Ahhhh, now I can size that pump!
So, if you ONLY need 50 kW of heat duty to warm your extract to the boiling point of ethanol you can use the following equation for the water used to heat:
Q = mc(Th-Tc)
We know Q = 50kW = 50 kJ/s = 50000 J/s
m = mass flow rate of water (g/s)
C = Heat capacity of water = 4.184 J/g*K
Th = 60C = 333 K
Tc = 35C = 308 K
m = 478 g/s. Let’s assume 500 g/s
Water’s density = 1 g/mL
478 mL/s = 1,720 L / hr of water flowing through your boiler.
You were basically right on the money with 1.7 m3/hr pump capacity but I would size on the higher end.
Determining your exchange area will be a much more difficult task especially considering the loss of ethanol you will experience