Hairpin heat exchanger design

  • Thread starter Seta
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Summary:: I need to know the preheat temperature of the water entering the gas boiler (which i think is the outlet temperature of the cold water Tc,o leaving the heat exchanger)

Hi, i'm doing a design project of a hairpin heat exchanger. The problem is:
In a public shower, the city water is heated in a gas boiler from 15 °C to 43 °C at an average flow rate of 2 kg/s. In an attempt to save energy consumption, it is proposed to collect the lukewarm water drained at the temperature of 38 ° C and pass it through a heat exchanger to preheat the incoming cold water (which is at 15 °C). If the dimensions of the space in which the heat exchanger will be installed are 1m x 3m x 1m, design a hairpin-type heat exchanger which must properly perform this task.
From here, i have to do a schematic diagram of the water circuit with different devices (position of the HX, pumps, valves, boiler,...), and do the different calculations.

My problem is, i couldn't calculate the preheat temperature of the water leaving the HX.

I've done the schematic diagram, is this the right way to do it, and should i add or remove something from it ?


schematic diagram of the water circuit.png
 

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  • #2
BvU
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Hello @Seta , :welcome: !

Did you read the guidelines ? We are not allowed to help if you don't post your own attempt at solution.

Also, it would help if you provide some context and information to help us assist you at the proper level: what do you know about HX design , do you have access to the VDI Heat Atlas, ever seen Q = UA*LMTD etc. And who made the choice of a hairpin-type instead of a plate HX -- teacher ?
 
  • #3
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Sorry, it's my first time using this forum.
What i want to do is to find the number of hairpins needed with : Q = UA_hp *LMTD*N_hp where A_hp is the area of the inner tube of one hairpin HX, and N_hp is the number of hairpins.

The problem here is that to find Q i have to know either the outlet temperature of the lukewarmed water Th,o or the preheated water Tc,o with : Q = m_c * Cp_c * ( Tc,o - Tc,i ) = m_h * Cp_h * ( Th,i - Th,o ).
Same thing goes for LMTD.

Then I'll calculate U by calculating the sum of the resistances, and to do so I'll have to find the convection coefficients inside the tube and in the annulus by calculating the reynolds number, nusselt number ....

The main problem here is the lack of information about the temperatures, that's what kept me stuck.

And yes, the teacher decided to use a hairpin type HX.

Thank you.
 
  • #4
BvU
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Good info!

The usual path in designing such a contraption is that you make a starting guess of U and if necessary refine in subsequent iterations. From Coulson and Richardson Vol 6: water/water shell and tube: 800-1500 W/(m2 ##\cdot## K). Or here 150-1200. So perhaps try 1000. (low flow water at low P allows thin pipes).

The main problem here is the lack of information about the temperatures, that's what kept me stuck.
You have approximately equal flowrates, so with an infinitely large exchanger you can preheat to the 38 ##^\circ##C (if you believe that number) and cool down to 15. The limit is size and in 3 m3 you can stuff a huge area of pipe. The tradeoff is investment versus efficiency: adding area doesn't give a good return on investment after a while.

Do the calculations with an equation solver (see engineeringtoolbox), or better: iterate in excel for good understanding or use the ##\varepsilon-##NTU method.

Check how sensitive (or rather insensitive) heat recovery percentage is under changes in UA once you are in the range 80-90%.

Allow some fouling and there you are !

Nice exercise; let us know how it goes :smile:
 
  • #5
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From what i understand, i have to take Th,o = 15 °C and Tc,o = 38 °C , but if i do so, which value should i take for LMTD since ( Th,i - Tc,o ) = ( Th,o - Tc,i ) = 0 ?
On the other hand, when i asked my teacher about the preaheat temperature Tc,o , he said that it has to be calculated.
 
  • #6
BvU
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From what i understand, i have to take Th,o = 15 °C and Tc,o = 38 °C
That's with an infinitely big HX. No wonder LMTD breaks down !
So take e.g. 90% recovery: LMTD still breaks down but we know that in the limit ##\Delta T_A \rightarrow \Delta T_B## the LMTD goes like ##LMTD \rightarrow \Delta T_A## (don't we ? :smile: )

he said that it has to be calculated.
Sure. It follows from Q = UA LMTD
 
  • #7
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Alright, thank you so much !
 
  • #8
BvU
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Sheer curiosity: how is it going ? I scribbled a few calculations and can imagine this is a meaningful and challenging exercise !
Also raises a few questions: the schematic suggests a single-shower application, but the 3 m3 seems a bit hefty for that ?

With 10 l/min and 23 ##^\circ##C ##\Delta##T, a maximum of 16 kW can be recovered, but then there is no ##\Delta##T left over (I think the professinals call that the 'approach').
If we aim for 80% that leaves ##\Delta##T = 4.6 ##^\circ##C meaning we need a UA of 3.5 kW/##^\circ##C

It can be done. Compare notes ?
 
  • #9
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Yes that's true, this isn't for one shower only, i should've added more.

Honestly, i am a little bit lost, i started calculations with 80% recovery :

q = 0.8 * 2 * 4.1786 * (43 - 15) = 187.2 kW // I took Cp_c at T = (43+15)/2 = 29 °C

Now i'll start to work with NTU-eps method:

q_max = m_c * Cp_c * (Th,i - Tc,i) = 192.234 kW // i took Cp_c at T = (38+15)/2 = 26.5 °C

==> eps = q / q_max

==> NTU = eps / (1-eps) // I took the heat capacity ratio equals 1 because the flowrates and the specific heat capacities are aproximately the same.

==> Uo * A_hp * N_hp = Cmin * NTU = m_c * Cp_c * NTU

And now, by first guessing Uo i can find the proper geometry, or i choose the dimensions first, then i'll calculate Uo.

I don't know if this the right way to do it, but i strongly doubt the results.
 
  • #10
BvU
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I don't see the 43##^\circ## anywhere near the heat exchanger ?!?!?

Other than that:

I suppose you want to run 12 showers simultaneously ?

i strongly doubt the results
Why ? Not sure I follow this NTU method -- maybe more ime tomoow.

I took the heat capacity ratio equals 1 because the flowrates and the specific heat capacities are aproximately the same.
I agree.

A_hp ? N_hp ?
 
  • #11
BvU
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Had to rehash my NTU stuff (almost never use it):
q = 0.8 * 2 * 4.1786 * (43 - 15) = 187.2 kW // I took Cp_c at T = (43+15)/2 = 29 °C ##\qquad## (1)

Now i'll start to work with NTU-eps method:

q_max = m_c * Cp_c * (Th,i - Tc,i) = 192.234 kW // i took Cp_c at T = (38+15)/2 = 26.5 °C ##\qquad## (2)

==> eps = q / q_max##\qquad## ##\qquad## ##\qquad## (3)

==> NTU = eps / (1-eps) // I took the heat capacity ratio equals 1 because the flowrates and the specific heat capacities are aproximately the same. ##\qquad## (4)

==> Uo * A_hp * N_hp = Cmin * NTU = m_c * Cp_c * NTU ##\qquad## (5)
ad 1:
The HX sees one input at 38, the other at 15 C. The 43 C is elsewhere, so it can't appear here​
I suppose the 0.8 is ##\varepsilon##​
I suppose the 2 is mass flow, 2 m3/s = 120 L/min, so 12 showers​
ad 2: I agree (but 3 digits is sufficient accuracy here ...)

ad 3:
No, ##q =\varepsilon\,q_{\,\text{max}}\ ##, so some 154 kW !

ad 4: Fully agree

ad 5:
I guess A_hp is a hairpin area and N_hp is their number :rolleyes:


And now, by first guessing Uo i can find the proper geometry, or i choose the dimensions first, then i'll calculate Uo.

I don't know if this the right way to do it, but i strongly doubt the results.

You have an initial guess for U. That gives you an area.
Pick a pipe diameter ##\Rightarrow## number of pipes. A check on the 3 m3.
Flow speed in the pipes, pressure drop, pipe wall thickness etc.

Then you might look at U again

good luck!

##\ ##
 

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