Question 2 of 6: Length of a Cube-Packed Gas Heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam; one textbook and any non-communicating calculator permitted. Format: two parts of three problems each (Part A — Heat Transfer, Part B — Mass Transfer), all problems worth 25 points; candidates attempt at least two per part and only the first two per part are marked. The six problems are renumbered 1–6 and all are solved below for completeness. Property data are taken from the problem statements and standard open-book tables (the paper supplies the air-property table A-9); values used are stated in each Given block.
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — conduction resistances, critical radius, fins, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — the packed-bed, distillation and sublimation problems of this style; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential distillation and gas adsorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — packed-bed and convective mass-transfer coefficients; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and the paper's Table A-9 (properties of air).
Part A — Heat Transfer
Question 2: Length of a Cube-Packed Gas Heater (25 points)
Given. Air $\dot m = 18\ \text{kg/hr} = 0.005\ \text{kg/s}$ heated $40 \to 360\ \text{°C}$; pipe diameter $D = 0.10$ m packed with cubes of side $a = 6.45$ mm; the pipe wall is held at $400\ \text{°C}$. Air properties at the mean bulk temperature $200\ \text{°C}$ (Table A-9): $c_p = 1023\ \text{J/kg}\cdot\text{K}$, $k = 0.03779\ \text{W/m}\cdot\text{K}$, $\mu = 2.577\times10^{-5}\ \text{kg/m}\cdot\text{s}$, $Pr = 0.697$.
Quantity
Value
Air rate $\dot m$
0.005 kg/s (18 kg/hr)
Inlet / outlet air T
40 °C → 360 °C
Pipe wall T
400 °C
Pipe diameter / cube side
0.10 m / 6.45 mm
Find. The length $L$ of packed pipe that delivers the required duty.
Figure 2 — Air is heated as it threads the packed bed. All of the heat enters through the 400 °C pipe wall; the cubes carry no heat source of their own, but they scour the wall film and mix the gas, which is what raises the wall coefficient.
Check
Only the pipe wall is heated. The cubes have no heat source of their own, so at steady state they sit near the local gas temperature and every watt must cross the wall area $\pi D_t L$. The packing surface is not a 400 °C heating surface. The wall coefficient uses Leva's correlation for a gas heated in a packed tube, with $d_p$ taken as the diameter of the equal-volume sphere (8.00 mm for a 6.45 mm cube). Taking $d_p$ as the cube side instead gives $L \approx 1.8$ m, so the answer is about 1.6 to 1.8 m depending on that choice, stated here as the paper's Note 1 asks.
Approach. Fix the duty from a sensible-heat balance, get the wall-to-gas coefficient of the packed tube from Leva's correlation, then size the wall area (hence the tube length) with the log-mean temperature difference against the constant-temperature wall.
Duty from the air-side energy balance.
$$Q = \dot m\,c_p\,(T_{out}-T_{in}) = 0.005\,(1023)(360-40) = \boxed{1637\ \text{W}}.$$
Equivalent particle diameter. A cube of side $a$ has the volume of a sphere of diameter
$$d_p = \left(\frac{6a^3}{\pi}\right)^{1/3} = \left(\frac{6(6.45)^3}{\pi}\right)^{1/3} = 8.00\ \text{mm},$$
so $d_p/D_t = 0.080$, inside the 0.05 to 0.3 range of the correlation.
Superficial mass velocity and particle Reynolds number. The empty-pipe area is $A_c = \tfrac{\pi}{4}(0.10)^2 = 7.854\times10^{-3}\ \text{m}^2$, so $G = \dot m/A_c = 0.6366\ \text{kg/m}^2\text{s}$ and
$$Re_p = \frac{d_p G}{\mu} = \frac{(0.00800)(0.6366)}{2.577\times10^{-5}} = 198.$$
Wall-to-gas coefficient (Leva, gas heated).
$$\frac{h_w D_t}{k} = 0.813\,Re_p^{0.90}\,e^{-6d_p/D_t} = 0.813(198)^{0.90}e^{-0.48} = 58.6,$$
$$h_w = \frac{58.6(0.03779)}{0.10} = \boxed{22.1\ \text{W/m}^2\text{K}}.$$
The exponential factor (0.619 here) accounts for the looser packing and bypassing next to the wall when the particles are large relative to the tube.
Log-mean temperature difference (constant wall temperature). With $\Delta T_1 = 400-40 = 360$ and $\Delta T_2 = 400-360 = 40$,
$$\Delta T_{lm} = \frac{360-40}{\ln(360/40)} = 145.6\ \text{K}.$$
Wall area and tube length. The wall area needed is $A_w = Q/(h_w\,\Delta T_{lm}) = 1637/(22.1\times145.6) = 0.507\ \text{m}^2$, and since $A_w = \pi D_t L$,
$$L = \frac{A_w}{\pi D_t} = \frac{0.507}{\pi(0.10)} = \boxed{1.62\ \text{m} \approx 1.6\ \text{m}}.$$
A little over a metre and a half of packed 10 cm pipe does the job. The length is set by the gas film at the wall, and the packing is what makes that film thin enough: the particles keep breaking up the boundary layer that would otherwise build along a bare wall at this low gas flow. The result is only as good as the correlation, so a candidate should name the one used and the particle diameter fed into it.
