Condensor Power thermodynamics

1. Jan 31, 2005

knarl

Gday,

Currently I have the task of producing specifications for a water cooled condenser used in distilling liquids. Initially I thought it would be a simple thermodynamics problem. But it has proved quite complex, and I need pointing in the right direction.

The problem is, I need to figure out how big to make the condenser to be able to condense all liquid at the same rate as it is boiled. That is, the condenser needs to dissipate the same amount of heat as the boiler element produces, P.

The condenser will consist of copper tubbing, length L and inner radius r1 and outer radius r2. The water will flow through the pipe at a rate F with an initial temperature of Tw. The temperature of the vapour outside the tube will be T.

My aim is to be able to assign a value to all the variables except for L, therefore work out how long the copper tube needs to be.

To begin, I looked up the thermal conductivity of copper (390 W.m^-1.K^-1), and assumed I could work out the power conducted by the copper by knowing the surface area and temperature inside and outside. For the temperature inside, I used an average of the input water temp and output water temp. I worked out the output water temp by using the specific heat of water and flow rate to match the power needed.

Does using that average water temperature invalidate my results because of the non linearity of heating the water??

I worked out I need approx 2.7m of 3.8mm inner diameter, 5mm outer diameter tube with a water flow rate of 2L/min to dissipate 2000W from 78C vapour using 35C input water. Does this add up?

I'd appreciate if someone with a bit more knowledge of thermodynamics could have a go at the problem and tell me what they get and how they went about it.

Thanks for reading if you got this far.

Neil Stephens.

2. Feb 1, 2005

Clausius2

That seems quite a serious problem. You should read some advanced books on it. It is not a simple problem, and the formulae is not simple too. Just look for someone nearby who knows about heat transfer processes, an engineer or so. I would need to know the geometry exactly because heat transfer depends upon it a lot. And I think this is not the best place to do such a hard calculation. People here could give you some initial estimation, but it probably will be far away from the real value.

Good luck.

3. Feb 1, 2005

da_willem

If I understand correctly the cooling water flows through a single cilindrical pipe while the vapour that needs cooling is outside that pipe? I think it is best to work with dimensionless numbers like the Reynolds number and Nusselt number. There exist empirical dimensionless correlations for the heat transport depending on the geometry. By first estimating, by means of these dimensionless numbers, the character of the involved flows (is the pipeflow laminar or turbulent, how important is free convection compared to forced convection, etc.) you can than select the important empirical correlations and find out the resistance to heat transport.

4. Feb 1, 2005

knarl

Thanks guys. I feared this was a super complex problem

Correct

Is it possible for you to point me towards a source of information about these correlations?

I'm not affraid of some research. But knowing where to start is why I came to this forum

Thanks

5. Feb 2, 2005

da_willem

I tried to find something using Google, but I couldn't find much. Maybe you can purchase a book on '(heat) transport phenomena'. I could give you some of the correlations that might be involved but if you want to tackle this problem seriously and accurately you need much more research I guess.

Well as I see it there are three regions wich show some resistance to heat transport:

1) the heat flow from cooling water to the cylinder wall.
2) the conduction of heat through the copper cylinder wall
3) the flow of vapour around the cilinder

All of these can be expressed using a 'heat tranfer coefficient' h. So the resitcance to heat transport is proportional to 1/h. The resistances above mentioned are 'in series' so you can find the total h (lets call it U) using:

$$\frac{1}{U}=\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}$$

How do you find the heat flux from this U:

For pure conduction in 1D Fouriers law relates a temperature gradient via the heat conduction coefficient ($\lambda[/tex]) to the heat flux [itex]\phi_q[/tex] (J/s): $$\phi_q '' = -\lambda \frac{dT}{dx}$$ (where the '' denotes 'per unit area') Engineers usually use Newtons law of cooling: $$\phi_q ''= h \Delta T$$ Wich is in my opinion only the definition of h. This is the realtion you need. So first find out all the h's and the use this relation to find the heat flux from the driving force: the temperature difference. Now what are these h's? Usually a dimensionless relation is given in terms of the Nusselt number: $$Nu=\frac{D \lambda}{h}$$ From wich you can find h if you know the characteristic dimension D (in case of a cilinder the diameter) and the conductivity coefficient of the appropiate substance. 1) For this you need to know the character of the flow. As a criterium you can use the Reynolds number: $$Re = \frac{ \rho v D}{ \mu}$$ the density times velocity times diameter divided by the dynamic viscosity. For the numbers you gave the velocity is 2,94 m/s and the Reynolds number comes down to ~10^4 wich is pretty turbulent (the transition from laminar to turbulent takes place around 2000-2500). So you might say depending on your desired accuracy: by all these vortices in the flow it is mixed pretty good and use a temperature of the flow independent of the radial coordinate (so [itex]1/h_2 = 0$). Or you can use the empirical relation for turbulent pipe flow:

$$Nu=0.027Re^{0.8} Pr^{0.33}$$ with $$Pr=\frac{\nu}{a}$$

So the 'Prandtl number' is the kinematic viscosity ($\mu / \rho$) divided by the thermal diffusivity ($a=\lambda / \rho c_p$)

If your flow turns out to be laminar there are other relations you need to use.

2) This one is the easiest as it only involves conduction. Fouriers law in cilindrical coordinates reads:

$$\phi_q '' = -\lambda \frac{dT}{dr}$$
$$\phi_q = 2 \pi r L \phi_q ''$$

Solving this using an energy balance for a cilinder yields: $$h=\frac{2 \pi \lambda L}{ln(r_2 / r_1)}$$

3) This one is the most complicated. It crucially depends on the geometry and conditions. Are you gonna lead the vapours around the cylinder by forced convection? With what vleocity are you gonna force this flow? And is this flow perpendicular to the pipe? Or is there only free convection? All the information about the boundry layer that evolves around the cylinder can be expressed in a dimensionless correlation, but you need to know exacly the geometry and conditions of the situation.

For forced convection perpendicular to a long pipe:

$$Nu=0.332Re^{1/2}Pr^{1/3}$$

Note that Re and Pr now correspond to the flow of the vapours around the pipe, and may have very different numerical values than the ones you calculate for the flow inside the pipe.

Ok, now I have U, now what?
As I stated before you can find the heat flux from this U using Newtons law of cooling:

$$\phi_q ''= h\Delta T$$ With A the area.

Now ofcourse this temperature difference isn't constant. At the beginning of the pipe the water is cool and the vapours are pretty hot and at the end of the pipe your cooling water has heated up. So what to use for $\Delta T$? You used the mean temperature difference, but I think it is better to use the 'logarithmic mean temperature difference':

$$(\Delta T)_{ln} = \frac{\Delta T_L - \Delta T_0}{ln(\frac{\Delta T_L}{\Delta T_0})}$$
With the subscript denoting the location of the temperature difference at the beginning (0) or at the end (position L).

Now: the final result: $$\phi_q = UA(\Delta T)_{ln}$$

NOTE: If the inner surface area is substantially different from the outer surface area you might want to incoorporate the areas in the resistances and use hA as your conduction coefficients so:

$$\frac{1}{U'}=\frac{1}{h_1 2\pi r_1 L}+\frac{1}{h_2 2\pi r_1 L}+\frac{1}{h_3 2\pi r_2 L}$$ so $$\phi_q = U'(\Delta T)_{ln}$$

Last edited: Feb 2, 2005
6. Feb 2, 2005

Clausius2

1) Be careful because the correlation you posted is not valid with condensation phenomena. Our friend will have to research a bit in order to obtain a better correlation as a function of new adimensional numbers (Jackob).

2) The Nusselt Number is not what you posted. In fact:

$$Nu_D=\frac{hD}{\lambda}$$

3) I advice Karnl to consult the book "Heat Transfer Fundamentals" of Incropera & DeWitt.

7. Feb 2, 2005

You're right

8. Feb 2, 2005

Clausius2

Anyway, your effort trying to solve the problem is worth of mentioning, DaWillem. Good job.

9. Feb 2, 2005

da_willem

Thanks, now lets hope it helps him... :tongue:

10. Feb 2, 2005

knarl

Thanks very much! You've given me a flying start. Thanks da_willem for the huge effort, and thanks Clausius2 for the correction and book reference. I'm on my way to learning alot.

Regards,
knarl.