/*! 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 20 Water is boiling in a 25 -cm-dia... [FREE SOLUTION] | 91Ó°ÊÓ

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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

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

Steam in a heating system flows through tubes whose outer diameter is \(5 \mathrm{~cm}\) and whose walls are maintained at a temperature of \(180^{\circ} \mathrm{C}\). Circular aluminum alloy 2024-T6 fins \((k=186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of outer diameter \(6 \mathrm{~cm}\) and constant thickness \(1 \mathrm{~mm}\) are attached to the tube. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.

What is a conduction shape factor? How is it related to the thermal resistance?

A 6-m-diameter spherical tank is filled with liquid oxygen \(\left(\rho=1141 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1.71 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) at \(-184^{\circ} \mathrm{C}\). It is observed that the temperature of oxygen increases to \(-183^{\circ} \mathrm{C}\) in a 144-hour period. The average rate of heat transfer to the tank is (a) \(249 \mathrm{~W}\) (b) \(426 \mathrm{~W}\) (c) \(570 \mathrm{~W}\) (d) \(1640 \mathrm{~W}\) (e) \(2207 \mathrm{~W}\)

Hot water is flowing at an average velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) through a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(3 \mathrm{~cm}\) and \(3.5 \mathrm{~cm}\), respectively. The pipe passes through a \(15-\mathrm{m}\)-long section of a basement whose temperature is \(15^{\circ} \mathrm{C}\). If the temperature of the water drops from \(70^{\circ} \mathrm{C}\) to \(67^{\circ} \mathrm{C}\) as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. Answer: \(272.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

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