/*! 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 58 The brake shoe and steel drum of... [FREE SOLUTION] | 91Ó°ÊÓ

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The brake shoe and steel drum of a car continuously absorb \(75 \mathrm{~W}\) as the car slows down. Assume a total outside surface area of \(0.1 \mathrm{~m}^{2}\) with a convective heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) to the air at \(20^{\circ} \mathrm{C}\). How hot does the outside brake and drum surface become when steady conditions are reached?

Short Answer

Expert verified
The outside surface temperature reaches \( 95^{\circ} \mathrm{C} \) at steady state.

Step by step solution

01

Identify Energy Balance Equation

In steady conditions, the power input into the brake system by friction, which is absorbed as heat, must equal the power dissipated into the surrounding air by convection. This means: \( Q_{in} = Q_{out} \). Where \( Q_{in} = 75 \mathrm{~W} \) and \( Q_{out} = h A (T_{s} - T_{ ext{air}}) \). Here, \( h \) is the convective heat transfer coefficient, \( A \) is the surface area, \( T_{s} \) is the surface temperature, and \( T_{ ext{air}} \) is the air temperature.
02

Set Heat Input Equal to Heat Loss

Use the equation for power lost by convection: \[ Q_{out} = hA(T_s - T_{air}) \]. Set \( Q_{in} \) as equal to \( Q_{out} \): \( 75 = 10 \times 0.1 \times (T_s - 20) \).
03

Solve for Temperature \( T_s \)

Rearrange the equation from Step 2: \( T_s - 20 = \frac{75}{1} = 75 \). Therefore, \( T_s = 75 + 20 = 95 \).
04

Conclude with Outside Surface Temperature

The solution shows that the outside surface temperature \( T_s \) of the brake and drum, when in steady conditions, becomes \( 95^{\circ} \mathrm{C} \).

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Key Concepts

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

Energy Balance Equation
The concept of the Energy Balance Equation is crucial when dealing with systems like the braking system of a car. Imagine this equation as a balance scale where the energy entering the system must exactly match the energy leaving it, especially in steady-state situations. For our exercise, the braking system absorbs energy due to friction; let's call this energy absorption "Power Input" (\(Q_{in}\)). This power input heats the system every time the car slows down.
In order for the system to remain in equilibrium, this heat must be released into the surrounding air. This process of releasing energy is termed "Power Dissipation" (\(Q_{out}\)). Our balance equation thus reads: \(Q_{in} = Q_{out}\).
  • \(Q_{in}\) represents the power absorbed, or energy input, which is \(75 \text{ W}\) in our example.
  • \(Q_{out}\) represents the power dissipated, or energy output, via cooling down.
By ensuring these two equal in steady conditions, we can determine other characteristics of the system like external temperatures.
Surface Temperature
Surface temperature measures how hot the brake shoe and steel drum's surface become as energy is continually transferred to the surrounding air. In our case, the surface temperature depends on several factors, such as the convective heat transfer coefficient \(h\), the surface area \(A\), and the ambient air temperature \(T_{air}\).
To find the surface temperature, the equation used is: \[Q_{out} = hA(T_s - T_{air})\]
  • Where \(T_s\) is the surface temperature we want to find.
  • \(T_{air}\) is the air temperature, given as \(20^{\circ} \text{C}\) in the problem.
Our task is to rearrange this equation to solve for \(T_s\). Solving step-by-step, we can see that \[75 = 10 \times 0.1 \times (T_s - 20)\]This process tells us that the surface, at steady state, reaches \(95^{\circ} \text{C}\). This method shows how surface temperature is a point of equilibrium.
Power Dissipation
Power dissipation highlights how energy is expended from the system, particularly how the heat from brakes passes to the air. In simple terms, the energy absorbed by the braking mechanism is shed to prevent the system from overheating, maintaining safety and efficiency.
In our scenario, the power dissipation happens through a process called convection. Convection involves the transfer of heat between the surface (break system) and the fluid surrounding it (air) without direct contact. The efficacy of this energy transfer is encapsulated in the convective heat transfer coefficient \(h\), which influences how effectively the surface becomes cooler.
Convective heat transfer not only defines the energy transformation from the break to the air but also affects steady-state temperatures. By ensuring the calculated \(Q_{out}\) matches the \(Q_{in}\), we maintain a desired surface condition. This illustrates that competent heat dissipation can prevent potential overmounting temperatures in engineered systems.

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Most popular questions from this chapter

Two rigid insulated tanks are connected with a pipe and valve. One tank has \(0.5 \mathrm{~kg}\) air at \(200 \mathrm{kPa}\), \(300 \mathrm{~K}\) and the other has \(0.75 \mathrm{~kg}\) air at \(100 \mathrm{kPa}\) \(400 \mathrm{~K}\). The valve is opened and the air comes to a single uniform state without any heat transfer. How do you determine the final temperature and pressure?

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