Question

For heat transfer purposes, a standing man can be modeled as a 30 -cm diameter, 175 -cm long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of \(34^{\circ} \mathrm{C} .\) For a convection heat transfer coefficient of \(10 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C},\) determine the rate of heat loss from this man by convection in an environment at \(20^{\circ} \mathrm{C}\).

Short answer

Answer: The rate of heat loss from the man by convection is approximately 231 W.

Step-by-step solution

Identify the given information

We are given the following information: - Diameter of the cylinder: \(30 \,\text{cm}\) - Length (height) of the cylinder: \(175 \,\text{cm}\) - Side surface average temperature: \(34^{\circ} \mathrm{C}\) - Convection heat transfer coefficient: \(10 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) - Environment temperature: \(20^{\circ} \mathrm{C}\) We need to determine the rate of heat loss from the side surface of the cylinder by convection.

Calculate the surface area of the cylinder

The formula for the lateral surface area of a cylinder is: \(A = 2\pi rh\) Where \(r\) is the radius and \(h\) is the height. First, convert the diameter and length to meters: - Diameter of the cylinder: \(0.3 \,\text{m}\) - Length of the cylinder: \(1.75 \,\text{m}\) Now calculate the radius of the cylinder: \(r = \frac{0.3}{2} = 0.15 \,\text{m}\) Then, we can find the surface area of the cylinder: \(A = 2\pi (0.15) (1.75) \approx 1.65 \,\text{m}^{2}\)

Calculate the rate of heat loss

The formula for convective heat transfer is: \(Q = hA(T_{s} - T_{\infty})\) Where \(Q\) is the rate of heat loss, \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, \(T_{s}\) is the side surface temperature, and \(T_{\infty}\) is the environment temperature. Using the given values, \(Q = (10 \,\mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C})(1.65 \,\text{m}^{2})(34^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C})\) \(Q \approx 231 \,\mathrm{W}\) The rate of heat loss from the man by convection is approximately 231 W.

Key concepts

Heat Transfer Coefficient

The heat transfer coefficient is a measure of a material's ability to conduct heat. It is denoted by the symbol 'h' and expressed in units of watts per square meter per degree Celsius (W/m²·°C). This coefficient quantifies how well heat is transferred between a surface and a fluid moving past that surface. In our example, the heat transfer coefficient of 10 W/m²·°C indicates how effectively heat is being lost from the man modeled as a cylinder to the cooler surrounding air. The higher the heat transfer coefficient, the more efficient the heat transfer process is, and vice versa. This coefficient is essential in calculations as it directly affects the rate of heat loss or gain through convective processes.

Understanding the heat transfer coefficient is essential because it involves material properties, surface characteristics, and the dynamics of the fluid flow. In practical scenarios, this coefficient can vary depending on several factors, including fluid velocity, viscosity, and the nature of the surface (smooth or rough).

Surface Area Calculation

The surface area calculation is crucial in determining how much area is available for heat exchange between the body (in our case, the 'man' modeled as a cylinder) and the environment. The formula used for calculating the lateral surface area of a cylinder is \[ A = 2\times\pi\times r\times h \]where 'A' stands for the area, 'r' for the radius, and 'h' for the height of the cylinder. In heat transfer problems, it is important to convert all measurements to the same units, typically meters, before calculation. It is the surface area that directly interacts with the surrounding air, and therefore it plays a pivotal role in the rate of convective heat exchange. By multiplying this surface area with the heat transfer coefficient and the difference in temperature between the surface and the surrounding air, we can determine the heat energy transferred per unit of time.

Thermal Energy Loss

Thermal energy loss, often referred to simply as heat loss, is the process by which thermal energy is transferred from a warmer object to a cooler environment. The rate of heat loss, represented by 'Q', measures the amount of heat energy (in watts, W) that the body loses over time. This calculation is particularly important in understanding energy efficiency and thermal comfort.

In our example, the formula for convective heat transfer,\[ Q = h\times A\times (T_{s} - T_{\infty}) \]is used, where 'T_s' is the surface temperature, and 'T_∞' is the ambient temperature. This formula enables us to calculate the thermal energy loss of the man modeled as a cylinder, giving us insights into how much heat he is losing to his surroundings and how this might impact his thermal comfort. The concept of thermal energy loss is widely applied in various fields such as building design, clothing insulation, and even in medical applications like hypothermia treatment.

Cylinder Surface Area

The surface area of a cylinder is a vital aspect when analyzing convective heat transfer in cylindrical objects. It refers to the area of the cylinder's outside surface that is exposed to the surrounding environment. For vertical cylinders with insulated top and bottom like the model of the standing man, we only consider the lateral surface area for heat exchange.

The lateral surface area 'A' of a cylinder is found by the formula:\[ A = 2\pi rh \]This is derived from the fact that when a cylinder is 'unrolled', the side surface forms a rectangle with 'h' as the height and the circumference of the base, \( 2\pi r \), as the length. Calculating the cylinder's surface area accurately is crucial for determining the thermal energy loss in heat transfer problems. This area acts as the 'gateway' for heat moving from the cylinder to the surrounding air and therefore directly influences how much heat is lost over time.