1. **Problem statement:** Find the effective thermal conductivity $k_{eff}$ in the vertical direction for a composite earth section made of stone, soil, and iron ore with given dimensions and conductivities. 2. **Given data:** - Widths: soil left $0.25'$, soil middle $2.5'$, soil right $0.25'$, total width $3'$ - Heights: stone $0.5'$, soil $4'$ - Thermal conductivities: $k_1=0.3$ (soil), $k_2=25$ (iron ore), $k_3=1.6$ (stone) 3. **Approach:** Since the geometry is two-dimensional and vertical heat flow is considered, the layers are in series vertically and parallel horizontally. 4. **Calculate vertical thermal resistance for each vertical section:** - Left soil column (soil + stone): $$R_{left} = \frac{0.5}{k_3 \times 0.25} + \frac{4}{k_1 \times 0.25} = \frac{0.5}{1.6 \times 0.25} + \frac{4}{0.3 \times 0.25}$$ $$= \frac{0.5}{0.4} + \frac{4}{0.075} = 1.25 + 53.33 = 54.58$$ - Middle iron ore column (iron ore + stone): $$R_{middle} = \frac{0.5}{k_3 \times 2.5} + \frac{4}{k_2 \times 2.5} = \frac{0.5}{1.6 \times 2.5} + \frac{4}{25 \times 2.5}$$ $$= \frac{0.5}{4} + \frac{4}{62.5} = 0.125 + 0.064 = 0.189$$ - Right soil column (soil + stone): same as left $$R_{right} = 54.58$$ 5. **Calculate total conductance (inverse of resistance) for each column:** $$G_{left} = \frac{1}{R_{left}} = \frac{1}{54.58} = 0.0183$$ $$G_{middle} = \frac{1}{R_{middle}} = \frac{1}{0.189} = 5.29$$ $$G_{right} = 0.0183$$ 6. **Total conductance for the entire width (parallel columns):** $$G_{total} = G_{left} + G_{middle} + G_{right} = 0.0183 + 5.29 + 0.0183 = 5.3266$$ 7. **Effective thermal resistance for entire width:** $$R_{eff} = \frac{1}{G_{total}} = \frac{1}{5.3266} = 0.1877$$ 8. **Effective thermal conductivity $k_{eff}$:** Using total height $H = 4.5'$ and total width $W = 3'$: $$k_{eff} = \frac{H}{R_{eff} \times W} = \frac{4.5}{0.1877 \times 3} = \frac{4.5}{0.563} = 7.99$$ **Final answer:** $$\boxed{k_{eff} \approx 8.0 \text{ Btu/h} \cdot \text{ft} \cdot \text{R}}$$