steam generator project

[salvatore87]

Hello everyone are struggling with the sizing of a steam generator, for a concentration plant with 10mw peak parabolic mirrors. the thermovector fluid is diathermic oil, while the input temperature in steam turbine is 400°c. based on this data I determined the temperature of the oil input and output from the generator, respectively of 415c and 330c, and the water input of 53 c. I calculated the power to trade between the two fluids
25,88mw. at this point the only missing data to calculate the necessary surface is the overall coefficient of heat exchange. to do this I separated the transformations in the steam generator in three stages:
1) Liquid heating up to the p=30 bar(inlet pressure in turbine)
2)evaporization t=235c
3)superheater
for each of these phases I calculated the global heat exchange coefficient, hypothesized the only presence of the convection.
the determined values were as follows: 53.9w/m^2 k ;58.6 w/m^2 58.9w/m^2 I would like to know if these values are plausible thanks!

 

[paulpaul]

Hi.
I would have expected a slightly lower k on the heater, since one of the two fluids is in steam phase, and one higher on the evaporator, since one of the two fluids is undergoing a phase transformation. How did you calculate k?

 

[salvatore87]

ks were calculated through the theory of natural convection, neglecting the effects of conduction through the walls of the tube. In practice I set a speed for each fluid, and from that knowing the flow I calculated the number of reynolds that in all cases was turbulent. after I got the number of nusselt through an empirical formula for nu=f (re,pr) turbulent flows and finally k=nu* thermal conductivity/dh (hydraulic diameter,derive from the passage area).
n.b the k relative to the evaporator I calculated it, considering only the convection for the outer side (oil side) because inside there is a phase passage and therefore hi (convection coefficient) is very large. This implies that k=1/(1/he+1/hi) is about equal to he.

dataort oil=115 kg/s, water flow=8.6 kg/s, v_olio=1m/s, v_water=0.5 m/s

 

[salvatore87]

However diathermic oil at those temperatures (415-330) is in steam phase,
That's why k is low!

 

[paulpaul]

Bye!
responding to you last night I didn't look well at the oil drills, since typically when using as a thermovector fluid it tends not to evaporate, since its main advantage over water is its highest boiling point even in unpressed circuits: "typically" you probably have a different application.
to understand better (I'm also curious :smile I would like to know how this steam generator is made: If I didn't realize it was a "water pipe" configuration with forced crossing, with diathermic oil outside tubes with low speed? but does the oil have a phase transformation in the generator or enters and comes out as steam?

 

[salvatore87]

the generator in question has differences regarding the one with burner of a classic steam system.in this situation, as is specified at the beginning, we are in the presence of a plant with parabolic mirrors concentration. therefore there are no combustion gases, but diathermic oil that gives the heat to the water. the oil in question has a boiling point of 250c therefore entering to 330 degrees is already to the state of steam. Things change when molten salts are used instead of oil, which can reach higher temperatures (550 c) and thus allow to achieve greater yield.

 

[salvatore87]

in this pdf you will find the scheme of a generator used for this application.

 

[paulpaul]

Thank you! I'll be gone until tomorrow then I look at us!

 

[paulpaul]

Come back. . .

summing up to see if I understand (this information I need to recalculate k):

- the oil always remains in steam phase
- the three exchangers (economizer, evaporator and overheater) are all tube beam with water side pipes
- the flow in the exchangers is always crossed
- How is the oil vapour circulated?
- the speeds you wrote about what base you assigned? Are they the same for all heat exchangers?
- Is the evaporator in natural circulation with vertical tubes?

I tried to track the exchange diagrams on the three exchangers: do you confirm them?

- economizer:
= 6338 kw
water: tin = 53 °c tout = 235 °c
oil: tin = 351 °C - tout = 330 °C

- evaporator:
q = 15446 kw
water: tin = tout = 235 °C
oil: tin = 402 °c - tout = 351 °c

- overheating
= 3646 kw
water: dye = 235 °C - tout = 400 °C
oil: tin = 415 °C - tout = 402 °C

- should not replace oil with one having a higher boiling point (so that there are) and have one of the two always liquid fluids? I do not know the system upstream however and any needs;

- the coefficient on the outer side tubes calculated it following an empirical correlation nu = f(re, pr) for pipe counters?

We update!

 

[salvatore87]

perfect on everything.
ps.
-the oil steam almost certainly circulates thanks to a pump inserted in the solar system.

-The speeds I assigned with the aim of reducing the load losses, but I have not consulted any manual on this. One thing is certain though, if I increase the speed I will certainly have an increase in convection (flow even more turbulent = higher thermal exchange) but I will have to provide a greater prevalence to the fluid. The speeds are the same for the three exchangers.

I can't tell you much about the evaporator, the only pattern I can rely on is what I sent you. In that case the evaporator is horizontal and natural circulation I believe.

- The only compatible oil I found is that! (I send you the technical sheet) even if the maximum recommended is 400c. I looked for other oils with higher working temperatures but the only fluids I found are molten salts (in the specification of the project is said to use oil).

-the formula is nu=0.023*(re)^0.8*(pr)^0.4 used when the flow is turbulent (in all cases examined).

Thank you so much for your news! !

 

[paulpaul]

Hello! Unfortunately in these days I don't have much time but I bring you here under some of my considerations that come to mind immediately:

1. if the oil always remains in steam phase, you must cmq predict a compressor to circulate it, and then bring it to a certain pressure; if it is liquid, a pump is required. I don't think natural circulation is feasible within pipes connecting the solar field with the gv: you have to have enough pressure to win load losses. You would also improve the heat exchange.

2. reading in your d/s, "basta" elevate oil pressure to 11,32 bar and get a saturation temperature of 405 °c. It's a little low for the steam cycle you've designed, but maybe looking on the net you can find fluids that at pressures around the 5-10 bars are still liquids at 415 °C or lower the steam temperature at the entrance of the turbine.

3. if you hold it liquid, you have a simpler machine (pompa) that "expresses" less energy than a compressor (I would say that a "vite" in this case could be suitable) to equal flow rates and prevails, you would improve convective exchange coefficients and simplify the exchangers (minor volume).

said this, we remain in the field of oil in steam phase: if the boiler is at natural circulation, they would change the proportions of the thermal powers between evaporator and economizer.

there are external flow reports on pipes (cf. kays & london, incropera.. .oppure searches on google "flow across tube banks" or even correlation of zhukauskas) that allow you to treat the external convection on pipe counters and find he in this configuration.
for the internal flow is well a simple relationship type dittus boelter. determines the number of tubes considering having an average speed in them of 2-3 m/s if you have liquid water or I would say max 20 m/s if in steam phase. then you have to play on the length to have the heat exchange considered.
in both cases (internal and external influence) I would say that you are in forced convection and forced.

Let me know something, I can only connect occasionally!

 

[paulpaul]

better a point: if the boiler is at natural circulation, the movement of the biphase water/vapor mixture inside the evaporator is due to the natural convection (density differences) and therefore it is difficult to impose a speed. It's a two-phasic flow that's hard to shape. I would limit myself to calculate the k (driven by he and possibly conduction of pipe wall) and therefore the surface of exchange of the evaporator considering the delta t medium log.
I think the size of the evaporator is too specialized (at least for me!) :smile:
for the heater instead in all cases I would consider forced convection, monophase flow, as mentioned above.

 

[salvatore87]

then using your indications I came to the following results:
-economizer: he=196,3(w/m^2/k) hi=17038
-evaporator: he=277,02(consider only the external side)
-heater: he=289,7 hi=8,070(this last value seems a little low!! I used a pr=1,969)

n.b I fixed an external diameter of the pipes of 51mm and thickness of 7mm (for all 3 components), and a oil velocity of 1m/s, as regards the hydraulic diameter on the oil side I hypothesized a step of 76mm between the pipe files, and I used formula 4(pass area)/perimeter

 

[salvatore87]

rectifying!! I have done the accounts and get a hi coefficient of the superheater equal to 1002.5!!! I send you the file with the thermodynamic properties that I have considered, in the case of the heater.

 

[paulpaul]

Hi.
the tubes seem great and above all too thick.. .you should not need these thicknesses (you only have 30 bars, and worse running).
I tried to make some accounts on the economizer and hypothesize a tube beam I get a very low number of reynolds (unless calculation errors) in full laminar flow (to avoid). Perhaps it is better to pass to other geometry: your hi = 17038 w/m2°c how you calculated it? How many pipes? What speed?

 

[salvatore87]

Maybe I figured out where I was wrong in the hi calculation. the hydraulic diameter must be calculated compared to the total passage area,(tube area*n.tubi) right?

 

[paulpaul]

Yes, total area = ntubi*atubo. corrections should however be applied to the dtubo.

In practice you have to go for attempts. a low speed reduces your load losses but at the same time makes your heat exchange worse. We must seek, as always, a compromise between exchange, losses, encumbrance and economics.

a tube beam normally consists of some dozen tubes which in this case would be extremely small, with laminar flow inside. probably a tube solution in the tube could be preferable, at least for the economizer. I did not do more thorough calculations. You could try to see with about twenty tubes what the super-exchange is coming to you, that is the beam length.

 

[salvatore87]

then with de=51mm, 50 tubes and flow =8,598 kg/s speed 4m/s comes out a length of 25m of the beam, maybe it is too!

 

[paulpaul]

for the first calculation I had wrong an account: I had inserted the viscosity of the water wrong with three orders of magnitude! :smile:

I've done things again, assuming I have 7 tubes 32x3. mi is hi = 17500 w/m2°c, and a load loss of 0.017 bar/m.

I attach xls where if you want you can do some tests with different geometries: I extracted it from another xls sheet I already had, I hope I didn't make mistakes!

As for he, the correlation of zhukaskas is valid as perhaps I had already told you by flow perfectly crossed on the tube beam, without "collateral" flows: I think that for what you have to determine, it can be okay. you can find in the literature of correlations that take into account the real geometry of the exchanger: For example, on the manual of mechanical engineer hoepli the topic is well treated, I had implemented the matlab calculation scheme long ago and I used it to design three water/gas tube bundles with excellent results regarding the prediction of output temperatures.

 

[salvatore87]

excuse but with your data, would exit a beam length of 383 m only for the economizer!! is feasible as what?