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I'm a student of Integrated Water Management, and looking at a geothermal bore that is producing potable water for a town's water supply.

The bore water reaches the surface at $50$ $^oC$, and flows at 1.6 million liters per day.

It's suggested a vapour expander engine could convert this heat to useful work, but such an engine may not commercially exist. Nevertheless, I'd like to calculate the energy potential.

Can anyone help with the Excel formula I'd need to calculate this?

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    $\begingroup$ This SOUNDS like homework cast, more cleverly than usual, in the form of a real world problem. Is it? Homework, assignments and their ilk ARE acceptable but certain rules apply. Saying so is the 1st rule. $\endgroup$ Aug 1, 2015 at 9:35
  • $\begingroup$ It's my student research project (so homework I guess), but it is a real world case study. I really appreciate your reply below. $\endgroup$ Aug 1, 2015 at 10:27

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Any engine that works at $50$ $^oC$ on the hot side is going to have a low efficiency. This may not matter but needs to be noted.

The theoretical max possible $\eta$ (efficiency) is that defined by the Carnot cycle. Depending on the actual implementation the cycle used may be different, but this sets an upper possible theoretical limit.

Assume $T_{cold}$ = $25$ $^oC$. Colder may be possible but cold side thermal sinking can be challenging - as will be seen, an enormous amount of "waste" heat energy will be present. Not all of this will flow via the cold sink - but a significant amount will.

$$\eta_{carnot} = \frac{\triangle T}{T_{hot}}$$

Here $T_{hot}$ at $50$ $^oC$ = $273$ $+$ $50$ $=$ $323$ $K$.

$\triangle T$ $=$ $(50-25)$ $=$ $25$ $K$.

$\eta_{carnot}$ $=$ $25/323$ $=$ $7.7\%$.

At high $T_{hot}$/$T_{cold}$ ratios it is possible in practice to get > $50\%$ $\eta_{carnot}$ but as $\triangle T$ drops and $T_{hot}/T_{cold}$ is small, only a small fraction of the Carnot efficiency is possible. Here with $\eta_{carnot}$ $=$ $7.7\%$ I'd guess that much more than $2\%$ efficiency would be hard to achieve.

$1.6$ x $10^6$ $l$ x $sg$ $=$ $1$ x $4.2$ $kJ/kg$ x $\triangle T$ x $2\%$ =~~~ $1.3$ x $10^8$ $J$ $=$ $3,360$ $MJ/day$

$Power$ $=$ $E/day /24 hours /360$ $=$ $~$ $39$ $kW$
E&OE.

That's certainly useful - but it comes from a lot of water - - 18.5 litre/second!
It's possible that better alternatives may exist.

Home heating, agriculture, "fish raising" - eg prawns, mixed aquaponics and more.

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  • $\begingroup$ Am I right in calculating the energy available as: Q = 1,600,000 * 0.001163 (kWh/kg/°K) * (50-25) = 46,520.00 kWh/day $\endgroup$ Aug 1, 2015 at 10:31
  • $\begingroup$ Something is wrong for me... 1,600,000 l x sg=1 x 4.2 kJ/kg x 2% = 134,400 Joules (not 134,400,000) $\endgroup$ Aug 1, 2015 at 11:27
  • $\begingroup$ @user1745767 KILOJoule per kg. so: 1,600,000 l x 1 kg/l x 4200 J/kg x 2% = .... . $\endgroup$ Aug 1, 2015 at 13:19
  • $\begingroup$ @user1745767 I edited my answer to take account of my having omitted the 25C delta T in the energy content calculation. The 39 kW is "useful". At retail cost in NZ for electricity that's worth about $230. In some countries far less. For interest - what country is this in? $\endgroup$ Aug 1, 2015 at 13:38
  • $\begingroup$ Thanks Russell, I appreciate you pointing that out, I see my mistake with the kilojoules vs joules. This is in Australia. If it's 39kW of electricity, then your right it's not insignificant. 39kw x 24 hours = 936kWh x 365 days = 341,640 kWh (36 MWh).... at say AUD 0.18 = AUD $61,000 $\endgroup$ Aug 2, 2015 at 11:16

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