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.