Question

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses alternating layers of the storage material and the flow passage. Each layer of the storage material is an aluminum slab of width W = 0.05 m, which is at an initial temperature of 25°C. Consider conditions for which the storage unit is charged by passing a hot gas through the passages, with the gas temperature and the convection coefficient assumed to have constant values of To = 600°C and h = 100 W/m2K throughout the channel. Hot gas -AI T(x,o) - Th 25°C T-600°C h-100W/m2K L=0.025m ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible radiation exchange with surroundings. PROPERTIES: Table 4-1, Aluminum, pure (T - 600K = 327°c): k = 231 W/m K, C = 1033 Jkg K. p=2702 kg/m a) Can you use the lumped capacitance method? If so, show the reason. b) Derive the relation of temperature vs time by applying energy balance on the system c) Derive the relation to calculate the amount of energy transferred for a certain time t. d) How long will it take to achieve 75% of the maximum possible energy storage? e) What is the temperature of the aluminum at this time?

Answer 1

SOLUTION Step 2 AI 70 - 2 Trio =T; = 25C Too = 600°c h=100 L=1025m please consider given. 988umptions Part A characteristic length B; HL Қ 100 * 0.025 931 = 0.0108 Bi 51 Hence the lumped capacitance methods we can use, pvc 0;[ı- explained D] Q - 4 Est Dest max cove) o; sest - exp[-+/) DEst, mex 0.75 (75%)

Ith Slc h nove 2702 * 0.025 * 1033 100 697.79 Ithn 698 5 from equation ® -- exp (t/698 ) = -0.75 - exp (+/698)=-17 0.75 exP ( 22 ) - 25 It 968S CD tep 2 :- temperature of the auminum at this time calculated from equation CYP (4/2) Ti - Tao T- Too T- Too + (Te-Too) exp(-4/2n)

600 +65–600) exp (–968/698) 11 600 - 575* 0.28 98 600-143.63 456.36 456°C Hence Selecting properties awminum is reasonable for system, at Gook this