Question 12.54: KNOWN: Furnace wall temperature and aperture diameter. Dista......

ANALYSIS: (a) From Eq. 12.5, the heat rate leaving the furnace aperture and intercepted by the detector is

q = I _{ e } A _{ a } \cos \theta_{ 1 } \omega_{ s – a }.

From Eqs. 12.14 and 12.28

I _{ e }=\frac{ E _{ b }\left( T _{ f }\right)}{\pi}=\frac{\sigma T _{ f }^4}{\pi}=\frac{5.67 \times 10^{-8}(1500)^4}{\pi}=9.14 \times 10^4  W / m ^2 \cdot sr .

From Eq. 12.2,

\omega_{ s – a }=\frac{ A _{ n }}{ r ^2}=\frac{ A _{ s } \cdot \cos \theta_2}{ r ^2}=\frac{10^{-5}  m ^2 \times \cos 45^{\circ}}{(1  m )^2}=0.707 \times 10^{-5} sr.

Hence

q =9.14 \times 10^4  W / m ^2 \cdot sr \left[\pi(0.02  m )^2 / 4\right] \cos 30^{\circ} \times 0.707 \times 10^{-5}  sr =1.76 \times 10^{-4}  W.

(b) With the window, the heat rate is

q =\tau\left( I _{ e } A _{ a } \cos \theta_1 \omega_{ s – a }\right)

where τ is the transmissivity of the window to radiation emitted by the furnace wall. From Eq. 12.55,

\tau=\frac{\int_0^{\infty} \tau_\lambda G _\lambda d \lambda}{\int_0^{\infty} G _\lambda d \lambda}=\frac{\int_0^{\infty} \tau_\lambda E _{\lambda, b }\left( T _{ f }\right) d \lambda}{\int_0^{\infty} E _{\lambda, b } d \lambda}=0.8 \int_0^2\left( E _{\lambda, b } / E _{ b }\right) d \lambda=0.8 F _{(0 \rightarrow 2  \mu m )}.

With λT = 2 μm × 1500 K = 3000 μm·K, Table 12.1 gives F _{(0 \rightarrow 2  \mu m )}=0.273. Hence, with τ = 0.273 × 0.8 = 0.218, find

q =0.218 \times 1.76 \times 10^{-4}  W =0.384 \times 10^{-4}  W.