A parabolic trough field runs at geometric concentration under a direct-beam…
PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · Earth, Climate & Energy Physics · The laws of thermodynamics
Energy from Physics, Ch The solar resource resolved: geometry, spectrum and the limits of concentration
Problem
A parabolic trough field runs at geometric concentration C=80 under a direct-beam irradiance on the aperture (mirror) plane of FDN=900Wm−2. Its performance test, referenced to the aperture area, is fitted by the collector efficiency equation with FR(τα)=0.70 and FRUL=0.65Wm−2K−1. The heat-transfer fluid enters at Ti=620K, and the ambient temperature, taken as the dead state for exergy, is T0=Ta=300K.
((a)) Use the collector efficiency equation to find the field's thermal efficiency and the useful heat delivered per unit aperture area.
((b)) Treating the delivered heat as if it were all available at the single temperature Ti, find its exergy per unit aperture area and express it as a fraction of FDN: under that approximation, this is the fraction of the incoming direct beam that a reversible downstream engine could turn into work.
((c)) A blackbody receiver at this same concentration, radiating only to space and driving a reversible engine against the same T0, has receiver-engine efficiency η(T)=(1−CFDNσT4)(1−TT0). By evaluating η(T) on a grid of trial temperatures, find the optimum receiver temperature Topt and the corresponding maximum efficiency, and compare it with the fraction found in part (b). Split the difference into three bookkeeping terms: the shortfall of the fitted intercept below 1, the excess of the fitted loss over the ideal receiver's emission, and the cost of running at 620K rather than Topt. Say why the first two are not by themselves a measured optical loss and a measured heat loss.