Question 12.55: KNOWN: Thermocouple is irradiated by a blackbody furnace at ......

ANALYSIS: (a) The irradiation on the detector due to emission from the furnace which passes through the filter is defined as

G _{ d }= q _{ f \rightarrow d } / A _{ d }=50  W / m ^2                  (1)

where the power leaving the furnace and intercepted at the detector is

q _{ f \rightarrow d }=\left[ I _{ f } \cdot A _{ f } \cos \theta_{ f } \cdot \omega_{ d – f }\right] \tau_{\text{filter}}=\left[\frac{\sigma T ^4}{\pi} \cdot A _{ f } \cos \theta_{ f } \cdot \frac{ A _{ d }}{ L ^2}\right] \tau_{\text{filter}}.                       (2)

The transmittance of the filter is

\tau_{\text {filter }}=\tau_{\lambda 1} F _{0-\lambda T }+\tau_{\lambda 2}\left(1- F _{0-\lambda T }\right)=0 \times 0.4036+0.8(1-0.4036)=0.477               (3)

where F _{0-\lambda T }=0.4036 with λT = 2.4 × 1500 = 3600 µm·K from Table 12.1. Combining Eqs. (1) and (2) and substituting numerical values,

G _{ d }=(1 / \pi) 5.67 \times 10^{-8}  W / m ^2 \cdot K ^4 \times(1500  K )^4\left(25 \times 10^{-6}  m ^2 \times 1\right)\left( A _{ d } / L ^2\right) \times 0.477 / A _{ d }=50  W / m ^2
find

L = 147 mm.

(b) Using the foregoing equations in the IHT workspace along with the IHT Radiation Tool, Band Emission Factor, G was computed and plotted as a function of L for selected blackbody temperatures.

The irradiation decreases with increasing separation distance x as the inverse square of the distance. At any fixed separation distance, the irradiation increases as \mathrm T_{\mathrm f} increases. In what manner will G depend upon \mathrm T _{\mathrm f}? Is \mathrm G \sim \mathrm T _{\mathrm f}^4? Why not?

COMMENTS: The IHT workspace used to generate the above plot is shown below.

// Irradiation, Eq (2):
G = qfd / Ad
qfd = Ief * Af * omegadf * tauf
omegadf = Ad / L^2
Ief = Ebf / pi
Ebf = sigma * Tf^4
sigma = 5.67e-8
// Transmittance, Eq (3):
tauf = tau1 * FL1Tf + tau2 * ( 1 – FL1Tf )
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1Tf = F_lambda_T(lambda1,Tf) // Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
// Assigned Variables:
G = 50 // Irradiation on detector, W/m^2
Tf = 1000 // Furnace temperature, K
//Tf = 1500
//Tf = 2000
Af = 25 * 1e-6 // Furnace aperture, m^2
Ad = 0.003 * 0.007 // Detector area, m^2
tau1 = 0 // Spectral transmittance, <= lambda1
tau2 = 0.8 // Spectral transmittance, >= lambda2
lambda1 = 2.4 // Wavelength, mum
//L = 0.194 // Separation distance, m
L_mm = L * 1000 // Separation distance, mm
/* Data Browser Results – Part (a)
Ebf FL1Tf Ief L omegadf qfd tauf Ad Af Df G
Tf lambda1 sigma tau1 tau2
2.87E5 0.4036 9.137E4 0.1476 0.0009634 0.00105 0.4771 2.1E-5 2.5E-5 0.025
50 1500 2.4 5.67E-8 0 0.8 */