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In an experiment, the temperature of a hot gas stream is to be measured by a thermocouple with a spherical junction. Due to the nature of this experiment, the response time of the thermocouple to register 42 percent of the initial temperature difference must be within 5 s. The properties of the thermocouple junction are k = 35 W/m•K, rho = 8500 kg/m3, and cp = 320 J/kg•K. If the heat transfer coefficient between the thermocouple junction and the gas is 250 W/m2•K, determine the diameter of the junction.

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raphealnwobi

Answer:

The diameter of the junction is 5.06 mm

Explanation:

Given that:

ρ = 8500 kg/m³

k = 35 W/m.k

[tex]c_p=320J/kg.K[/tex]

the heat transfer coefficient between the thermocouple junction and the gas  = [tex]hA_s=250W/m^2.k[/tex]

The initial temperature difference (t₀) = 42% = 0.42

t = 5 s

[tex]\frac{T(t)-T_\alpha }{T_i-T_\alpha } =1-t_0 = 1-0.42=0.58[/tex]

[tex]t=-\frac{\rho c_pL_c}{hA_s}ln\frac{T(t)-T_\alpha}{T_i-T_\alpha}[/tex]

Substituting values:

[tex]5=-\frac{8500* 320*L_c}{250}ln(0.58)\\L_c=8.436*10^{-4[/tex]

[tex]L_c=\frac{V}{A}=\frac{\frac{4\pi r_o^3}{3} }{4\pi r_o^2} = \frac{r_o}{3}=\frac{D}{6}\\ Therefore, D=6L_c\\D=6*8.436*10^{-4}=5.06mm[/tex]

The diameter of the junction is 5.06 mm

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