Sample Cooler Design

Your design could be something like this....

https://www.watco-group.co/surface-condenser-in-thermal-power-plant/

https://www.farad.gr/products/steam-condensers-drain-coolers

https://www.htri.net/xist

With the low flow (volume) rate, I would recommend a plate type heat exchanger...

http://marineinfobox.blogspot.com/2017/04/comparison-between-shell-and-plate-type.html

A few things here.

Do you have a budget to work with?

140c condensate... is that under pressure?

The final approach temperature Δ is pretty tight being 10c... cooling would be difficult but not impossible., and by difficult it would require some expense. And with youR flow rate, some recirculating. It good to keep a night flow to retain turbulence.

<...140c condensate... is that under pressure?...>

In order for it to be liquid water at that temperature, the pressure needs to be north of 3.6barg, according to Mayhew & Rogers, "Thermodynamic Properties of fluids", any edition.

Vapour pressure = 3.64 bara at 140°C. That's why I said 3 barg = 4 bara, including a tad safety, in #4.

That’s why I asked the question... and Codemaster followed up with the required details.

At 140°C the primary circuit must be at ~ 3barg or higher.

First calculate the kW heat transferred.

I would try a cooling water outlet temperature, and calculate the log mean ΔT. Higher cooling water outlet temperature means lower cooling water flow but lower log mean ΔT, hence greater exchange area required, and vice versa. There may be a limit on cooling water flow available.

If it's a shell and tube exchanger you can find data on kW/(m²K) in eg Perry, hence the exchange area. But assuming both fluids are clean water a plate heat exchanger should be OK, and likely more compact. Best to speak to a supplier for sizing.

Then you could try a couple more cooling water outlet temperatures, and see what seems to give the optimum design.

Just re-read the original post and see cooling water flow is specified as 20m³/h, 10x the condensate flow. So temp rise is 0.1 x condensate temp drop., ie 10K to 40°C.

Oh, any number of ways.

• The most popular these days is to talk the problem through with companies that supply heat exchangers - one can find all their phone numbers on their websites.
• Another is to bounce the problem off a local Chemical Engineer.
• A third is to use Perry, “The Chemical Engineer’s Handbook”, any edition.
• etc.

Dear Sam6516

The given data by you, seen as follows:

i) Hot condensate Flow Rate: 2 M^3/Hr = 2000 Kg./Hr. taken as Wc. (The density of condensate at Temp 140 Deg., is 0.975, which will be 975 Kg./M^3 and can be ignored for simplicity) However the Weight of 2 M^3 hot condensate = 2000 x (0.975) = 1950 Kgs./Hr

Ii] Condensate Temp., In and Out: 140 deg.C and 40 deg.C , taken as Tcin and Tcout respectively, a difference is 100 Deg.C

iii) Cooling Water Flow Rate: 20 M^3/Hr., with inlet Temp. of 30 deg.C (Here density of water considered as 1.0 for simplicity) taken as Tw-in and Tw-out. The Outlet Temp. of Cooling water circulated is to be calculated in order to find out the Log Mean Temp.

iv) You have not mentioned the pressure of the hot condensate and this is required to decide the Thickness of the Tube to be used in the Heat Exchanger.

CALCULATIONS:

1. The Basis is HEAT LOST BY CONDENSATE = HEAT GAINED BY COOLING WATER, which is also equal to U x A x (LGMT) where U is the Over-All Heat Transfer Co.eff., A is the Area of the Heat Exchanger in M^2, LGMT is Log Mean Temp. in Deg.C

2. Wc X (Tc in – Tc out) = Wcw x (Tw-out - Tw-in) = U x A x (LGMT)

3. The Tube Material considered in my estimate is SS Tube of 20 MM dia., and Length is 2000MM. Thickness will have to be decided based on the Pressure of the Condensate to be cooled.

If you substitute the values, you get the Outlet Tem.of Cooling Water as 40 Deg. 2000 x (140 – 40) = (20 x 1000) x (Tout-30) this will give a cooling water outlet Temp = 40 Deg.C which is NOT possible since the APPROACH TEMP. for Heat Transfer IS ZERO and your Heating surface required will be infinity.

It is recommended to have an APPROACH TEMP., of 10 Degrees for optimum design. Hence the Outlet Temp. of Condensate will be 50 Deg.C, the Temp. drop of Condensate will be 90 Deg.C and Cooling Water outlet temp. will be 40 Deg.C., rise of 10 Deg.C You can also have an Approach Temp., of 5Deg.C but surface area will go up and hence cost will go up cost-benefit ratio will decide the Approach Temp.

For these temperatures as shown or arrived at, by me will give you a Log Mean Temp. of 41.695 Deg.C The Value for the Over-all Heat transfer for this system will be 75 K.Cal/Hr/Deg.C and hence U = 75. Substituting this value in the equations shownabbove you will get a Heating Surface or Cooling Surface as 56.120 M^2 I am not giving the full calculations as you can also calculate/work out. ADD 10% on this as margin for scaling effect/fouling factor an area of 61.732 Sq. M will be sufficient.

Based on 20 MM Dia, 2000 MM length tube you can calculate the number of Tubes required. The Ligament can be taken as 8 MM and then you can arrive at the dia of the Heat Exchanger Shell. Thickness will depend upon the pressure of condensate which you know, but not mentioned in your posting.

I hope it is clear to you and any doubt, inform me through personal messaging system available in the CR4

Thanks.

DHAYANNADHAN.S

"...If you substitute the values, you get the Outlet Tem.of Cooling Water as 40 Deg. 2000 x (140 – 40) = (20 x 1000) x (Tout-30) this will give a cooling water outlet Temp = 40 Deg.C which is NOT possible since the APPROACH TEMP. for Heat Transfer IS ZERO and your Heating surface required will be infinity. ..."

...or a counter flow arrangement could be used.

Using countercurrent heat exchanger, condensate 140 to 40°, cooling water 30 to 40°. I make the LMDT 39.1°.

You said Over-all Heat transfer for this system will be 75 K.Cal/Hr/Deg. Is that per m²? If so it seems very low. A quick look suggests something more like 1000 K.Cal/Hr/Deg/m².

But anyway the OP appears to have lost interest.

Thank you so much. I'll get back to you with the pressure value of the condensate.

This matter is best dealt with by local design practices, and not via the forum.