/*! 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 27 The temperature controller for a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wall-mounted insulation pad. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The required heater power when the desired dryer air temperature is \(50^{\circ} \mathrm{C}\) is \(P_e = 0.015~\text{W}\).

Step by step solution

01

Calculate the Heat Transfer Rate Between Air and Exposed Switch Surface

First, calculate the heat transfer rate (\(Q\)) between the air and the exposed switch surface using the convection heat transfer equation: \[ Q = hA(T_\text{set} - T_\infty) \] Where: \(Q\) is the heat transfer rate, \(h\) is the convection heat transfer coefficient (\(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), \(A\) is the exposed surface area of the switch (\(30 \cdot 10^{-6} \mathrm{~m}^{2}\)), \(T_\text{set}\) is the switch set temperature (\(70^{\circ} \mathrm{C}\)), \(T_\infty\) is the desired dryer air temperature (\(50^{\circ} \mathrm{C}\)). Plug in the values to get the heat transfer rate: \[ Q = 25 \cdot 30 \cdot 10^{-6} (70 - 50) \]
02

Equate the Heat Transfer Rate with the Heater Power to Find the Required Heater Power

Next, equate the heat transfer rate calculated in step 1 with the heater power (\(P_e\)), since the heat transferred from the switch to air must be equal to the power supplied to the heater: \[ P_e = Q \] Now, plug in the value for the heat transfer rate: \[ P_e = 25 \cdot 30 \cdot 10^{-6} (70 - 50) \] Calculate the heater power: \[ P_e = 25 \cdot 30 \cdot 10^{-6} (20) \] \[ P_e = 0.015 ~\text{W} \] Therefore, the required heater power when the desired dryer air temperature is \(50^{\circ} \mathrm{C}\) is \(P_e = 0.015~\text{W}\).

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.

Understanding a Bimetallic Switch
A bimetallic switch is a crucial component in temperature control systems, like the one found in a clothes dryer. It consists of two different metals, fused together, each with a distinct coefficient of thermal expansion. When the temperature changes, the metals expand at different rates, causing the strip to bend. This bending action can open or close a circuit, controlling the operation of appliances.

Bimetallic switches are designed to respond at specific temperatures, making them effective in maintaining a set temperature range. In the context of a clothes dryer, the switch opens at the desired maximum air temperature, ensuring the dryer does not exceed this limit. As temperature is a vital factor in drying processes, understanding how bimetallic switches function helps in controlling and optimizing appliance performance.

By regulating appliance operation the switch offers safety and efficiency in temperature-sensitive environments, preventing overheating. These switches provide a simple yet reliable method of temperature control that is utilized in various devices beyond clothes dryers.
What is Heat Transfer Coefficient?
The heat transfer coefficient, often denoted as "h," measures a material's capacity to transfer heat through convection. It is a crucial factor in determining how quickly heat is exchanged between a surface and the fluid (such as air) moving over it. The unit of measurement is typically \(\text{W/m}^2 \cdot \text{K}\), and it varies depending on factors like fluid properties, flow characteristics, and the nature of the surface.

In the exercise about the dryer switch, the heat transfer coefficient is given as 25 \(\text{W/m}^2 \cdot \text{K}\). This indicates how well heat is transferred between the air and the switch surface. Knowing this coefficient is crucial for calculating the required heater power; it helps determine the rate at which heat needs to be added or removed to maintain desired temperatures.

A higher heat transfer coefficient suggests better heat transfer capabilities, meaning quicker temperature changes. In practical applications, understanding the heat transfer coefficient contributes to designing more efficient heating and cooling systems, influencing everything from household appliances to industrial operations.
Heater Power Calculation Explained
Calculating the heater power required to achieve a particular temperature involves using the convection heat transfer equation. This ensures that the heat produced by the heater compensates for heat loss to maintain the desired temperature setting.

The formula used is:
  • \(Q = hA(T_\text{set} - T_\infty)\)
Where:
  • \(Q\) is the heat transfer rate,
  • \(h\) is the heat transfer coefficient,
  • \(A\) is the surface area,
  • \(T_\text{set}\) is the desired switch temperature,
  • \(T_\infty\) is the current air temperature.
In the given exercise, equating this to the heater power forms the basis for the calculation:
  • \(P_e = Q\)
Plugging in the appropriate values for area, temperature differences, and the heat transfer coefficient, we derive the power needed by the heater. In this case, calculations show that only a small amount of power, 0.015 Watts, is necessary.
Understanding heater power calculations ensures effective energy management, allowing devices to operate at optimal efficiency while minimizing energy waste. This is invaluable in energy conservation efforts, reducing operational costs and environmental impact.

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

Most of the energy we consume as food is converted to thermal energy in the process of performing all our bodily functions and is ultimately lost as heat from our bodies. Consider a person who consumes \(2100 \mathrm{kcal}\) per day (note that what are commonly referred to as food calories are actually kilocalories), of which \(2000 \mathrm{kcal}\) is converted to thermal energy. (The remaining \(100 \mathrm{kcal}\) is used to do work on the environment.) The person has a surface area of \(1.8 \mathrm{~m}^{2}\) and is dressed in a bathing suit. (a) The person is in a room at \(20^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). At this air temperature, the person is not perspiring much. Estimate the person's average skin temperature. (b) If the temperature of the environment were \(33^{\circ} \mathrm{C}\), what rate of perspiration would be needed to maintain a comfortable skin temperature of \(33^{\circ} \mathrm{C}\) ?

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

An overhead 25-m-long, uninsulated industrial steam pipe of \(100-\mathrm{mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{\circ} \mathrm{C}\), and the coefficient associated with natural convection is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steam is generated in a gas-fired boiler operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02\) per \(\mathrm{MJ}\), what is the annual cost of heat loss from the line?

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

A \(50 \mathrm{~mm} \times 45 \mathrm{~mm} \times 20 \mathrm{~mm}\) cell phone charger has a surface temperature of \(T_{s}=33^{\circ} \mathrm{C}\) when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity \(\varepsilon=0.92\) and is subject to a free convection heat transfer coefficient of \(h=4.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The room air and wall temperatures are \(T_{\infty}=22^{\circ} \mathrm{C}\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), respectively. If electricity costs \(C=\$ 0.18 / \mathrm{kW} \cdot \mathrm{h}\), determine the daily cost of leaving the charger plugged in when not in use.

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.