heat exchanger verification

[Peppe.Giuli]

Good evening community,

I am struggling with the choice of a heat exchanger (balanced plates or inspectable plates) for a cooler, after performing all the calculations of the case I have arisen a doubt about a temperature that at the moment I hypothesized (and I hope to calculate the correct value! )

I attached an image to be more understandable, inside the cylinder in aisi 304 of diameter 1400 mm I find of the material at the input temperature of 110°c and I would like it to cool up to 50 °c, through passage of water between the diameters 1420 and 1444 mm (t= 20°c and flow of 35 mc/h).
on the warm side, I hypothesized a water input temperature of 80 °C and exit at 50 °C but this follows that I have a high deltat and I have difficulty with the balanced models.
so I ask, how can I verify the temperature of the water in contact with the wall (is it right to consider it equal to 80°c)? hypothesize an average value is correct? how to calculate the water temperature in contact with the wall for heat transmission?

thanks for the answers

Giuseppe

 

[paulpaul]

I don't understand. First you talk about a plate heat exchanger and then you show the section of a tube heat exchanger. Can I miss something?

 

[Peppe.Giuli]

I don't understand. First you talk about a plate heat exchanger and then you show the section of a tube heat exchanger. Can I miss something?

the photo illustrates the question that I did, the calculation of the water temperature "heated" by transmission of heat between the material at 100 °C and water at 15 °C. maybe I have stretched too much in the case and title of the incorrect discussion. Excuse me.

 

[paulpaul]

Forgive me, but there's still something that doesn't come back, at least to me:

- I am not yet clear what a plate exchanger has to do with the image (refer to the image and forget the plate exchanger)
- it is not clear whether the "material" to be carried from 110 °C (in the second message to 100 °C) to 50 °C inside is still or has a certain speed (hypotice is firm)
- it is not clear the water input temperature in the enclosure (in the first message say 20 °c, in the second 15 °c) - little changes in the concept, but just for accuracy;
- then speak of "hot side" and hypothesizes a water input/output temperature of 80 and 50 °c: but the hot side was not the one with "material" inside and the cold side that with water at 15 (or 20 °c) incoming?
- How long is this pipe in the pipe?

for now are these minimum info to try to make some reasoning

 

[Peppe.Giuli]

- the image is linked to the exchanger since the entry on side 1 of the exchanger is linked to the amount of water passing between the cylinders.
- the material at 110 °c is firm
- the entrance temperature to the enclosure is at 15°c
- the tube is 1400 mm long
- the hot side is that of the material that incoming has a temperature of 100 °c and exit I would like it to arrive at 50 °c but the "problem" is that it is not a liquid. so I hypothesized that at 15°C full water intercapedine and the material at 110 °C it could be assumed a temperature of 80 °C to be inserted inside the calculations for my heat exchanger.
the question is precisely this: what is the temperature of water exchange (15°c) after being in contact with the material at 110°c?

Thank you.

 

[paulpaul]

I would say that first, if you haven't already done it, you need to understand the heat you have to subtract to the material: q = mcdt, where is the mass of the material, and its specific heat (which I think you know) and dt the difference between the initial and final temperature of the material (110 - 50 = 60 °c).
then you need to know how long (t) you need to cool it, and get the thermal power p = q/t.
from here, calculate the required flow of water with an energy balance: m = p/cdt, where this time is the specific heat of water and dt the temperature difference between output and water input. the input one you know (15 °c), the output one depends on how then you cool the water to have it again at 15 °c (I assume you have a closed circuit with a exchanger): usually on a exchanger you hypothesize a dt of 10-15 °c.
have you followed this reasoning to determine the flow of water? or do you have different constraints? Is the water outlet temperature free?

after these energy considerations that apply regardless of the method of exchange but which serve to set the reasoning, you must go to see if this exchanger will be efficient: At first sight I think there is a lot of material that will exchange very little heat, being far from the walls of exchange, exchange that will depend however on the coefficient of conducting the material (if solid) and also on that of convection (if fluid). What kind of material is it? Do you know these coefficients?

we then see if it is possible to answer your question, which seems to me to understand if it is lawful to consider the wall at a "media" temperature.

 

[Peppe.Giuli]

I started with these considerations you described, then I also got design lanes and going ahead with calculations I had the following results:

the constraints I have are:
- the cooling water flow of 35 m3/h;
- cylinder mass of 4000 kg
- the outgoing water temperature of the exchanger is 50 °C;
- the specific heat of the material is 1400 j/kg k

My results are:
- temperature of water heat exchange from 15 arrives at18°c
- power of 120 kw
- conduction coeff of 4600 w/m2 k
- convection coeff 1110 w/m2 k

thank you very much for the availability

 

[paulpaul]

I started with these considerations you described, then I also got design lanes and going ahead with calculations I had the following results:

the constraints I have are:
- the cooling water flow of 35 m3/h;
- cylinder mass of 4000 kg
- the outgoing water temperature of the exchanger is 50 °C;
- the specific heat of the material is 1400 j/kg k

would be a heat to exchange of q = 4000 x 1400 x 60 = 336.000 kj, correct? I do not understand the 50 °c of water outlet temperature if more under affirmations that are 18 °c...

My results are:
- temperature of water heat exchange from 15 arrives at18°c
- power of 120 kw

these data instead come back, considering the water flow of 35 m3/h and the thermal jump of 3 °c.
we say that admitting not to disperse heat, you would exchange the above heat in 2800 s, that is just less than an hour.

- conduction coeff of 4600 w/m2 k
- convection coeff 1110 w/m2 k

the first is the conducting coefficient of solid material to cool, correct? eye that there is an error in the units of measurement, since the conduction coefficient always has w/mk units.
the second presumption is the coefficient of convective exchange water side.

Let me know if I interpreted correctly, that then maybe we proceed with the other arguments.

 

[Peppe.Giuli]

would be a heat to exchange of q = 4000 x 1400 x 60 = 336.000 kj, correct? I do not understand the 50 °c of water outlet temperature if more under affirmations that are 18 °c...


these data instead come back, considering the water flow of 35 m3/h and the thermal jump of 3 °c.
we say that admitting not to disperse heat, you would exchange the above heat in 2800 s, that is just less than an hour.


the first is the conducting coefficient of solid material to cool, correct? eye that there is an error in the units of measurement, since the conduction coefficient always has w/mk units.
the second presumption is the coefficient of convective exchange water side.

Let me know if I interpreted correctly, then maybe we proceed with other reasonings

consider 50 °c, the output temperature of the material; i.e. it enters inside at 110°c and cools up to 50°c by thermal exchange with water at 15 °c (and then rises up until 18°c when it enters contact with the wall ).

 

[Peppe.Giuli]

would be a heat to exchange of q = 4000 x 1400 x 60 = 336.000 kj, correct? I do not understand the 50 °c of water outlet temperature if more under affirmations that are 18 °c...


these data instead come back, considering the water flow of 35 m3/h and the thermal jump of 3 °c.
we say that admitting not to disperse heat, you would exchange the above heat in 2800 s, that is just less than an hour.


the first is the conducting coefficient of solid material to cool, correct? eye that there is an error in the units of measurement, since the conduction coefficient always has w/mk units.
the second presumption is the coefficient of convective exchange water side.

Let me know if I interpreted correctly, that then maybe we proceed with the other arguments.

the second is precisely that of water, the first value is the convective thermal exchange value of the material, in addition I took into consideration the coeff. of steel: 15 w/ m °c

 

[paulpaul]

ok but if the material to cool is solid (as I think I have understood) how does it have a convection coefficient, which by definition is defined only for fluids? is not a solid and has a natural convection? in natural convection it really seems very high as value.

 

[Peppe.Giuli]

you are right, it is a granulous type material with thermal conductivity equal to 0.4 w/m k.

 

[paulpaul]

It is clearly a non-stationary conduction problem, where your material will take time to get to the desired temperature. You can not even use the simplified method of concentrated thermal capacity, given the low conductivity of the material.
exist in literature of analytical solutions for the case of cylinder (what is this), but with a much greater l/d ratio. otherwise you must approach the problem with the method of finite differences, by hand or by making a simulation with a dedicated software.
However, the average wall temperature can be assumed equal to the average temperature between water and material temperature (the latter varies over time however), and independent from the cylinder height (the water temperature varies very little and the material seems to me homogeneously distributed).
I made a quick calculation of the convective coefficient water side, and I get a little lower than your (about 900 w/m2 k).

 

[Peppe.Giuli]

Thanks for the advice! !