In the incandescent lamp, the electrical energy converted into the Joule heating in the thin-wire filament is again converted into thermal radiation by surface emission. This is shown in Figure (a). The filament is at T_{2}= 2, 900 K and has a total emissivity \epsilon _{r,2} = 0.8 . This thermal radiation emission is mostly absorbed in the glass bulb (approximated as a sphere of diameter D_{1} = 6 cm) and the remainder is transmitted through the bulb. In problem 4.13 at T_{2} = 2,900 K , the calculated fraction of the radiation in the visible range is F_{\Delta λT} (visible) = 0.1012 . Assume that the remainder of the emission is absorbed in the glass bulb, i.e., \dot{S}_{ e,\sigma } = A_{r,2}(1 − F_{\Delta λT} )\epsilon _{r,2} \sigma _{SB}T ^{4} _{2} . Here the filament is a wire with D_{2} = 0.2 mm, L_{2} = 1 m . This heat absorbed by the glass bulb is removed by a thermobuoyantflow surface convection and by surface radiation. The ambient and the surrounding surfaces are at T_{f,∞} = T_{s,∞} = 300 K . The bulb emissivity is that of smooth glass listed in Table
Question 6.11: In the incandescent lamp, the electrical energy converted in...
(a) Figure (b) shows the thermal circuit diagram.
(b) Applying the integral-volume energy equation, and assuming a steadystate heat transfer, we have
= A_{r,2}(1 − F_{\Delta λT })\epsilon _{r,2}\sigma _{SB}T^{ 4}_{ 2} .
Here Q_{1} = 0, and for the surface convection we have
Q_{ku,1-∞} =\frac{T_{1}-T_{f,\infty }}{\left\langle R_{ku}\right\rangle _{D}}\left\langle R_{ku}\right\rangle _{D}=\frac{D}{A_{ku}\left\langle Nu\right\rangle _{D}k_{f}}
From Table , for the sphere we have
\left\langle Nu\right\rangle _{D}=(\left\langle Nu_{D,l}\right\rangle ^{6}+\left\langle Nu_{D,t}\right\rangle ^{6})^{1/6}\left\langle Nu_{D,l}\right\rangle=2.0 + 0.878a_{1}Ra^{1/4}_{D}
\left\langle Nu_{D,t}\right\rangle=\frac{0.13Pr^{0.22}}{(1+0.61Pr^{0.81})^{0.42}Ra^{1/2}_{D}}
a_{1}=\frac{4}{3}\frac{0.503}{[1+(0.492/Pr)^{9/16}]^{4/19}}
Ra_{D}=\frac{g\beta _{f}(T_{1}-T_{f,\infty })D^{3}}{v_{f}\alpha _{f}}
For radiation, we have a two-surface enclosure with both surfaces being gray, diffuse opaque surfaces with A_{r,∞} \gg A_{r,2}. Then we use Q_{r,1-2}=A_{r,1}\epsilon _{r,1}[E_{b,1}(T_{1})-E_{b,2}(T_{2})] for F_{1-2}=1 with \frac{A_{r,1}}{A_{r,2}} \simeq 0 or \epsilon _{r,2}=1 , because we also have F_{1-∞} = 1, i.e.,
Q_{r,1}=Q_{r,1-\infty }=A_{r,1}\epsilon _{r,1}[E_{b,1}(T_{1})-E_{b,\infty }(T_{\infty })] A_{r,1}=A_{ku}We now substitute for Q_{r,1-∞} and Q_{ku,1-∞} in the energy equation, i.e.,
A_{ku}\left\langle Nu\right\rangle _{D}\frac{k_{f}}{D}(T_{1}-T_{f,\infty })+A_{r,1}\epsilon _{r,1}\sigma _{SB}(T_{1}^{4}-T_{s,\infty }^{4}) =A_{r,2}(1-F_{\Delta \lambda T})\epsilon _{r,2}\sigma _{SB}T^{4}_{2}This equation is solved for T_{1} and is nonlinear in T_{1}. A software is used for the solution. The numerical values for the parameters are
\epsilon _{r,1} = 0.94 Table
\epsilon _{r,2} = 0.8, T_{2} = 2,900 K, \sigma _{SB} = 5.67 × 10^{−6} W/m^{2}-K^{4} Table
D_{1} = 0.06 m ,T_{f,∞} = T_{s,∞} = 300 K
k_{f} = 0.0267 W/m-K Table
Pr = 0.69 Table
β_{f} = 1/T_{f,∞} = 1/300 1/K ideal gas \beta _{f}=\frac{1}{T_{f}} ,p =\rho _{f}\frac{R_{g}}{M} T_{f}
ν_{f} = 1.566 × 10^{−5} m^{2}/s Table
α_{f} = 2.257 × 10^{−5} m^{2}/s Table
g = 9.807 m/s^{2}
A_{r,2} = πD_{2}L_{2} = π × 2 × 10^{−4} (m) × 10^{−2} (m) = 6.280 × 10^{−6} m^{2}
F_{\Delta λT} = 0.1012.
From the solver, we have the results
T_{1} = 410.2 K
= 137.1^{\circ }C.
The intermediate quantities computed are
\dot{S}_{ e,σ} = 21.28 W , Ra_{D} = 2.202 × 10^{6} , \left\langle Nu\right\rangle _{D} = 19.68,
Q_{r,1-∞} = 10.37 W , Q_{ku,1-2} = 10.91 W
