/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A hair dryer is basically a duct... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.

Solve this system of two equations with two unknowns using EES: $$ \begin{aligned} x^{3}-y^{2} &=10.5 \\ 3 x y+y &=4.6 \end{aligned} $$

A hollow spherical iron container with outer diameter \(20 \mathrm{~cm}\) and thickness \(0.2 \mathrm{~cm}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). If the outer surface temperature is \(5^{\circ} \mathrm{C}\), determine the approximate rate of heat loss from the sphere, in \(\mathrm{kW}\), and the rate at which ice melts in the container. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

Air at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) blows over a pond. The surface temperature of the pond is at \(40^{\circ} \mathrm{C}\). Determine the heat flux between the surface of the pond and the air.

A 2.1-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at \(20^{\circ} \mathrm{C}\). Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be \(180^{\circ} \mathrm{C}\) in steady operation. Also, the voltage drop and electric current through the wire are measured to be \(110 \mathrm{~V}\) and \(3 \mathrm{~A}\), respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: \(156 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.