Air at Ti = 25 °C enters a bed of diameter Db = 0.05 m with a mass flux of G = 2 kg/m^2s. Two types of bed particles are under consideration viz. (1) Spherical particles of diameter Dp = 0.003 m, void fraction equal to ε = 0.4 and (2) Cylindrical particles of diameter Dcyl equal to length Lcyl

Question

Air at [latex]T_{i}=25^{\circ} C[/latex] enters a bed of diameter [latex]D_{b}=0.05 m[/latex] with a mass flux of [latex]G=2 kg / m ^{2} s[/latex]. Two types of bed particles are under consideration viz. (1) Spherical particles of diameter [latex]D_{p}=0.003 m[/latex], void fraction equal to [latex]\epsilon=0.4[/latex] and (2) Cylindrical particles of diameter [latex]D_{ ext {cyl }}[/latex] equal to length [latex]L_{ ext {cyl }}[/latex] of 0.002 m, void fraction equal to [latex]\epsilon=0.45[/latex]. The bed heights are to be chosen such that the exiting air is at [latex]T_{e}=55^{\circ} C[/latex] when the bed wall is steam jacketed and maintained at [latex]T_{w}=100^{\circ} C[/latex]. Evaluate the two options and decide the better one.

Accepted Answer

Air properties at a mean temperature of [latex]T_{m}=\frac{T_{i}+T_{e}}{2}=\frac{25+55}{2}=40{ }^{\circ} C[/latex] are read off air tables

[latex] ho=1.117 kg / m ^{3}[/latex] , [latex]\mu=1.91 imes 10^{-5} kg / m s[/latex] , [latex]C_{p}=1005 J / kg K[/latex] , k = 0.0274 W/m K , [latex]\operatorname{Pr}=0.699[/latex]

Case 1: Bed with spherical particles:

Mass flux through the bed is given as [latex]G=2 kg / m ^{2} s[/latex]. The superficial air velocity is then given by

[latex]U_{s}=\frac{G}{ ho}=\frac{2}{1.117}=1.791 m / s[/latex]

The Reynolds number based on particle diameter is then given by

[latex]R e_{p}=\frac{ ho U_{s} D_{p}}{\mu}=\frac{1.117 imes 1.791 imes imes 0.003}{1.91 imes 10^{-5}}=314.63[/latex]

Use Eq. 17.125 to obtain the wall to gas heat transfer coefficient as

[latex]\frac{h_{c} D_{p}}{k}=0.203 \operatorname{Re}_{p}^{\frac{1}{3}} \operatorname{Pr}^{\frac{1}{3}}+0.220 \operatorname{Re}_{p}^{0.8} \operatorname{Pr}^{0.4}[/latex] (17.125)

[latex]h_{c}=\left(0.203 \operatorname{Re}_{p}^{\frac{1}{3}} \operatorname{Pr}^{\frac{1}{3}}+0.220 \operatorname{Re}_{p}^{0.8} \operatorname{Pr}^{0.4} ight) \frac{k}{D_{p}}=\left(0.203 imes 314.63^{\frac{1}{3}} imes 0.699^{\frac{1}{3}} ight.[/latex] [latex]\left.+0.220 imes 314.63^{0.8} imes 0.699^{0.4} ight) \frac{0.0263}{0.003}=182.1 W / m ^{2}{ }^{\circ} C[/latex]

The Stanton number [latex]S t_{b}[/latex] is then calculated as

[latex]S t_{b}=\frac{h_{c}}{ ho U_{s} C_{p}}=\frac{182.1}{1.117 imes 1.791 imes 1005}=0.0906[/latex]

With all the temperatures specified, the LMTD is calculated as

[latex]L M T D=\frac{\left(T_{w}-T_{i} ight)-\left(T_{w}-T_{e} ight)}{\ln \frac{\left(T_{w}-T_{i} ight)}{\left(T_{w}-T_{e} ight)}}=\frac{(100-25)-(100-55)}{\ln \left(\frac{100-25}{100-55} ight)}=58.7{ }^{\circ} C[/latex]

Total heat transferred to air is calculated as

[latex]Q=G A_{b} C_{p}\left(T_{e}-T_{i} ight)=2 imes \frac{\pi imes 0.05^{2}}{4} imes 1005 imes(55-25)=118.4 W[/latex]

Noting that [latex]Q=h_{c}\left(\pi D_{b} H ight) L M T D[/latex], we solve for bed height H as

[latex]H=\frac{Q}{h_{c} \pi D_{b} L M T D}=\frac{118.4}{182.1 imes 0.05 imes 58.7}=0.0705 m[/latex]

Performance parameters are now calculated. The friction factor is calculated using Ergun equation as

[latex]f=\frac{1-\epsilon}{\epsilon^{3}}\left(1.75+150 \frac{1-\epsilon}{\operatorname{Re}_{p}} ight)=\frac{1-0.4}{0.4^{3}}\left(1.75+150 \frac{1-0.4}{314.63} ight)=19.09[/latex]

Colburn j factor is calculated as

[latex]j=S t_{b} P r^{\frac{2}{3}}=0.0906 imes 0.699^{\frac{2}{3}}=0.0714[/latex]

Performance parameter of the fixed bed heat exahnger is calculated as

[latex]\frac{J}{F^{\frac{1}{3}}}=\frac{j}{f^{\frac{1}{3}}}=\frac{0.0714}{19.09^{\frac{1}{3}}}=0.0267[/latex]

Case 2: Bed with cylindrical particles:

Since the particles are not spherical, we make use of equivalent diameter for the particles given by [latex]D_{h}=\sqrt{\frac{A_{s}}{\pi}}[/latex] where [latex]A_{s}[/latex] is the particle surface area given by [latex]A_{s}=\pi D_{ cyl } L cyl +2 \frac{\pi D_{ cyl }^{2}}{4}[/latex] With [latex]D_{ ext {cyl }}=L_{ ext {cyl }}=0.002 m[/latex] . we have [latex]A_{s}=1.5 \pi D_{c y l}^{2}=1.5 imes[/latex] [latex]\pi imes 0.002^{2}=1.885 imes 10^{-5} m ^{2}[/latex]. Hence, the equivalent diameter is given by [latex]D_{h}=\sqrt{\frac{1.885 imes 10^{-5}}{\pi}}=0.00245 m[/latex] The rest of the calculations follow the same procedure as was usedwith the spherical particale case except that Eq. 17.124 is used for obtaining the heat transfer coefficient. Leaving the details to the reader, the results are tabulated below

[latex]\frac{h_{c} D_{p}}{k}=2.58 \operatorname{Re}_{p}^{\frac{1}{3}} \operatorname{Pr}^{\frac{1}{3}}+0.094 \operatorname{Re}_{p}^{0.8} \operatorname{Pr}^{0.4}[/latex] (17.124)

Results for cylindrical bed particles
[latex]U_{s}:[/latex]	1.791 m/s	[latex]\operatorname{Re}_{h}:[/latex]	256.9
[latex]N u_{c}:[/latex]	21.45	[latex]h_{c}:[/latex]	[latex]239.98 W / m ^{2}{ }^{\circ} C[/latex]
H:	0.0535 m	[latex] ext { St: }[/latex]	0.1194
j :	0.094	f :	12.5
[latex]j / f^{\frac{1}{3}}:[/latex]	0.0405

Comparisons: Bed with cylindrical particles is more compact since it requires a smaller bed height. It is also the better of the two since the heat transfer performance parameter is larger of the two values. Hence, it is recommended that the bed with cylindrical particles be chosen for the application.

Question 17.16: Air at Ti = 25 °C enters a bed of diameter Db = 0.05 m with ...

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