Question

The heat transfer rate due to free convection from a vertical surface, 1 m high and 0.6 m wide, to quiescent air that is 20 K colder than the surface is known. What is the ratio of the heat transfer rate for that situation to the rate corresponding to a vertical surface, 0.6 m high and 1 m wide, when the quiescent air is 20 K warmer than the surface

Answer 1

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

$R_a = \frac{g\beta TL^3P_r}{v^2}$

$R_a = \frac{9.8*0.00333* 20*1^3*0.69}{(15.89x10^{-6})2} \\\\R_a = 1.784 *10^9$

Case 2, L = 0.6 m

$R_a = \frac{g\beta TL^3P_r}{v^2} \\\\R_a = \frac{9.8*0.00333* 20*0.6^3*0.69}{(15.89*10^{-6})^2}\\\\ R_a = 3.853 *10^8$

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:

$\frac{q_1}{q_2} = \frac{h_1 \delta T}{h_2 \delta T} \\\\\frac{q_1}{q_2} = \frac{h_1}{h_2} \\\\But, \frac{hL}{k} = CR_a^n L, \ \ h=\frac{k}{L}(CR_a^n L)\\\\\frac{q_1}{q_2} = \frac{C_1R_a_1^n L_2}{C_2R_a_2^n L_1} = \frac{0.1(1.784*10^9)^{\frac{1}{3}} *0.6}{0.59(3.853*10^8)^{\frac{1}{4}} *1} \\\\\frac{q_1}{q_2} = \frac{72.76}{82.66} = 0.88$

Therefore, the ratio of heat transfer rate is 0.88