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 air flow. 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 \(100^{\circ} \mathrm{F}\).

Since the person is generating heat at a rate of 300 Btu/h, we can convert this to Watts (W) by using the conversion factor 1 Btu/h = 0.293071 W. Thus, the heat transfer rate, Q, will be: Q = 300 Btu/h * 0.293071 W/(Btu/h) = 87.9213 W

We know that the convective heat transfer can be expressed as: Q = h * A * (T_outer - T_air), where h is the heat transfer coefficient, A is the exposed surface area, T_outer is the average outer surface temperature, and T_air is the air temperature. The heat transfer coefficient (h) can be calculated using empirical correlations for forced convective heat transfer, such as the Dittus-Boelter correlation: Nu = 0.023 * Re^(0.8) * Pr^(0.3), where Nu is the Nusselt number, Re is the Reynolds number, and Pr is the Prandtl number. We are given the exposed surface area (A = 18 ft^2) and the air temperature (T_air = 85°F = 29.44°C). We will need to find the Reynolds number (Re), the Prandtl number (Pr), and the Nusselt number (Nu) to calculate the heat transfer coefficient (h).

To calculate the Reynolds number (Re), we need to use the equation: Re = (ρ * V * D) / μ, where ρ is the air density, V is the air velocity, D is the diameter of the cylinder (representing the person), and μ is the dynamic viscosity. We are given the air velocity (V = 6 ft/s = 1.8288 m/s) and the diameter (D = 1 ft = 0.3048 m). We need to find the density (ρ) and dynamic viscosity (μ) of the air at 100°F (37.78°C) from standard air properties tables or online resources. At 100°F (37.78°C), the properties of air are approximately: ρ = 1.146 kg/m^3 μ = 1.963 * 10^-5 kg/(m * s) Now we can calculate the Reynolds number (Re): Re = (1.146 * 1.8288 * 0.3048) / (1.963 * 10^-5) = 3,915.11 For the Prandtl number (Pr), we can use the specific heat (c_p), the dynamic viscosity (μ), and the thermal conductivity (k) of the air. Pr = (c_p * μ) / k At 100°F (37.78°C), the properties of air are approximately: c_p = 1006.43 J/(kg*K) k = 0.0281 W/(m*K) Thus, the Prandtl number (Pr) is: Pr = (1006.43 * 1.963 * 10^-5) / 0.0281 = 0.7098

Now we can calculate the Nusselt number (Nu) using the Dittus-Boelter correlation: Nu = 0.023 * Re^(0.8) * Pr^(0.3) = 0.023 * (3915.11)^(0.8) * (0.7098)^(0.3) = 94.26 The heat transfer coefficient (h) can be calculated by multiplying the Nusselt number (Nu) by the thermal conductivity (k) and dividing by the diameter (D): h = (Nu * k) / D = (94.26 * 0.0281) / 0.3048 = 8.7144 W/(m^2*K)

Now we can use the convective heat transfer equation to calculate the average outer surface temperature (T_outer): Q = h * A * (T_outer - T_air) T_outer = (Q / (h * A)) + T_air = (87.9213 / (8.7144 * 1.6729)) + 29.44 = 34.11°C So, the average temperature of the outer surface is 34.11°C.

If the air velocity is doubled (V = 3.6576 m/s), we will need to recalculate the Reynolds number, Nusselt number, and heat transfer coefficient. New Reynolds number: Re_new = (1.146 * 3.6576 * 0.3048) / (1.963 * 10^-5) = 7,830.22 New Nusselt number: Nu_new = 0.023 * Re_new^(0.8) * Pr^(0.3) = 0.023 * (7830.22)^(0.8) * (0.7098)^(0.3) = 175.13 New heat transfer coefficient: h_new = (Nu_new * k) / D = (175.13 * 0.0281) / 0.3048 = 16.2238 W/(m^2*K) New average outer surface temperature: T_outer_new = (Q / (h_new * A)) + T_air = (87.9213 / (16.2238 * 1.6729)) + 29.44 = 31.09°C If the air velocity is doubled, the new average temperature of the outer surface would be 31.09°C.