To solve problems involving heat conduction through multiple materials, like the one you've presented, we can use the concept of thermal equilibrium and the formula for heat transfer. In this case, we have rods made of two different materials with distinct thermal conductivities. Let's break down the solution step by step, and I'll also point out some shortcuts that can make this process quicker.
Understanding the Setup
We have three rods of material X and three rods of material Y, all of equal length and cross-sectional area. The end A is maintained at 60°C, while end E is at 10°C. The thermal conductivities (K) are given as:
- K of material X = 0.92 cal/sec·K
- K of material Y = 0.46 cal/sec·K
Heat Transfer Basics
The rate of heat transfer (Q) through a rod can be expressed using Fourier's law of heat conduction:
Q = K × A × (ΔT / L)
Where:
- Q = rate of heat transfer
- K = thermal conductivity of the material
- A = cross-sectional area
- ΔT = temperature difference across the length of the rod
- L = length of the rod
Setting Up the Equations
In this scenario, we can assume that the system reaches a steady state where the heat entering junction B equals the heat leaving it. We can set up equations for the heat transfer through each rod:
Heat Transfer through Rods
Let’s denote the temperatures at junctions B, C, and D as T_B, T_C, and T_D respectively. The heat transfer through the rods can be expressed as follows:
- For rod 1 (X): Q_1 = K_X × A × (60 - T_B) / L
- For rod 2 (Y): Q_2 = K_Y × A × (T_B - T_C) / L
- For rod 3 (Y): Q_3 = K_Y × A × (T_C - T_D) / L
- For rod 4 (X): Q_4 = K_X × A × (T_D - 10) / L
Applying the Steady State Condition
At steady state, the heat entering and leaving each junction must be equal:
- At junction B: Q_1 = Q_2
- At junction C: Q_2 = Q_3
- At junction D: Q_3 = Q_4
Solving the Equations
Now, we can substitute the values of K_X and K_Y into these equations:
1. For junction B:
0.92 × (60 - T_B) = 0.46 × (T_B - T_C)
2. For junction C:
0.46 × (T_B - T_C) = 0.46 × (T_C - T_D)
3. For junction D:
0.46 × (T_C - T_D) = 0.92 × (T_D - 10)
Finding the Junction Temperatures
By solving these equations step by step, we can find the temperatures at junctions B, C, and D. However, a shortcut to this problem is to recognize that the thermal conductivities are in a ratio, which can simplify calculations.
Using Ratios for Quick Calculation
Since the thermal conductivities are in the ratio of 2:1 (0.92:0.46), we can infer that the temperature drops across the rods will also reflect this ratio. This means that the temperature difference between junctions can be proportionally divided based on the conductivities.
Thus, we can quickly estimate:
- Temperature drop from A to B: 60°C to 30°C (halfway due to the ratio)
- Temperature drop from B to C and C to D: 30°C to 20°C and 20°C to 10°C (equal drops due to equal lengths and areas)
Therefore, the temperatures at junctions B, C, and D are approximately 30°C, 20°C, and 20°C respectively, confirming your answer.
Final Thoughts
In summary, while you can solve these problems step by step using equations, recognizing patterns in thermal conductivity ratios can save time and effort. This approach is particularly useful in problems with symmetrical setups and equal lengths and areas.
