/*! 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 63 A \(50 \mathrm{~mm} \times 45 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
The daily cost of leaving the charger plugged in when not in use can be calculated by finding the total surface area (A) of the charger, then determining the heat loss by convection (Q_c) and heat loss via radiation (Q_r). Next, calculate the total power usage (P) by adding heat losses. Finally, calculate the energy usage in kilowatt-hours per day and multiply it by the cost of electricity ($0.18/kWh) to find the daily cost: \[\text{Daily cost} = \frac{P \times 24}{1000} \times 0.18\]

Step by step solution

01

Calculate the total surface area of the charger

To calculate the heat loss, first, we need to determine the total surface area of the charger. The charger has dimensions \(50 \mathrm{~mm} \times 45 \mathrm{~mm} \times 20 \mathrm{~mm}\), and it has six faces. Total surface area (A) = 2 × (Length × Width + Length × Height + Width × Height) \[A = 2 \times ((50 \times 45) + (50 \times 20) + (45 \times 20 ))\] Convert the area from \(\mathrm{mm^2}\) to \(\mathrm{m^2}\), by dividing by \(10^6\). \[A = \frac{2 \times ((50 \times 45) + (50 \times 20) + (45 \times 20 ))}{10^6}\]
02

Calculate the heat loss by convection from the charger surface

Calculate the heat loss by convection using the formula, \(Q_c = hA(T_s - T_\infty)\). Given, \(T_s = 33^{\circ} \mathrm{C}\), \(T_\infty = 22^{\circ} \mathrm{C}\), and \(h = 4.5 \mathrm{~W/m^2 \cdot K}\). \[Q_c = hA(T_s - T_\infty)\]
03

Calculate the heat loss via radiation from the charger surface

Calculate the heat loss by radiation using the formula, \(Q_r = \varepsilon \sigma A(T_s^4 - T_{sur}^4)\). Given, \(\varepsilon = 0.92\), \(T_{sur} = 20^{\circ} \mathrm{C}\), and \(\sigma = 5.67 \times 10^{-8} \mathrm{~W/m^2 \cdot K^4}\) (Stefan-Boltzmann constant). \[Q_r = \varepsilon \sigma A(T_s^4 - T_{sur}^4)\]
04

Calculate the total power usage of the charger

The total power usage by the charger (P) is equal to the sum of heat losses due to convection and radiation. \(P = Q_c + Q_r\)
05

Calculate the energy usage in kilowatt-hours per day

Calculate the energy usage in kilowatt-hours per day. Energy usage = \(\frac{Power \times Time}{1000}\), as 1 kilowatt = 1000 watts Given, time = 24 hours. Energy usage = \(\frac{P \times 24}{1000}\) kWh/day
06

Calculate the daily cost of leaving the charger plugged in

Given the cost of electricity, \(C = \$0.18/\mathrm{kWh}\), we can find the daily cost by multiplying the energy usage in kilowatt-hours per day with the cost of electricity. Daily cost = Energy usage × Cost of electricity Daily cost = \(\frac{P \times 24}{1000} \times 0.18)\) After calculating the above expression, we will get the daily cost of leaving the charger plugged in when not in use.

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.

Convection Heat Transfer
Convection heat transfer involves the movement of heat through a fluid, which could be a liquid or a gas. This transfer happens because of the fluid's motion across a solid surface, such as air moving over the surface of a cell phone charger.
This type of heat transfer is often characterized by a convection heat transfer coefficient, symbolized by the letter \(h\). The convection heat transfer coefficient depends on properties like fluid velocity and its viscosity. In many practical problems, including our exercise, it is provided as a constant value, simplifying calculations.

The mathematical expression used to calculate heat loss due to convection is:
  • \(Q_c = h \, A (T_s - T_\infty)\)
Where:
  • \(Q_c\) is the heat loss due to convection.
  • \(A\) is the surface area.
  • \(T_s\) is the surface temperature.
  • \(T_\infty\) is the ambient temperature.
By using this formula, we can determine how much heat is being transferred from the charger's surface to the surrounding air.
Radiation Heat Transfer
Radiation heat transfer is a process where heat energy is emitted from a surface in the form of electromagnetic waves.
This type of transfer can occur in the absence of a medium, meaning heat can be transferred through a vacuum. This is different from conduction and convection, which require a medium.

In our exercise, heat loss through radiation is calculated using the formula:
  • \(Q_r = \varepsilon \sigma \ A (T_s^4 - T_{sur}^4)\)
Where:
  • \(Q_r\) is the heat loss by radiation.
  • \(\varepsilon\) is the emissivity of the surface.
  • \(\sigma\) is the Stefan-Boltzmann constant.
  • \(T_s\) and \(T_{sur}\) are the temperature of the surface and the surroundings, respectively.
The formula highlights the relation between the power emitted and the temperature difference elevated to the power of four, emphasizing that even small changes in temperature can significantly affect radiation heat transfer.
Surface Emissivity
Surface emissivity, denoted as \(\varepsilon\), is a measure of a material's ability to emit energy by radiation. Values of emissivity range between 0 and 1, where 1 represents a perfect black body that emits all incident energy.

Real-world surfaces often have emissivity values less than 1, depending on their material properties, surface texture, and temperature. For instance, in our exercise, the emissivity of the cell phone charger is given as 0.92.

A high emissivity value means that the surface is a good emitter of radiation, releasing more heat than surfaces with lower emissivity under similar conditions. The value of emissivity plays a crucial role in calculating radiation heat loss using the equation for \(Q_r\). Consequently, selecting a material with appropriate emissivity is vital in heat management applications like electronics cooling.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant, \(\sigma\), is a crucial element in radiation heat transfer calculations. It represents the proportionality constant in the Stefan-Boltzmann Law, which relates the total energy radiated per unit area of a black body to the fourth power of its temperature.
The value of \(\sigma\) is approximately \(5.67 \times 10^{-8} \, \mathrm{W/m^2\cdot K^4}\), and it helps quantify the energy emitted in terms of radiative heat loss.

This law and its constant are fundamental in understanding how bodies exchange heat through radiation. In our exercise, \(\sigma\) enables us to calculate the impact of temperature differences on the heat radiated by the cell phone charger. Understanding this constant allows engineers and scientists to predict how objects will behave thermally under various conditions, improving the design and efficiency of electronic devices.

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

A wall has inner and outer surface temperatures of 16 and \(6^{\circ} \mathrm{C}\), respectively. The interior and exterior air temperatures are 20 and \(5^{\circ} \mathrm{C}\), respectively. The inner and outer convection heat transfer coefficients are 5 and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

The concrete slab of a basement is \(11 \mathrm{~m}\) long, \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02 / \mathrm{MJ}\), what is the daily cost of the heat loss?

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.

Consider a carton of milk that is refrigerated at a temperature of \(T_{m \mathrm{r}}=5^{\circ} \mathrm{C}\). The kitchen temperature on a hot summer day is \(T_{\infty}=30^{\circ} \mathrm{C}\). If the four sides of the carton are of height and width \(L=200 \mathrm{~mm}\) and \(w=100 \mathrm{~mm}\), respectively, determine the heat transferred to the milk carton as it sits on the kitchen counter for durations of \(t=10 \mathrm{~s}, 60 \mathrm{~s}\), and \(300 \mathrm{~s}\) before it is returned to the refrigerator. The convection coefficient associated with natural convection on the sides of the carton is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(0.90\). Assume the milk carton temperature remains at \(5^{\circ} \mathrm{C}\) during the process. Your parents have taught you the importance of refrigerating certain foods from the food safety perspective. Comment on the importance of quickly returning the milk carton to the refrigerator from an energy conservation point of view.

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the roof under steadystate conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the same ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airflow conditions over the roof, increasing with increasing air speed. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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.