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Chapter 1: Problem 132

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.

Short Answer

Expert verified
Answer: To find the convection heat transfer coefficients at the beginning and the end of the heating process, use the formula: \(h =\frac{P_\text{heater}}{(T_\text{heater} - T_\text{water}) \cdot A}\) For the beginning of the heating process, plug in the initial surface temperature of the heating element (120掳C), the initial temperature of the water (20掳C), and the surface area calculated in Step 3: \(h_\text{start} = \frac{0.8 \mathrm{\ kW}}{(120^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) \cdot A}\) For the end of the heating process, use the final temperature of the water (80掳C): \(h_\text{end} =\frac{0.8 \mathrm{\ kW}}{(120^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C}) \cdot A}\) Calculate h_start and h_end to find the convection heat transfer coefficients at the beginning and the end of the heating process, respectively.

Step by step solution

01

Calculate the energy required to heat the water

First, we need to calculate the energy required to raise the temperature of the water from 20掳C to 80掳C. We can use the specific heat capacity of water (c_water = 4.18 kJ/kg掳C) and the mass of the water (m_water = 75 kg) to do this. The energy required (Q) can be calculated using the formula: \(Q = m_\text{water} \cdot c_\text{water} \cdot \Delta T\) where 螖T is the temperature difference (80掳C - 20掳C = 60掳C). Plug in the values and solve for Q: \(Q = 75 \mathrm{\ kg} \cdot 4.18 \mathrm{\ kJ/kg^{\circ} C} \cdot 60^{\circ} C\)
02

Calculate the time it takes to heat the water

Next, we need to determine how long it takes for the heater to transfer the required energy into the water. The heater has a power of 800 W (which is equal to 0.8 kW), so we can use the formula: time = \(\frac{Q}{P_\text{heater}}\) Plug in the values and solve for time: time = \(\frac{75 \mathrm{\ kg} \cdot 4.18 \mathrm{\ kJ/kg^{\circ} C} \cdot 60^{\circ} C}{0.8 \mathrm{\ kW}}\)
03

Calculate the surface area of the heating element

Since we need to find the convection heat transfer coefficients at the beginning and at the end of the heating process, we should first calculate the surface area of the heating element. The heating element is a cylinder, and its surface area (A) can be calculated using the formula: For a cylinder of length L and radius r, \(A = 2 \pi r L\) Plug in the given values: \(A = 2 \pi \cdot 0.0025 \mathrm{\ m} \cdot 0.4 \mathrm{\ m}\)
04

Find the convection heat transfer coefficient at the beginning and the end of the heating process

We can use the formula for the convection heat transfer coefficient (h) related to the heat exchanged by the heating element and the rate of temperature increase: \(h =\frac{P_\text{heater}}{(T_\text{heater} - T_\text{water}) \cdot A}\) Using this formula, find the convection heat transfer coefficient at the beginning of the heating process by plugging in the initial surface temperature of the heating element (120掳C), the initial temperature of the water (20掳C), and the surface area calculated in Step 3: \(h_\text{start} = \frac{0.8 \mathrm{\ kW}}{(120^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) \cdot A}\) Repeat the steps for the end of the heating process with the final temperature of the water (80掳C): \(h_\text{end} =\frac{0.8 \mathrm{\ kW}}{(120^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C}) \cdot A}\)

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Key Concepts

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

Convection Heat Transfer Coefficient
Understanding the convection heat transfer coefficient is vital in determining the efficiency of thermal energy transfer from the heating element to water. This coefficient, often denoted as \( h \), quantifies the convective heat transfer process between a solid surface and a fluid, in this case, water.
To calculate \( h \) at different times, use the formula:
\( h = \frac{P_{\text{heater}}}{(T_{\text{heater}} - T_{\text{water}}) \cdot A} \)
Here's how it works:
  • \( P_{\text{heater}} \) is the power of the heater, which is 0.8 kW in this example.
  • \( T_{\text{heater}} \) is the surface temperature of the heating element, which remains constant at \(120^{\circ} \text{C}\).
  • \( T_{\text{water}} \) varies from the initial \(20^{\circ} \text{C}\) to \(80^{\circ} \text{C}\) during heating.
  • \( A \) is the surface area of the heater, calculated previously.
The coefficient changes as the temperature difference \( (T_{\text{heater}} - T_{\text{water}}) \) changes. By understanding \( h \), you can evaluate how quickly and efficiently heat is transferred.
Electric Resistance Heating
Electric resistance heating utilizes the natural resistance found in materials to convert electrical energy into thermal energy. This process is widely used in various heating applications due to its simplicity and efficiency.
In this example, an 800-W electric resistance heating element is used. This implies that:
  • 800 watts of electrical power is consistently converted into heat energy.
  • The heating capability can be directly calculated by the power output, i.e., 0.8 kW.
The heating element releases heat continuously at a consistent rate, making it straightforward to predict how long it takes to achieve desired temperature changes. Electric resistance heating elements are made from materials with high resistivity, ensuring efficient heat generation.
Specific Heat Capacity
Specific heat capacity is a material's intrinsic property that measures how much heat energy is required to change the temperature of a unit mass by one degree Celsius. In the context of water, which has a high specific heat capacity of 4.18 kJ/kg掳C, it takes considerable energy to raise its temperature.

The formula to calculate the energy needed is:
\( Q = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \)
Where:
  • \( Q \) is the total energy needed to heat the water.
  • \( m_{\text{water}} \) is the mass of the water, which, in this case, is 75 kg.
  • \( \Delta T \) is the temperature change, from \(20^{\circ} \text{C}\) to \(80^{\circ} \text{C}\), which gives \(60^{\circ} \text{C}\).
Understanding the specific heat capacity helps estimate the amount of energy required to achieve temperature changes in materials, especially those like water with a higher capacity to store heat.
Surface Area Calculation
Calculating the surface area of the heating element is crucial for determining the heat transfer rate through convection. The heating element is modeled as a cylinder, and its surface area can be determined using the cylindrical surface area formula.
For a cylinder, the surface area \( A \) is given by:
\( A = 2 \pi r L \)
Where:
  • \( r \) is the radius of the cylinder. Given a diameter of 0.5 cm, we convert to radius by dividing by 2, which equals 0.25 cm or 0.0025 m when converted to meters.
  • \( L \) is the length of the cylinder, which is 40 cm or 0.4 m when converted to meters.
Substitute these values into the formula to calculate \( A \). This value plays a significant role in calculating the convection heat transfer coefficient, as it influences the area over which heat transfer occurs.

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Most popular questions from this chapter

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

While driving down a highway early in the evening, the air flow over an automobile establishes an overall heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The passenger cabin of this automobile exposes \(9 \mathrm{~m}^{2}\) of surface to the moving ambient air. On a day when the ambient temperature is \(33^{\circ} \mathrm{C}\), how much cooling must the air conditioning system supply to maintain a temperature of \(20^{\circ} \mathrm{C}\) in the passenger cabin? (a) \(670 \mathrm{~W}\) (b) \(1284 \mathrm{~W}\) (c) \(2106 \mathrm{~W}\) (d) \(2565 \mathrm{~W}\) (e) \(3210 \mathrm{~W}\)

The critical heat flux (CHF) is a thermal limit at which a boiling crisis occurs whereby an abrupt rise in temperature causes overheating on fuel rod surface that leads to damage. A cylindrical fuel rod of \(2 \mathrm{~cm}\) in diameter is encased in a concentric tube and cooled by water. The fuel generates heat uniformly at a rate of \(150 \mathrm{MW} / \mathrm{m}^{3}\). The average temperature of the cooling water, sufficiently far from the fuel rod, is \(80^{\circ} \mathrm{C}\). The operating pressure of the cooling water is such that the surface temperature of the fuel rod must be kept below \(300^{\circ} \mathrm{C}\) to avoid the cooling water from reaching the critical heat flux. Determine the necessary convection heat transfer coefficient to avoid the critical heat flux from occurring.

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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