/*! 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 28 A house has an electric heating ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 28

A house has an electric heating system that consists of a \(300-W\) fan and an electric resistance heating element placed in a duct. Air flows steadily through the duct at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\) and experiences a temperature rise of \(5^{\circ} \mathrm{C}\). The rate of heat loss from the air in the duct is estimated to be \(250 \mathrm{~W}\). Determine the power rating of the electric resistance heating element.

Short Answer

Expert verified
Answer: The power rating of the electric resistance heating element is 2450 W.

Step by step solution

01

Calculate the heat gained by the air

First, let's find the rate of heat gain by the air. The specific heat capacity of air is approximately \(1000 \mathrm{~J/(kg \cdot K)}\). Using the mass flow rate (\(0.6 \mathrm{~kg/s}\)) and the temperature rise (\(5^{\circ} \mathrm{C}\)), we can calculate the rate of heat gain as follows: Rate of heat gain = mass flow rate 脳 specific heat capacity 脳 temperature rise Rate of heat gain = \((0.6 \mathrm{~kg/s}) \times (1000 \mathrm{~J/(kg \cdot K)}) \times (5 \mathrm{~K})\)
02

Calculate the rate of heat gain

Now, let's compute the rate of heat gain: Rate of heat gain = \((0.6 \mathrm{~kg/s}) \times (1000 \mathrm{~J/(kg \cdot K)}) \times (5 \mathrm{~K})\) Rate of heat gain = \(3000 \mathrm{~W}\)
03

Consider the rate of heat loss and power consumed by the fan

Next, we have to take into account the rate of heat loss from the air in the duct (\(250 \mathrm{~W}\)) as well as the power consumed by the fan (\(300 \mathrm{~W}\)). The sum of these two is the total power not being used for heating the air. Total_power_not_used_for_heating = Rate of heat loss + Power consumed by the fan Total_power_not_used_for_heating = \(250 \mathrm{~W} + 300\mathrm{~W}\)
04

Calculate the total_power_not_used_for_heating

Now, let's compute the total_power_not_used_for_heating: Total_power_not_used_for_heating = \(250 \mathrm{~W} + 300\mathrm{~W}\) Total_power_not_used_for_heating = \(550 \mathrm{~W}\)
05

Determine the power rating of the electric resistance heating element

Finally, let's determine the power rating of the electric resistance heating element, which is the difference between the rate of heat gain and the total power not used for heating: Power_rating_of_heating_element = Rate of heat gain - Total_power_not_used_for_heating Power_rating_of_heating_element = \(3000 \mathrm{~W} - 550 \mathrm{~W}\)
06

Calculate the power_rating_of_heating_element

Now, let's compute the power_rating_of_heating_element: Power_rating_of_heating_element = \(3000 \mathrm{~W} - 550 \mathrm{~W}\) Power_rating_of_heating_element = \(2450 \mathrm{~W}\) Therefore, the power rating of the electric resistance heating element is \(2450 \mathrm{~W}\).

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Key Concepts

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

Electric Heating System
An electric heating system is a common method used to warm indoor spaces. This method transfers electric energy into thermal energy through heating elements. In our example, the system includes a fan that expels air through a duct, encountering a resistance heating element.

The main goal of this system is to ensure the air is heated by the resistance element, even as it flows continuously. The efficiency of this process depends on several factors, such as the power of the heating element, the operation of the fan, and potential heat losses.

To optimize the performance, it鈥檚 crucial to account for the energy input (from electricity) and make sure it's mixed with minimal energy loss. The heat gain that occurs as air passes through the duct results in a rise in temperature, demonstrating that the system efficiently converts electrical energy into heat.
Specific Heat Capacity
Specific heat capacity is a measure of how much energy is needed to raise the temperature of a substance. In simple terms, it tells us how heat-resistant a substance is.

Air has a specific heat capacity of approximately 1000 J/(kg路K), which means you need 1000 Joules of energy to increase the temperature of 1 kilogram of air by 1 Kelvin.

In the context of electric heating systems, knowing the specific heat capacity of air helps in determining the energy required to achieve a desired temperature rise.
  • Mass flow rate: This quantifies the amount of air passing through the system.
  • Temperature rise: This is the increase in temperature we observe as the air absorbs heat.
By multiplying specific heat capacity with mass flow and temperature rise, you determine the heat energy (in Watts) needed for the air.

This knowledge enables engineers to design more effective heating systems, ensuring proper energy use without excessive power consumption.
Heat Loss
Heat loss is an important aspect in the analysis of heating systems. It refers to the amount of energy that escapes from the system without contributing to the heating of the air. Effectively managing and minimizing heat loss is vital for energy efficiency.

In our discussion, there is a known heat loss of 250 W from the air in the duct. This means that out of all the energy input into the system, 250 W is lost. Some common reasons for heat loss include:
  • Poor insulation.
  • Conduction through duct materials.
  • Incomplete sealing of duct connections allowing hot air to escape.


Addressing these issues can significantly improve the system's efficiency. Once heat loss is identified, it must be subtracted from the total power input to calculate the effective power contributing to actual heating. By focusing on reducing heat loss, you allow more of the electric power consumed to be used for heating, saving energy and costs in the process.

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Most popular questions from this chapter

The deep human body temperature of a healthy person remains constant at \(37^{\circ} \mathrm{C}\) while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter.

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is \(10^{\circ} \mathrm{C}\), while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.

A 4-m \(\times 5-\mathrm{m} \times 6-\mathrm{m}\) room is to be heated by one ton ( \(1000 \mathrm{~kg}\) ) of liquid water contained in a tank placed in the room. The room is losing heat to the outside at an average rate of \(10,000 \mathrm{~kJ} / \mathrm{h}\). The room is initially at \(20^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), and is maintained at an average temperature of \(20^{\circ} \mathrm{C}\) at all times. If the hot water is to meet the heating requirements of this room for a 24-h period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. Answer: \(77.4^{\circ} \mathrm{C}\)

What is the physical mechanism of heat conduction in a solid, a liquid, and a gas?

Steady heat conduction occurs through a \(0.3\)-m-thick \(9 \mathrm{~m} \times 3 \mathrm{~m}\) composite wall at a rate of \(1.2 \mathrm{~kW}\). If the inner and outer surface temperatures of the wall are \(15^{\circ} \mathrm{C}\) and \(7^{\circ} \mathrm{C}\), the effective thermal conductivity of the wall is (a) \(0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(0.83 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(2.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(5.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)
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