THERMAL BYU 12

Concept: In a steady state, the rate of heat generation inside a system must equal the rate of heat transfer out through its boundaries.

Step 1: Analyzing Figure I
Let the power of each identical heater be $P$. In Figure I, there are three heaters, so the total power generated inside the chamber is $3P$. The heat flows out through a single membrane with thermal conductance $K$.
Using the law of thermal conduction, heat current $H = K \Delta T$: $$3P = K(T_{inside} – T_{outside})$$ $$3P = K(25 – 10) = 15K$$ $$P = 5K \quad \dots \text{(Equation 1)}$$

Step 2: Analyzing Figure II
The chamber is divided into three compartments (A, B, C) by identical membranes. Heat flows sequentially from C $\to$ B $\to$ A $\to$ Outside.

• Compartment A (Outermost): The heat passing through the membrane at A is the sum of heat generated in all three compartments (C, B, and A), which is $3P$. $$3P = K(T_A – 10)$$ Substituting $P = 5K$: $$15K = K(T_A – 10) \Rightarrow 15 = T_A – 10 \Rightarrow \mathbf{T_A = 25^\circ \text{C}}$$
• Compartment B (Middle): The heat passing through the membrane between B and A corresponds to the heat generated in C and B, which is $2P$. $$2P = K(T_B – T_A)$$ $$2(5K) = K(T_B – 25) \Rightarrow 10 = T_B – 25 \Rightarrow \mathbf{T_B = 35^\circ \text{C}}$$
• Compartment C (Innermost): The heat passing through the membrane between C and B corresponds to the heat generated in C only, which is $P$. $$P = K(T_C – T_B)$$ $$5K = K(T_C – 35) \Rightarrow 5 = T_C – 35 \Rightarrow \mathbf{T_C = 40^\circ \text{C}}$$