Question

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas.

Answer 1

Answer:

as the exercise is incomplete, I add the information that is missing:

Consider a packed bed of 75-mm-diameter aluminum spheres ? = 2700 kg/m3, and c = 950 J/kg

Answer: The copper can store 34.1% more thermal energy at 272.42ºC

Explanation:

Given:

Diameter packed bed = 75 mm

Density = 2700 kg/m³

specific heat = 950 J/kg K

initial temperature sphere = 25ºC

thermal conductivity = 240 W/m K

temperature of gas = 300ºC

convection coefficient = 75 W/m²K

The characteristic length is

$L_{c} =\frac{r}{3} =\frac{0.0375}{3} =0.0125m$

The Biot number is equal to

$Bi=\frac{hL_{c} }{k} =\frac{75*0.0125}{240} =3.9x10^{-6}$

Bi < 1, we will use the lumped capacitance method

The energy transfer is equal to:

$\frac{Q}{Q_{max} } =(1-exp(\frac{t}{t_{t} } ))$ (eq. 1)

$t_{t} =\frac{pV}{hA} =\frac{p\frac{\pi D^{3} }{6} c}{h\pi D^{2} } =\frac{pDc}{6h} \\t_{t}=\frac{2700*0.075*950}{6*75} =427.5s$

Replacing in eq. 1

$0.9=1-exp(\frac{-t}{t_{t} } )\\0.1=exp(\frac{-t}{t_{t} } )\\\frac{-t}{t_{t} } =ln(0.1)\\t=2.3t_{t}=2.3*427.5=983.25s$

The temperature at the center of sphere is

$T=T\alpha +(T_{i} -T\alpha )exp(\frac{-6ht}{pDc} )=300+(25-300)exp(\frac{-6*75*983.25}{2700*0.075*950} )=272.42C$

We obtain the percentage increase

Cpcu = 385 J/kg K

(pCp)cu = 8933 * 385 = 3439205 J/m³K

from water:

(pCp)wa = 2700 * 950 = 2565000 J/m³K

the percentage increase is

%P = (3439205 - 2565000)/2565000 = 34.1%

The copper can store 34.1% more thermal energy.