Question

2. A thermal energy storage unit is composed of rectangular channels and is well insulated from outer surface. It encloses alternating layers and flow passages (shown using the arrows in the image below). Each layer of the storage material is an aluminum slab of width (thickness) W=0.05 m, which is at initial temperature of 25 °C. The storage unit is charged by passing a hot gas through the passages, with the gas temperature and convection coefficient assumed to have constant values of To=600 °C and h= 100 W/m .K throughout the channel. Note: Aluminum properties: Aluminum @T-600 K, K-231 W/mK,c-1033 J/kg.K, density p = 2702 kg/m Storage Material (i) (ii) (iii) Under what conditions the lumped capacitance method can be used? Note: the characteristic length is the half thickness (3 marks) How long it will take to achieve 75% of the maximum possible energy storage? (15 marks) What is the value of the temperature of the aluminum slab at this time? (10 marks) Is there any advantage to using copper slabs (with density p=8900 kg/m and e = 400J/kg.K) of equivalent size instead of aluminum? (5 marks) (iv)

Answer 1

Here, in lumped heat analysis there is no advisable to use copper because of more time as compared to Aluminium.

4) As fe Biot Number Lumpeol teat System u seal /00 X0 05 Biot 2X23 Biot o.olo8 Thus Biot o. ,Sim ce Lume H Biotothan Heisler chout used.' sis usea. Ti 5t293= 298, T 600t2 = 873 K, 7- G00k T To tohr, e for slab. T- Too e 298-873 loo 2 G0083 2f02 X OROSX J033 e - (S75 O.0014 3309x - C213) e Ta king lnboth sides, An ( 0.00 14330 9AT T S19.80s At time T s9.so Tempenature T- As per (2) question, zq8-813 T-83 200 e 2to2x005xo33 19.80 T 873 Rath sidesWe get - S7S a.106 T-873 -273 T-873 T 600kfo Aluminium. P 0 = 400 J[17 then Loox 2 e Copper used 298-813 Goo 873 x400 - CS75 e - 273) tarng n both side s. We gat T-663 S./-* THea, iwe 1s moxe then tHat uminiun, so there 1s Calventage af used coppeL Slalo