Chapter 4: Problem 2
What is lumped system analysis? When is it applicable?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 4: Problem 2
What is lumped system analysis? When is it applicable?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeConsider a 7.6-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(\left.\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Such a meat chunk intially at \(2^{\circ} \mathrm{C}\) is dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the center temperature of the meat chunk to rise to \(75{ }^{\circ} \mathrm{C}\) is (a) \(136 \mathrm{~min}\) (b) \(21.2 \mathrm{~min}\) (c) \(13.6 \mathrm{~min}\) (d) \(11.0 \mathrm{~min}\) (e) \(8.5 \mathrm{~min}\)
An ordinary egg can be approximated as a \(5.5-\mathrm{cm}-\) diameter sphere whose properties are roughly \(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.14 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\). The egg is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is dropped into boiling water at \(97^{\circ} \mathrm{C}\). Taking the convection heat transfer coefficient to be \(h=\) \(1400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the center of the egg to reach \(70^{\circ} \mathrm{C}\). Solve this problem using analytical one-term approximation method (not the Heisler charts).
How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?
The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).
Aluminum wires, \(3 \mathrm{~mm}\) in diameter, are produced by extrusion. The wires leave the extruder at an average temperature of \(350^{\circ} \mathrm{C}\) and at a linear rate of \(10 \mathrm{~m} / \mathrm{min}\). Before leaving the extrusion room, the wires are cooled to an average temperature of \(50^{\circ} \mathrm{C}\) by transferring heat to the surrounding air at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the necessary length of the wire cooling section in the extrusion room.
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