Question 13.7: Consider an air heater consisting of a semicircular tube for...
Consider an air heater consisting of a semicircular tube for which the plane surface is maintained at 1000 K and the other surface is well insulated. The tube radius is 20 mm, and both surfaces have an emissivity of 0.8. If atmospheric air flows through the tube at 0.01 kg/s and T_{m} = 400 K, what is the rate at which heat must be supplied per unit length to maintain the plane surface at 1000 K? What is the temperature of the insulated surface?
Known: Airflow conditions in tubular heater and heater surface conditions.
Find: Rate at which heat must be supplied and temperature of insulated surface.
Schematic:
Assumptions:
1. Steady-state conditions.
2. Diffuse, gray surfaces experiencing uniform irradiation and radiosity.
3. Negligible tube end effects and axial variations in gas temperature.
4. Fully developed flow.
Properties: Table A.4, air (1 atm, 400 K): k = 0.0338 W/m · K, μ = 230 × 10^{-7} kg/s · m, c_{p} = 1014 J/kg · K, Pr = 0.69.
Analysis: Since the semicircular surface is well insulated and there is no external heat addition, a surface energy balance yields
-q_{2,rad} = q_{2,conv}
Since the tube constitutes a two-surface enclosure, the net radiation transfer to surface 2 may be evaluated from Equation 13.23. Hence
q_{12} = q_{1} = -q_{2} = \frac{σ(T_{1}^{4} – T_{2}^{4})}{\frac{1 – ε_{1}}{ε_{1}A_{1}} + \frac{1}{A_{1}F_{12}} + \frac{1 – ε_{2}}{ε_{2}A_{2}}} (13.23)
\frac{σ(T_{1}^{4} – T_{2}^{4})}{\frac{1 – ε_{1}}{ε_{1}A_{1}} + \frac{1}{A_{1}F_{12}} + \frac{1 – ε_{2}}{ε_{2}A_{2}}} = hA_{2}(T_{2} – T_{m})
where the view factor is F_{12} = 1 and, per unit length, the surface areas are A_{1} = 2r_{o} and A_{2} = \pi r_{o}. With
Re_{D} = \frac{ρu_{m}D_{h}}{\mu} = \frac{\dot{m}D_{h}}{A_{c}\mu} = \frac{\dot{m}D_{h}}{(\pi r_{o}^{2}/2)\mu}
the hydraulic diameter is
D_{h} = \frac{4A_{c}}{P} = \frac{2\pi r_{o}}{\pi + 2} = \frac{0.04\pi m}{\pi + 2} = 0.0244 m
Hence
Re_{D} = \frac{0.01 kg/s × 0.0244 m}{(\pi/2)(0.02 m)^{2} × 230 × 10^{-7} kg/s · m} = 16,900
From the Dittus–Boelter equation,
Nu_{D} = 0.023 Re_{D}^{4/5} Pr^{0.4}
Nu_{D} = 0.023(16,900)^{4/5}(0.69)^{0.4} = 47.8
h = \frac{k}{D_{h}}Nu_{D} = \frac{0.0338 W/m · K}{0.0244 m}47.8 = 66.2 W/m^{2} · K
Dividing both sides of the energy balance by A_{1}, it follows that
\frac{5.67 × 10^{-8} W/m^{2} · K^{4} [(1000)^{4} – T_{2}^{4}] K^{4}}{\frac{1 – 0.8}{0.8} + 1 + \frac{1 – 0.8}{0.8}\frac{2}{\pi}} = 66.2\frac{\pi}{2}(T_{2} – 400) W/m^{2}
or
5.67 × 10^{-8} T_{2}^{4} + 146.5T_{2} – 115,313 = 0
A trial-and-error solution yields
T_{2} = 696 K
From an energy balance at the heated surface,
q_{1,ext} = q_{1,rad} + q_{1,conv} = q_{2,conv} + q_{1,conv}
Hence, on a unit length basis,
q'_{1,ext} = h\pi r_{o}(T_{2} – T_{m}) + h2r_{o}(T_{1} – T_{m})
q'_{1,ext} = 66.2 × 0.02[\pi(696 – 400) + 2(1000 – 400)] W/m
q'_{1,ext} = (1231 + 1589) W/m = 2820 W/m
Comments:
1. The irradiation, radiosity, and convection heat flux distributions along the surfaces would not be uniform. As a consequence, we would expect the temperature of the insulated surface to be higher in the corner regions adjacent to surface 1 and lower in the crown of the enclosure. Determination of the irradiation, radiosity, temperature, and convection heat flux distributions along the various surfaces would require a more complex analysis incorporating many radiative surfaces.
2. Applying an energy balance to a differential control volume about the air, it follows that
\frac{dT_{m}}{dx} = \frac{q'_{1}}{\dot{m}c_{p}} = \frac{2820 W/m}{0.01 kg/s (1014 J/kg · K)} = 278 K/m
Hence the air temperature change is significant, and a more representative analysis would subdivide the tube into axial zones and would allow for variations in air and insulated surface temperatures between zones. Moreover, a two-surface analysis of radiation exchange would no longer be appropriate.
