Consider a bed consisting of particles of diameter Dp = 0.02 m which has been heated to a uniform temperature of Tb = 80 °C. Air at a temperature of Ti = 15 °C is blown through the bed with a mass velocity of G = 0.5 kg/s m^2. Immediately afterwards, it is seen that the air exiting the bed is

Question

Consider a bed consisting of particles of diameter [latex]D_{p}=0.02[/latex] which has been heated to a uniform temperature of [latex]T_{b}=80^{\circ} C[/latex]. Air at a temperature of [latex]T_{i}=15^{\circ} C[/latex] is blown through the bed with a mass velocity of [latex]G=0.5 kg / s m ^{2}[/latex]. Immediately afterwards, it is seen that the air exiting the bed is at [latex]T_{e}=65^{\circ} C[/latex]. Estimate the bed height using the correlations proposed by (a) Whitaker and (b) Lof and Hawley. Porosity of the bed is known to be [latex]\epsilon=0.5[/latex].

Accepted Answer

Step 1 Air properties are taken at the mean temperature of [latex]T_{m}=\frac{T_{i}+T_{o}}{2}=\frac{15+65}{2}=40^{\circ} C[/latex].

Density of air: [latex] ho=1.117 kg / m ^{3}[/latex]

Specific heat of air: [latex]C_{p}=1005 J / kg ^{\circ} C[/latex]

Kinematic viscosity of air: [latex] u=17.1 imes 10^{-6} m ^{2} / s[/latex]

Thermal conductivity of air: [latex]k=0.0274 W / m ^{\circ} C[/latex]

Prandtl number of air: [latex]\operatorname{Pr}=0.699[/latex]

Step 2 Superficial velocity of air is calculated as (based on unit bed cross section area) [latex]U_{s}=\frac{G}{ ho}=\frac{0.5}{1.092}=0.448 m / s[/latex]. Hence, the modified Reynolds number is given by

[latex]R e_{m}=\frac{U_{s} D_{p}}{ u(1-\epsilon)}=\frac{0.448 imes 0.02}{17.1 imes 10^{-6}(1-0.5)}=1047.1[/latex]

The volumetric heat transfer coefficient may then be calculated based on Whitaker correlation (17.122) [latex]h_{v(a)}[/latex] as

[latex]h_{v}=\frac{6(1-\epsilon)^{2} k}{\epsilon D_{p}^{2}}\left(0.5 \operatorname{R} e_{m}^{\frac{1}{2}}+0.2 \operatorname{Re}_{m}^{\frac{2}{3}} ight) \operatorname{Pr}^{\frac{1}{3}}[/latex] (17.122)

[latex]h_{v(a)}=\frac{6(1-0.5)^{2} imes 0.0274}{0.5 imes 0.02^{2}}\left(0.5 imes 1047.1^{\frac{1}{2}}+0.2 imes 1047.1^{\frac{2}{3}} ight) 0.0 .699^{\frac{1}{3}}[/latex] [latex]=6711.9 W / m ^{3}{ }^{\circ} C[/latex]

Step 3 Modeling heat transfer between bed and air: Consider a bed of height H and unit cross section area as shown in Fig. 17.28. The axis of the bed is vertical and oriented parallel to the z axis. Consider a bed element of thickness [latex]\Delta z[/latex] as shown. Heat balance requires that heat gained by air as it moves up through the element be equal to the heat lost by the bed particles. Thus

[latex]G C_{p} T(z+\Delta z)-G C_{p} T(z)=h_{v}\left(T_{b}-T ight) \Delta z[/latex]

Using Taylor expansion and retaining two terms and in the limit [latex]\Delta z ightarrow 0[/latex], the above simplifies to

[latex]G C_{p} \frac{ d T}{ d z}=h_{v}\left(T_{b}-T ight)[/latex]

This equation is in variable separable form and is easily integrated once with respect to z to get [latex]-\ln \left(T_{b}-T ight)=\frac{h_{v} z}{G C_{p}}+C[/latex] where C is a constant of integration. Using [latex]T=T_{i}[/latex] at z = 0 we have [latex]-\ln \left(T_{b}-T_{i} ight)=C[/latex] . Hence the air temperature at any z is given by

[latex]\ln \left(\frac{T_{b}-T_{i}}{T_{b}-T} ight)=\frac{h_{v} z}{G C_{p}}[/latex]

We are given that [latex]T=T_{e}[/latex] at z = H. Hence, we have

[latex]H=\frac{G C_{p}}{h_{v}} \ln \left(\frac{T_{b}-T_{i}}{T_{b}-T_{e}} ight)[/latex]

Step 4 Based on above analysis the bed height may be calculated.

Whitaker correlation: [latex]H_{(a)}=\frac{0.5 imes 1005}{6711.9} \ln \left(\frac{80-15}{80-65} ight)=0.110 m[/latex]

Lof and Hawley correlation: [latex]H_{(b)}=\frac{0.5 imes 1005}{6186.88} \ln \left(\frac{80-15}{80-65} ight)=0.119 m[/latex]

From a design point of view the bigger answer may be chosen.

Question 17.15: Consider a bed consisting of particles of diameter Dp = 0.02...

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