Respuesta :
Answer:
see explanation below
Explanation:
Given that,
[tex]T_1 =[/tex] 500Ā°C
[tex]T_2[/tex] = 25Ā°C
d = 0.2m
L = 10mm = 0.01m
Uā = 2m/s
Calculate average temperature
[tex]\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5[/tex]
262.5 + 273
= 535.5K
From properties of air table A-4 corresponding to [tex]T_{avg}[/tex] = 535.5K [tex]\approx 550K[/tex]
k = 43.9 Ć 10ā»Ā³W/m.k
v = 47.57 Ć 10ā»ā¶ mĀ²/s
[tex]P_r = 0.63[/tex]
A)
Number for the first strips is equal to
[tex]R_e_x = \frac{u_o.L}{v}[/tex]
[tex]R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4[/tex]
Calculating heat transfer coefficient from the first strip
[tex]h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3[/tex]
[tex]h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2[/tex]
The rate of convection heat transfer from the first strip is
[tex]q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W[/tex]
The rate of convection heat transfer from the fifth trip is equal to
[tex]q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)[/tex]
[tex]h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2[/tex]
Calculating [tex]h_o_-_4[/tex]
[tex]h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2[/tex]
The rate of convection heat transfer from the tenth strip is
[tex]q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)[/tex]
[tex]h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2[/tex]
Calculating
[tex]h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2[/tex]
Calculating the rate of convection heat transfer from the tenth strip
[tex]q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W[/tex]
The rate of convection heat transfer from 25th strip is equal to
[tex]q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)[/tex]
Calculating [tex]h_o_-_2_5[/tex]
[tex]h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2[/tex]
Calculating [tex]h_o_-_2_4[/tex]
[tex]h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2[/tex]
Calculating the rate of convection heat transfer from the tenth strip
[tex]q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W[/tex]
