Let's solve the problem step by step. --- ## Given Data- **Al ball:** - \( \rho C = 2.5 \) MJ/m³K = \( 2.5 \times 10^6 \) J/m³K - \( k = 240 \) W/mK - Radius \( r = 2 \) cm = .02 m - Initial temperature \( T_i = 320 \) K - Surrounding temperature \( T_\infty = 280 \) K - Heat transfer coefficient \( h = 50 \) W/m²K --- ## (a) **Time Constant and Lumped Capacitance Check** ### **Step 1: Characteristic Length and Biot Number** - **Volume of sphere:** \( V = \frac{4}{3} \pi r^3 \) - **Surface area of sphere:** \( A = 4 \pi r^2 \) - **Characteristic length (\( L_c \)):** \[ L_c = \frac{V}{A} = \frac{\frac{4}{3}\pi r^3}{4\pi r^2} = \frac{r}{3} \] \[ L_c = \frac{.02}{3} = .00667~\text{m} \] - **Biot number (\( Bi \)):** \[ Bi = \frac{h L_c}{k} \] \[ Bi = \frac{50 \times .00667}{240} = .00139 \] - Since \( Bi < .1 \), **the lumped capacitance model is appropriate**. ### **Step 2: Time Constant (\( \tau \))** \[ \tau = \frac{\rho C_p V}{hA} \] - \( V = \frac{4}{3} \pi (.02)^3 = 3.351 \times 10^{-5}~\text{m}^3 \) - \( A = 4\pi (.02)^2 = .00503~\text{m}^2 \) \[ \tau = \frac{2.5 \times 10^6 \times 3.351 \times 10^{-5}}{50 \times .00503} \] \[ \tau = \frac{83.775}{.2515} = 333~\text{s} \] --- ### **Summary (a):** - **Biot number** \( = .00139 \) (**lumped model is appropriate**) - **Time constant** \( \tau = 333~\text{s} \) --- ## (b) **Length of Fins for Infinite Model** Given: 6 fins, each with \( r_{fin} = 1 \) mm = .001 m, \( k = 240 \) W/mK. For a pin fin: - **Fin parameter** \( m = \sqrt{\frac{h P}{k A_c}} \) - For a cylinder, \( P = 2\pi r_{fin} \), \( A_c = \pi r_{fin}2 \): \[ m = \sqrt{\frac{h \cdot 2\pi r_{fin}}{k \cdot \pi r_{fin}^2}} = \sqrt{\frac{2h}{k r_{fin}}} \] \[ m = \sqrt{\frac{2 \cdot 50}{240 \cdot .001}} = \sqrt{\frac{100}{.24}} = \sqrt{416.67} = 20.4~\text{m}^{-1} \] For an "infinite" fin, \( mL \geq 2.5 \): \[ L \geq \frac{2.5}{m} = \frac{2.5}{20.4} = .123~\text{m} \] --- ### **Summary (b):** - **Fin length for infinite model:** **at least .123 m (12.3 cm)** --- ## (c) **With 6 Infinitely Long Fins** - **Fins are infinitely long, same temperature as the ball** - **Neglect fin thermal capacity** ### **Step 1: Fin Resistance** - **Fin base area:** \( A_{fin,base} = \pi r_{fin}^2 \) - **Fin resistance:** For infinite fin, \[ R_{fin} = \frac{1}{m k A_{fin,base}} \] - Number of fins: 6 ### **Step 2: Total Thermal Resistance** - **Sphere convection resistance:** \( R_{sphere} = \frac{1}{hA_{sphere}} \) - **Total resistance (parallel):** \[ \frac{1}{R_{total}} = \frac{1}{R_{sphere}} + 6 \cdot \frac{1}{R_{fin}} \] Calculate each: #### 1. **Fin base area** \[ A_{fin,base} = \pi (.001)^2 = 3.14 \times 10^{-6}~\text{m}^2 \] #### 2. **Fin resistance** \[ R_{fin} = \frac{1}{20.4 \cdot 240 \cdot 3.14 \times 10^{-6}} = \frac{1}{.0154} = 64.9~\text{K/W} \] #### 3. **Sphere convection resistance** \[ R_{sphere} = \frac{1}{50 \cdot .00503} = \frac{1}{.2515} = 3.98~\text{K/W} \] #### 4. **Total resistance** \[ \frac{1}{R_{total}} = \frac{1}{3.98} + \frac{6}{64.9} = .251 + .0924 = .343 \] \[ R_{total} = \frac{1}{.343} = 2.92~\text{K/W} \] ### **Step 3: New Time Constant** \[ \tau = (\rho C_p V) \cdot R_{total} \] \[ \tau = 83.775 \cdot 2.92 = 244.6~\text{s} \] ### **Step 4: Check Biot Number** \[ Bi = \frac{(\rho C_p V)}{A_{total} \cdot R_{total} \cdot k} \] But since the added fins are assumed to be at the same temperature as the ball and their capacity is neglected, the lumped model is still valid. --- ### **Summary (c):** - **Lumped model is still appropriate** - **New time constant with 6 infinite fins:** \( \boxed{245~\text{s}} \) (rounded) --- ## **Final Answers** ### (a) **Time constant:** \( \boxed{333~\text{s}} \); **Lumped model is appropriate** ### (b) **Fin length for infinite:** \( \boxed{.123~\text{m}} \) (12.3 cm) ### (c) **With 6 infinite fins:** Lumped model is appropriate; **Time constant:** \( \boxed{245~\text{s}} \) --- Let me know if you want any clarifications or deeper explanations!