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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 × specific heat capacity × 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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Most popular questions from this chapter

A 40-cm-long, 800-W electric resistance heating element with diameter \(0.5 \mathrm{~cm}\) and surface temperature \(120^{\circ} \mathrm{C}\) is immersed in \(75 \mathrm{~kg}\) of water initially at \(20^{\circ} \mathrm{C}\). Determine how long it will take for this heater to raise the water temperature to \(80^{\circ} \mathrm{C}\). Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process.

A 300-ft-long section of a steam pipe whose outer diameter is 4 in passes through an open space at \(50^{\circ} \mathrm{F}\). The average temperature of the outer surface of the pipe is measured to be \(280^{\circ} \mathrm{F}\), and the average heat transfer coefficient on that surface is determined to be \(6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine \((a)\) the rate of heat loss from the steam pipe and (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent, and the price of natural gas is $$\$ 1.10 /$$ therm ( 1 therm \(=100,000\) Btu).

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{~kg} / \mathrm{s}\) and is heated to \(60^{\circ} \mathrm{C}\). The rate of heat transfer to the water is (a) \(20 \mathrm{~kW}\) (b) \(42 \mathrm{~kW}\) (c) \(84 \mathrm{~kW}\) (d) \(126 \mathrm{~kW}\) (e) \(334 \mathrm{~kW}\)

\(80^{\circ} \mathrm{C}\). Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process. 1-133 It is well known that wind makes the cold air feel much colder as a result of the wind chill effect that is due to the increase in the convection heat transfer coefficient with increasing air velocity. The wind chill effect is usually expressed in terms of the wind chill temperature (WCT), which is the apparent temperature felt by exposed skin. For outdoor air temperature of \(0^{\circ} \mathrm{C}\), for example, the wind chill temperature is \(-5^{\circ} \mathrm{C}\) at \(20 \mathrm{~km} / \mathrm{h}\) winds and \(-9^{\circ} \mathrm{C}\) at \(60 \mathrm{~km} / \mathrm{h}\) winds. That is, a person exposed to \(0^{\circ} \mathrm{C}\) windy air at \(20 \mathrm{~km} / \mathrm{h}\) will feel as cold as a person exposed to \(-5^{\circ} \mathrm{C}\) calm air (air motion under \(5 \mathrm{~km} / \mathrm{h}\) ). For heat transfer purposes, a standing man can be modeled as a 30 -cm- diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of \(34^{\circ} \mathrm{C}\). For a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from this man by convection in still air at \(20^{\circ} \mathrm{C}\). What would your answer be if the convection heat transfer coefficient is increased to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as a result of winds? What is the wind chill temperature in this case?

Consider two walls of a house that are identical except that one is made of 10 -cm-thick wood, while the other is made of 25 -cm-thick brick. Through which wall will the house lose more heat in winter?
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