/*! 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 49 In Problem 3.48, the electrical ... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problem 3.48, the electrical power required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C}\) depends on the thermal conductivity of the wall material \(k\), the thermal contact resistance \(R_{t, c}^{\prime}\) and the convection coefficient \(h\). Compute and plot the separate effect of changes in \(k\) \((1 \leq k \leq 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), \quad R_{t, c}^{\prime} \quad\left(0 \leq R_{t, c}^{\prime} \leq 0.1 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\right)\), and \(h\left(10 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) on the total heater power requirement, as well as the rate of heat transfer to the inner surface of the tube and to the fluid.

Short Answer

Expert verified
In order to study the separate effects of changes in thermal conductivity $k$, thermal contact resistance $R_{t, c}^{\prime}$, and convection coefficient $h$ on the total heater power requirement (P), the rate of heat transfer to the inner surface of the tube (q_w), and the rate of heat transfer to the fluid (q_f), the following steps must be followed: 1. Obtain the equations governing the heat transfer rates, considering heat conduction, thermal contact resistance, and convection. 2. Calculate the rates of heat transfer to the inner surface (q_w) and to the fluid (q_f) using the obtained equations. 3. Determine the total heater power requirement (P) by equating it to the heat transfer rates. 4. Calculate P, q_w, and q_f for the given ranges of $k$, $R_{t, c}^{\prime}$, and $h$ values. 5. Plot the separate effects of changes in $k$, $R_{t, c}^{\prime}$, and $h$ on P, q_w, and q_f by varying each parameter individually while keeping the others constant. This will provide a visual representation of how these changes affect heating requirements and heat transfer rates.

Step by step solution

01

Find the equations governing the electrical power required, the heat transfer rate to the inner surface, and the heat transfer rate to the fluid

The exercise involves heat transfer through a wall material and fluid, with thermal conductivity k, thermal contact resistance Rₜ,ₒ, and the convection coefficient h. To find the equations governing the electrical power required P, the heat transfer rate to the inner surface, and the heat transfer rate to the fluid, we first need to apply the steady-state, one-dimensional heat conduction equation: \[q = k \frac{T_o - T_w}{x}\] where q is the heat transfer rate, T_o is the outer temperature, T_w is the inner (wall) temperature, and x is the wall thickness. We also need to consider the heat transfer from the wall to the fluid due to convection: \[q = h (T_w - T_f)\] where T_f is the fluid temperature. Considering thermal contact resistance Rₜ,ₒ in the wall, we have: \[q = \frac{T_o - T_w}{R_{t, c}^{\prime}}\] Since the heater must supply the same heat transfer rate to maintain the steady-state, all the heat transfer rates are equal (q_conduction = q_convection = q_contact_resistance). Now we can create equations for the rates of heat transfer to the inner surface and to the fluid.
02

Find the rate of heat transfer to the inner surface and to the fluid

To find the rate of heat transfer to the inner surface (q_w) and to the fluid (q_f), we can substitute and solve the heat transfer equations in Step 1, considering all three heat transfer rates equal: 1. For heat transfer to the inner surface (q_w): \[q_w = k \frac{T_o - T_w}{x}\] 2. For heat transfer to the fluid (q_f): \[q_f = h (T_w - T_f)\]
03

Prepare the equations to compute the total heater power requirement

Since the total heat transfer rate supplied by the heater is equal to the electrical power required for the heater (P), we can write: \[P = q_w = q_f\]
04

Calculate P, q_w, and q_f for the given k, Rₜ,ₒ, and h

Given the range of k, Rₜ,ₒ, and h values, we will now compute the total heater power requirement (P), the rate of heat transfer to the inner surface (q_w), and the rate of heat transfer to the fluid (q_f): 1. Compute the corresponding T_w and T_f for the given k, Rₜ,ₒ, and h values by substituting them into the equations from Steps 2 and 3. 2. Calculate the heat transfer rates q_w and q_f using the obtained T_w and T_f values. 3. Compute the total heater power requirement P by equating it to q_w and q_f.
05

Plot the separate effects of k, Rₜ,ₒ, and h on P, q_w, and q_f

Now, we will plot the separate effects of changes in k, Rₜ,ₒ, and h on the total heater power requirement (P), the rate of heat transfer to the inner surface (q_w), and the rate of heat transfer to the fluid (q_f): 1. For varying k values (1 to 200 W/m*K), keep Rₜ,ₒ, and h constant, and compute P, q_w, and q_f for each k value. Plot k on the x-axis and P, q_w, and q_f on the y-axis on separate graphs. 2. For varying Rₜ,ₒ values (0 to 0.1 m*K/W), keep k and h constant, and compute P, q_w, and q_f for each Rₜ,ₒ. Plot Rₜ,ₒ on the x-axis and P, q_w, and q_f on the y-axis on separate graphs. 3. For varying h values (10 to 1000 W/m²*K), keep k and Rₜ,ₒ constant, and compute P, q_w, and q_f for each h value. Plot h on the x-axis and P, q_w, and q_f on the y-axis on separate graphs. This will provide you with a visual representation of how changes in k, Rₜ,ₒ, and h affect the total heater power requirement, the rate of heat transfer to the inner surface of the tube, and the rate of heat transfer to the fluid.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity \(k\) is a material property that measures a material's ability to conduct heat. It is a key element in determining how well or poorly heat will flow through an object. In this context, it is measured in watts per meter-kelvin (W/m·K). The higher the thermal conductivity of a material, the better it is at transferring heat. Metals generally have high thermal conductivity, while insulating materials have low thermal conductivity.

In heat transfer, thermal conductivity plays a crucial role in the conduction process. When you have a temperature difference across a material, heat will flow from the hot side to the cold side, driven by this temperature gradient. The rate of heat flow (or heat transfer rate \(q\)) through a wall can be expressed as follows:

\[q = k \frac{T_o - T_w}{x}\]

Here, \(T_o\) is the outer surface temperature, \(T_w\) is the inner surface temperature, and \(x\) is the wall thickness. Higher thermal conductivity means more efficient heat flow, impacting the power requirement of a heater to maintain specific temperature conditions.
Convection Coefficient
The convection coefficient \(h\) represents the effectiveness of heat transfer from a solid surface to a fluid, or vice versa. This parameter plays a vital role in convection heat transfer processes, which occur when a fluid (liquid or gas) flows over a surface. The units of \(h\) are watts per square meter-Kelvin (W/m²·K).

Convection occurs naturally due to the buoyancy effect, or it can be forced, like using a fan. The rate of heat transfer via convection from the wall to the fluid is described mathematically as:

\[q = h (T_w - T_f)\]

Where \(T_w\) is the wall temperature, and \(T_f\) is the fluid temperature. The difference \((T_w - T_f)\) drives the heat transfer. A high convection coefficient indicates more efficient heat transfer, meaning that heat is carried away more effectively by the fluid. This can significantly affect both the heat transfer rate and the power required by a heater in thermal management systems.
Thermal Contact Resistance
Thermal contact resistance \(R_{t, c}^{\prime}\) is the resistance to heat flow across the interface between two materials in contact. It occurs because of surface roughness, material differences, or presence of intermediate layers like air gaps. Like electrical resistance, higher thermal contact resistance hinders heat flow, represented in units of meter-square Kelvin per watt (m²K/W).

In thermal analysis, it is crucial to account for this resistance when calculating the total thermal resistance of a system. The relationship is given by:

\[q = \frac{T_o - T_w}{R_{t, c}^{\prime}}\]

Where \(T_o\) and \(T_w\) are the temperatures at the outer and inner surfaces, respectively. A higher \(R_{t, c}^{\prime}\) indicates a greater hindrance to heat flow, necessitating more energy (or power) to maintain thermal equilibrium across contacts. Reducing thermal contact resistance, such as improving surface finish or using thermal interface materials, can improve the efficiency of heat transfer systems.

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Most popular questions from this chapter

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(\rho_{e}=10^{-6} \Omega \cdot \mathrm{m}\), thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and emissivity \(\varepsilon=0.20\). The heater is designed to deliver air at a temperature of \(T_{\infty}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {sur }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable electric current \(I\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

A nanolaminated material is fabricated with an atomic layer deposition process, resulting in a series of stacked, alternating layers of tungsten and aluminum oxide, each layer being \(\delta=0.5 \mathrm{~nm}\) thick. Each tungsten-aluminum oxide interface is associated with a thermal resistance of \(R_{t, i}^{\prime \prime}=3.85 \times 10^{-9} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The theoretical values of the thermal conductivities of the thin aluminum oxide and tungsten layers are \(k_{\mathrm{A}}=1.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{T}}=6.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The properties are evaluated at \(T=300 \mathrm{~K}\). (a) Determine the effective thermal conductivity of the nanolaminated material. Compare the value of the effective thermal conductivity to the bulk thermal conductivities of aluminum oxide and tungsten, given in Tables A.1 and A.2. (b) Determine the effective thermal conductivity of the nanolaminated material assuming that the thermal conductivities of the tungsten and aluminum oxide layers are equal to their bulk values.

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