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Water is boiling in a 25 -cm-diameter aluminum pan \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) at \(95^{\circ} \mathrm{C}\). Heat is transferred steadily to the boiling water in the pan through its \(0.5-\mathrm{cm}\)-thick flat bottom at a rate of \(800 \mathrm{~W}\). If the inner surface temperature of the bottom of the pan is \(108^{\circ} \mathrm{C}\), determine \((a)\) the boiling heat transfer coefficient on the inner surface of the pan and \((b)\) the outer surface temperature of the bottom of the pan.

Short Answer

Expert verified
The boiling heat transfer coefficient on the inner surface of the pan is 3824.83 W/m²K, and the outer surface temperature of the bottom of the pan is 103.75°C.

Step by step solution

01

Calculate the temperature difference across the pan's bottom

In this step, we find the temperature difference between the inner surface and boiling water in the pan. Given the inner surface temperature is \(108^{\circ} \mathrm{C}\) and the boiling water temperature is \(95^{\circ} \mathrm{C}\), the temperature difference is: \(\Delta T = (108 - 95)\mathrm{C} = 13\mathrm{C}\)
02

Calculate the heat transfer rate across the pan's bottom

We are given that the heat transfer rate, \(q\), is \(800 \mathrm{~W}\). We can now use this value in the conduction equation to find the boiling heat transfer coefficient, \(h_{f}\).
03

Apply the conduction equation through the pan's bottom

The conduction equation is given by: \(q = kA\frac{\Delta T}{\Delta x}\) where \(k = 237 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) is the thermal conductivity of aluminum, \(A\) is the area of the pan's bottom, and \(\Delta x = 0.5\,\mathrm{cm}\) is the thickness of the bottom. The area of the pan's bottom can be calculated as: \(A = \pi r^2\), where \(r = 12.5\,\mathrm{cm} = 0.125\,\mathrm{m}\) is the radius of the pan. So, \(A = \pi(0.125)^2 = 0.0491\,\mathrm{m^2}\). We can now rearrange the conduction equation to find the boiling heat transfer coefficient, \(h_{f}\): \(h_{f} = \frac{k\Delta T}{q\Delta x}A\)
04

Calculate the boiling heat transfer coefficient

Plugging in the values into the equation, we get: \(h_{f} = \frac{(237\frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(13\mathrm{K})}{(800\,\mathrm{W})(0.005\,\mathrm{m})}(0.0491\,\mathrm{m^2}) = 3824.83\,\frac{\mathrm{W}}{\mathrm{m^2} \cdot \mathrm{K}}\) Therefore, the boiling heat transfer coefficient on the inner surface of the pan is \(3824.83\,\frac{\mathrm{W}}{\mathrm{m^2} \cdot \mathrm{K}}\).
05

Calculate the outer surface temperature of the pan's bottom

We know that the inner surface temperature of the pan is \(108^{\circ} C\), so the temperature difference through the bottom will be the same as the one through the boiling waster to inner surface. The conduction equation for the outer surface is: \(q = h_{f}A\Delta T_{out}\) \(\Delta T_{out} = \frac{q}{h_{f}A}\) Plugging in the values, we get: \(\Delta T_{out} = \frac{(800\,\mathrm{W})}{(3824.83\,\frac{\mathrm{W}}{\mathrm{m^2} \cdot \mathrm{K}})(0.0491\,\mathrm{m^2})} = 4.25\,\mathrm{K}\) So, the outer surface temperature of the pan's bottom is \(108^{\circ} \mathrm{C} - 4.25\,\mathrm{K} = 103.75^{\circ} \mathrm{C}\). In summary, \((a)\) the boiling heat transfer coefficient on the inner surface of the pan is \(3824.83\,\frac{\mathrm{W}}{\mathrm{m^2} \cdot \mathrm{K}}\) and \((b)\) the outer surface temperature of the bottom of the pan is \(103.75^{\circ} \mathrm{C}\).

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Key Concepts

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

Conduction Equation
Understanding heat transfer is crucial in many practical applications, ranging from cooking to industrial processes. The conduction equation is one such fundamental concept that predicts the rate of heat transfer through materials due to temperature differences. In its simplest form, the conduction equation is expressed as:
\[\begin{equation}q = kA\frac{\Delta T}{\Delta x}\end{equation}\]
Here, \(q\) represents the rate of heat transfer in watts (\(\mathrm{W}\)). The \(k\) is the material’s thermal conductivity, indicating how well the material conducts heat. It is measured in watts per meter-kelvin (\(\frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\)). The area through which the heat is being transferred is denoted as \(A\) and is measured in square meters (\(\mathrm{m}^2\)), and \(\Delta T\) is the temperature difference across the material in Kelvin or degrees Celsius. Finally, \(\Delta x\) represents the thickness of the material through which heat is conducted, measured in meters (\(\mathrm{m}\)).

Applying the Conduction Equation

In the given exercise, we applied the conduction equation to determine the boiling heat transfer coefficient of the water in the aluminum pan. By rearranging the equation and inserting the known values, we calculated the coefficient effectively. This process is a perfect example of how theoretical principles are pivotal in resolving practical problems accurately.
Thermal Conductivity
Thermal conductivity, represented by the symbol \(k\), is a material property that signifies the rate at which heat is transferred through the material due to a temperature gradient. Higher values of thermal conductivity represent a better ability to conduct heat. For example, metals typically have high thermal conductivity, which is why they feel colder to the touch compared to materials like wood, even if they are at the same temperature.
In the exercise, we used the thermal conductivity value of aluminum, \(k=237\frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\), which signifies that aluminum is an excellent conductor of heat. This property is one reason why metals like aluminum are commonly used in cookware; they distribute heat quickly and evenly across the surface. Understanding thermal conductivity is critical not just in daily tasks such as cooking, but also in designing heating systems, insulation, and even electronic devices where heat dissipation is key.
Boiling Heat Transfer
Boiling heat transfer refers to the process of heat exchange that occurs when a liquid reaches its boiling point and undergoes a phase change to gas. This type of heat transfer is complex because it involves both heat conduction and convection, plus the additional energy required to change the phase of the liquid, known as the latent heat of vaporization.
The boiling heat transfer coefficient, which we calculated as \(3824.83\,\frac{\mathrm{W}}{\mathrm{m^2} \cdot \mathrm{K}}\) in the exercise, is a measure of how effectively heat is delivered to the boiling liquid. This coefficient is influenced by many factors, including the properties of the liquid, the surface condition of the heating surface, and the pressure in the environment. The exercise demonstrated that even a slight change in surface temperature could significantly impact the heat transfer rate. Boiling heat transfer plays a crucial role in many industries, including power generation and food processing, and requires careful analysis to optimize systems for safety, efficiency, and cost-effectiveness.

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Most popular questions from this chapter

Can the thermal resistance concept be used for a solid cylinder or sphere in steady operation? Explain.

A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20 -cm-thick concrete walls and ceiling \((k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The kiln is maintained at \(40^{\circ} \mathrm{C}\) by injecting hot steam into it. The two ends of the kiln, \(4 \mathrm{~m} \times 5 \mathrm{~m}\) in size, are made of a 3 -mm-thick sheet metal covered with 2 -cm-thick Styrofoam \((k=0.033 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The convection heat transfer coefficients on the inner and the outer surfaces of the kiln are \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Disregarding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at \(-4^{\circ} \mathrm{C}\).

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To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield with thickness of \(5 \mathrm{~mm}\) and thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outside ambient temperature is \(-10^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature inside the automobile is \(25^{\circ} \mathrm{C}\). Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting.

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

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