Speed in evaporator tube

[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.

[mad fridgie]

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!!!!!!!

[Peter_1]

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.

[desA]

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.

[mad fridgie]

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

[desA]

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.

[mad fridgie]

Carrying on now we have more data,
your evaporating pressure, is this your outlet pressure leaving the pipe.

[Aik]

Peter_1, do you need to know results or theoretical calculations?


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 :)

[Peter_1]

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.

[mad fridgie]

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 !!!!!!!

[Peter_1]

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.

[Peter_1]

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.

[mad fridgie]

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

[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

[DTLarca]

See Spread Sheet attached in message further down the thread....

[Peter_1]

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.

[Peter_1]

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.

[desA]

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.

[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.

[Peter_1]

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.

[desA]

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...

[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.

[Aik]

Happy New Year.

Where can I find different viscosities of refrigerant at different conditions?
Solkane will help you :)

http://www.solvay-fluor.com/docroot/fluor/static_files/attachments/download.htm

[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...

[DTLarca]

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.

[Peter_1]

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.

[DTLarca]

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.

[mad fridgie]

What is the heat source for the evap?

[desA]

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.

[nike123]

Some good software:
http://www.nist.gov/el/building_environment/evapcond_software.cfm

I think it wont tell you speed, but it will tell you pressure drop.;)

[Peter_1]

What is the heat source for the evap?

The ground MadFridgie

[mad fridgie]

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

[Peter_1]

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.

[desA]

The info I quoted comes from a commercial evap coil design program. Their pressure drops seem credible.

[Peter_1]

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.

[mad fridgie]

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)

[Peter_1]

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.

[mad fridgie]

You will have a bigger tempertaure difference with R407C, approx 2.5C.
+2.5Cevap at inlet and 0C sst at outlet.

[DTLarca]

Testing testing testing testing...

[DTLarca]

Testing testing testing testing - am I being limited to the number of words I can post?

[DTLarca]

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[DTLarca]

Thus - if you had 1bar pressure drop along a single length (70m + lots of bends) then you would have 0.25Bar pressure drop along two parallel 70m lengths. But then you would also have twice as much pipe surface with two 70m lengths. So you half the lengths each to 35m. Now your pressure drop will be 0.125Bar.

[DTLarca]

Testing testing testing ...

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Cool, editing worked twice :)

[DTLarca]

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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.

Testing other users names...

DesA

Testing reply...

DesA's coil calcs have bends - I assume?

My spreadsheet can be altered with the addition of a 6th section of pipe length for superheating but then the superheated vapour's density and viscosity are also required.

Testing...

So far able to post some stuff but not able to go back and edit those posts.

[DTLarca]

For 1/2" pipe add about 0.75 meters for every 180° U-Bend per attached image.