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steam generator project

  • Thread starter Thread starter salvatore87
  • Start date Start date
salvatore87 said:
excuse but with your data, would exit a beam length of 383 m only for the economizer!! is feasible as what?
Click to expand...
I threw in some numbers just to show you the hi calculation. the xls sheet must be completed with the calculation of he, of the conducting resistance and with the calculation of total k.
the calculation of the total surface I have never even set it :) so you have to change some parameters (e.g. number and diameter of the tubes, pipe pitch, etc.) looking for an excellent between load losses, heat exchange and dimensions. excel is convenient for that, since the project of a exchanger is very often iterative.

you can also have hundreds of tubes and diameters of the tube beam of the size order of the meters, just to have an idea.

The scheme I would follow is This is it.Classic.
 
perfect but in any case, the convection on the water side has very high values compared to the oil side, so its influence on the k, and reflex on the total surface is poor. Do you agree?
 
salvatore87 said:
perfect but in any case, the convection on the water side has very high values compared to the oil side, so its influence on the k, and reflex on the total surface is poor. Do you agree?
Click to expand...
agree
However I am used to always calculate everything, calculate a heat exchanger from some hundreds or thousands of kw without knowing what happens side tubes is not recommended (we also consider calculation of load losses).
build a xls sheet for he calculation starting from correlations for pipe counters...
 
ok already done... I had also sent it in a previous message :) now they perfect it taking into account also the load losses side tubes and coat side. ...
 
ok, perfect it also taking into account the influence of tube configuration on speed: if you have tubes online (I would limit myself for simplicity to this case, look qui two examples, one of tubes online the other of staggered tubes) you have the minimum area on which to calculate the external velocity given by

(d)

in which d is the diameter of the circle that circumscribes the beam of pipes, n the maximum number of tubes per row that the fluid meets transversally (the thread usually placed on the middle plane of the exchanger) and dd horizontal distance between the diaphragms (the sects that reverse the external flow).

the mass flow divided by am and the density of the fluid is the maximum speed you need to calculate king, which is called max king in correlations.
In this way you can calculate the speed to consider in the correlations starting from the flow, exploiting both the transverse step and the distance between the diaphragms (as well as the diameter of the beam).
 
excuse but the distance between the sects and beam length are not necessarily equal then?? I can have more sets distributed along the entire length of the right beam? and dd represents the distance between two successive sects?
 
exactly, dd is the distance between the setts, constant between one sett and the other and between the ends sets and the tube plates: this distance is less than the total length of the beam, with the objective of deviating the external flow and crossing as far as possible the two flows.
decreasing this distance favours thermal exchange (equal to all the rest) but load losses are increased.
 
I apologize if I ask you some questions but from the practical point of view I have never seen a exchanger. . . then between the sects and the cloak must exist a certain free area to allow the passage of the external flow?
 
after making changes to the program I came to a solution:
n.tubi=10
de = 40 mm
d mantle= 4m (circular circle)
length tube beam= 139m (I can only one pass? ? )
pipe load loss=0,0019 bar/m
load loss mantle side=3,1 bar(the formula is n*f*(density/2*v^2)*x)n=number of ranks
f=factor of friction obtained from moody
x=correct factor between 1 and 1.2
 

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Yes, sects or diaphragms serve to reverse the flow and therefore there is a part of the beam where the flow is not crossed. in the first scheme I had turned you can understand how they are arranged. there may also be other geometries but that is the most typical.
you can't ignore the geometry of the external flow if you have to calculate the external coefficient: I tried to give you only the minimum information to determine it with sufficient precision.
There are a thousand other constructive details of a heat exchanger on which you might dwell but 1) is not my first job so surely there are people more competent than me to do it and 2) I think you do not need to solve your problem!

So they gave you a problem that you never faced in class?
 
salvatore87 said:
after making changes to the program I came to a solution:
n.tubi=10
de = 40 mm
d mantle= 4m (circular circle)
length tube beam= 139m (I can only one pass? ? )
pipe load loss=0,0019 bar/m
load loss mantle side=3,1 bar(the formula is n*f*(density/2*v^2)*x)n=number of ranks
f=factor of friction obtained from moody
x=correct factor between 1 and 1.2
Click to expand...
There's something wrong... now I can't look at it. I hope I can do it tonight, but I don't guarantee you! the diameter of the bandage should be around 2 m the length is excessive (the tube rods are available asmrcially in lengths from 6 and 12 m. the tubes I think are too few, I would increase the number of tubes by reducing the diameter.
 
paulpaul said:
Yes, sects or diaphragms serve to reverse the flow and therefore there is a part of the beam where the flow is not crossed. in the first scheme I had turned you can understand how they are arranged. there may also be other geometries but that is the most typical.
you can't ignore the geometry of the external flow if you have to calculate the external coefficient: I tried to give you only the minimum information to determine it with sufficient precision.
There are a thousand other constructive details of a heat exchanger on which you might dwell but 1) is not my first job so surely there are people more competent than me to do it and 2) I think you do not need to solve your problem!

So they gave you a problem that you never faced in class?
Click to expand...
no the novelty was only in the fact that the generator must be integrated with a concentration plant. the track is the following and nothing special is required for optimization:

- sizing of a concentrated solar system with parabolic solar collectors for a nominal power of 10 mwp

data: turbine input temperature t0= 400°c, p0=30 bar, condenser with evaporative tower or air condenser, diathermic oil thermovector fluid, bari location.
calculate: surface collectors, total energy produced during the year, sizing steam generator and capacitor.
calculation of maximum thermal accumulation.
 
What do you think of this?
n.tubi=139
♪
d mantle=1.8m (circuit circle)
length beam tube = 12m
pipe load loss=5,19*10^-6 bar/m
load loss mantle side=6,11 bar(the formula is n*f*(density/2*v^2)*x)n=number of ranks
f=factor of friction obtained from moody
x=correct factor between 1 and 1.2
 
Forget about the latest version where there were mistakes, in summary this is what I got for the economizer.
-n.tubi=100
-Length beam 12m
-Heat diameter 2m
-de=40mm
the losses through the whole system (economizer+evaporator+heater) are:
0,0023 bar(in the cloak)
0,0033 bars(in tubes)

the total area of exchange is 662 m^2 which seems to me a good result, do you agree?
 

Attachments

I can't check all your excel's accounts, but the pdc side bands look a little high, limiting me to the economizer. I see that you used a rectangular tube configuration with transverse step = longitudinal step = 76 mm, correct? You also neglected the thickness of the pipes, right?

I noticed for the moment a mistake on the number of reynolds side bandme: must be

♪

vmax seems to me correctly calculated, but I saw that you used a hydraulic diameter corresponding to a section between 4 tubes: Instead, you have to use the simple outer diameter of the pipes, the number of reynolds in the external flows does not require the calculation of the hydraulic diameter. Remember that the number of reynolds calls a characteristic length, which only in the internal flow is the hydraulic diameter but in the external flows can be the outer diameter of a cylinder (if the flow is on a cylinder, as in this case) or the length of a plate, if the flow is parallel to a flat plate.

If the beam comes along, you can use two tube side steps, multiplying the average logarithmic deltat of the case in countercurrent for a coefficient found in literature (usually called "f") that takes into account the different geometry of the flows. and their temperature.
the friction coefficient on the external side is however not calculated using moody (which is valid for internal flows) but with other graphs in which you have max king in the ascisse and where you have to position yourself on the curve corresponding to the correct step.
the formula for calculating pdc seems to me correct: if you have used the configuration of tubes "quadrata" then x = 1 (but it can also be much higher than 1 and also less than 1 for relationships between the two steps different from 1!).
 
paulpaul said:
I really had bed bar here! ;)

in the xls file I didn't look if you used conversion factors in the pdc, I stopped at reynolds since it didn't come back.
Click to expand...
in fact I was wrong was not in bar but in pascal excuse me :) then excuse if to calculate re use de, then also in the calculation of he I have to use it (he=nu*conducibility/de) right?
 

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