Let's solve the problem step by step. --- ## **Given Data** - Width of base, \( W = 20 \) mm = **0.02 m** - Length of base, \( L_b = 3 \) mm = **0.003 m** - Length of fin, \( L_f = 15 \) mm = **0.015 m** - Number of fins, \( N = 6 \) - Fin thickness, \( \delta = 1.8 \) mm = **0.0018 m** - Chip temperature, \( T_c = 85^\circ \mathrm{C} \) - Ambient temperature, \( T_\infty = 20^\circ \mathrm{C} \) - Convection coefficient, \( h = 100 \) W/m²·K - Thermal conductivity (aluminum), \( k = 180 \) W/m·K - Contact resistance, \( R''_{t,c} = 2 \times 10^{-6} \) m²·K/W --- ## **(a) What is the chip power when \( N = 6 \), \( \delta = 1.8 \) mm?** ### **1. Calculate the Area** #### **Fin Geometry** - Number of spaces between fins: \( N-1 \) - Each fin: thickness \( \delta \) - Each space: width \( s \) - Total width: \( N\delta + (N-1)s = W \) - Solve for \( s \): \[ s = \frac{W - N\delta}{N-1} \] \[ s = \frac{0.02 - 6 \times 0.0018}{5} = \frac{0.02 - 0.0108}{5} = \frac{0.0092}{5} = 0.00184\ \text{m} \] #### **Fin Surface Area** Each fin has 2 sides, and a tip (but tips are adiabatic, so ignore). - **Each fin's side area:** \( 2 \times (L_f \times W) \) - **Each fin's base area:** \( \delta \times W \) (but heat transfer from base is included separately) - **Total area from fin sides:** \( A_\text{fin sides} = N \times 2 \times (L_f \times \delta) \) - **Area from exposed base:** \( A_\text{base} = (N-1) \times s \times W \) But for heat transfer, use: \[ A_\text{fin, one} = 2 \times L_f \times W \] \[ A_\text{fin, total} = N \times 2 \times L_f \times W \] \[ A_\text{base} = (N-1) \times s \times W \] Let's calculate: - \( N = 6 \) - \( L_f = 0.015 \) m - \( W = 0.02 \) m - \( s = 0.00184 \) m \[ A_\text{fin, total} = 6 \times 2 \times 0.015 \times 0.02 = 0.036 \text{ m}^2 \] \[ A_\text{base} = 5 \times 0.00184 \times 0.02 = 0.000184 \text{ m}^2 \] --- ### **2. Calculate the Fin Efficiency (\( \eta_f \))** Fin type: straight, adiabatic tip. \[ m = \sqrt{\frac{2h}{k\delta}} \] \[ m = \sqrt{\frac{2 \times 100}{180 \times 0.0018}} \] \[ m = \sqrt{\frac{200}{0.324}} = \sqrt{617.28} \approx 24.84 \] \[ mL_f = 24.84 \times 0.015 = 0.373 \] Fin efficiency for adiabatic tip: \[ \eta_f = \frac{\tanh(mL_f)}{mL_f} \] \[ \tanh(0.373) \approx 0.356 \] \[ \eta_f = \frac{0.356}{0.373} = 0.954 \] --- ### **3. Overall Surface Efficiency (\( \eta_o \))** \[ \eta_o = \frac{N \cdot \eta_f \cdot 2 L_f W + (N-1)sW}{N \cdot 2 L_f W + (N-1)sW} \] Plug in values: - \( N = 6 \) - \( \eta_f \approx 0.954 \) - \( 2L_f W = 2 \times 0.015 \times 0.02 = 0.0006 \) - \( (N-1)sW = 0.000184 \) Numerator: \[ 6 \times 0.954 \times 0.0006 + 0.000184 = 0.003432 + 0.000184 = 0.003616 \] Denominator: \[ 6 \times 0.0006 + 0.000184 = 0.0036 + 0.000184 = 0.003784 \] \[ \eta_o = \frac{0.003616}{0.003784} = 0.956 \] --- ### **4. Total Heat Sink Surface Area** \[ A_{tot} = N \cdot 2L_f W + (N-1)sW = 0.003784\ \text{m}^2 \] --- ### **5. Heat Transfer from Sink to Air** \[ Q = \frac{\eta_o h A_{tot} (T_c - T_\infty)}{1 + \eta_o h A_{tot} R''_{t,c}} \] \[ Q = \frac{0.956 \times 100 \times 0.003784 \times (85 - 20)}{1 + 0.956 \times 100 \times 0.003784 \times 2 \times 10^{-6}} \] \[ Q = \frac{0.956 \times 100 \times 0.003784 \times 65}{1 + 0.956 \times 100 \times 0.003784 \times 2 \times 10^{-6}} \] \[ = \frac{0.956 \times 0.3784 \times 65}{1 + 0.000723} \] \[ = \frac{0.3616 \times 65}{1.000723} \] \[ = \frac{23.504}{1.000723} \] \[ Q \approx 23.487\ \text{W} \] --- ### **Final Answer for (a):** \[ \boxed{Q \approx 23.5\ \text{W}} \] --- ## **(b) Effect of Increasing N while Holding \( \delta \) Constant** - As \( N \) increases (with fixed \( \delta \)), spacing \( s \) between fins decreases, but more fin surface area is exposed for heat transfer. - More fins increases total heat transfer area and thus increases power dissipated. - However, if \( s \) becomes very small, air flow between fins may be restricted, reducing the convective heat transfer coefficient \( h \). - So, **increasing N increases chip power up to a point**, after which gains are limited by air flow restrictions. --- ### **Summary Table** | Parameter | Value | |-----------|-------| | Chip Power (\( N=6, \delta=1.8\,\text{mm} \)) | **23.5 W** | | Effect of Increasing N | **Increases power, but with diminishing returns as spacing decreases** | --- ## **Boxed Final Answers** **(a) Chip power for \( N = 6 \), \( \delta = 1.8 \) mm:** \[ \boxed{Q \approx 23.5\ \mathrm{W}} \] **(b) Effect of increasing number of fins while holding \( \delta \) constant:** \[ \boxed{\text{Power increases as number of fins increases, but eventually plateaus due to reduced air flow between fins.}} \]

# Step-by-Step Solution for Heat Sink Analysis ## Given Data - **Width of base, \( W \)**: 20 mm = **0.02 m** - **Length of base, \( L_b \)**: 3 mm = **0.003 m** - **Length of fin, \( L_f \)**: 15 mm = **0.015 m** - **Number of fins, \( N \)**: 6 - **Fin thickness, \( \delta \)**: 1.8 mm = **0.0018 m** - **Chip temperature, \( T_c \)**: 85°C - **Ambient temperature, \( T_\infty \)**: 20°C - **Convection coefficient, \( h \)**: 100 W/m²·K - **Thermal conductivity (aluminum), \( k \)**: 180 W/m·K - **Contact resistance, \( R''_{t,c} \)**: \( 2 \times 10^{-6} \) m²·K/W --- ## (a) Chip Power Calculation ### 1. Calculate the Area #### Fin Geometry To compute the effective surface area for heat transfer, we first need to determine the spacing \( s \) between the fins. - The total width equation for the fins and spaces is given by: \[ s = \frac{W - N\delta}{N-1} \] Substituting the values: \[ s = \frac{0.02 - 6 \times 0.0018}{5} = 0.00184\ \text{m} \] #### Fin Surface Area Each fin has two sides contributing to heat transfer (ignoring adiabatic tips). - **Area of one fin's sides**: \[ A_\text{fin, one} = 2 \times (L_f \times \delta) \] - **Total area from fin sides**: \[ A_\text{fin, total} = N \times A_\text{fin, one} \] Calculating: \[ A_\text{fin, total} = 6 \times 2 \times 0.015 \times 0.02 = 0.036\ \text{m}^2 \] - **Area from exposed base**: \[ A_\text{base} = (N-1) \times s \times W \] Calculating: \[ A_\text{base} = 5 \times 0.00184 \times 0.02 = 0.000184\ \text{m}^2 \] --- ### 2. Calculate the Fin Efficiency (\( \eta_f \)) We determine the efficiency of the fin using the following formula, which accounts for the geometry and material properties. \[ m = \sqrt{\frac{2h}{k\delta}} \] Substituting the values: \[ m = \sqrt{\frac{200}{0.324}} \approx 24.84 \] Next, calculate \( mL_f \): \[ mL_f = 24.84 \times 0.015 \approx 0.373 \] Using the efficiency formula for fins with adiabatic tips: \[ \eta_f = \frac{\tanh(mL_f)}{mL_f} \] \[ \eta_f \approx 0.954 \] --- ### 3. Overall Surface Efficiency (\( \eta_o \)) The overall efficiency combines the contributions from all fins and the base area. \[ \eta_o = \frac{N \cdot \eta_f \cdot 2 L_f W + (N-1)sW}{N \cdot 2 L_f W + (N-1)sW} \] Calculating both numerator and denominator: - **Numerator**: \[ 0.003616 \] - **Denominator**: \[ 0.003784 \] Thus, \[ \eta_o \approx 0.956 \] --- ### 4. Total Heat Sink Surface Area The total effective area for heat transfer is: \[ A_{tot} = N \cdot 2L_f W + (N-1)sW \approx 0.003784\ \text{m}^2 \] --- ### 5. Heat Transfer from Sink to Air The heat transfer \( Q \) can be calculated using the formula: \[ Q = \frac{\eta_o h A_{tot} (T_c - T_\infty)}{1 + \eta_o h A_{tot} R''_{t,c}} \] Substituting in the values: \[ Q \approx 23.5\ \text{W} \] --- ### Final Answer for (a): \[ \boxed{Q \approx 23.5\ \text{W}} \] --- ## (b) Effect of Increasing N while Holding \( \delta \) Constant Increasing the number of fins while keeping the thickness \( \delta \) constant leads to several outcomes: - **Increased Surface Area**: More fins mean more surface area for heat dissipation, which generally increases the power transferred away from the chip. - **Spacing and Airflow**: As \( s \) decreases, airflow between fins may become restricted, potentially reducing the convective heat transfer coefficient \( h \). - **Diminishing Returns**: There is a point where adding more fins yields less additional power due to reduced airflow. **Conclusion**: Increasing \( N \) increases chip power but with diminishing returns as fin spacing decreases. --- ### Summary Table | Parameter | Value | |-----------|-------| | Chip Power (\( N=6, \delta=1.8\,\text{mm} \)) | **23.5 W** | | Effect of Increasing N | **Increases power, but with diminishing returns as spacing decreases** | --- ## Boxed Final Answers **(a) Chip power for \( N = 6 \), \( \delta = 1.8 \) mm:** \[ \boxed{Q \approx 23.5\ \mathrm{W}} \] **(b) Effect of increasing number of fins while holding \( \delta \) constant:** \[ \boxed{\text{Power increases as number of fins increases, but eventually plateaus due to reduced air flow between fins.}} \]