Let's solve the problem step by step! --- # Problem Understanding You are **asked to determine the temperature at the contact point of two very long fins ("aletas muy largas")**, one circular and one square, given their properties and operating conditions in the table. ## Table Values (translated and unified to SI units where needed) | Shape | Dimension | Conductivity (k) | h (conv. coeff.) | T(, tip) | T∞ (ambient) | α (diffusivity) | |----------|---------------------|-------------------------|----------------------------|-------------------|--------------|---------------------------| | Circular | r = 5 cm = .05 m | 80 W/m·K | 60 W/m²·K | T(,1) = 150°C | 25°C | 1.172 ×10⁻⁵ m²/s | | Square | l = 2 in = .0508 m | 30 Btu/h·ft·°F ≈ 51.9 W/m·K | 18 Btu/h·ft²·°F ≈ 102.2 W/m²·K | T(.4) = 302°F ≈ 150°C | 82.4°F ≈ 28°C | .182 in²/s ≈ 1.174×10⁻⁵ m²/s | > For simplicity, let's work with the **circular fin** values in SI units. --- # Step 1: Governing Equation for "Very Long" Fins For a **very long fin**, the end temperature is given by: \[ T(x) = T_\infty + (T_b - T_\infty) e^{-mx} \] - \( T(x) \): temperature at distance x from the base - \( T_\infty \): ambient temperature - \( T_b \): base (contact) temperature (**what we need to find**) - \( m = \sqrt{\frac{hP}{kA_c}} \) - h: convection coefficient - P: perimeter - \( A_c \): cross-sectional area - k: thermal conductivity --- # Step 2: Set Up Equations for Both Fins At the **contact point x= for both fins** (since they touch), the temperatures must be equal: \[ T_{circular}() = T_{square}() = T_{contact} \] But we don't have both base temperatures; instead, we have temperatures at certain points from the base. For the **circular fin**: - At \( x_1 = .1 \) m, \( T_1 = 150^\circ C \) For the **square fin**: - At \( x_2 = .4 \) m, \( T_2 = 150^\circ C \) Let the unknown contact temperature be \( T_ \). --- # Step 3: Write the Two Fin Equations at Their Measured Points ## Circular Fin \[ T_1 = T_\infty + (T_ - T_\infty) e^{-m_1 x_1} \] Where: - \( T_1 = 150^\circ C \) - \( T_\infty = 25^\circ C \) - \( x_1 = .1 \) m ### Calculate \( m_1 \): - \( h = 60 \) W/m²·K - \( k = 80 \) W/m·K - For a rod: \( r = .05 \) m - \( P = 2\pi r = 2\pi(.05) = .314 \) m - \( A_c = \pi r^2 = .00785 \) m² \[ m_1 = \sqrt{ \frac{hP}{kA_c} } = \sqrt{ \frac{ 60 \times .314 }{ 80 \times .00785 } } \] Calculate numerator: - \( 60 \times .314 = 18.84 \) Calculate denominator: - \( 80 \times .00785 = .628 \) So, \[ m_1 = \sqrt{ \frac{18.84}{.628} } = \sqrt{30} \approx 5.477 \ \text{m}^{-1} \] --- ## Square Fin - \( T_2 = 150^\circ C \) - \( T_\infty = 28^\circ C \) - \( x_2 = .4 \) m ### Calculate \( m_2 \): - \( h = 102.2 \) W/m²·K - \( k = 51.9 \) W/m·K - For a square: \( l = .0508 \) m - \( P = 4l = .2032 \) m - \( A_c = l^2 = .00258 \) m² \[ m_2 = \sqrt{ \frac{ hP }{ kA_c } } = \sqrt{ \frac{ 102.2 \times .2032 }{ 51.9 \times .00258 } } \] Numerator: - \( 102.2 \times .2032 = 20.78 \) Denominator: - \( 51.9 \times .00258 = .1338 \) So, \[ m_2 = \sqrt{ \frac{20.78}{.1338} } = \sqrt{155.3} \approx 12.46 \ \text{m}^{-1} \] --- # Step 4: Write Both Equations **Circular fin:** \[ 150 = 25 + (T_ - 25) e^{-5.477 \times .1} \] **Square fin:** \[ 150 = 28 + (T_ - 28) e^{-12.46 \times .4} \] --- # Step 5: Solve for \( T_ \) Let's solve each exponent: - \( 5.477 \times .1 = .5477 \) - \( e^{-.5477} \approx .578 \) - \( 12.46 \times .4 = 4.984 \) - \( e^{-4.984} \approx .00682 \) Now plug these in: ### Circular: \[ 150 = 25 + (T_ - 25) \times .578 \] \[ 150 - 25 = (T_ - 25) \times .578 \] \[ 125 = (T_ - 25) \times .578 \] \[ T_ - 25 = \frac{125}{.578} \approx 216.3 \] \[ T_ = 216.3 + 25 = 241.3^\circ C \] --- ### Square: \[ 150 = 28 + (T_ - 28) \times .00682 \] \[ 150 - 28 = (T_ - 28) \times .00682 \] \[ 122 = (T_ - 28) \times .00682 \] \[ T_ - 28 = \frac{122}{.00682} \approx 17897 \] \[ T_ = 17897 + 28 = 17925^\circ C \] This value is **unphysically high**. Let's check the numbers: - \( e^{-4.984} \) should be about \( .00683 \), which matches. - The main issue is that the square fin's ambient temperature is only 3°C higher than the circular fin, but the decay is much more severe (higher m, longer distance). --- # Step 6: Conclusion **The two equations produce very different answers, suggesting either the units or initial data for the square fin are inconsistent.** However, for the **circular fin**, the process and calculation can be relied upon: --- ## **Final Answer (Circular Fin):** \[ \boxed{T_{contact} = 241.3^\circ C} \] This is the temperature at the contact point, calculated based on the circular fin data. --- ### **Summary of Steps** 1. **Write the fin temperature distribution formula for a very long fin.** 2. **Calculate the fin parameter \( m \) for each fin.** 3. **Set up the temperature equation at the given distance for each fin.** 4. **Solve for the unknown base/contact temperature.** --- ## **If you want to check the square fin, double-check all units and input data.** But for the circular fin, the answer is **241.3°C**. If you need the square fin calculation in detail or want to check unit conversions, let me know!