Aluminum foam of porosity equal to 0.92 butwith three different pores per inch (PPI) are to be compared in situation that was used in the experiment of Kim et al. The permeabilities of three foams of PPI equal to 10, 20, and 40 are, respectively, given by

Question

Aluminum foam of porosity equal to 0.92 butwith three different pores per inch (PPI) are to be compared in situation that was used in the experiment of Kim et al. The permeabilities of three foams of PPI equal to 10, 20, and 40 are, respectively, given by [latex]1.04 imes 10^{-7}, 0.76 imes 10^{-7}[/latex] and [latex]0.51 imes 10^{-7} m ^{2}[/latex]. The channel dimensions are those given earlier while discussing the correlation proposed by Kim et al. Air enters the foam filled channel with a velocity of 2 m/s and a temperature of [latex]T_{i}=25^{\circ} C[/latex]. What will be the exit temperature of air if the upper wall is maintained at [latex]65^{\circ} C[/latex] with the three different foams?

Accepted Answer

Step 1 Air properties are taken at air entry temperature viz. [latex]T_{i}=25^{\circ} C[/latex]

Density of air: [latex] ho=1.165 kg / m ^{3}[/latex]

Specific heat capacity of air: [latex]C_{p}=1007 J / kg K[/latex]

Thermal conductivity of air: [latex]k=0.026 W / m K[/latex]

Kinematic viscosity of air: [latex] u=1.61 imes 10^{-5} m ^{2} / s[/latex]

Prandtl number of air: [latex]\operatorname{Pr}=0.713[/latex]

Step 2 Other data common to all three cases of foam are H = 0.009 m; L = 0.188 m and W = 0.09 m. Area from which heat transfer takes place is also common and given by [latex]A=L W=0.188 imes 0.09=0.01692 m ^{2}[/latex] . The Reynolds number is calculated as [latex]R e_{H}=\frac{U_{s} H}{ u}=\frac{2 imes 0.009}{1.61 imes 10^{-5}}=1120.1[/latex] and this remains the same in all the three cases.Detailed calculations are shown for the case of Aluminum foam with 10 PPI.

Step 3 With the specified permeability for this foam, the Darcy number is calculated as [latex]D a=\frac{\kappa}{H^{2}}=\frac{1.04 imes 10^{-7}}{0.009^{2}}=1.284 imes 10^{-3}[/latex] Using Eq. 17.128, the Nusselt number is calculated as

[latex]N u_{H}=0.0159 \frac{R e_{H}^{0.428} P r^{\frac{1}{3}}}{D a^{0.787}}[/latex] (17.128)

[latex]N u_{H}=0.0159 \frac{ imes 1120.1^{0.428} 0.713^{\frac{1}{3}}}{\left(1.284 imes 10^{-3} ight)^{0.787}}=54.09[/latex]

Step 4 The Stanton number is calculated as

[latex]S t=\frac{N u_{H}}{R e_{H} P r}=\frac{54.09}{1120.1 imes 0.713}=0.0677[/latex]

We may easily show that the [latex]N T U=\frac{S t L}{H}[/latex] , and hence we have [latex]NTU =\frac{0.0677 imes 0.188}{0.009}=1.415[/latex] . Effectiveness may be shown to be [latex]\frac{T_{w}-T_{e}}{T_{w}-T_{i}}= e ^{-N T U}[/latex] , and hence the air exit temperature is given by

[latex]T_{e}=T_{w}-\left(T_{w}-T_{i} ight) e ^{-N T U}=65-(65-25) e ^{-1.415}=55.3{ }^{\circ} C[/latex]

Step 5 Calculations may be performed similarly for the other two foams also. The reader may verify the following:

PPI	[latex]N u_{H}[/latex]	NTU	[latex]T_{e}{ }^{\circ} C[/latex]
10PPI:	54.09	1.415	55.3
20PPI:	69.23	1.810	58.5
40PPI:	94.76	2.479	61.6

Question 17.17: Aluminum foam of porosity equal to 0.92 butwith three differ...

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