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Chapter 2: Problem 158

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 bearings 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 quenching process is 1615 W.

Step by step solution

01

Find the volume of the steel ball

To find the mass of the steel ball-bearing, we first need to find its volume. The volume of a sphere can be calculated using the formula: \(V = \dfrac{4}{3} \pi r^{3}\) where V is the volume, and r is the radius. The diameter of the steel ball bearing is given as 25mm (0.025m), so the radius is half of that, which is 12.5mm (0.0125m). Now let's plug the values into the equation: \(V = \dfrac{4}{3} \pi(0.0125)^{3} ≈ 8.18 × 10^{-6} \mathrm{m}^3\)
02

Determine the mass of the steel ball

To find the mass of the steel ball, we will use the formula: \(mass = \rho × V \) where ρ is the density and V is the volume. We have already calculated the volume of the steel ball and we are also given the density of the material (ρ = 7900 kg/m³). Plug these values into the equation: \(mass = 7900 × 8.18 × 10^{-6} ≈ 0.0646 \mathrm{kg}\)
03

Calculate the rate of heat loss

Now that we have the mass of the steel ball, we can use this information to calculate the rate of heat loss using the equation mentioned before: Rate of heat loss = mass × specific heat × rate of temperature change We have all the necessary values: mass = 0.0646 kg, specific heat (c) = 500 J/(kg·K), and the rate of temperature change = -50 K/s (the negative sign indicates a decrease in temperature). Plug these values into the equation: Rate of heat loss = \(0.0646 × 500 × (-50) ≈ -1615 \mathrm{W}\) The negative sign indicates that the heat is being lost from the steel ball bearing, which is expected during the quenching process. So, the rate of heat loss during the quenching process is 1615 W.

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Most popular questions from this chapter

The heat conduction equation in a medium is given in its simplest form as $$ \frac{1}{r} \frac{d}{d r}\left(r k \frac{d T}{d r}\right)+\hat{e}_{\mathrm{gen}}=0 $$ Select the wrong statement below. (a) The medium is of cylindrical shape. (b) The thermal conductivity of the medium is constant. (c) Heat transfer through the medium is steady. (d) There is heat generation within the medium. (e) Heat conduction through the medium is one-dimensional.

How do you recognize a linear homogeneous differential equation? Give an example and explain why it is linear and homogeneous.

The temperatures at the inner and outer surfaces of a \(15-\mathrm{cm}\)-thick plane wall are measured to be \(40^{\circ} \mathrm{C}\) and $28^{\circ} \mathrm{C}$, respectively. The expression for steady, one-dimensional variation of temperature in the wall is (a) \(T(x)=28 x+40\) (b) \(T(x)=-40 x+28\) (c) \(T(x)=40 x+28\) (d) \(T(x)=-80 x+40\) (e) \(T(x)=40 x-80\)

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}$. The 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.

A large plane wall has a thickness \(L=50 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the left surface \((x=0)\), it is subjected to a uniform heat flux \(\dot{q}_{0}\) while the surface temperature \(T_{0}\) is constant. On the right surface, it experiences convection and radiation heat transfer while the surface temperature is \(T_{L}=225^{\circ} \mathrm{C}\) and the surrounding temperature is \(25^{\circ} \mathrm{C}\). The emissivity and the convection heat transfer coefficient on the right surface are \(0.7\) and $15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. (a) Show that the variation of temperature in the wall can be expressed as $T(x)=\left(\dot{q}_{0} / k\right)(L-x)+T_{L},(b)\( calculate the heat flux \)\dot{q}_{0}$ on the left face of the wall, and (c) determine the temperature of the left surface of the wall at \(x=0\).
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