to create a wind water heater that when the wind is blowing can output 2 litres per minute at 45 degrees C.

i had calculated with the liquid friction equation in my original post-

area - 2 m^2

water viscosity at 10 C - 1.3

distance between plate -0.002

velocity - 2

power = viscosity x area x speed / distance between plates

power = 1.3 x 2 x 2 / 0.002

power = 2600 watts

i thought this was alot of energy and am now not sure that i understand the equation

my best guestimate at heat loss on an insulated tank this size is 272 - 680 watts

i read somewhere that 2092 watts will boil one litre of water in 2 seconds

please correct any and all of my guestimates

guest

Approaching the problem from my electrical pov and assuming a 12 degree C water temperature to start...

********flow***************
"output 2 litres per minute at 45 degrees C"
(45 [C] - 12 [C]) * 4.18 [J/g-C] * 2000 [g] = 275880 [J]
275880 [J] / 60 [s] = 4598 [J/s]
That's 4.6 [kW] !

*********loss*************
"my best guestimate at heat loss on an insulated tank this size is 272 - 680 watts"
Depends on size and insulation, but your estimates here may be reasonable.
I've been taking measurements om my HOME domestic hot water tank for over a year now. It is a 92 eff rated, 50 gallon, electric unit. Average continuous heat LOSS is about 90 Watts.

**********boil*************
"i read somewhere that 2092 watts will boil one litre of water in 2 seconds"
Not quite...
(100 [C] - 12 [C]) * 4.18 [J/g-C] * 1000 [g] = 367840 [J]
367840 [J] / 2092 [J/s] = 176 [s]
just under 3 minutes!

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

If this is a "home" wind project, best wishes. Do your research thoroughly. A TRUE 5 kW wind turbine is quite a large home project.

thanks for the replies they are very helpful

i found this displacement pump and restrictor friction heater design http://www.freepatentsonline.com/4596209.pdf

can anyone point me to orific friction calculations/explanations or if you're really bored outline a pump and restrictor friction heater that can output 2 litres per minute at 45 degrees C

i also found this wind water heater design http://www.raunvis.hi.is/~ornh/Links-Research/Windmill.pdf

i see now that both these designs are superior to my original idea as they have inbuilt wing tip speed effieciency controls.

please hyperlink anything relevant.

guest

If the energy is used to increase the flow temperature then the balance is:

Power at the restrictor:

Ph= Q*Δp [W] and

Thermal power:

Pth=Q*Cp*ρ*Δθ since Ph=Pth we obtain that Δp / Δθ = Cp*ρ where

Q= fluid flow [m^3/s]

Δp = pressure drop at restrictor [N/m^2]

Cp = heat capacity of fluid [J/kg°K]

Ρ = specific mass of fluid [kg/m^3]

If the fluid is water then Cp=4187 [j/kg°K] and ρ = 994.4 [kg/m^3]

This leads to a pressure drop of 4.164 [N/mm^2°K]

This means that for 35°K temperature increase the pressure drop should be 145.7 [N/mm^2].

Since 1 psi = 0.00689 N/mm^2 the pressure drop in psi will be 145.7/0.00689=21.15 ksi !

Quite difficult to have a simple pump to generate such a pressure and especially for water.

If the fluid is a mineral oil (hydraulic oil for instance) the parameters do change:

Cp= 1880 [J/kg°K] and ρ= 915 [kg/m^3] so that Δp / Δθ = 1.72 [N/mm^2°K] value which is 2.42 times smaller as the value for water. Even under such positive changes the direct heating will ask for high pressure drops and thus for heavy pumps which are expensive and require a serious maintenance, very fine filtering a.s.o.

The solution is to heat a water volume with a smaller pressure drop and transfer the heat via heat exchanger to the user's circuit. This is the way mentioned patent suggests. Gear pumps can work for long times in pressure ranges from 6 to 21 N/mm^2. Oil filters are still required in order to obtain a high MTBF but the system is more tolerant to dirt due to lower pressure.

The regulation proposed is but not optimal and there are better ways to do it according to what is the requirement.

In fact there are 2 parameters to optimize: energy harvest and temperature. For the longevity of the system it is interesting to maintain a temperature as low as possible (around 60..70°C) for a reduced heat transfer area and a reserve with respect to vapor generation and possible cavitation. A reduced temperature will also reduce the investment for the thermal isolation of the reservoir.

Same reservoir will work as a heat accumulator giving the possibility to supply hot water even if for a limited time wind is low.

It is, for corrosion reasons, better to have a 2 fluids system since hot water could destroy seals and corrode the pump. Oil will protect the components from primary circuit.

Since between the pressure drop and the temperature rise there is a direct linear relationship it is enough to control the restrictor via the generated pressure drop and introduce a limit function of reservoir temperature. Those controls do not need electronics.

If the temperature goes over a threshold it is possible to still use the pump as brake in order to avoid an over-speed but use a by-pass leading the hot oil to an exchanger to atmosphere an so limit the ingression of heat to the reservoir. The availability of pressurized oil offers a lot of possibilities for controls with components available of the shelf.

thanks nick name

related forum http://www.eng-tips.com/viewthread.cfm?qid=257265&page=1reat

quote -

If you have a closed loop (with no heat loss) with a pump and a orifice plate and insert the power P[kW] as mechanical power in the pump this will increase the temperature as:

P * t = m * cp * dT

P = Power [kW]
t = Time you run the pump [s]
m = mass of fluid [kg]
cp = specific heat [kJ/kgK]
dT = temperature increase [K]

guest

Exactly what I wrote, as you see physics are all over the world the same at least for some of us.

Here are the schematics for the system as described in last mail:

You can determine the reservoir dimensions taking into account the hot water flow and the time it is needed and the average power from the wind mill. If you need a helping ha,d I can give you the support. The resistor is in fact a piloted valves to maintain the pressure drop constant (or almost constant), the valve will by-pass hot oil to an external heat exchanger if temperature goes to high. Control can be also more complex and better optimized this being only an example of possibilities. It is not a final solution.

Great answer, Nick Name.

I hope the original poster appreciates the value of your input, and the time you've spent.

(Your nice clear drawing contrasts with the one starting this thread. I think you might not have seen the thread, another in the series of "other world" physics posts. In this case, I really thought the original poster did the whole thing as a joke, -- and my first responses were in that joking spirit... but as the thread developed, it started to look like he might have been serious??!!)

nickname i would appreciate any support i can get

i dont really have any understanding of physics and working out how your water heater worked was facinating

so lets design it!

my water heater is for a bath that will continually be fed with water(from the heater) and so will be continually overflowing

the first thing i need to work out is just how much water i need?

all circumstances where the heater isnt heating the bathwater to 42 C are ignored

what is the minimum flow rate at 42 C to compensate for the heat loss and keep the bath temp at 42 C?

the bath dimensions = 0.5m x 0.45m x 0.4m(deep)

side wall surface area = 0.76 m^2

side wall - concrete - thermal conductivity = 1.7 w/mk

side wall thickness = 0.2

heat loss from side wall = 206 w

water surface heat loss; outdoor-sheltered = 5000 w/m^2

bath water surface area = 0.225 m^2

bath water surface heat loss = 1125 w

total heat loss = 1331 w

this is as far as i dare to go and even this is a stretch

is the heat loss through the side walls and water surface even relevant if hot water is continually entering and leaving the bath?

i have stated in the above question to work out the flow rate at 42 C for a bath temp af 42 C. how will higher temps at less flowrate to keep the bath at 42 C effect the overall heater design?

please help!

1- do you need over flow or do you only need a constant temperature bath?

2- In case that you do not need for any reason a continuous over flow it is only necessary to compensate the heat losses. As you noticed the surface losses are big in comparison to the other. It is thus recommended to insulate the best possible way the water surface for a reduction of those losses (93%!). In this case the sketch I made will satisfy the requirements -as principle- and with some corrections. For the water circuit it will be necessary to provide a small pump which could be actuated by the wind energy as well. Water should come at the lowest bath level in and leave it at the highest.

3- If you need an over flow the losses have to be increased with the energy corresponding to the water over flow at the bath temperature, in case of need you shall define the water flow, it can be regulated by the connection to cold water supply.

4- Do you have any information about wind velocity probabilistic distribution at your location (especially during the cold period of the year)? Do you expect outer temperatures under 10°C? for how long? How is temperature at night?

To "design" the system as you say information is too scare.

The system has to be designed for the maximal energy request under the worse working conditions so both have to be defined very well from the start if not the project will be a failure.

1- do you need over flow or do you only need a constant temperature bath?

yes. well i would like to design it to compare it to non continuous flow designs. using this technology http://www.missouriwindandsolar.com/Water_Heater_Elements.html (bottom of page) it would take a day or two to heat the water for one bath.

2- In case that you do not need for any reason a continuous over flow it is only necessary to compensate the heat losses. As you noticed the surface losses are big in comparison to the other. It is thus recommended to insulate the best possible way the water surface for a reduction of those losses (93%!). In this case the sketch I made will satisfy the requirements -as principle- and with some corrections. For the water circuit it will be necessary to provide a small pump which could be actuated by the wind energy as well. Water should come at the lowest bath level in and leave it at the highest.

to reduce surface heat loss i would use a bath lid which would still allow for continuos feed and continuos overflow. surface water heat loss with lid would then be about 600 w

3- If you need an over flow the losses have to be increased with the energy corresponding to the water over flow at the bath temperature, in case of need you shall define the water flow, it can be regulated by the connection to cold water supply.

so just so i understand correctly-

if the bath walls and water surface loss is x and the over flow is y the input flow has to be x + y

i was under the impression that the oil temp was regulated, and so thought that the regulated oil temp would regulate water temp at constant flowrate

4- Do you have any information about wind velocity probabilistic distribution at your location (especially during the cold period of the year)? Do you expect outer temperatures under 10°C? for how long? How is temperature at night?

here is the climatic data for the area http://www.bom.gov.au/climate/averages/tables/cw_097000.shtml

i think the area has a mean yearly wind speed of 8 m/s

i am investigating wind power first as it will influence what kind of property i end up purchasing.

so to create a bath that operates at 42 C 25% of the time maybe a windspeed parameter of 10 m/s

I do not like to go into details with a (ghost sorry!) guest. So that following step is to register and then we can discuss more about how to go on.

nickname and everyone i was the guest

ill start the ball rolling by stating the steps i think the design should take and the parts i think i can or cant complete

1- heat and flow rate of water out of heat exchange to keep the bath at 42 C. i can work out bath heat loss for a full still bath but am lost when trying to understand what is required for a continuous feed and continuous overflow bath.

2-length and dimension of pipe submerged in oil pool. i have found this equation

The overall heat transfer coefficient for a wall or heat exchanger can be calculated as:

1 / U A = 1 / h₁ A₁ + dx_w / k A + 1 / h₂ A₂ (1)

where

U = the overall heat transfer coefficient (W/m²K)

A = the contact area for each fluid side (m²)

k = the thermal conductivity of the material (W/mK)

h = the individual convection heat transfer coefficient for each fluid (W/m²K)

dx_w = the wall thickness (m)

i was thinking of using this equation but have had trouble finding information on h = the individual convection heat transfer coefficient for each fluid(oil and water).should i use copper pipe or stainless steel pipe?

3- selection of gear pump and piloted valve. i havent researched this enough to know what i can or cant do yet, but will get on it.

4-turbine selection. at the moment i am unsure if i will use wind or hydro so perhaps step 3 could be the final step stating the power required at the gear pump.

as i have mentioned i dont really know what i am doing so this approach maybe completely misguided or back to front.

thanks for the assistance.

maitland -- This replies to just one question of yours. For piping in an oil-water heat exchanger copper is the best bet. It has better heat transfer characteristics and will be easier to assemble into the desired configuration.

For oil carrying piping away from water use carbon steel or copper pipe, as well as bronze /brass valves anywhere the pressure is under 250psi whichever material better suits you. If the fluid is an oil and the rotary pump is likely made of carbon steel materials the expense of anything stainless steel in the plumbing is unnecessary.

Ed Weldon

ed - i only questioned the use of copper when i read somewhere that oil reacts to the copper in such a way that over time it reduces its heat transfer abilities. the article was suggesting using stainless steel piping instead.

it isnt a big issue and i think i will just use the copper

thanks for clarifying

it seams this thread is dead

i will finish off by stating some guidelines i have come up with

- i have chosen hydro power

- i am happy that the total bath heat loss will be around 1300 w

- so i need 1300 w at the hydro turbine(havent worked out heat loss fr exchanger and pump yet)

- gear pump max rpm above hydro turbine

- orifice plate size ? seams irrelevant exept that it cannot create an upstream pressure that the gear pump/hydro turbine cannot handle

- copper tube length ?

- oil filter ?

- when completed will be able to fine tune by increasing/decreasing hydro turbine speed and by throttling the water flow through the heat exchanger

- keep the oil temp at about 60 C

one last question - does anyone have any ideas for an off the shelf reservoir for the oil?

all coments welcomed thanks

There are reservoirs for the fluids sprayed in agriculture made from reinforced plastics. They are very low cost but I do not know up to how many degrees they can be used.

Now to come to some of your questions:

- the 1300W does it cover ALL losses? including heating of continuous flow?

- to choose the pump speed you should consider

1- the turbine torque and the pump torque

2- the power developed by turbine and the power absorbed by the pump

Ppump = η* Power turbine = Q*p [flow x pressure] η= transmission efficiency

Tpump= η* Tturbine/i [ i= transmission ration >1 since the pump will turn faster than the turbine for several reasons]

T pump= V*p/(2* pi) [ V= pump displacement /revolution]

Qpump= V* npump= V*nturbine*i [n number of revolutions/time unit]

All those equations have to be fulfilled at same time.

3- Orifice you should for many reasons avoid cavitation after the orifice so that in fact you should have a series of orifices with a drop/orifice so that the pressures ratio will be less 2. When you are so far I shall give you indications how to design the resistor.

4- In a 1st approximation you can consider that the convection coefficients are on the oil side around 300 W/m²°K and 2500 W/m²°K on the water side the wall thermal resistance can be neglected. This will give you the transfer area required for the power you want to transfer to water. Of course a coefficient > 1 has to be considered in order to avoid due to dirt or other reasons a limitation of the transfer capacity.

After this you choose your tube and compute the oil velocity and recalculate the oil "h" value for the oil side and with it again the required area. You may use the mean area A= L* pi*(de+di)/2. You have the interest to have a higher velocity for the oil in order to have a higher "h"oil (h≈ v^0.8 where v= oil velocity in the tube).

I think for the moment it is enough to work with. For the computations you can use the book of Liennard (I think) which free on the net and a very good one with every think one needs to solve problems as yours.

Let me know how it goes on and I shall correct your computations if anything seams to need it.

As you see it is still not dead your thread.

Nick

nick

i dont know if the 1300 w covers all losses and i cant seam to find any answers

i dont really understand physics and maths

i was just going to do some rough experiments to guestimate what all the losses would be as i got closer to purchasing the parts

would a shell and tube heat exchanger be suitable instead of an oil resovior and copper coil ? placing the pump and orifice in between the oil out and oil in ?

my calculations for copper pipe in heat exchanger

neglecting the wall thermal resistance i came up with

u = 1/1/oil convection coeffiecient+1/water convection coeffiecient

u = 1/1/300 + 1/2500

u = 267.85

to find the area required

power = u x area x temp difference

1300 = 267.85 x area x 35

1300 = 9375 x area

area = 1300 / 9375

area = .1386 m squared

using online calculator with a 22mm diameter tube i would need a length of 200 mm

this length seams too short

this maybe due to no flow being considered?

correction to the above calculation

power = u x area x log mean temp difference

1300 = 267.85 x area x 27

1300 = 7232 x area

area = .179 m squared

using online calculator with a 22mm diameter tube i would need a length of 258 mm

Heat exchanger Design

Data:

Water flow 2l/min = 3.33E-5 m³/s

Initial temperature of water 10°C

Expected average temperature of bath 45°C

Losses in bath (according to your computations 1300 W

Estimated losses in the heat exchanger 10% of net power

1- Temperature losses in bath between in and outlet

Δθ_WB = P_LB/(Q*ρ*c_P)_W 1300/(3.33E-5*990*4181)= 9.4 °K

Inlet temperature of water = mean+0.5* Δθ_WB = 45+9.4/2= 49.72 °C ≈ 50°C

2- Heating power requirement for water

P_HW =(Q*ρ*c_P)_W* Δθ_W = 3.33E-5*990*4181*(50-10)=5.51E3 W

Remark: if no continuous flow then power hast o compensate losses = 1300 W.

3- Oil flow

Minimal temperature of oil > maximal temperature of water =50°C choice θ_{Oil min}=65°C

Power in front of exchanger P_HE = 1.1* P_HW=6065 W

This power carried by the oil is P_HOil =(Q*ρ*c_P)_Oil* Δθ_Oil = P_HE

Δθ_Oil depends on the pressure drop generated by the pump(s) at the orifice:

Δθ_Oil = Δp/(ρ*c_P)_Oil The value of Δp depends on the pump type, for cost reasons I make the choice of a gear pump, for general purpose those pumps offer a maximal log-life pressure of 14…21 MPa. If we compute with the average 17 MPa then we obtain:

Δθ_Oil = Δp/(ρ*c_P)_Oil = 17E6/(915*1880)≈10°K

This result shows that for a higher temperature difference several pumps in series have to be used.

Q _Oil = P_HE /(ρ _Oil *c_POil*) = 6065/(915*1880*10)=3.53E-4 m³/s = 21.16 m³/min.

This is the "net flow" expected at the exchanger, the pump has an internal leak estimated after a longer use at 5% so that the theoretical flow of the pump at the imposed rpm by the turbine should be 21.16/0.95=22.3 l/min, rounded up≈23 l/min

If pumps are connected in series the results can be different but the choice is a problem of available funds.

1 pump Flow 23 l/min θ_{Oil min}=65°C θ_{Oil max}=75°C

2 pumps 11.8 65 85

3 pumps 8.24 65 95

4- Heat transfer

4.1- Oil side

The more turbulent the flow is the higher the transfer intensity for this reason oil will flow in the tube and water around.

Computation will be done for an oil flow of 8.24 l/min and with same model can be repeated for the other flows. Q _Oil = 8.24l/min = 1.37 E-4 m³/s.

The Re value should be if possible ≥2300 let us compute the tube diameter for the flow and the Re=2300. Hydraulic diameter of tube = d_i Re= w*d_i/ν w= Q/A A= π*di^2/4

Re= 4*Q/(π*di*ν)→d_i= (4/π)*Q/(Re*ν) = (4/π)* 1.37 E-4 /(2300 *2.3E-4) =3.30E-04 m= =0.33mm. Even the flow if only a pump is considered will lead to a d_i≈1mm.

It is clear that with such flows of oil a turbulent flow cannot be obtained. We have to choose a tube which can be bend and is not so big.

We choose d_i = 6mm = 6E-3 m and d_e= 8mm= 8E-3m. If for different reasons other tubes will be preferred the computations can be done using this one as model. For a laminar flow the local Nusselt value depends on the Gz value = Re*Pr*d_H/X where X= distance to tube inlet.

For values of Gz ≥0.1 the local Nusselt stays constant at 4.364. For oil Pr≈5

Re= 4*Q/(π*di*ν) = 4*1.37 E-4/( π*6E-3 *2.3E-4)= 126.4 still in the laminar domain.

For Gz= 0.4 we obtain X= Re*Pr*d_H/Gz =126.4*5*6E-3/0.1= 37.92 m.

This means that under given conditions the full section is not yet in laminar conditions for a tube < 9.5 m. This also means that the heat transfer coefficient will be more important.

The Nusselt value is ≈ 7.5 and leads to a convection coefficient :

h_Oil =Nu*λ_Oil/d_i = 7.5*0.15/6E-3 = 187.5 W/m²°K

Thermal resistance parameter will be 1/ h_Oil= 5.33E-03

4.2 Tube wall

Material cooper dimensions 6/8 mm The resistance parameter will be

ln(d_e/d_i)/λ_cooper= 7.23E-04

4.3 Water side

Water will flow in a quasi laminar manner so that same law for Nusselt can be accepted

4.4 Global thermal resistance

Rth=5.33E-3+7.23E-4+3.14E-3 =9.19E-2 the equivalent heat transfer coefficient will thus be

H total= 108.8 W/m²°K

4.5 Heat transfer area

S= Q/ (Htotal* Δθ_HE)

Δθ_HE = (Δθ_HEmax- Δθ_HEmin)/ln(Δθ_HEmax/ Δθ_HEmin) = ((65-10)- (95-50))/ln(55/ 45) = 49.8 °K

S= 6065/(108.8*49.8) = 1.12 m²

Length of tube = L= S/ (π*d_i)=1.12/(π*6E-3)=59.5 m.

If only losses are to be compensated the required tube length will be 59.5*1300/6065=12.8≈13m.

The values I indicated were from a table computed a long time ago and considered a turbulent flow which is not the case in this application.

I hope that now you have all what you need.

May I make a comment you wanted to develop a system for which you did not have the basics but were not willing to ask a professional. In fact I went a lot over the usual limits since you were stuck but it is not a fair approach to ask for such a detailed support which requires a lot of work.

I hope now you can build up your system and I at least expect you to communicate the goal and the results.

Have a happy new year

Nick

thanks nickname

worked examples help make understanding things so much clearer

i am sure that this thread will help not only me and i will attempt to give it some finality by posting my design and finished product

thanks again

maitland