Question

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

Answer: The lumped system analysis is more likely to be applicable for the solid that is allowed to cool naturally in the air.

Step-by-step solution

Find the heat transfer resistance in the environment

The heat transfer resistance in the environment can be represented by the heat transfer coefficients, \(h\). As water has a higher heat transfer coefficient than air, we can assume that \(h_{water} > h_{air}\).

Analyze the Biot number

The Biot number is given as \(Bi = \frac{hL}{k}\), where \(L\) is the characteristic length and \(k\) is the thermal conductivity of the solid. Since the solids are identical, we assume that the characteristic lengths and thermal conductivities are equal. Therefore, we can compare the Biot numbers using the heat transfer coefficients we determined in Step 1. Since \(h_{water} > h_{air}\), the Biot number for the solid in water will be larger than the Biot number for the solid in air, i.e., \(Bi_{water} > Bi_{air}\).

Determine the applicability of the lumped system analysis

For lumped system analysis to be applicable, the Biot number must be much less than 1 (Bi << 1). Since \(Bi_{water} > Bi_{air}\), the Biot number for the solid in the air is more likely to be much smaller than 1. Therefore, the lumped system analysis is more likely to be applicable for the solid that is allowed to cool naturally in the air.