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

Short answer

Answer: The new average temperature of the outer surface would be 31.09°C if the air velocity is doubled.

Step-by-step solution

Calculate Heat Transfer Rate

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

Write the Convective Heat Transfer Equation

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).

Calculate the Reynolds and Prandtl Numbers

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

Calculate the Nusselt Number and Heat Transfer Coefficient

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)

Calculate the Average Outer Surface Temperature

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.

Calculate the New Average Outer Surface Temperature if the Air Velocity is Doubled

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.

Key concepts

Convective Heat Transfer

Convective heat transfer is a fundamental concept in the study of thermodynamics and fluid dynamics, involving the movement of heat due to the bulk motion of fluids. In everyday language, it's like feeling the breeze on your skin when standing next to a fan on a warm day. The faster the air moves, the more heat it carries away from your skin, making you feel cooler. This process relies heavily on the heat transfer coefficient, denoted as 'h', which quantifies how effectively heat is transferred from the surface to the air.

In practical scenarios, like the one described in the exercise, the convective heat transfer equation is used: \[ Q = h \times A \times (T_{\text{outer}} - T_{\text{air}}) \]where:

• \(Q\) is the heat transfer rate,
• \(A\) is the surface area,
• \(T_{\text{outer}}\) and \(T_{\text{air}}\) are the outer surface and air temperatures, respectively.

This equation shows the relation between several physical factors and convective heat transfer, highlighting how both the area exposed to the air and the temperature difference between a surface and the air contribute to how effectively heat is transported away from the body.

Reynolds Number

The Reynolds number, often abbreviated as \(Re\), is a dimensionless quantity that helps determine the flow regime of a fluid. It provides insight into whether the flow will be laminar (smooth and orderly) or turbulent (chaotic). In the context of the problem, it is calculated using the equation:\[ Re = \frac{\rho \times V \times D}{\mu} \]where:

• \(\rho\) is the fluid density,
• \(V\) is the fluid velocity,
• \(D\) is the characteristic length (often diameter for cylindrical objects),
• \(\mu\) is the dynamic viscosity.

A low Reynolds number implies laminar flow, while a high Reynolds number suggests turbulent flow. In our exercise, increasing the air velocity leads to a higher Reynolds number. This transition indicates more mixing of the air layers, enhancing the convective heat transfer effectiveness.

Nusselt Number

The Nusselt number (\(Nu\)) is another crucial dimensionless number in heat transfer, representing the enhancement of heat transfer through convection as compared to conduction. It acts as a bridge between the physical behavior of the problem and the empirical correlations used to predict heat transfer rates.The equation used for estimating \(Nu\) in forced convection problems like ours is often the Dittus-Boelter correlation:\[ Nu = 0.023 \times Re^{0.8} \times Pr^{0.3} \]The Nusselt number helps determine the heat transfer coefficient \(h\), connecting the convective heat transfer with easily measurable quantities such as air velocity and temperature. A higher Nusselt number corresponds to increased convective heat transfer, confirming the fan's effectiveness in cooling the person's body when the air velocity is increased.

Prandtl Number

The Prandtl number (\(Pr\)) is a dimensionless number that indicates the relative thickness of the velocity and thermal boundary layers in a fluid flow. It is defined by the relation:\[ Pr = \frac{c_{\text{p}} \times \mu}{k} \]where:

• \(c_{\text{p}}\) is the specific heat capacity at constant pressure,
• \(\mu\) is the dynamic viscosity,
• \(k\) is the thermal conductivity of the fluid.

In simpler terms, the Prandtl number tells how fast thermal diffusivity compares with momentum diffusivity (or viscosity). It plays a significant role in the Nusselt number calculation, serving as an essential factor in determining the heat transfer coefficient. For most gases, including air, the Prandtl number remains constant over a wide range of temperatures, making it a handy value for simplifying heat transfer calculations in practical applications like the one in the exercise.