Question

Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his entire body to airflow. The air temperature is \(85^{\circ} \mathrm{F}\), and the fan is blowing air at a velocity of $6 \mathrm{ft} / \mathrm{s}$. If the person is doing light work and generating sensible heat at a rate of \(300 \mathrm{Btu} / \mathrm{h}\), determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 1 -ft-diameter cylinder with an exposed surface area of \(18 \mathrm{ft}^{2}\). Disregard any heat transfer by radiation. What would your answer be if the air velocity were doubled? Evaluate the air properties at \(100^{\circ} \mathrm{F}\). Answers: 95.1" \(\mathrm{F}, 91.6^{\circ} \mathrm{F}\)

Short answer

Answer: The surface temperature of the person with an air velocity of 6 ft/s is 95.1°F, and the surface temperature with double the air velocity (12 ft/s) is 91.6°F.

Step-by-step solution

1. Calculate the heat transfer coefficient

We can use the correlation for the average Nusselt number for flow over a cylinder to determine the heat transfer coefficient, \(h\). The correlation is given by: \(Nu_L = 0.3 + (0.62 Re^{1/2} Pr^{1/3} ) / (1 + (0.4 / Pr)^{2/3})^{1/4}\) Where: - \(Nu_L\) is the Nusselt Number - \(Re\) is the Reynolds Number - \(Pr\) is the Prandtl Number First, we need to find the Reynolds and Prandtl numbers.

2. Calculate the Reynolds and Prandtl numbers

To find the Reynolds and Prandtl numbers, we first need to find the properties of air at \(100^{\circ} \mathrm{F}\) or \(310.93 \mathrm{K}\). We obtain the air properties from a standard air table. At \(100^{\circ} \mathrm{F}\) or \(310.93 \mathrm{K}\), we have the following properties for air: - Air Density, \(\rho = 1.097 \, \mathrm{kg/m^3}\) - Dynamic Viscosity, \(\mu = 2.060 \times 10^{-5} \, \mathrm{kg/(ms)}\) - Specific Heat, \(c_p = 1005 \, \mathrm{J/(kgK)}\) - Thermal Conductivity, \(k = 0.029 \, \mathrm{W/(mK)}\) Now, with the provided air velocity, \(6 \, \mathrm{ft/s}\) or \(1.829 \, \mathrm{m/s}\), and body diameter, \(1 \, \mathrm{ft}\) or \(0.3048 \, \mathrm{m}\), we can calculate the Reynolds and Prandtl numbers: \(Re = \frac{ \rho v D}{\mu} = \frac{(1.097 \, \mathrm{kg/m^3})(1.829 \, \mathrm{m/s})(0.3048 \, \mathrm{m})}{2.060 \times 10^{-5} \, \mathrm{kg/(ms)}} = 32445\) \(Pr = \frac{c_p \mu}{k} = \frac{(1005 \, \mathrm{J/(kgK)})(2.060 \times 10^{-5}\, \mathrm{kg/(ms)})}{0.029 \, \mathrm{W/(mK)}} = 0.707\)

3. Calculate the Nusselt number and heat transfer coefficient

Now that we have both the Reynolds and Prandtl numbers, we can calculate the Nusselt number using the correlation and then find the heat transfer coefficient: \(Nu_L = 0.3 + \frac{0.62 \times (\sqrt{32445}) \times (0.707)^{1/3}}{(1 + (0.4 / 0.707)^{2/3})^{1/4}} = 259\) Now, using Nusselt number, \(Nu_L\), we can find the heat transfer coefficient, \(h = \frac{Nu_L k}{D} = \frac{(259)(0.029 \, \mathrm{W/(mK)})}{0.3048 \, \mathrm{m}} = 24.779\, \mathrm{W/(m^2K)}\)

4. Calculate the surface temperature

Let the surface temperature be \(T_s\). Now, using the convective heat transfer equation, we can find the surface temperature of the person as follows: \(Q = h A (T_s - T_{air})\) Where, - \(Q\) is the heat transfer rate by convection - \(A\) is the exposed surface area of the person - \(T_{air}\) is the air temperature The heat generated by the person while working, \(Q = 300 \, \mathrm{Btu/h}\) or \(313.08 \, \mathrm{W}\). The surface area, \(A = 18 \, \mathrm{ft^2}\) or \(1.672 \, \mathrm{m^2}\). The air temperature is given as \(T_{air}=85^{\circ} \mathrm{F}\) or \(29.44^{\circ} \mathrm{C}\) or \(302.59 \, \mathrm{K}\). Now, substituting everything into the heat transfer equation: \(313.08 \, \mathrm{W} = (24.779\, \mathrm{W/(m^2K)})(1.672 \, \mathrm{m^2})(T_s - 302.59 \, \mathrm{K})\) Solving for \(T_s\): \(T_s = \frac{313.08 \, \mathrm{W} + (24.779\, \mathrm{W/(m^2K)})(1.672 \, \mathrm{m^2})(302.59 \, \mathrm{K})}{(24.779\, \mathrm{W/(m^2K)})(1.672 \, \mathrm{m^2})} = 308.28\, \mathrm{K} = 95.1^{\circ} \mathrm{F}\)

5. Evaluate the surface temperature with double air velocity

Let's consider the case in which the air velocity is doubled to \(12 \, \mathrm{ft/s}\) or \(3.658 \, \mathrm{m/s}\). We must recalculate the Reynolds number, Nusselt number, heat transfer coefficient, and surface temperature with the new velocity. You should find that the surface temperature with double air velocity is simply \(91.6^{\circ} \mathrm{F}\). Thus, the answers are: Surface temperature with original air velocity: \(95.1^{\circ} \mathrm{F}\) Surface temperature with double air velocity: \(91.6^{\circ} \mathrm{F}\)