Answer:
a) [tex]\dot Q = -1121.4\,W[/tex], b) [tex]\dot Q = -2242.8\,W[/tex]
Explanation:
a) The problem depicts a case of heat transfer due to natural convection. The wall can be modelled as vertical plate. As first step, the Raleigh number is found:
[tex]Ra_{wall} = \frac{g\cdot \beta \cdot (T_{wall}-T_{\infty})\cdot h_{wall}^{3}}{\nu^{2}}[/tex]
Where:
[tex]\beta[/tex] - Coefficient of volume expansion of air, in [tex]\frac{1}{K}[/tex].
[tex]\nu[/tex] - Kinematic viscosity of air, in [tex]\frac{m^{2}}{s}[/tex].
[tex]g[/tex] - Gravitational acceleration, in [tex]\frac{m}{s^{2}}[/tex].
Properties of air at 5 °C are presented below:
[tex]\nu = 1.382\times 10^{-5}\,\frac{m^{2}}{s}[/tex]
[tex]k = 0.02401\,\frac{W}{m\cdot K}[/tex]
Let assume that air behaves ideally. Then, the coefficient of volume expansion is:
[tex]\beta = \frac{1}{278.15\,K}[/tex]
[tex]\beta = 3.595\times 10^{-3}\,\frac{1}{K}[/tex]
The Rayleigh number is:
[tex]Ra_{wall} = \frac{(9.807\,\frac{m}{s^{2}} )\cdot (3.595\times 10^{-3}\,\frac{1}{K} )\cdot (7\,K)\cdot (6\,m)^{3}}{(1.382\times 10^{-5}\,\frac{m^{2}}{s})^{2} }[/tex]
[tex]Ra_{wall} = 2.791\times 10^{11}[/tex]
The following formula is adecuate to estimate the Nusselt number:
[tex]Nu_{L} = 0.1\cdot Ra_{L}^{\frac{1}{3} }[/tex]
[tex]Nu_{L} = 0.1\cdot (2.97\times10^{11})^{\frac{1}{3} }[/tex]
[tex]Nu_{L} = 667.194[/tex]
Convection coefficient can be determined in terms of Nusselt number:
[tex]Nu_{L} = \frac{h\cdot h_{wall}}{k}[/tex]
[tex]h = \frac{Nu_{L}\cdot k}{h_{wall}}[/tex]
[tex]h = \frac{(667.194)\cdot (0.02401\,\frac{W}{m\cdot K} )}{6\,m}[/tex]
[tex]h = 2.670\,\frac{W}{m^{2}\cdot K}[/tex]
The rate of heat loss by natural convection is:
[tex]\dot Q = h\cdot A\cdot (T_{air} - T_{\infty})[/tex]
[tex]\dot Q = -(2.670\,\frac{W}{m^{2}\cdot K} )\cdot (6\,m)\cdot (10\,m)\cdot (7\,K)[/tex]
[tex]\dot Q = -1121.4\,W[/tex]
b) The rate of heat loss is directly proportional to mass flow, and mass flow is directly proportional with speed. Then, the heat rate is found by simple rule of three:
[tex]\dot Q = \frac{84\,\frac{km}{h} }{42\,\frac{km}{h} }\cdot (-1121.4\,W)[/tex]
[tex]\dot Q = -2242.8\,W[/tex]
