Answer to Question #233016 in Molecular Physics | Thermodynamics for Unknown346307

Question

A furnace wall is made up of three layers of thicknesses 250 mm, 100 mm

and 150 mm with thermal conductivities of 1.65, k and 9.2 W/m°C respectively. The inside is

exposed to gases at 1250°C with a convection coefficient of 25 W/m2°C and the inside surface is

at 1100°C, the outside surface is exposed air at 25°C with convection coefficient of 12 W/m2°C.

Determine :

(i) The unknown thermal conductivity ‘k’ ;

(ii) The overall heat transfer coefficient ;

(iii) All surface temperatures.

Answer 1

$L_A = 250\; mm = 0.25 \;m \\ L_B= 100 \;mm = 0.1 \;m \\ L_C = 150 \;mm = 0.15\; m \\ k_A = 1.65 \;W/m°C \\ k_C=9.2 \;W/m°C \\ t_{hf}= 1250 \;°C \\ t_1= 1100 \;°C \\ h_{hf}= 25 \; W/m^2°C \\ h_{cf}= 12 \;W/m^2 °C$

(i) Thermal conductivity, k (= k B ) :

The rate of heat transfer per unit area of the furnace wall,

$q=h_{hf}(t_{hf}-t_1) \\ = 25(1250-1100) = 3750 \;W/m^2 \\ q= \frac{(ΔT)_{overall}}{(R_{th})_{total}} \\ q= \frac{(t_{hf}-t_{cf})}{(R_{th})_{conv-hf} -R_{th-A}+ R_{th-B} +R_{th-C}+(R_{th})_{conv-cf}} \\ 3750 = \frac{1250-25}{\frac{1}{h_{hf}} + \frac{L_A}{k_A} + \frac{L_B}{k_B} + \frac{L_C}{k_C} + \frac{1}{h_{cf}} } \\ 3750 = \frac{1225}{\frac{1}{25} + \frac{0.25}{1.65} + \frac{0.1}{k_B} + \frac{0.15}{9.2} + \frac{1}{12} } \\ = \frac{1225}{0.04 + 0.1515 + \frac{0.1}{k_B} + 0.0163 + 0.0833 } \\ = \frac{1225}{0.2911 + \frac{0.1}{k_B}} \\ 3750 (0.289 + \frac{0.1}{k_B}) = 1225 \\ \frac{0.1}{k_B}= \frac{1225}{3750} -0.2911 = 0.0355 \\ k_B= k= \frac{0.1}{0.0355} = 2.817 \; W/m^2°C$

(ii) The overall transfer coefficient, U :

The overall heat transfer coefficient,

$U= \frac{1}{(R_{th})_{total}} \\ (R_{th})_{total} = \frac{1}{25} + \frac{0.25}{1.65} + \frac{0.1}{2.817} + \frac{0.15}{9.2} + \frac{1}{12} \\ = 0.04 + 0.1515+0.0355+0.0163+0.0833 = 0.3266 \;°C \;m^2/W \\ U= \frac{1}{0.3266}= 3.06 \;W/m^2 °C$

(iii) All surface temperature ; t₁ , t₂ , t₃ , t₄ :

$q=q_A=q_B=q_c \\ 3750 = \frac{t_1-t_2}{L_A/k_A} = \frac{t_2-t_3}{L_B/k_B} = \frac{t_3-t_4}{L_C/k_C} \\ 3750 = \frac{1110 -t_2}{0.25/1.65} \\ t_2= 1100 -3750 \times \frac{0.25}{1.65} = 531.8 \;°C \\ 3750 = \frac{531.8 -t_3}{0.1/2.817} \\ t_3= 531.8 -3750 \times \frac{0.1}{2.817} = 398.6 \;°C \\ 3750 = \frac{398.6 -t_4}{0.15/9.2} \\ t_4= 398.6 -3750 \times \frac{0.15}{9.2} = 337.5 \;°C$

Check using outside conversion,

$q= \frac{337.5-25}{1/h_{cf}}= \frac{337.5-25}{1/12}=3750 \;W/m^2$