just case b 1. As more and more components are placed on a single integrated circuit (chip), the amount of heat that is dissipated continues to increase. However, this increase is limited by the maximum allowable chip operating temperature. Consider a 15 mm x 15 mm square chip. In order to improve the heat dissipation, a five by five array of copper pin fins are to be joined to the outer surface of the chip. The fins have a diameter of 1.5 mm and a length of 15 mm. Cooling air can be supplied at T∞ = 25°C, and the maximum allowable chip base temperature is T₀ = 75°C. The convective heat transfer coefficient is h = 250 W/m²·K, and the copper has a thermal conductivity of k = 401 W/m·K.Consider two cases: Case A: Convecting Tip, and Case B: Adiabatic TipAnd for each calculate the following at steady state:(a) Heat transfer rate from a single fin.(b) Efficiency of a single fin.(c) Total heat transfer rate of the array. ### Heat Transfer from Extended Surfaces#### Equation for Temperature Distribution:\[ \theta = c_1 e^{mx} + c_2 e^{-mx} \]#### Case B Boundary Conditions – Adiabatic Tip- **At $ x = 0 $**: \[ T = T_0 \]- **At $ x = L $**: \[ \left( \frac{dT}{dx} \right)_{x=L} = 0 \]#### Temperature Profile Solution\[ \frac{\theta}{\theta_0} = \frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{e^{mx} + e^{-mx}}{1 + e^{-2mL}} \quad \text{(17-38)} \]**or**\[ \frac{\theta}{\theta_0} = \frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{\cosh[m(L - x)]}{\cosh mL} \quad \text{(17-39)} \]#### Heat Transfer Rate\[ q = kAm\theta_0 \tanh mL \quad \text{(17-45)} \]#### Diagram ExplanationThe diagram depicts a cylindrical rod extended from a surface, illustrating the heat transfer scenario. An adiabatic tip is shown, indicating no heat flow at the tip, which relates to the boundary conditions described above. The solution equations provide the temperature distribution and heat transfer rate for this type of setup.

a) Heat transfer from a fin with insulated tipq= Mtank(mL)q = hPkA θs tan h (mL)m = hPkA Cross…

Answered: As more and more components are placed…

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter2: Steady Heat Conduction

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Problem 2.33P

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just case b

1. As more and more components are placed on a single integrated circuit (chip), the amount of heat that is dissipated continues to increase. However, this increase is limited by the maximum allowable chip operating temperature. Consider a 15 mm x 15 mm square chip. In order to improve the heat dissipation, a five by five array of copper pin fins are to be joined to the outer surface of the chip. The fins have a diameter of 1.5 mm and a length of 15 mm. Cooling air can be supplied at T∞ = 25°C, and the maximum allowable chip base temperature is T₀ = 75°C. The convective heat transfer coefficient is h = 250 W/m²·K, and the copper has a thermal conductivity of k = 401 W/m·K.

Consider two cases: Case A: Convecting Tip, and Case B: Adiabatic Tip

And for each calculate the following at steady state:

(a) Heat transfer rate from a single fin.
(b) Efficiency of a single fin.
(c) Total heat transfer rate of the array.

$### Heat Transfer from Extended Surfaces #### Equation for Temperature Distribution: \[ \theta = c_1 e^{mx} + c_2 e^{-mx} \] #### Case B Boundary Conditions – Adiabatic Tip - **At $ x = 0 $**: \[ T = T_0 \] - **At $ x = L $**: \[ \left( \frac{dT}{dx} \right)_{x=L} = 0 \] #### Temperature Profile Solution \[ \frac{\theta}{\theta_0} = \frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{e^{mx} + e^{-mx}}{1 + e^{-2mL}} \quad \text{(17-38)} \] **or** \[ \frac{\theta}{\theta_0} = \frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{\cosh[m(L - x)]}{\cosh mL} \quad \text{(17-39)} \] #### Heat Transfer Rate \[ q = kAm\theta_0 \tanh mL \quad \text{(17-45)} \] #### Diagram Explanation The diagram depicts a cylindrical rod extended from a surface, illustrating the heat transfer scenario. An adiabatic tip is shown, indicating no heat flow at the tip, which relates to the boundary conditions described above. The solution equations provide the temperature distribution and heat transfer rate for this type of setup.$

Transcribed Image Text:### Heat Transfer from Extended Surfaces #### Equation for Temperature Distribution: \[ \theta = c_1 e^{mx} + c_2 e^{-mx} \] #### Case B Boundary Conditions – Adiabatic Tip - **At $ x = 0 $**: \[ T = T_0 \] - **At $ x = L $**: \[ \left( \frac{dT}{dx} \right)_{x=L} = 0 \] #### Temperature Profile Solution \[ \frac{\theta}{\theta_0} = \frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{e^{mx} + e^{-mx}}{1 + e^{-2mL}} \quad \text{(17-38)} \] **or** \[ \frac{\theta}{\theta_0} = \frac{T - T_{\infty}}{T_0 - T_{\infty}} = \frac{\cosh[m(L - x)]}{\cosh mL} \quad \text{(17-39)} \] #### Heat Transfer Rate \[ q = kAm\theta_0 \tanh mL \quad \text{(17-45)} \] #### Diagram Explanation The diagram depicts a cylindrical rod extended from a surface, illustrating the heat transfer scenario. An adiabatic tip is shown, indicating no heat flow at the tip, which relates to the boundary conditions described above. The solution equations provide the temperature distribution and heat transfer rate for this type of setup.

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