You may be right. Back of the envelope kinds of computation though, 6 miles of 12" diameter bore hole is 99,525 sq ft of surface area for thermal transfer. If y

You may be right. Back of the envelope kinds of computation though, 6 miles of 12" diameter bore hole is 99,525 sq ft of surface area for thermal transfer. If you think of the lower half of that as the heat injection surface (so 3 miles -> 6 miles and then back up to 3 miles on the way out) call it 100k sqft of rock surface. Now some people have spent a lot of time studying rock (https://pubs.usgs.gov/of/1988/0441/report.pdf https://pubs.usgs.gov/of/1988/0441/report.pdf) I'm not one of them (more of a casual geologist because I like hiking in the mountains) but if I'm reading the heat transfer equations from pg 91 of that report it still seems like you'd get a decent amount of heat transfer (order of 10's of MJ) into the water.