# Problem Statement Two coaxial parallel disks of equal diameter \( D = 1\,\mathrm{m} \) are placed \( 1\,\mathrm{m} \) apart. Both behave as black bodies. **Question:** At what new distance should the disks be placed so that the radiation heat transfer between them is reduced by 75% (i.e., only 25% of the original value remains)? --- # Step 1: Radiative Heat Exchange Between Two Parallel Disks The net radiative heat exchange between two black parallel disks is given by: \[ Q = \sigma (T_1^4 - T_2^4) \cdot A_1 \cdot F_{1 \rightarrow 2} \] Where: - \( \sigma \): Stefan–Boltzmann constant - \( A_1 \): Area of disk 1 - \( F_{1 \rightarrow 2} \): View factor from disk 1 to disk 2 Since both disks are equal and coaxial: \[ A_1 = A_2 = \frac{\pi D^2}{4} \] --- # Step 2: View Factor for Coaxial Parallel Disks For two coaxial parallel disks of diameter \( D \) and separation \( L \): \[ F_{1 \rightarrow 2} = \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{1 + (D/L)^2}} \right] \] --- # Step 3: Setting Up the Reduction Condition Let \( Q_1 \) be the heat transfer at \( L_1 = 1\,\mathrm{m} \), and \( Q_2 \) at new distance \( L_2 \): \[ Q_2 = .25 Q_1 \] All other terms are unchanged except for the view factor, so: \[ \frac{F_{2}}{F_{1}} = \frac{Q_2}{Q_1} = .25 \] Let \( x = D / L \), so \( x_1 = 1/1 = 1 \), \( x_2 = 1/L_2 \). Compute \( F_1 \): \[ F_1 = \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{1 + 1^2}} \right] = \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{2}} \right] \] \[ \sqrt{2} \approx 1.4142 \] \[ \frac{1}{\sqrt{2}} \approx .7071 \] \[ F_1 = \frac{1}{2} \left[ 1 - .7071 \right] = \frac{1}{2} \times .2929 = .14645 \] Let \( F_2 \) be the view factor at new distance \( L_2 \): \[ F_2 = \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{1 + (1/L_2)^2}} \right] \] From the reduction condition: \[ \frac{F_2}{F_1} = .25 \] \[ F_2 = .25 F_1 = .25 \times .14645 = .0366125 \] --- # Step 4: Solve for the New Distance \( L_2 \) Set \( F_2 \) to the value above and solve for \( L_2 \): \[ .0366125 = \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{1 + (1/L_2)^2}} \right] \] \[ .073225 = 1 - \frac{1}{\sqrt{1 + (1/L_2)^2}} \] \[ \frac{1}{\sqrt{1 + (1/L_2)^2}} = 1 - .073225 = .926775 \] \[ \sqrt{1 + (1/L_2)^2} = \frac{1}{.926775} = 1.07896 \] \[ 1 + (1/L_2)^2 = (1.07896)^2 = 1.16312 \] \[ (1/L_2)^2 = 1.16312 - 1 = .16312 \] \[ 1/L_2 = \sqrt{.16312} = .40389 \] \[ L_2 = \frac{1}{.40389} \approx 2.475 \] --- # Final Answer ## New Distance \[ \boxed{L_2 \approx 2.48~\mathrm{m}} \] **Therefore, the disks should be placed approximately \( 2.48\,\mathrm{m} \) apart for a 75% reduction in radiation heat transfer.** --- ## Summary of Steps 1. **Write the view factor equation for two coaxial disks.** 2. **Plug in initial and required reduction to find the new view factor.** 3. **Solve for the new distance.**