Question

A thermal storage is made with a rock (granite) bed of 70 ft³ which is heated to 720 R using solar energy. A heat engine receives a Q_H from the bed and rejects heat to the ambient at 520 R. The rock bed therefore cools down and as it reaches 520 R the process stops. Find the energy the rock bed can give out. What is the heat engine efficiency at the beginning of the process and what is it at the end of the process? 

Answer 1

Assume the whole setup is reversible and that the heat engine operates in a Carnot cycle. The total change in the energy of the rock bed is 

 u₂ − u₁ = q = C ∆T = 0.21 (720 - 520) = 42 Btu/lbm

m = ρV = 172 × 70 = 12040 lbm; Q = mq = 505 680 Btu

To get the efficiency assume a Carnot cycle device

η = 1 - T_o / T_H = 1 - 520/720 = 0.28 at the beginning of process  

η = 1 - To / TH = 1 - 520/520 = 0 at the end of process