Electrical efficiency ? ? R(load)/R(source ("TEG"))
? = R(load)/[R(load) + R(source)]
It's unbelieveable such a fundamental error could make it past the editors and be published in EETimes: A cursory glance would show that if the source resistance is less than or equal the load resistance, the efficiency ? would be =1.0
Editor, The Hearing Blog
For the record, when the source resistance is zero, the efficiency ? is 100%; while when in maximum power transfer conditions, the source resistance = the load resistance, the efficiency ? = 50% i.e. the voltage drop and hence the i˛R power dissipation is equal across the source and load.
Also this must use true waste heat as a source.
if not then increasing thermal conductivity will lower the mother ships efficiency. This would negate any power produced by the parasitc system running of the exaust heat.
This is great for legacy systems, but given the cost of energy, it is a poorly planned new design that could effectively utilize this technology.
Real energy savings can be had with heat pump water heaters (cool the house, create hot water simultaneously), or the University of Michigan using the Great Lakes for a heat sink and cutting their air conditioning bills by about $500K per month.
Solar electric energy is electric power generation from sunlight. It could be direct to PV - Photovoltaic, or it could be indirect. An instance of the indirect type is concentration of solar power, where the energy of the sun is concentrated to boil water used to generate power.
A Book For All Reasons Bernard Cole1 Comment Robert Oshana's recent book "Software Engineering for Embedded Systems (Newnes/Elsevier)," written and edited with Mark Kraeling, is a 'book for all reasons.' At almost 1,200 pages, it ...