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I'm thinking about a system that would replace water in a swimming pool with clean water from an Artesian well.

The water in the well is about 1C and the swimming pool is around 20C. Reheating the whole pool is obviously very costly, so I'm looking to use a counter-current heat exchanger. What kind of efficiency can I expect from it?

This is just a concept idea for now, just looking at possible angles.

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    $\begingroup$ Depends on the materials and design... First define exactly what you want to achieve - flow rate, outlet temperature etc then you can find a suitable heat exchanger. $\endgroup$
    – Solar Mike
    May 25 '19 at 6:27
  • $\begingroup$ Have a look at engineering.stackexchange.com/a/13743/10902 $\endgroup$
    – Solar Mike
    May 25 '19 at 6:29
  • $\begingroup$ I'm looking at reasonable materials (copper, steel, brass, etc.) and a rate of about 1L/sec. The inlet temperature is 1C and the outlet temperature is 21C. $\endgroup$
    – Cyberax
    May 26 '19 at 7:04
  • $\begingroup$ So, flow rates are both the same or likely to be different? $\endgroup$
    – Solar Mike
    May 26 '19 at 8:46
  • $\begingroup$ Yes, since it's meant for water replacement the flow rates will be identical. $\endgroup$
    – Cyberax
    May 27 '19 at 18:53
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Unless you are going to pump water removed, true counter flow will be difficult due to low pressure drop to support flow velocity on discharge. Perhaps you could coil up thin gauge tubing in an overflow bucket or drum. Run the supply water through the tubing. For heat transfer you can get good heat transfer on tube side but on return side it would be low due to low velocity unless you are going to pump the return as well.

For overall heat transfer coefficient, it's the inverse of the sum of the inner and outer convective resistances conduction resistance and any fouling factor. For convective losses there are correlations called Nusselt number correlations. It is a non-trivial effort to figure out but can be done with some assumptions.

The big obstacle is return flow is not the same as supply flow due to evaporation losses so this may make heat recovery impractical.

Without having done the calculation of the overall heat transfer coefficient for this problem I wonder whether there is much savings to be had

As for efficiency, the commenter above is correct: it depends on surface area, and I think also on ambient heat losses to environment for hot/returned fluid.

For $1 \frac{\mathrm{l}}{\mathrm{s}}$ flow, l from the author comment above, neglecting any heat recovery the heat load would be $$ \begin{align} Q &= \dot{m} \, C_{\text{p}} \, \Delta T \\[10px] &= 1 \frac{\mathrm{l}}{\mathrm{s}} \times \left(20\sideset{^{\circ}}{}{\mathrm{C}}-1 \sideset{^{\circ}}{}{\mathrm{C}}\right) \times 1 \frac{\mathrm{kg}}{\mathrm{l}} \times 4.18 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \\[10px] &= 79 \, \mathrm{Watts} \end{align} \,.$$ This is the heat requirement before any heat recovery.

Assuming you are saving electric heat in the United States, this is worth about $$ 0.13 \frac{\mathrm{USD}}{\mathrm{kW} \cdot \mathrm{hr}} \times .079 \, \mathrm{kW} ~~=~~ .103 \frac{\mathrm{USD}}{\mathrm{hr}} ~~\approx~~ 900 \frac{\mathrm{USD}}{\mathrm{year}} \,,$$if you could recover 100 percent.

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  • $\begingroup$ I suggested an edit for some formatting. I didn't include this change in the suggestion, but I suspect that you meant $\approx 90 \frac{\mathrm{USD}}{\mathrm{year}}$ rather than $\approx 900 \frac{\mathrm{USD}}{\mathrm{year}} ;$ looks like an extra zero snuck in there. $\endgroup$
    – Nat
    Jun 2 '19 at 17:57
  • $\begingroup$ Looks like the heat capacity of water, $C_{\text{p}} ,$ had a unit error in it. I think you meant $4.18\frac{\mathrm{kJ}}{\mathrm{kg}\cdot\mathrm{K}}$ rather than $4.18\frac{\mathrm{J}}{\mathrm{kg}\cdot\mathrm{K}} ,$ which would've given you $\sim 79 \, \mathrm{kW}$ rather than $\sim 79 \, \mathrm{W} .$ Then, assuming we also correct the extra-zero, this should've led to an estimated electrical heating cost of $\sim 90,000 \frac{\mathrm{USD}}{\mathrm{year}} ,$ which is pretty close to what I got. $\endgroup$
    – Nat
    Jun 2 '19 at 19:42
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tl;dr- Sounds like you'll need to spend about $84,000 \frac{\mathrm{USD}}{\mathrm{year}}$ to heat the water for your pool. The proposed heat exchanger might realistically give you about $50 \%$ savings, or $42,000 \frac{\mathrm{USD}}{\mathrm{year}},$ if we neglect its own operational expenses, capital cost, depreciation, etc..


If you've raising $1 \frac{\mathrm{L}}{\mathrm{s}}$ of water from $1\sideset{^{\circ}}{}{\mathrm{C}}$ to $20\sideset{^{\circ}}{}{\mathrm{C}} ,$ then that requires a heat duty of about $$ \begin{align} Q &~~=~~ \frac{\mathrm{d}m}{\mathrm{d}t} ~~ C_{\text{p}} ~~ \Delta T \tag{1} \\[10px] &~~\approx~~ \left(1 \frac{\mathrm{L}}{\mathrm{s}} \times \frac{1000\,\mathrm{g}}{1 \, \mathrm{L}}\right) \left(4.2 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\right) \left(20\sideset{^{\circ}}{}{\mathrm{C}} - 1\sideset{^{\circ}}{}{\mathrm{C}}\right) \\[10px] &~~=~~ 79.8 \, \mathrm{kW} \,. \end{align} $$

Math check

If you want to check my math, then you can query WolframAlpha with the following:

(heat capacity of water) * ((20 - 1) Kelvin) * (molar density of water) * (1 liter) / (1 second)

, which, checking just now, yields a result of $78.84 \, \mathrm{kW} .$

Then, if we assume that your cost of electricity is $0.12 \frac{\mathrm{USD}}{\mathrm{W} \cdot \mathrm{hr}},$ that's about $9.58 \frac{\mathrm{USD}}{\mathrm{hr}}$ or $83,943 \frac{\mathrm{USD}}{\mathrm{year}}$ worth of electricity to heat the water.

Since you're looking at burning more than $80,000 \frac{\mathrm{USD}}{\mathrm{year}}$ on just heating a pool, it does seem reasonable to try to find a way to reduce that cost.

Next, as you've noted, we can't generally get ideal heat exchange. The closer we try to approach it, the more costly the heat exchanger, plus the greater other operational expenses become.

While it's really hard to give an exact figure, especially we don't have many details, the heuristic that folks in industry have previously recommended to me is that a decent heat exchanger can have a temperature approach of about $$ \left.\Delta T\right|_{\text{approach}} ~=~ T_\text{hot-in} - T_\text{cold-out} ~\approx~ 10\,\mathrm{K} \tag{2} \,,$$ while a more robust heat exchanger might try for a $\left.\Delta T\right|_{\text{approach}}$ of about $5 \, \mathrm{K} .$

Since the numbers above were all pretty linear (this is a nice, simple system!), we can basically say that you'll have a savings-factor of $$ \xi_{\text{savings}} ~\approx~ \frac{19\,\mathrm{K} - \left.\Delta T\right|_{\text{approach}}}{19\,\mathrm{K}} ~=~ 1 - \frac{\left.\Delta T\right|_{\text{approach}}}{19\,\mathrm{K}} \tag{3} \,.$$ So, $$ \xi_{\text{savings}} ~\approx~ \left\{ \begin{array}{cl} 47.4 \% & \text{if decent heat exchanger with}~\left.\Delta T\right|_{\text{approach}} \approx 10\,\mathrm{K} \\ 73.9 \% & \text{if good heat exchanger with}~\left.\Delta T\right|_{\text{approach}} \approx 5\,\mathrm{K} \end{array} \right. \tag{4} \,. $$ We're definitely roughly guesstimating here, but it looks like the heat exchanger might realistically save you $50 \%$ on heating costs – though, to be clear, it may offset some of those savings with additional maintenance costs, depreciation, etc..

So, in short, I'm guessing maybe $50 \%$ efficiency. Given the prior estimate of $83,943 \frac{\mathrm{USD}}{\mathrm{year}}$ for heating, that'd seem to be about $42,000 \frac{\mathrm{USD}}{\mathrm{year}}$ in savings on your electric bill, before factoring in stuff like potential increases in electrical costs to run pumps, maintenance fees, capital costs, etc..

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  • $\begingroup$ Thanks! This heuristic was what I've been looking for. The system should probably pay off even if the duty cycle is reduced from 100% (3.6 tons of water per hours is quite a bit, after all). $\endgroup$
    – Cyberax
    Jun 3 '19 at 1:26

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