In a manufacturing plant, a quench hardening process is used to treat steel ball bearings \((c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\) \(60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) of \(25 \mathrm{~mm}\) in diameter. After being heated to a prescribed temperature, the steel ball bearings are quenched. Determine the rate of heat loss if the rate of temperature decrease in the ball bearing at a given instant during the quenching process is \(50 \mathrm{~K} / \mathrm{s}\).

Short Answer

Expert verified
Answer: The rate of heat loss during the quench hardening process for the given steel ball bearing is 1625 W.

Step by step solution

01

Calculate volume and mass of steel ball bearing

To determine the mass of the steel ball, first, calculate its volume: Volume (V) = \(\frac{4}{3}\pi (\frac{d}{2})^3\) Where d = diameter of the ball bearing Plug in the values: d = 25mm = 0.025m V = \(\frac{4}{3}\pi (\frac{0.025}{2})^3 = 8.18 \times 10^{-6} \mathrm{m}^3\) Now, calculate mass using the steel's density formula: Mass (m) = Volume × Density m = \(8.18 \times 10^{-6} \mathrm{m}^3 \times 7900 \mathrm{~kg} / \mathrm{m}^{3} = 0.065 \mathrm{~kg}\)
02

Calculate the rate of heat loss

To calculate the rate of heat loss (Q_dot), use the following formula: Q_dot = \(m c \frac{dT}{dt}\) Plug in the values: Q_dot = \((0.065 \mathrm{~kg})(500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(-50 \mathrm{~K} / \mathrm{s}) = -1625 \mathrm{~W}\) Note that the negative sign indicates heat being lost from the steel ball bearing. The rate of heat loss during the quench hardening process is \(1625 \mathrm{W}\).

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Key Concepts

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

Quench Hardening
Quench hardening is a metalworking process used to increase the hardness and strength of steel and other alloys. During this heat treatment, a high-temperature metal part is rapidly cooled, or 'quenched,' in a liquid such as water, oil, or brine. This rapid cooling transforms the metal's microstructure, causing it to become much harder.

However, this rapid temperature change can induce stresses and distortions in the material, which is why precise control of quenching conditions, including the cooling rate and the temperature of the quenching medium, is crucial to achieving the desired qualities in the final product.

Understanding the rate at which heat is removed from the material during quenching is key to optimizing the hardening process without causing unwanted effects such as cracking.
Thermal Conductivity
Thermal conductivity, denoted by the symbol 'k', is a measure of a material's ability to conduct heat. It represents how easily heat can pass through a material. When a metal like steel is quenched, its thermal conductivity affects how rapidly heat is transferred from the metal to the surrounding cooling medium.

In terms of quench hardening, a higher thermal conductivity means that heat can escape more quickly, which can influence the cooling rate and, consequently, the metal's final microstructure. Knowing the thermal conductivity of steel is essential for predicting and controlling heat loss during the quench hardening process.
Specific Heat Capacity
The specific heat capacity, commonly represented by the symbol 'c', is a physical property that indicates how much heat energy is required to raise the temperature of a unit mass of a substance by one-degree Celsius or Kelvin. In the context of our steel ball bearings, the specific heat capacity affects how much heat needs to be removed to lower the temperature of the material.

A higher specific heat capacity means that more energy must be extracted to produce the same change in temperature, which directly impacts the rate of cooling and the resulting hardness of the quenched metal. It's an important factor in calculating heat loss and managing the heat treatment process.
Rate of Temperature Change
The rate of temperature change, denoted as \(\frac{dT}{dt}\), is the speed at which the temperature of an object changes over time, typically expressed in degrees per second or minute. This rate is crucial in quench hardening since it determines how quickly the object cools down and solidifies into its hardened state.

A rapid rate of temperature change, as in quenching, is necessary to achieve certain material properties, like increased toughness and strength. By understanding the relationship between the specific heat capacity, the mass of the object, and the desired temperature change rate, engineers can calculate the rate of heat loss. This calculation is fundamental in designing and regulating the quenching process to match specific metallurgical outcomes.

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Most popular questions from this chapter

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

A spherical communication satellite with a diameter of \(2.5 \mathrm{~m}\) is orbiting around the earth. The outer surface of the satellite in space has an emissivity of \(0.75\) and a solar absorptivity of \(0.10\), while solar radiation is incident on the spacecraft at a rate of \(1000 \mathrm{~W} / \mathrm{m}^{2}\). If the satellite is made of material with an average thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the midpoint temperature is \(0^{\circ} \mathrm{C}\), determine the heat generation rate and the surface temperature of the satellite.

How do differential equations with constant coefficients differ from those with variable coefficients? Give an example for each type.

Exhaust gases from a manufacturing plant are being discharged through a 10 - \(\mathrm{m}\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\), wall thickness of \(10 \mathrm{~cm}\), and thermal conductivity of \(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The exhaust gases are discharged at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure specific heat of the exhaust gasses is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular day, the outer surface of the exhaust stack experiences radiation with the surrounding at \(27^{\circ} \mathrm{C}\), and convection with the ambient air at \(27^{\circ} \mathrm{C}\) also, with an average convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are 0.9. Assuming steady one-dimensional heat transfer, (a) obtain the variation of temperature in the exhaust stack wall and (b) determine the inner surface temperature of the exhaust stack.

The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

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