Question

A very large metal plate of thickness d and thermal conductivity k is cooled by a stream of air at temperature \(T_\infty = 300\) K with a heat transfer coefficient h, as shown in the figure. The centerline temperature of the plate is \(T_P\). In which of the following case(s) can the lumped parameter model be used to study the heat transfer in the metal plate? 

• A. \(h = 10\) Wm⁻²K⁻¹, \(k = 100\) Wm⁻¹K⁻¹, \(d = 1\) mm, \(T_P = 350\) K
• B. \(h = 100\) Wm⁻²K⁻¹, \(k = 100\) Wm⁻¹K⁻¹, \(d = 1\) m, \(T_P = 325\) K
• C. \(h = 100\) Wm⁻²K⁻¹, \(k = 1000\) Wm⁻¹K⁻¹, \(d = 1\) mm, \(T_P = 325\) K
• D. \(h = 1000\) Wm⁻²K⁻¹, \(k = 1\) Wm⁻¹K⁻¹, \(d = 1\) m, \(T_P = 350\) K

Hint:

The Biot number \(\text{Bi} = \frac{h L_c}{k}\) can be thought of as the ratio of resistances: \(\frac{\text{Internal Conduction Resistance}}{\text{External Convection Resistance}}\). The lumped model works when this ratio is small, meaning the internal resistance is negligible, and the temperature inside the body is uniform. For a plane wall of thickness \(d\), the characteristic length \(L_c\) is \(d/2\).

Answer 1

The Correct Option is A, C

Step 1: Understanding the Concept:
The lumped parameter model (or lumped capacitance method) is a simplification used in transient heat conduction analysis. It is valid when the internal resistance to heat conduction within the object is negligible compared to the external resistance to heat convection at the surface. This condition is quantified by the Biot number (Bi).
Step 2: Key Formula or Approach:
The lumped parameter model is considered valid if the Biot number is small, typically: \[ \text{Bi} \le 0.1 \] The Biot number is defined as: \[ \text{Bi} = \frac{h L_c}{k} \] where:
- \(h\) is the convection heat transfer coefficient.
- \(k\) is the thermal conductivity of the solid.
- \(L_c\) is the characteristic length of the object. For a large plate cooled from both sides, the characteristic length is half the thickness: \(L_c = d/2\).
The temperatures \(T_P\) and \(T_\infty\) are not needed to check the validity of the model, only to know that heat transfer is occurring.
Step 3: Detailed Calculation:
Let's calculate the Biot number for each case. Remember to convert \(d\) to meters.
(A) \(h = 10, k = 100, d = 1\) mm = 0.001 m:
- \(L_c = d/2 = 0.001 / 2 = 0.0005\) m.
- \(\text{Bi} = \frac{h L_c}{k} = \frac{10 \times 0.0005}{100} = \frac{0.005}{100} = 0.00005\).
- Since \(0.00005 \le 0.1\), the lumped parameter model is valid.
(B) \(h = 100, k = 100, d = 1\) m:
- \(L_c = d/2 = 1 / 2 = 0.5\) m.
- \(\text{Bi} = \frac{h L_c}{k} = \frac{100 \times 0.5}{100} = 0.5\).
- Since \(0.5>0.1\), the model is not valid.
(C) \(h = 100, k = 1000, d = 1\) mm = 0.001 m:
- \(L_c = d/2 = 0.001 / 2 = 0.0005\) m.
- \(\text{Bi} = \frac{h L_c}{k} = \frac{100 \times 0.0005}{1000} = \frac{0.05}{1000} = 0.00005\).
- Since \(0.00005 \le 0.1\), the lumped parameter model is valid.
(D) \(h = 1000, k = 1, d = 1\) m:
- \(L_c = d/2 = 1 / 2 = 0.5\) m.
- \(\text{Bi} = \frac{h L_c}{k} = \frac{1000 \times 0.5}{1} = 500\).
- Since \(500>0.1\), the model is not valid.
Step 4: Final Answer:
The lumped parameter model can be used in cases (A) and (C).
Step 5: Why This is Correct:
The validity of the lumped capacitance model depends solely on the Biot number being less than or equal to 0.1. The calculations show that only cases (A) and (C) satisfy this criterion. Physically, these cases represent situations with high internal conduction (high \(k\)) and/or low external convection (low \(h\)) and/or small dimensions (small \(d\)), which leads to a uniform temperature distribution within the plate.