Problem 4
Question
On a hot day, a fan is blowing warm air over a person. The air is at a temperature of \(45^{\circ} \mathrm{C}\) and the person can be assumed to be represented as a flat plate with dimensions of \(1 \mathrm{~m} \times 0.2 \mathrm{~m}\) having a surface temperature of \(28^{\circ} \mathrm{C}\). If the convection heat transfer coefficient is found to be \(100 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), determine the rate of heat transfer from the hot air to the person.
Step-by-Step Solution
Verified Answer
The rate of heat transfer is 340 W.
1Step 1: Identify Known Values
First, let's identify and list all the known values from the problem. The air temperature \(T_{\infty}\) is \(45^{\circ} \mathrm{C}\), the surface temperature \(T_s\) is \(28^{\circ} \mathrm{C}\), and the convection heat transfer coefficient \(h\) is \(100 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). The surface area of the person (flat plate) is \(1 \mathrm{~m} \times 0.2 \mathrm{~m} = 0.2 \mathrm{~m}^2\).
2Step 2: Determine the Temperature Difference
The temperature difference between the hot air and the surface of the person is the driving force for heat transfer. We calculate this as \(\Delta T = T_{\infty} - T_s = 45^{\circ} \mathrm{C} - 28^{\circ} \mathrm{C} = 17^{\circ} \mathrm{C}\).
3Step 3: Calculate the Rate of Heat Transfer
The rate of heat transfer \(\dot{Q}\) can be calculated using the formula for convective heat transfer: \[ \dot{Q} = h \cdot A \cdot \Delta T \] Substituting the given values, we get:\[ \dot{Q} = 100 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} \times 0.2 \mathrm{~m}^2 \times 17^{\circ} \mathrm{C} = 340 \mathrm{~W} \]
Key Concepts
Heat Transfer CalculationTemperature DifferenceHeat Transfer Coefficient
Heat Transfer Calculation
Heat transfer calculation is essential to understanding how thermal energy moves from one object to another. When it comes to convection heat transfer, there's a particular equation that governs the process. This equation is \[ \dot{Q} = h \cdot A \cdot \Delta T \]where \(\dot{Q}\) is the rate of heat transfer, \(h\) is the heat transfer coefficient, \(A\) is the area through which heat is being transferred, and \(\Delta T\) is the temperature difference across that area.
- To perform these calculations, start by identifying the known values such as the heat transfer coefficient, the surface area, and the temperature difference.
- Substitute these into the equation and solve for the rate of heat transfer.
Temperature Difference
Temperature difference (\(\Delta T\)) is a key player in heat transfer scenarios. It represents the thermal gradient driving the transfer of heat. The greater the difference in temperature between two objects or environments, the more heat will flow between them naturally. It's important to remember:
- Temperature difference is calculated simply by subtracting the surface temperature from the temperature of the surrounding air or fluid.
- This difference is often expressed in Celsius or Kelvin.
Heat Transfer Coefficient
The heat transfer coefficient ( \(h\) ) is a measure of the convective heat transfer ability. Essentially, it signifies how readily heat transfers between an object and a fluid or air flowing around it.
- The larger the heat transfer coefficient, the more efficient the heat transfer will be.
- This coefficient is dependent on the properties of the fluid (such as viscosity and thermal conductivity) and the nature of the surface interaction.
Other exercises in this chapter
Problem 2
A person who has experienced hypothermia is heated steadily from an external source. The thermal conductivity of skin \(0.4 \mathrm{~W} / \mathrm{m}^{\circ} \ma
View solution Problem 3
A skin sample has a heat flux meter attached to it in a laboratory. The meter indicates that a heat flux of \(45 \mathrm{~W} / \mathrm{m}^{2}\) passes through a
View solution Problem 5
Consider an average person standing in the middle of a room. The various surfaces are found to be at a temperature of \(15^{\circ} \mathrm{C}\) in the winter an
View solution Problem 9
A sunbather who loses no heat through the surface pointing away from the sun is exposed to solar radiation at a rate of \(750 \mathrm{~W} / \mathrm{m}^{2}\). Th
View solution