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Chapter 3: Problem 127

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.

Short Answer

Expert verified
In summary, to find the friction coefficient \(\mu\), we first obtain an expression relating torque, contact pressure, and the friction coefficient as \(\mu = \frac{\tau}{P \cdot r}\). Next, we determine the radial temperature distribution \(T(r)\) in the disk using the energy balance and obtain the expression \(T(r) = \frac{-\omega \mu P}{2k} r^2 + C_1 r + C_2\). Finally, we use the given test conditions to calculate the contact pressure \(P\), the friction coefficient \(\mu\), and the maximum disk temperature by evaluating \(T(r_2)\).

Step by step solution

01

Relating Torque, Contact Pressure, and Friction Coefficient

To find an expression for the friction coefficient, we need to relate it to the torque and the known force F. According to the friction force formula, the friction force between the disks can be expressed as: \(F_r = \mu P\) Where \(F_r\) is the friction force, \(\mu\) is the friction coefficient, and P is the contact pressure between the rotating and stationary disks. The torque 蟿 can be calculated using force and its radial distance: \(\tau = F_r \cdot r\) Now, combining both equations, we get: \(\tau = \mu P \cdot r\) We can now find the expression for the friction coefficient: \(\mu = \frac{\tau}{P \cdot r}\) Now that we have the expression for \(\mu\), we can use it later to find the value of the friction coefficient for the given test conditions.
02

Determining the Radial Temperature Distribution

To find the radial temperature distribution, we need to use the energy balance. We know that the energy generation in the disk is due to the friction force, and this energy has to be distributed evenly across the disk to maintain a constant temperature. The heat generation per unit volume can be given by: \(q'' = \mu P \cdot \omega\) Where \(q''\) is the heat generation per unit volume, \(\omega\) is the angular velocity, and \(\mu P\) is the friction force acting between the disks. The energy balance equation can be expressed as: \(q'' = k \frac{d^2T}{dr^2}\) Where k is the thermal conductivity of the disk material and \(\frac{d^2T}{dr^2}\) is the second-order derivative of temperature with respect to the radial distance r. Now we can solve this differential equation for temperature distribution T(r): \(k \frac{d^2T}{dr^2} = \mu P \cdot \omega\) Integrating twice, we get: \(T(r) = \frac{-\omega \mu P}{2k} r^2 + C_1 r + C_2\) Now, using the given boundary conditons, we can find the constants \(C_1\) and \(C_2\). The conditions are: when r = \(r_1\), \(T(r_1) = T_1\); and when r = \(r_2\), \(T(r_2) = T_2\), where \(T_1\) is known, and \(T_2\) represents the temperature at the outer radius. This gives us the expression for the radial temperature distribution T(r).
03

Evaluating the Friction Coefficient and the Maximum Disk Temperature

Now, we can find the value of the friction coefficient using the expression derived in Step 1: \(\mu = \frac{\tau}{P \cdot r}\) For given test conditions, F = 200N, 蠅 = 40rad/s, 蟿 = 8Nm. We can find the contact pressure P using the given force F: \(P = \frac{F}{\pi(r_2^2 - r_1^2)}\) Calculate P and then substitute the values into the equation for \(\mu\), to get the value of the friction coefficient. Next, we evaluate the maximum disk temperature. It is noteworthy that the maximum temperature occurs at the outer radius \(r_2\). Therefore, we calculate the temperature at the outer radius \(r_2\) using the expression for T(r) derived in Step 2: \(T(r_2) = \frac{-\omega \mu P}{2k} r_2^2 + C_1 r_2 + C_2\) Substitute the values of 蠅, \(\mu\), P, \(r_2\), \(T_1\), k, and \(C_1\), \(C_2\) obtained from the boundary conditions in the equation to evaluate the maximum disk temperature.

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Key Concepts

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

Friction Coefficient
The friction coefficient, often symbolized as \( \mu \), is a crucial parameter in understanding how two surfaces interact. In the context of a disk brake system, it represents the ease with which the rotating disk slides against the stationary disk. This measurement is vital for determining the effectiveness and safety of brake systems.
To calculate \( \mu \), you need to know the torque \( \tau \) required to maintain rotation and the contact pressure \( P \) between the disks. The relationship is given by the formula:
  • \( \mu = \frac{\tau}{P \cdot r} \)
Where:
  • \( \tau \) is the torque
  • \( P \) is the contact pressure
  • \( r \) is the radial distance
With this formula, you can compute the friction coefficient once you know the force applied and the resulting torque. This understanding is essential for engineers to design efficient braking systems that provide reliable stopping power. By knowing \( \mu \), adjustments can be made to the brake design to enhance performance and safety.
Radial Temperature Distribution
The radial temperature distribution \( T(r) \) in a disk brake is a representation of how temperature changes from the center of the disk to its outer edge. This is crucial because temperature variation affects the material's thermal stress and efficiency.To find \( T(r) \), we use the energy balance equation linked to frictional heat generation:
  • \( q'' = k \frac{d^2T}{dr^2} \)
Where:
  • \( q'' \) is the heat generation per unit volume
  • \( k \) is the thermal conductivity
  • \( \frac{d^2T}{dr^2} \) is the second-order radial temperature derivative
Integrating this equation gives:
  • \( T(r) = \frac{-\omega \mu P}{2k} r^2 + C_1 r + C_2 \)
The boundary conditions鈥攖emperature at the inner and outer radii鈥攈elp determine constants \( C_1 \) and \( C_2 \). This expression for \( T(r) \) lets us predict the temperature across the disk, helping ensure the material will withstand the thermal loads it encounters during braking. Understanding \( T(r) \) is essential for maintaining structural integrity and performance under thermal stress.
Thermal Conductivity
Thermal conductivity \( k \) is a property that indicates how well a material conducts heat. In the context of the disk brake problem, it's a critical factor in determining how heat moves through the disk material due to friction.In the energy balance equation:
  • \( k \frac{d^2T}{dr^2} = q'' \)
\( k \) appears as a constant that affects how quickly temperature changes can propagate through the disk. The higher the \( k \), the more efficiently heat is conducted, distributing the temperature more evenly across the disk.This property is essential for preventing localized overheating, which can lead to warping or failure in brake systems. By selecting disk materials with appropriate thermal conductivity, engineers can optimize heat dissipation and ensure stable performance even under high stress or repeated use conditions.
Choosing the right \( k \) enables ideal thermal performance, balancing between too rapid cooling or retaining excessive heat.

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Most popular questions from this chapter

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is \(3 \mathrm{~mm}\) thick and whose interior surface is maintained at a temperature of \(36^{\circ} \mathrm{C}\). On a calm day the convection heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), but with \(30 \mathrm{~km} / \mathrm{h}\) winds it reaches \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In both cases the ambient air temperature is \(-15^{\circ} \mathrm{C}\). (a) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at \(-15^{\circ} \mathrm{C}\) on the windy day?

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{i}=0.046 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{i}=50 \mathrm{~mm}\) and steel panels, each of thermal conductivity \(k_{p}=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{p}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{\infty, i}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as \(h_{i}=h_{o}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?
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