Phil et al. --
I have tried to find a good answer for you (since I would like to know for myself) -- but I never arrived at a definitive answer. I was looking for a paper that stated "Lake E may be estimated as calculated PET", perhaps with a coefficient, but I could not find such paper.
The closest I came was in a statement in Ward and Trimble (2004), p. 92: "As discussed, lake evaporation is a good surrogate for PET, and the closest surrogate for a lake, and thus for PET, is the buried pan." That is, pan measurements converted to lake evaporation (with coefficients differing from 0.6-0.8 across the USA) are better than lake-E and PET calculations. They don't directly say that "PET estimates may be used to estimate lake E", but I guess we may infer that this may be the case (perhaps friends at Ohio State Univ. could walk down the hall and ask Andy Ward if he'd agree or not...).
Ward, A. D., and S. W. Trimble. Environmental Hydrology. CRC, 2004.
A very good review of methods is given by McMahon et al. (2013):
McMahon, T. A., M. C. Peel, L. Lowe, R. Srikanthan, and T. R. McVicar. “Estimating Actual, Potential, Reference Crop and Pan Evaporation Using Standard Meteorological Data: A Pragmatic Synthesis.” Hydrology and Earth System Sciences 17, no. 4 (2013): 1331–63.
They cite work by Morton:
Morton, Fred I. “Operational Estimates of Lake Evaporation.” Journal of Hydrology 66, no. 1 (1983): 77–100.
where differences between potential evaporation over land and water bodies is discussed, but I couldn't find a definitive statement regarding the use of PET to estimate lake E. Their definition of PET allowed for greater humidity and lower temperatures over water bodies, which may not be appropriate for the way SWAT deals with things. It could be construed to infer that open water E was less than PET, if PET were calculated based on higher T and much lower humidity values over land than water -- but SWAT makes no such differentiation, and land-based T and humidity values are applied across the subbasin without regard to the surface wetness of HRUs. (Actual ET, not PET, takes into account soil moisture, and I suppose crop coefficients, too.)
In any case I found nothing in the literature (yet) that stated that lake E = 0.6*PET. I'm not sure why that's the default in SWAT. I've generally had a positive water balance in my models and I've changed the evaporation coefficients from 0.6 to 1 to maximize my evaporation from reservoirs, ponds, and wetlands. My first presumption was that the 0.6 coefficient looked suspiciously like a pan coefficient, and that if so, it wasn't appropriate for converting PET to lake E.
All for now,
-- Jim