A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors where it is heated. Air enters a \(900-\mathrm{W}\) hair dryer at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\), and leaves at \(50^{\circ} \mathrm{C}\). The cross-sectional area of the hair dryer at the exit is \(60 \mathrm{~cm}^{2}\). Neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer, determine \((a)\) the volume flow rate of air at the inlet and \((b)\) the velocity of the air at the exit.

Short Answer

Expert verified
The temperature at the inlet is 25°C and the exit is 50°C, and the cross-sectional area at the exit is 60 cm². The pressure at the inlet is 100 kPa. **Answer:** To find the volume flow rate at the inlet and the exit velocity of the air in the hair dryer, follow these steps: 1. Calculate the mass flow rate of the air using the power of the hair dryer, specific heat of air, and temperature difference between the inlet and exit points. 2. Determine the air density at the inlet using the ideal gas equation and then calculate the volume flow rate at the inlet using the mass flow rate and air density. 3. Find the exit velocity of the air using the cross-sectional area of the hair dryer at the exit and the volume flow rate at the inlet.

Step by step solution

01

Find the mass flow rate of air

Using the power of the hair dryer and the specific heat of air, we can find the mass flow rate of the air passing through the electric resistors. We will use the energy balance equation in the form of power to find the mass flow rate: $$ P = \dot{m} \cdot c_p \cdot (T_{out} - T_{in}) $$ Where, \(P = 900\,\text{W}\), \(\dot{m}\) is the mass flow rate, \(c_p = 1005\,\text{J/kgK}\) is the specific heat at constant pressure for air, \(T_{in} = 25^{\circ}\text{C}\) , \(T_{out} = 50^{\circ}\text{C}\). Solving for \(\dot{m}\), we get: $$ \dot{m} = \frac{P}{c_p \cdot (T_{out} - T_{in})} $$
02

Calculate the volume flow rate at the inlet

Next, we will find the volume flow rate at the inlet using the mass flow rate and the air density. The ideal gas equation will be used to find the density at the inlet: $$ \rho_{in} = \frac{P_{in}}{R \cdot T_{in}} $$ Where, \(P_{in} = 100\,\text{kPa}\), \(R = 287\,\text{J/kgK}\) is the specific gas constant for air, \(T_{in} = 25^{\circ}\text{C} = 298\,\text{K}\). After calculating the density, we can find the volume flow rate at the inlet as follows: $$ \dot{V}_{in} = \frac{\dot{m}}{\rho_{in}} $$
03

Calculate the exit velocity of air

Finally, using the cross-sectional area at the exit, we will find the velocity of air at the exit. The exit velocity can be found using the following equation: $$ v_{out} = \frac{\dot{V}_{in}}{A_{out}} $$ Where, \(A_{out} = 60\,\text{cm}^2 = 0.006\,\text{m}^2\) is the cross-sectional area of the hair dryer at the exit. After calculating these values, we will have the volume flow rate of air at the inlet and the velocity of air at the exit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Balance Equation
The energy balance equation is fundamental in understanding how energy is conserved in a physical system. It's an expression of the first law of thermodynamics, which states that energy cannot be created or destroyed, only converted from one form to another. In the context of the hair dryer example, we apply the energy balance to calculate the mass flow rate of air.

An electric hair dryer converts electrical energy into heat, which is transferred to the air passing over the resistors. The equation for power in terms of the energy balance is:
\[\begin{equation} P = \.{m} \cdot c_p \cdot (T_{\text{out}} - T_{\text{in}})\end{equation}\]
where:
  • P represents the power input (in Watts),
  • \dot{m} is the mass flow rate of the air (in kg/s),
  • c_p is the specific heat at constant pressure (in J/kg K), indicates how much energy is needed to raise the temperature of 1 kg of air by 1 degree Kelvin,
  • T_{\text{in}} and T_{\text{out}} are the inlet and exit temperatures of the air (in degrees Celsius), converted to Kelvin in calculations.
It's important to note that neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer means we assume all the electrical energy is being used to heat the air. This simplified assumption helps focus on understanding the relationship between power and the mass flow rate of air, which is crucial for determining the correct operation of thermal systems like the hair dryer.
Mass Flow Rate
Mass flow rate is a measure of how much mass of a substance passes through a given surface per unit time. It is an essential concept in both engineering and physics, particularly when analyzing fluid systems like our hair dryer scenario. In the equation given in the energy balance section, mass flow rate (\(\dot{m}\)) can be determined when we have the power input and can calculate the temperature change:
\[\begin{equation} \dot{m} = \frac{P}{c_p \cdot (T_{\text{out}} - T_{\text{in}})}\end{equation}\]
In practical terms, this value tells us how much air the hair dryer moves and heats up on a continuous basis. A higher mass flow rate would imply more air is being heated, thus potentially drying hair faster. By manipulating the mass flow rate, manufacturers can design hair dryers for different use cases, such as a quicker dry or a slower, more controlled heat application.

Understanding mass flow rate is also key in various applications beyond hair dryers. It is used in designing cooling systems, calculating fuel consumption in engines, and even determining the dosage in drug delivery systems. It links directly to the conservation of mass, another principle that, along with energy conservation, dictates the behavior of physical systems.
Ideal Gas Equation
The ideal gas equation is a cornerstone of thermodynamics, providing a link between the pressure, volume, temperature, and amount of gas. While real gases do not always behave ideally, the ideal gas law is a useful approximation for gases at low pressure and high temperature, similar to the conditions in our hair dryer example:
\[\begin{equation} PV = nRT\end{equation}\]
This relationship simplifies for a constant amount of gas to:
\[\begin{equation} \rho = \frac{P}{RT}\end{equation}\]
where \(\rho\) represents the density of the gas, \(P\) is the pressure, \(R\) is the specific gas constant, and \(T\) is the absolute temperature in Kelvin.

By using the ideal gas equation, we can determine the air density at the inlet of the hair dryer, which, combined with the mass flow rate, helps us find the volume flow rate. Here's the density calculation based on the equation:\[\begin{equation} \rho_{\text{in}} = \frac{P_{\text{in}}}{R \cdot T_{\text{in}}}\end{equation}\]
With this equation, we have a powerful tool to analyze not only heating and cooling systems but also to understand atmospheric conditions, calculate buoyancy, and even design pneumatic systems. Whether studying the weather or designing an airplane's cabin pressure system, the ideal gas law is widely applied and critical to predicting gas behavior.

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Most popular questions from this chapter

Heat treatment is common in processing of semiconductor material. A 200-mm- diameter silicon wafer with thickness of \(725 \mu \mathrm{m}\) is being heat treated in a vacuum chamber by infrared heater. The surrounding walls of the chamber have a uniform temperature of \(310 \mathrm{~K}\). The infrared heater provides an incident radiation flux of \(200 \mathrm{~kW} / \mathrm{m}^{2}\) on the upper surface of the wafer, and the emissivity and absorptivity of the wafer surface are \(0.70\). Using a pyrometer, the lower surface temperature of the wafer is measured to be \(1000 \mathrm{~K}\). Assuming there is no radiation exchange between the lower surface of the wafer and the surroundings, determine the upper surface temperature of the wafer. (Note: A pyrometer is a non-contacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.

A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -hour period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\).

What is asymmetric thermal radiation? How does it cause thermal discomfort in the occupants of a room?

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