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Chapter 4: Problem 13

An electrical heater \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Estimate the temperature reached by the heater when dissipating \(50 \mathrm{~W}\) with the surface of the block at a temperature of \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
To estimate the temperature reached by the heater, we first calculate the thermal resistance using the formula \(R_{\text{th}} = \frac{\ln \frac{r_{2}}{r_{1}}}{2 \pi k L}\), which gives us \(R_{\text{th}}\). Then, we find the heat transfer through the cylindrical path using \(\Delta Q = P \cdot R_{\text{th}}\). Finally, we determine the temperature difference using \(\Delta T = \Delta Q \cdot R_{\text{th}}\) and add it to the surface temperature of \(25^{\circ} \mathrm{C}\) to estimate the temperature reached by the heater.

Step by step solution

01

Calculate the thermal resistance for heat transfer

The heat transfer in this case occurs through conduction. The thermal resistance for conduction through a cylinder is given by the formula: \[ R_{\text{th}} = \frac{\ln \frac{r_{2}}{r_{1}}}{2 \pi k L} \] where \(R_{\text{th}}\) is the thermal resistance, \(r_{1}\) is the inner radius of the heater, \(r_{2}\) is the outer radius of the heater, \(L\) is the length of the heater, and \(k\) is the thermal conductivity of the material. Here, the heater's diameter is \(5 \mathrm{~mm}\), so its radius \(r_1 =2.5 \mathrm{~mm}\), and the hole's radius is equal to the heater's radius, which means \(r_2 = 2.5 \mathrm{~mm}\). Since the hole is drilled normal to the surface, the length remains the same at \(100 \mathrm{~mm}\). The thermal conductivity of the material is \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
02

Find the heat transfer through the cylindrical path

Now that we have calculated the thermal resistance, we can find the heat transfer through the cylindrical path using the formula: \[ \Delta Q = P \cdot R_{\text{th}} \] where \(\Delta Q\) is the heat transfer, and \(P\) is the power dissipated by the heater. In this problem, the power dissipated by the heater is given as \(50 \mathrm{~W}\).
03

Determine the temperature difference between the heater and the surface

Finally, we can determine the temperature difference between the heater and the surface using the heat transfer formula: \[ \Delta T = \Delta Q \cdot R_{\text{th}} \] where \(\Delta T\) is the temperature difference. Once we have found the temperature difference, we can add it to the surface temperature to estimate the temperature reached by the heater. The surface of the block is at a temperature of \(25^{\circ} \mathrm{C}\). Following these steps, we can solve the problem and estimate the temperature reached by the heater when dissipating \(50 \mathrm{~W}\).

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Key Concepts

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

Heat Transfer
Understanding heat transfer is crucial when analyzing how heat moves from one object to another. In the case of the electrical heater scenario, heat transfer occurs through a process called conduction. Conduction is the transfer of heat through a solid material without any actual movement of the material itself. When you hold a metal rod in a fire, the heat travels up the rod, making it warm to the touch. This is conduction at work.
In this exercise, heat moves from the heater into the surrounding block. The efficient movement of this heat is vital to ensuring the heater reaches its desired operating temperature. Heat transfer relies heavily on the concept of thermal conductivity, which measures a material's ability to conduct heat. A high thermal conductivity indicates that a material can transfer heat easily, while a low value denotes the opposite.
  • Heat Transfer Type: Conduction
  • Material: Large block with thermal conductivity of 5 W/(m路K)
  • Power Dissipated: 50 W by the heater
Understanding how heat is transferred to surrounding materials will help in predicting the temperature that the heater will reach.
Thermal Resistance
Thermal resistance defines how difficult it is for heat to pass through a material. It's much like electrical resistance but for heat instead of electricity. When calculating thermal resistance especially for cylindrical objects like our heater, you use a specific formula. In this case, the formula to calculate the resistance through a cylindrical path is
\[ R_{\text{th}} = \frac{\ln \frac{r_{2}}{r_{1}}}{2 \pi k L} \]
where:
  • \( R_{\text{th}} \) is the thermal resistance
  • \( r_{1} \) and \( r_{2} \) are the inner and outer radii of the heater
  • \( k \) is the thermal conductivity of the material (5 W/m路K in this problem)
  • \( L \) is the length of the heater
For our exercise, since \( r_1 \) equals \( r_2 \) then \( R_{\text{th}} \) becomes zero due to the equation's logarithmic term. This complex idea tells us how much the heat will be resisted as it tries to move through the cylinder. Simply put, a high thermal resistance means less heat transfer, keeping the heater warmer.
Temperature Difference
To determine the temperature difference between the heater and the block surface, you have to consider how much heat is transferred through the thermal resistance. Calculating the temperature difference \( \Delta T \) is essential for finding out the actual temperature of the heater. The basic formula used is:
\[ \Delta T = \Delta Q \cdot R_{\text{th}} \]
where
  • \( \Delta T \) is the temperature difference
  • \( \Delta Q \) is the heat transfer which is equivalent to power dissipated (in this case, 50 W)
  • \( R_{\text{th}} \) is thermal resistance (calculated previously)
In our exercise, once you determine the temperature difference, add it to the given surface temperature of the block, which is 25掳C. This sum gives an estimate of the heater's temperature. Accurately calculating \( \Delta T \) ensures that you comprehend how different temperatures interact in conduction, allowing for practical applications in engineering and thermal management.

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Most popular questions from this chapter

Derive the nodal finite-difference equations for the following configurations. (a) Node \((m, n)\) on a diagonal boundary subjected to convection with a fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\). (b) Node \((m, n)\) at the tip of a cutting tool with the upper surface exposed to a constant heat flux \(q_{o}^{\prime \prime}\), and the diagonal surface exposed to a convection cooling process with the fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\).

Consider heat transfer in a one-dimensional (radial) cylindrical coordinate system under steady-state conditions with volumetric heat generation. (a) Derive the finite-difference equation for any interior node \(m\). (b) Derive the finite-difference equation for the node \(n\) located at the external boundary subjected to the convection process \(\left(T_{\infty 0}, h\right)\).

A small device is used to measure the surface temperature of an object. A thermocouple bead of diameter \(D=120 \mu \mathrm{m}\) is positioned a distance \(z=100 \mu \mathrm{m}\) from the surface of interest. The two thermocouple wires, each of diameter \(d=25 \mu \mathrm{m}\) and length \(L=300 \mu \mathrm{m}\), are held by a large manipulator that is at a temperature of \(T_{m}=23^{\circ} \mathrm{C}\). If the thermocouple registers a temperature of \(T_{\mathrm{tc}}=29^{\circ} \mathrm{C}\), what is the surface temperature? The thermal conductivities of the chromel and alumel thermocouple wires are \(k_{\mathrm{Ch}_{h}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. You may neglect radiation and convection effects.

For a small heat source attached to a large substrate, the spreading resistance associated with multidimensional conduction in the substrate may be approximated by the expression [Yovanovich, M. M., and V. W. Antonetti, in Adv. Thermal Modeling Elec. Comp. and Systems, Vol. 1, A. Bar-Cohen and A. D. Kraus, Eds., Hemisphere, \(\mathrm{NY}, 79-128,1988]\) \(R_{r(\mathrm{sp})}=\frac{1-1.410 A_{r}+0.344 A_{r}^{3}+0.043 A_{r}^{5}+0.034 A_{r}^{7}}{4 k_{\mathrm{sub}} A_{s, h}^{1 / 2}}\) where \(A_{r}=A_{s, h} / A_{s, \text { sub }}\) is the ratio of the heat source area to the substrate area. Consider application of the expression to an in- line array of square chips of width \(L_{h}=\) \(5 \mathrm{~mm}\) on a side and pitch \(S_{h}=10 \mathrm{~mm}\). The interface between the chips and a large substrate of thermal conductivity \(k_{\text {sub }}=80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=0.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If a convection heat transfer coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is associated with airflow \(\left(T_{\infty}=15^{\circ} \mathrm{C}\right)\) over the chips and substrate, what is the maximum allowable chip power dissipation if the chip temperature is not to exceed \(T_{h}=85^{\circ} \mathrm{C}\) ?

Free convection heat transfer is sometimes quantified by writing Equation \(4.20\) as \(q_{\text {coov }}=S k_{\text {eff }} \Delta T_{1-2}\), where \(k_{\text {eff }}\) is an effective thermal conductivity. The ratio \(k_{\text {eff }} / k\) is greater than unity because of fluid motion driven by buoyancy forces, as represented by the dashed streamlines. An experiment for the configuration shown yields a heat transfer rate per unit length of \(q_{\text {conv }}^{\prime}=110 \mathrm{~W} / \mathrm{m}\) for surface temperatures of \(T_{1}=53^{\circ} \mathrm{C}\) and \(T_{2}=15^{\circ} \mathrm{C}\), respectively. For inner and outer cylinders of diameters \(d=20 \mathrm{~mm}\) and \(D=60 \mathrm{~mm}\), and an eccentricity factor of \(z=10 \mathrm{~mm}\), determine the value of \(k_{\text {eff. }}\). The actual thermal conductivity of the fluid is \(k=0.255 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
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