Answer :
Answer:
The ratio of heat transfer rate is 0.88
Explanation:
Given;
Case1 :
height of vertical surface, L = 1 m
width of vertical surface, w = 0.6 m
Case 2:
height of vertical surface, L = 0.6 m
width of vertical surface, w = 1 m
At an assumed film temperature of air = 300 K
then, read off from heat transfer table, temperature inverse β, surface area flow rate v, and Pr, to determine Rayleigh number for the two cases.
β = 1/300 = 0.00333 K⁻¹
v = 15.89 x 10⁻⁶ m²/s
Pr = 0.69
Case 1, L = 1 m
[tex]R_a = \frac{g\beta TL^3P_r}{v^2}[/tex]
[tex]R_a = \frac{9.8*0.00333* 20*1^3*0.69}{(15.89x10^{-6})2} \\\\R_a = 1.784 *10^9[/tex]
Case 2, L = 0.6 m
[tex]R_a = \frac{g\beta TL^3P_r}{v^2} \\\\R_a = \frac{9.8*0.00333* 20*0.6^3*0.69}{(15.89*10^{-6})^2}\\\\ R_a = 3.853 *10^8[/tex]
From the values of Rayleigh numbers above, case 1 is Turbulent flow while case 2 is laminar flow
Thus: C₁ = 0.1, n₁ = ¹/₃
C₂ = 0.59, n₂ = 1/4
Ratio of heat transfer rate is given as:
[tex]\frac{q_1}{q_2} = \frac{h_1 \delta T}{h_2 \delta T} \\\\\frac{q_1}{q_2} = \frac{h_1}{h_2} \\\\But, \frac{hL}{k} = CR_a^n L, \ \ h=\frac{k}{L}(CR_a^n L)\\\\\frac{q_1}{q_2} = \frac{C_1R_a_1^n L_2}{C_2R_a_2^n L_1} = \frac{0.1(1.784*10^9)^{\frac{1}{3}} *0.6}{0.59(3.853*10^8)^{\frac{1}{4}} *1} \\\\\frac{q_1}{q_2} = \frac{72.76}{82.66} = 0.88[/tex]
Therefore, the ratio of heat transfer rate is 0.88