Respuesta :
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
[tex]L_{c} =\frac{r}{3} =\frac{0.0375}{3} =0.0125m[/tex]
The Biot number is equal to
[tex]Bi=\frac{hL_{c} }{k} =\frac{75*0.0125}{240} =3.9x10^{-6}[/tex]
Bi < 1, we will use the lumped capacitance method
The energy transfer is equal to:
[tex]\frac{Q}{Q_{max} } =(1-exp(\frac{t}{t_{t} } ))[/tex] (eq. 1)
[tex]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[/tex]
Replacing in eq. 1
[tex]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[/tex]
The temperature at the center of sphere is
[tex]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[/tex]
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.
