Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air. For which solid is the lumped system analysis more likely to be applicable? Why?

Short Answer

Expert verified
Answer: The lumped system analysis is more likely to be applicable for the solid body cooling in air, as the Biot number for the solid body in air is lower than the solid body in water, allowing the assumption of a uniform temperature within the body to be valid.

Step by step solution

01

Understand the heat transfer mechanisms

In both situations, heat transfer occurs between the hot solid body and its surroundings. For the solid submerged in water, convection is the dominant mechanism of heat transfer. In the case of the solid body exposed to air, convection and radiation both contribute to the heat transfer process.
02

Analyze heat transfer rate

The rate at which heat is transferred between the hot solid body and its surroundings depends on the heat transfer coefficient. The higher the heat transfer coefficient, the faster the temperature of the solid will decrease, and the less likely it is that the lumped system analysis will be applicable. In general, the heat transfer coefficient for convection in water is much higher than air due to the higher thermal conductivity of water.
03

Biot number

To determine the applicability of lumped system analysis, we consider the Biot number (Bi), which is a dimensionless number that characterizes the relative importance of internal conduction to external convection. It is given by the formula: Bi = \frac{h * L_c}{k} Where: - \(h\) is the convective heat transfer coefficient, - \(L_c\) is the characteristic length of the solid (usually, the volume divided by the surface area), and - \(k\) is the thermal conductivity of the solid. In the lumped system analysis, the Biot number must be much smaller than 1, typically smaller than 0.1, which means that the rate of internal heat conduction is much faster than the rate of heat transfer to the surroundings. In such a case, the temperature within the solid remains uniform at all times.
04

Comparison of the Biot number for two cases

Since the heat transfer coefficient for convection in water is much higher than convection in air, the Biot number for the solid in water will also be higher than the solid in air, given that the characteristic length and thermal conductivity of the solid are identical.
05

Conclusion

As the Biot number for the solid body in air is lower than the solid body in water, the lumped system analysis is more likely to be applicable for the solid body cooling in air. This is because, in this case, the rate of internal heat conduction is faster as compared to the heat transfer to surroundings, allowing the assumption of a uniform temperature within the body to be valid.

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Most popular questions from this chapter

Layers of 23 -cm-thick meat slabs \((k=0.47 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(7^{\circ} \mathrm{C}\) are to be frozen by refrigerated air at \(-30^{\circ} \mathrm{C}\) flowing at a velocity of \(1.4 \mathrm{~m} / \mathrm{s}\). The average heat transfer coefficient between the meat and the air is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to \(-18^{\circ} \mathrm{C}\). Also, determine the surface temperature of the meat slab at that time.

The 40-cm-thick roof of a large room made of concrete \(\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.88 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). After a heavy snow storm, the outer surface of the roof remains covered with snow at \(-5^{\circ} \mathrm{C}\). The roof temperature at \(18.2 \mathrm{~cm}\) distance from the outer surface after a period of 2 hours is (a) \(14^{\circ} \mathrm{C}\) (b) \(12.5^{\circ} \mathrm{C}\) (c) \(7.8^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-5^{\circ} \mathrm{C}\)

A hot dog can be considered to be a \(12-\mathrm{cm}-\mathrm{long}\) cylinder whose diameter is \(2 \mathrm{~cm}\) and whose properties are \(\rho=980 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.9 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.76 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(2 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). A hot dog initially at \(5^{\circ} \mathrm{C}\) is dropped into boiling water at \(100^{\circ} \mathrm{C}\). The heat transfer coefficient at the surface of the hot dog is estimated to be \(600 \mathrm{~W} / \mathrm{m}^{2}\). K. If the hot dog is considered cooked when its center temperature reaches \(80^{\circ} \mathrm{C}\), determine how long it will take to cook it in the boiling water.

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

Layers of 6-in-thick meat slabs \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(50^{\circ} \mathrm{F}\) are cooled by refrigerated air at \(23^{\circ} \mathrm{F}\) to a temperature of \(36^{\circ} \mathrm{F}\) at their center in \(12 \mathrm{~h}\). Estimate the average heat transfer coefficient during this cooling process. Solve this problem using the Heisler charts. Answer: \(1.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\)

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