Solution: Maximum Power of Heater
Step 1: Initial Conditions and Mole Distribution
For membrane to remain intact the pressure difference in the steady state must not cross the limit for membrane. In the limiting case it should be equal to the maximum pressure the membrane can withstand. The cylinder is divided into two parts with volumes $V_1$ and $V_2$ such that $V_2 = \eta V_1$. Since the gas was initially in equilibrium, the density is uniform. Therefore, the number of moles is proportional to volume. $$ n_1 = \frac{1}{1+\eta} n_{total}, \quad n_2 = \frac{\eta}{1+\eta} n_{total} $$ where $n_{total} = m/M$.
Step 2: Pressure Difference Relation
The difference in pressure across the membrane is given by ideal gas law: $$ \Delta P = P_1 – P_2 = \frac{n_1 R T_1}{V_1} – \frac{n_2 R T_2}{V_2} $$ Using the mole-volume proportionality ($n_1/V_1 = n_2/V_2 = n/V_{total}$): $$ \Delta P = \frac{n R}{V} (T_1 – T_2) \implies T_1 – T_2 = \frac{\Delta P V}{n R} $$
Step 3: Energy Balance in Quasi-Steady State
Let $H$ be the heat transfer rate coefficient such that power transferred is $P_{trans} = H(T_1 – T_2)$.
In the state where pressure difference is maximum and constant, the rate of temperature rise must be equal in both chambers ($\dot{T_1} = \dot{T_2} = \dot{T}$).
Energy Balance for Chamber 2:
$$ n_2 C_v \dot{T} = H(T_1 – T_2) $$
Energy Balance for Whole System:
$$ (n_1 + n_2) C_v \dot{T} = P_{max} \implies n C_v \dot{T} = P_{max} $$
Step 4: Deriving Max Power
Substitute $\dot{T}$ from the system equation into the Chamber 2 equation: $$ n_2 C_v \left( \frac{P_{max}}{n C_v} \right) = H(T_1 – T_2) $$ $$ \frac{n_2}{n} P_{max} = H \left( \frac{\Delta P V}{n R} \right) $$ Since $\frac{n_2}{n} = \frac{\eta}{\eta+1}$: $$ \frac{\eta}{\eta+1} P_{max} = \frac{H \Delta P V}{n R} $$ $$ P_{max} = \left( \frac{\eta+1}{\eta} \right) \frac{M V \Delta P H}{m R} $$