# Two-phase flow heat transfer coefficient

Gold Member
Hi, I'm looking to find a very very basic model to calculate the heat transfer coefficient in a tube with given heat power exchanged ##Q##. The fluid inside the tube starts sub-cooled (##T_{in}## is known) but exits the tube with a quality of steam around 0.15 (pressure is supposed constant). I'm willing to neglect all sort of effects that bubbles my induce on the heat transfer coefficients if they accumulate on the inner surface. I'd like just a simple rough estimation.

So I though to divide the problem in two part: first the sub-cooled water reaches saturation condition, and then it starts to boil. In the first part I could use ##Nu =0.023Re^{0.8}Pr^{0.3}##. As a first rough estimation I would chose an average temperature for the fluid, then I could refine the process in an iterative manner (considering the spatial dependence - but that is not my main issue here).

The problem is that I don't know how to treat the second part of the flow (where the water is partially boiling). I did quick research but the literature is very technical about this because it is a pretty studied but complicated problem. I was wondering if anyone knows a correlation used to get a very first rough estimated without getting too much into fluid dynamics (which I'm not an expert at).

Thanks Ric

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Mentor
Well, the boiling heat transfer coefficient is typically much larger than the heat transfer coefficient without boiling, isn't it. So, as a good approximation, I would just bound the answer as if the boiling did not take place. Also, 15% is a pretty low quality.

dRic2
Mentor
Here's a page on boiling heat transfer from Chemical Engineers Handbook, 5th Edition, by Perry and Chilton:

This edition has over 60 pages on heat transfer. The current edition is Perry's Chemical Engineer's Handbook, by Green and Southard: https://www.amazon.com/dp/0071834087/?tag=pfamazon01-20.

Have fun!

dRic2
Gold Member
Perry's Chemical Engineer's Handbook
long time, no see
The reason to this post was pretty much to avoid resorting that beast. 😆

Well, the boiling heat transfer coefficient is typically much larger than the heat transfer coefficient without boiling, isn't it. So, as a good approximation, I would just bound the answer as if the boiling did not take place. Also, 15% is a pretty low quality.
Would you consider this approximation to still be good at high pressure (like 70 bar)? I may be wrong but I think water properties changes quite a bit with pressure.

Mentor
Would you consider this approximation to still be good at high pressure (like 70 bar)? I may be wrong but I think water properties changes quite a bit with pressure.
Well, since the vapor specific volume is about 20 x the volume of liquid water at that pressure, I might consider also looking at the case of heat transfer with all gas present (using the viscosity and density of saturated water vapor at 70 bars , 285 C).

Gold Member
Well, since the vapor specific volume is about 20 x the volume of liquid water at that pressure, I might consider also looking at the case of heat transfer with all gas present (using the viscosity and density of saturated water vapor at 70 bars , 285 C).
Like the real behavior is given by some sort of interpolation between this two extrema ?

Mentor
Like the real behavior is given by some sort of interpolation between this two extrema ?
I would want to see how the two answers compare first.

Gold Member
I would want to see how the two answers compare first.
Thanks for the interest. I bit busy right now so I will take this very slowly. Sorry if I'll let you wait

Mentor
Thanks for the interest. I bit busy right now so I will take this very slowly. Sorry if I'll let you wait
No problem. Take your time. This type of thing requires some intuitive consideration.

dRic2
Gold Member
Hi @Chestermiller I have one more question before continuing. My final goal is to create a super-simplified model for a steam generator so where steam is created in the part outside of the tubes (I don't remember the name). In order to do that I have to consider the geometry. I have a triangular pitch. I struggle a bit to calculate the "free" area in order to calculate Reynolds number for the flow (I know the mass flow rate). For a triangular pitch I assumed it was given by this relation:
$$S_{\text{free}} = \frac {\sqrt{3}} {4} P^2 - \frac {\pi}{4} D^2$$
Where ##P## is the pitch of the lattice. But I'm not very sure. Can you confirm ?

Mentor
Hi @Chestermiller I have one more question before continuing. My final goal is to create a super-simplified model for a steam generator so where steam is created in the part outside of the tubes (I don't remember the name). In order to do that I have to consider the geometry. I have a triangular pitch. I struggle a bit to calculate the "free" area in order to calculate Reynolds number for the flow (I know the mass flow rate). For a triangular pitch I assumed it was given by this relation:
$$S_{\text{free}} = \frac {\sqrt{3}} {4} P^2 - \frac {\pi}{4} D^2$$
Where ##P## is the pitch of the lattice. But I'm not very sure. Can you confirm ?
So this is on the shell side of a shell-and-tube heat exchanger? Or is it a boiler with tubes on the bottom? You need to consult a handbook (like Perry's or Marks) to get the precise details for this. Or some text on heat exchanger design.

Homework Helper
My final goal is to create a super-simplified model for a steam generator so where steam is created in the part outside of the tubes (I don't remember the name)
The reason to this post was pretty much to avoid resorting that beast. 😆
I'm willing to neglect all sort of effects that bubbles my induce on the heat transfer coefficients if they accumulate on the inner surface. I'd like just a simple rough estimation.

What, precisely, is the purpose of your model ?

Gold Member
What, precisely, is the purpose of your model ?
Try to write a simple code to simulate a steam generator, just to learn something and getting better at programming.

Gold Member
So this is on the shell side of a shell-and-tube heat exchanger? Or is it a boiler with tubes on the bottom? You need to consult a handbook (like Perry's or Marks) to get the precise details for this. Or some text on heat exchanger design.
Yes. I was thinking about a simple cylinder filled with tubes in a triangular geometry. I would start by neglecting inlet and outlet geometry as a possible source of complication. Then the hot fluid will flow inside the tubes while the steam is generated on the shell side. The flow of the cold fluid I was initially thinking to be parallel to the tubes, but I know realize that there may be different options, so I don't really know. Since it's just something I would do to learn the process I would choose the simplest scenario.

Mentor
Yes. I was thinking about a simple cylinder filled with tubes in a triangular geometry. I would start by neglecting inlet and outlet geometry as a possible source of complication. Then the hot fluid will flow inside the tubes while the steam is generated on the shell side. The flow of the cold fluid I was initially thinking to be parallel to the tubes, but I know realize that there may be different options, so I don't really know. Since it's just something I would do to learn the process I would choose the simplest scenario.
Then if you are choosing the simplest scenario (parallel flow, with no baffling in the heat exchanger), you would probably work with the hydraulic diameter in the correlation equations, where the hydraulic diameter is equal to 4 times the cross sectional area for flow divided by the wetted perimeter.

Gold Member
you would probably work with the hydraulic diameter in the correlation equations, where the hydraulic diameter is equal to 4 times the cross sectional area for flow divided by the wetted perimeter.
Yes, so my question is about the cross sectional area (which I referred as ##S_{\text{free}}##) for a triangular pitch. Is my formula ok for that ?

Gold Member
Anyway I tried a very simple estimation of the heat transfer coefficient supposing a circular cross section. I used these values (together with tables for density, viscosity ecc):
##v_{l} = 4.5 \text{m/s}##
##d = 13 \text{mm}##
##P = 69 \text{bar}##

and I calculated the heat transfer for both cases of saturated steam and liquid:

* 100% saturated liquid, ## h \approx 33000 \frac{W} {m °C}##
* 100% saturated steam ## h \approx 1600 \frac{W} {m °C}##

PS: in the case of saturated steam I thought to keep the mass flow rate as a constant so, if ##v## was the velocity of the liquid I used for the vapor ##v_{vap} = v* \frac {rho_{l, sat}(69 \text{bar})}{rho_{v, sat}(69 \text{bar})}##

Do these results seems reasonable to you?

Mentor
Yes, so my question is about the cross sectional area (which I referred as ##S_{\text{free}}##) for a triangular pitch. Is my formula ok for that ?
For a unit cell, I would have half the 2nd term.

dRic2
Gold Member
Yes, my mistake. And if the flow rate ##\Gamma## is fixed can I calculate the velocity as
##v = \frac {\Gamma}{\rho*S*N}##
where N is te total number of pipes ?

PS: the order of magnitude for ##h## are the same as those I posted above. Do you think I should neglect the vapor or not to get a decent estimation ?

Mentor
Yes, my mistake. And if the flow rate ##\Gamma## is fixed can I calculate the velocity as
##v = \frac {\Gamma}{\rho*S*N}##
where N is te total number of pipes ?

PS: the order of magnitude for ##h## are the same as those I posted above. Do you think I should neglect the vapor or not to get a decent estimation ?
I would get the average velocity from $$\frac{\Gamma}{\rho(A-NA_T))}$$where A is the cross sectional area of the empty shell and ##A_T## is the cross sectional area of each tube.

Gold Member
Yes, seems better. Do you think my attempt was wrong ?

Anyway thanks a lot for your time.

Mentor
Anyway I tried a very simple estimation of the heat transfer coefficient supposing a circular cross section. I used these values (together with tables for density, viscosity ecc):
##v_{l} = 4.5 \text{m/s}##
##d = 13 \text{mm}##
##P = 69 \text{bar}##

and I calculated the heat transfer for both cases of saturated steam and liquid:

* 100% saturated liquid, ## h \approx 33000 \frac{W} {m °C}##
* 100% saturated steam ## h \approx 1600 \frac{W} {m °C}##

PS: in the case of saturated steam I thought to keep the mass flow rate as a constant so, if ##v## was the velocity of the liquid I used for the vapor ##v_{vap} = v* \frac {rho_{l, sat}(69 \text{bar})}{rho_{v, sat}(69 \text{bar})}##

Do these results seems reasonable to you?
Since the rho v's are the same, the only difference in Re is because of the viscosity difference. Does that cause such a big effect? Or is the Pr also that contributes?

Mentor
Yes, seems better. Do you think my attempt was wrong ?

Anyway thanks a lot for your time.
The only reason I would look at the unit cell would be to get the hydraulic diameter. That would be obtained by taking the unit cell area you determined, multiplying by 4, and dividing by the wetted perimeter ##\pi D/2##. I guess another way to get the hydraulic diameter is ##4(A-NA_T)/(N\pi D)##

Mentor
I used a MATLAB library called XSteam to get liquid and steam properties, but viscosity was not included so I had do do a quick search on the internet. As reference values at 300 °C I found for saturated conditions (Pa*s)

##\mu_V = 20*10^{-6}##
##\mu_L = 95*10^{-6}##

with those values I got

##Pr_L = 0.89##
##Pr_V = 1.66##

I also tried different cross-section sizes (with fixed flow rate) in order to get different velocities (from 4.5 to 16 m/s). The results obviously change but the order of magnitude is still 35000/40000 for saturated liquid and 2000 for saturated steam (in W/m/°C).
Given these values, what is it that causes the large difference between the liquid and vapor, the thermal conductivity? The Nu's should come out comparable, right?

Gold Member
Since the rho v's are the same, the only difference in Re is because of the viscosity difference. Does that cause such a big effect? Or is the Pr also that contributes?
I noticed that ##h## depends on the value of the flow rate. For low mass flow rate I get those values, but if I increase ##\Gamma## I get comparable results. I fear a mistake. Tomorrow I'm going to try again from scratch

Mentor

Gold Member
Hi, sorry for my late reply. Where can I find steam viscosity at 69 bar ? I tried on Perry's but didn't find it. Probably my fault.

Anyway I did again the calculations with those value of viscosity and I always get
##h_{l}/h_v \approx 2##.
I kept the mass flow rate fixed, but changed the dimensions in order to get different values for the velocity. For smaller liquid velocity (v = 4.5 m/s) I have ##h_l \approx 20000## and ##h_v \approx 10000## while for higher liquid velocities (up to 13 m/s) I get ##h_l \approx 40000## and ##h_v \approx 20000## (all in W/m^2/°C)

Mentor
Hi, sorry for my late reply. Where can I find steam viscosity at 69 bar ? I tried on Perry's but didn't find it. Probably my fault.

Anyway I did again the calculations with those value of viscosity and I always get
##h_{l}/h_v \approx 2##.
I kept the mass flow rate fixed, but changed the dimensions in order to get different values for the velocity. For smaller liquid velocity (v = 4.5 m/s) I have ##h_l \approx 20000## and ##h_v \approx 10000## while for higher liquid velocities (up to 13 m/s) I get ##h_l \approx 40000## and ##h_v \approx 20000## (all in W/m^2/°C)
https://www.engineeringtoolbox.com/steam-viscosity-d_770.html has liquid and vapor viscosities at saturation conditions. I also found a reference with Prantdl numbers for liquid water and water vapor as a function of temperature and pressure. According to the reference, at your conditions, they are both are very close to 1.0.

Please flesh out what Re and Pr you used for the different cases, and what k you used.

Gold Member
I checked the values for viscosity from that website and they agree perfectly. With those I calculated:
##Pr_l = 0.8963##
##Pr_v = 1.6594##

Using:
##k_l = 0.57##
##k_v = 0.063##

For Re number I get

*For ##v_l = 4.5## m/s -> ##v_v = v_l* \frac {\rho_l} {\rho_v} = 92## m/s
##Re_l = 8.35*10^6##
##Re_v = 3.97*10^7##

*For ##v_l = 13.23## m/s -> ##v_v = v_l* \frac {\rho_l} {\rho_v} = 273## m/s
##Re_l = 2.45*10^7##
##Re_v = 1.16*10^8##

Mentor
I checked the values for viscosity from that website and they agree perfectly. With those I calculated:
##Pr_l = 0.8963##
##Pr_v = 1.6594##

Using:
##k_l = 0.57##
##k_v = 0.063##

For Re number I get

*For ##v_l = 4.5## m/s -> ##v_v = v_l* \frac {\rho_l} {\rho_v} = 92## m/s
##Re_l = 8.35*10^6##
##Re_v = 3.97*10^7##

*For ##v_l = 13.23## m/s -> ##v_v = v_l* \frac {\rho_l} {\rho_v} = 273## m/s
##Re_l = 2.45*10^7##
##Re_v = 1.16*10^8##
This all seems pretty OK. So what do you want to do next?

Gold Member
So what do you want to do next?
I'd like to know what ##h## I should use in the case that the outlet flow rate has a quality of steam of 0.2 (20%). Thanks again for your time.

For example is 1st case (##v_l = 4.5## m/s) I get ##h_l \approx 20000## and ##h_v \approx 10000## W/m^2/°C

Mentor
Bird, Stewart, and Lightfoot give typical values for forced convection of 500-10000 for water. For gases, 50-500. Is it possible that you are using too small a diameter?

What fraction of the tube length is covered before the vapor begins forming. Also, the entire tube length after that is not going to be 20% quality.

Gold Member
Ok so my approach has changed. The final goal is to find a suitable value for ##h##. The steam generator I am considering is one from a nuclear power plant so I looked for papers about steam generators which met the data I have in mind and I reverse-engineered the problem and I calculated ##h## from ##Q = UA \Delta T_{ml}##. I have found a value of ##h## from 9000 to 12000 (slightly changing some parameters to see how much they weight). This looks ok, I think. Only thing I don't like is that I also get a velocity for the liquid of 0.34 m/s... I have really no idea whether this could be a reasonable value or not. I was expecting something around 3 or 4 m/s but I'm not sure

Mentor
Ok so my approach has changed. The final goal is to find a suitable value for ##h##. The steam generator I am considering is one from a nuclear power plant so I looked for papers about steam generators which met the data I have in mind and I reverse-engineered the problem and I calculated ##h## from ##Q = UA \Delta T_{ml}##. I have found a value of ##h## from 9000 to 12000 (slightly changing some parameters to see how much they weight). This looks ok, I think. Only thing I don't like is that I also get a velocity for the liquid of 0.34 m/s... I have really no idea whether this could be a reasonable value or not. I was expecting something around 3 or 4 m/s but I'm not sure
If you are aware of the design and operating conditions in the actual plant, are you doubting your ability to back out the flow velocity? That value doesn't seem very unreasonable to me.

Gold Member
are you doubting your ability to back out the flow velocity?
Well... I always blunder a lot of stuff when doing calculations... so I never actually trust myself if I don't have an idea of what the result might look like.

Anyway, assuming I didn't make a mistake I found the value which seems to agree with your predictions/considerations. Thanks again for your time.