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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 spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and dropped into a large body of water at \(T_{\infty}\) where it is cooled by convection with an average convection heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\), and determine net heat flux \(\dot{q}_{0}\) absorbed by the solar collector.

Consider steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly. The error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature is \((a)\) none, \((b)\) small, or \((c)\) significant.

In a food processing facility, a spherical container of inner radius \(r_{1}=40 \mathrm{~cm}\), outer radius \(r_{2}=41 \mathrm{~cm}\), and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a 800 -W electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(120^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container material by solving the differential equation, and \((c)\) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

What is heat generation? Give some examples.

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