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Speed in evaporator tube
Someone able to calculate these for me please?
I personally know who can do this - I can't - but i want to make this a general question.
What will be the theoretical speed through an 1/2" line - length 15, 30 and 70 m - of an evaporator, 2 kW, evaporating at 0°C for R410a and R407c, condensing at 35°C and 40°C and a SC of 10K and second case 20K, Sh of 5K?
We start with a saturated gas/liquid mixture in the beginning and ending with a superheated gas.
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Re: Speed in evaporator tube
Keeping with the theme of the question, we are missing some data, to calculate the next stage!
Data can be supplied in a number of ways!!!!!!!
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Re: Speed in evaporator tube
Edited, indeed, forgot some data, in fact most important data. Sorry :o
Reason for my question: a colleague/friend is installing DX heatpumps and he's using for this copper in lengths of 70m. I was suggesting that this is far too long and should be something in the range of 30 m. I say he could have pressure drops of more than 1 bar and he didn't believe this.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
Someone able to calculate these for me please?
I personally know who can do this - I can't - but i want to make this a general question.
What will be the theoretical speed through an 1/2" line - length 15, 30 and 70 m - of an evaporator, 2 kW, evaporating at 0°C for R410a and R407c, condensing at 35°C and 40°C and a SC of 10K and second case 20K, Sh of 5K?
We start with a saturated gas/liquid mixture in the beginning and ending with a superheated gas.
To try the simple approach, first.
The evaporator & condenser designers will probably use something along these lines:
1. Liquid : v,max ~ 1.5 m/s
2. Vapour : v,max ~ 15-25 m/s.
The major constraint is allowable pressure drop, with vapour erosion the next possibility.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
Edited, indeed, forgot some data :o
Reason for my question: a colleague/friend is installing DX heatpumps and he's using for this copper in lengths of 70m. I was suggesting that this is far too long and should be something in the range of 30 m. I say he could have pressure drops of more than 1 bar.
Still a bit more data required, if you want to know the velocity, your need to area, density and something else?:D
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Re: Speed in evaporator tube
Quote:
m' = r*Ac*v
Where :
m' = mass flowrate [kg/s]
r = density (vapour, or liquid, or 2-phase) [kg/m3]
Ac = cross-sectional area of pipe (ID-based) [m]
v = velocity [m/s]
You can work backwards to find 'v', if you have all the other data.
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Re: Speed in evaporator tube
Carrying on now we have more data,
your evaporating pressure, is this your outlet pressure leaving the pipe.
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1 Attachment(s)
Re: Speed in evaporator tube
Peter_1, do you need to know results or theoretical calculations?
Quote:
Originally Posted by
mad fridgie
Still a bit more data required, if you want to know the velocity, your need to area, density and something else?:D
Also pipe's location :)
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Re: Speed in evaporator tube
Quote:
Originally Posted by
mad fridgie
Carrying on now we have more data,
your evaporating pressure, is this your outlet pressure leaving the pipe.
Indeed MadFridgie. I know that a phenomena will happen in the tubes where liquid will be trapped between gas and vice-versa.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
Indeed MadFridgie. I know that a phenomena will happen in the tubes where liquid will be trapped between gas and vice-versa.
So you now need to work out the mass flow of each refrigerant/condition to meet your 2 KW, use PH chart, whilst looking at the chart look up the density of the refrigerant at the pipe exit conditions, then calculate as per DesA formular (maximum velocity/speed)
Or Just use Coolpack !!!!!!!
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Aik
Peter_1, do you need to know results or theoretical calculations?
Also pipe's location :)
Only 1 straight pipe Aik where the give evaporating pressure is the outlet pressure.
My colleague is now measuring 0°C (converted from pressure of 3.5 bar) for R407c (!) at the compressor inlet and he's thinking that the whole circuit of 14 kW (7 circuits of 2 kW) is evaporating at 0°C.
If there should be no pressure drop, then the whole circuit starts at -6°C (+/-)
Due to a possible pressure drop of 1 bar for R407c, the whole circuit stays at 0°C.
I advised him to reduce his lines to 30 m and that this should benefit the COP of the compressor due to the lower DP over the copper.
If I calculate Dp for a 1/2" line of 70m , R407c for the given conditions, then this gives me 1.2 bar DP (8.7K) for a gas line and a neglectable DP for the 1/2" liquid line.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
mad fridgie
So you now need to work out the mass flow of each refrigerant/condition to meet your 2 KW, use PH chart, whilst looking at the chart look up the density of the refrigerant at the pipe exit conditions, then calculate as per DesA formular (maximum velocity/speed)
Or Just use Coolpack !!!!!!!
The condition of the gas constantly changes while the refrigerant flows from the beginning to the end of the pipe. So I think we just can't assume outlet nor inlet conditions to calculate speed in this line. It's somehow an mathematical expression that needs to be used.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
The condition of the gas constantly changes while the refrigerant flows from the beginning to the end of the pipe. So I think we just can't assume outlet nor inlet conditions to calculate speed in this line. It's somehow an mathematical expression that needs to be used.
At those conditions the speed at the exit will remain constant (for each refrigerant and set of conditions)
Your speed will increase from the start to the end, due to change in density.
Have you worked out the amount of of surface area you require,
So if I understand it correctly, you have 7 circuits at 70 meters, you want lets say 14 circuits at 35 meters (same surface area, reduced pressure drop) but could reduced refrigerant heat transfer co-efficient
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Re: Speed in evaporator tube
Q=m'*dH - capacity [kW]
m' - mass flowrate [kg/s]
dH - difference of enthalpies of evaporating and suction refrigerant [kJ/kg]
Acording to this formula you can calculate m'. After with help DesA's formula you can calculate velocity. Entalpy you can find using CoolPack.
dP=lamda*(l/d)*(r*v^2)/2 - pressure drop on straight tube [Pa]
lamda - friction coefficient, for copper tubes lamda=0.03
l - tube's length [m]
d - inner diameter of tube [m]
r - density (vapour, or liquid, or 2-phase) [kg/m3]
v - velocity [m/s]
P.S. for more accuracy for 10^5< Re < 10^8
lamda=0.0032+(0.221/(Re^0.237))
Re - Reynolds number
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Re: Speed in evaporator tube
See Spread Sheet attached in message further down the thread....
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Re: Speed in evaporator tube
Quote:
Originally Posted by
mad fridgie
At those conditions the speed at the exit will remain constant (for each refrigerant and set of conditions)
Your speed will increase from the start to the end, due to change in density.
Have you worked out the amount of of surface area you require,
So if I understand it correctly, you have 7 circuits at 70 meters, you want lets say 14 circuits at 35 meters (same surface area, reduced pressure drop) but could reduced refrigerant heat transfer co-efficient
@Mad Fridgie, indeed, the speed will increase but I can imagine this will be not a linear function. So, that becomes a little bit, in fact a lot of a gray zone for me.
What I was saying was twofold:
1. sending 14 kW through 7 lines of 70 m gives a pressure drop of +/- 1 bar and reduces COP
2. Spreading the 490 m over 14 coils of 35 m will increase system performance.
But this is something I feel on my elbow but I can't prove t with figures.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
mad fridgie
So you now need to work out the mass flow of each refrigerant/condition to meet your 2 KW, use PH chart, whilst looking at the chart look up the density of the refrigerant at the pipe exit conditions, then calculate as per DesA formular (maximum velocity/speed)
Or Just use Coolpack !!!!!!!
I can calculate it manually with the refrigerant properties out of saturated tables or superheated valeus via the log p/h but then I'm starting with conditions which doesn't change that much while transporting the gas or liquid along the lines.
In the evaporator tubes, we have a very rapid and intense phase change an a huge expanding of the liquid to its gaseous state. So the characteristics changes continuously along the line of 70 m.
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Re: Speed in evaporator tube
A gift for Peter. Happy New Year for 2011.
http://i56.tinypic.com/1zmyy69.png
'Natural' circuit. 3c.
http://i51.tinypic.com/34do08x.png
7c. 2 unlinked tubes.
http://i53.tinypic.com/27y8hms.png
14c. 2 unlinked tubes.
http://i52.tinypic.com/14obr86.png
'Natural' circuit. 19c.
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Re: Speed in evaporator tube
desA, first of all, thanks for your help - as usual. For you and all the other colleagues, a Happy NewYear (in some parts of the world, it's already Newyear I guess)
desA, Let's take your coil consisting of 7 tubes, where the total coil DP is 7.66 kPa (or 0.076 bar...I'm used to work with bar)
Because the total pressure = 1/partial pressure losses (like a parallel connection of electrical resistances), can I then say that the DP over 1 tube of the coil is then 7 times 7.66 kPa or 53.62kPa (= 0.53 bar)?
Splitting the same capacity over 14 circuits reduces the DP with a factor of 7.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
DTLarca
Here's a spread sheet I produced, I'm thinking about 6 years ago maybe, for another discussion we had here.
You could just change the numbers for your conditions.
Thanks M..., DTLarca, I will try to figure out how this works. As usual if you made something, seems very impressing and professionally made. Something for tomorrow or Sunday.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
desA, first of all, thanks for your help - as usual. For you and all the other colleagues, a Happy NewYear (in some parts of the world, it's already Newyear I guess)
desA, Let's take your coil consisting of 7 tubes, where the total coil DP is 7.66 kPa (or 0.076 bar...I'm used to work with bar)
Because the total pressure = 1/partial pressure losses (like a parallel connection of electrical resistances), can I then say that the DP over 1 tube of the coil is then 7 times 7.66 kPa or 53.62kPa (= 0.53 bar)?
Splitting the same capacity over 14 circuits reduces the DP with a factor of 7.
As a first approximation, we could use the Bernoulli equation across each circuit.
dP,14/dP,7 ~ (Ac,7/Ac,14)^2 = (Ac,7/(2*Ac,7))^2 = 1/4
Backcheck:
dP,14/dP,7 = 0.0696/0.248 = 0.2806 ~ 0.25 = 1/4
The difference probably comes from his bend calculations, or the missing tubes. Each parallel pass in the evap same will have 'roughly' the same dP, since the system splits & meets again. Practically though, the end manifolds will really determine the true velocity distribution across each circuit. Also probably have to factor in the 2x length of the 7c versus the 14c - haven't gone through the hand calcs fully. You could plot the data I provided with circuits on horizontal axis & dP on vertical axis. Then curve-fit & estimate from there...
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1 Attachment(s)
Re: Speed in evaporator tube
DTLArca, Where can I find different viscosities of refrigerant at different conditions?
In attachment what I calculated but don't know if it's correct 85 kPa pressure drop.
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Re: Speed in evaporator tube
Happy New Year.
Quote:
Where can I find different viscosities of refrigerant at different conditions?
Solkane will help you :)
http://www.solvay-fluor.com/docroot/...s/download.htm
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Aik
Q=m'*dH - capacity [kW]
m' - mass flowrate [kg/s]
dH - difference of enthalpies of evaporating and suction refrigerant [kJ/kg]
Acording to this formula you can calculate m'. After with help DesA's formula you can calculate velocity. Entalpy you can find using CoolPack.
dP=lamda*(l/d)*(r*v^2)/2 - pressure drop on straight tube [Pa]
lamda - friction coefficient, for copper tubes lamda=0.03
l - tube's length [m]
d - inner diameter of tube [m]
r - density (vapour, or liquid, or 2-phase) [kg/m3]
v - velocity [m/s]
P.S. for more accuracy for 10^5< Re < 10^8
lamda=0.0032+(0.221/(Re^0.237))
Re - Reynolds number
Sorry, it's for single-phase substance. For two-phase states it must be integration is over velocity and density...
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Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
DTLArca, Where can I find different viscosities of refrigerant at different conditions?
In attachment what I calculated but don't know if it's correct 85 kPa pressure drop.
I'll modify the spread sheet a little later so that it can be used for this particular purpose.
I think I generated all my refrigerant data using coolpack or Klea Calc - I forget - will have a look later.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
desA
As a first approximation, we could use the Bernoulli equation across each circuit.
dP,14/dP,7 ~ (Ac,7/Ac,14)^2 = (Ac,7/(2*Ac,7))^2 = 1/4
Backcheck:
dP,14/dP,7 = 0.0696/0.248 = 0.2806 ~ 0.25 = 1/4
The difference probably comes from his bend calculations, or the missing tubes. Each parallel pass in the evap same will have 'roughly' the same dP, since the system splits & meets again. Practically though, the end manifolds will really determine the true velocity distribution across each circuit. Also probably have to factor in the 2x length of the 7c versus the 14c - haven't gone through the hand calcs fully. You could plot the data I provided with circuits on horizontal axis & dP on vertical axis. Then curve-fit & estimate from there...
desA, was my assumption then correct about the dp over 1 circuit (7 times total DP over the coil)?
I think your explanation was about the change from a 7 to 14 circuit evaporator.
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1 Attachment(s)
Re: Speed in evaporator tube
Quote:
Originally Posted by
Peter_1
DTLArca, Where can I find different viscosities of refrigerant at different conditions?
In attachment what I calculated but don't know if it's correct 85 kPa pressure drop.
I think you have used a vapour density 1/10th of what it actually will be, so, your vapour velocity is of course 10 times faster than it actually will be. I get a total pressure drop of 0.2Bar along the 70m. Your liquid calcs look okay though. Have a look at the modified spread sheet attached...
There are 4 sheets now:- Copper Tube Dimension Choices
- Refrigerant and tube values converted to velocity if all liquid and then again velocity if all vapour and thereafter, based on average X (kg/kg) values for each 5th pipe length, a total pressure drop is calculated based on the implied proportions of liquid versus vapour.
- R410A values
- R407C values.
But these calcs will be, seriously, estimates only. Very simple assumptions are being made.
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Re: Speed in evaporator tube
What is the heat source for the evap?
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Re: Speed in evaporator tube
http://i53.tinypic.com/119c8xj.png
Single circuit. 1c. dP/L=8.77 kPa/m ; dP=1896 kPa = 18.96 bar
http://i51.tinypic.com/34do08x.png
7c. 2 unlinked tubes. dP/L=0.248 kPa/m ; dP=7.66 kPa = 0.0766 bar
http://i53.tinypic.com/27y8hms.png
14c. 2 unlinked tubes. dP/L=0.0696 kPa/m ; dP=1.07 kPa = 0.0107 bar
------------
The useful feature here is to compare the dP/L values i.e. the pressure loss per meter of pipe run, in either the 1c, 7c, or 14c configurations.
dP = (dP/L)*L ; where L=pipe length.
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Re: Speed in evaporator tube
Some good software:
http://www.nist.gov/el/building_envi...d_software.cfm
I think it wont tell you speed, but it will tell you pressure drop.;)
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Re: Speed in evaporator tube
Quote:
Originally Posted by
mad fridgie
What is the heat source for the evap?
The ground MadFridgie
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Re: Speed in evaporator tube
Your load is not steady state, so massive differences in working pressures occur, I now understand the need to know your velocity through out the evap, I am not sure what velocities you require to ensure that oil logging/filming is limited (when you are changing phase), maybe some others have this required data
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Re: Speed in evaporator tube
Quote:
Originally Posted by
DTLarca
I think you have used a vapour density 1/10th of what it actually will be, so, your vapour velocity is of course 10 times faster than it actually will be. I get a total pressure drop of 0.2Bar along the 70m. Your liquid calcs look okay though. Have a look at the modified spread sheet attached...
There are 4 sheets now:
- Copper Tube Dimension Choices
- Refrigerant and tube values converted to velocity if all liquid and then again velocity if all vapour and thereafter, based on average X (kg/kg) values for each 5th pipe length, a total pressure drop is calculated based on the implied proportions of liquid versus vapour.
- R410A values
- R407C values.
But these calcs will be, seriously, estimates only. Very simple assumptions are being made.
I think DTLarca enlightened a little bit my confusion. The entire length of 70 m was divided in 5 different sections where in each section we have each time a new 'mixture' due to the ever increasing gas phase (and decreasing liquid phase) of the refrigerant along the copper tube.
But, I see a big difference between DTLarca and desA. I think I will make a practical setup for this and test this once.
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Re: Speed in evaporator tube
The info I quoted comes from a commercial evap coil design program. Their pressure drops seem credible.
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Re: Speed in evaporator tube
Quote:
Originally Posted by
desA
The info I quoted comes from a commercial evap coil design program. Their pressure drops seem credible.
To be not misunderstood, I don't doubt any of the posts here, this is all something which can't be found in common refrigeration books. This is very learning for me.
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Re: Speed in evaporator tube
With R410a the pressure drop on either case is pretty minimal, less than 1C on your sst, I would be more worried about oil logging, at these lower velocities and pipe volume (will the compressor run out of oil over time)
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Re: Speed in evaporator tube
Well, for the moment, these systems are running on R407c but we want to change this to R410a. We then can install smaller copper, 3/8" will then be our choice.
But the R407c is a special case, especially if you should encounter a pressure drop of around 1 bar along the line . You then should measure the same temperature (not same pressure) at the end and beginning of the copper.
My colleague is now measuring only at the outlet and he's measuring 4.5 bar ( equals 0°C ) and he's thinking that this is the same all over the line, from beginning till the end.
I know think he's right and I'm wrong. I thought he should face much larger DP's over 70 m copper.
To be 100% sure and to convince myself, I will make a practical setup in my evening classes and connected different lengths of copper to a speed regulated compressor.
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Re: Speed in evaporator tube
You will have a bigger tempertaure difference with R407C, approx 2.5C.
+2.5Cevap at inlet and 0C sst at outlet.
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Re: Speed in evaporator tube
Testing testing testing testing...
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Re: Speed in evaporator tube
Testing testing testing testing - am I being limited to the number of words I can post?