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Zepler [3.9K]
3 years ago
6

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb

ient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Engineering
1 answer:
Minchanka [31]3 years ago
7 0

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

   = √( 140*2 ) / ( 12 * 10^-3 * 55 )

   = √ 280 / 0.66 = 20.60

Attached below is the remaining answers

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All of these are true about steel EXCEPT that:
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Answer:B) heat tends to strengthen high-strength steel.

Explanation: You are right it is B

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According to the eNotes, a program that eliminates sales and promotions in an effort to minimize the bullwhip effect would be ca
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Options: True or false

Answer:True, it will be called CRMWRONGEDLP.

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3 years ago
(a) Consider a germanium semiconductor at T 300 K. Calculate the thermal equilibrium electron and hole concentrations for (i) Nd
padilas [110]

Answer:

a.

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2 × 10¹¹ cm⁻³

ii. electron concentration, n₀ = 1.33 × 10¹¹ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

b.  

i. electron concentration, n₀ = 2 × 10¹⁵ cm⁻³, hole concentration, p₀ = 2.205 × 10⁻³ cm⁻³

ii. electron concentration, n₀ = 1.47 × 10⁻³ cm⁻³, hole concentration, p₀ = 3 × 10¹⁵ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

Explanation:

a. For Germanium, intrinsic concentration n₁ = 2 × 10¹³ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ = acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ]      since N₁ > N₂

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ - 0) +√[(2 × 10¹⁵ cm⁻³ - 0)² + 4(2 × 10¹³ cm⁻³)²] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ +√[(4.0016 × 10³⁰ cm⁻⁶] ]

n₀ = 1/2[(2 × 10¹⁵ cm⁻³ + 2.0004 × 10¹⁵ cm⁻³ ]

n₀ = 1/2[4.0004 × 10¹⁵ cm⁻³ ]

n₀ = 2.0002 × 10¹⁵ cm⁻³ ≅ 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2 × 10¹³ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2 × 10¹¹ cm⁻³

ii.  For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ =  acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ]      since N₂ > N₁

p₀ = 1/2[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³) +√[(10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³)² + 4(2 × 10¹³ cm⁻³)²] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9 × 10³⁰ cm⁻⁶ + 16 × 10²⁶ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ +√[(9.0016 × 10³⁰ cm⁻⁶] ]

p₀ = 1/2[(3 × 10¹⁵ cm⁻³ + 3.0003 × 10¹⁵ cm⁻³ ]

p₀ = 1/2[6.0003 × 10¹⁵ cm⁻³ ]

p₀ = 3.00015 × 10¹⁵ cm⁻³ ≅ 3 × 10¹⁵ cm⁻³

Te electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2 × 10¹³ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4 × 10²⁶ cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.33 × 10¹¹ cm⁻³

b. For GaAs, intrinsic concentration n₁ = 2 × 10⁶ cm⁻³.

i. For the electron concentration, n₀ wit N₁ = donor concentration = 2 × 10¹⁵ cm⁻³ and N₂ =  acceptor concentration = 0,

n₀ = 1/2[(N₁ - N₂) +√[(N₁ - N₂)² + 4n₁²] ]      since N₁ > N₂   and N₁ - N₂ = 2 × 10¹⁵ cm⁻³ >> n₁ = 2 × 10⁶ cm⁻³

n₀ = (N₁ - N₂) = 2 × 10¹⁵ cm⁻³ - 0 = 2 × 10¹⁵ cm⁻³

The hole concentration p₀ is gotten from

n₀p₀ = n₁²

p₀ = n₁²/n₀ = (2.1 × 10⁶ cm⁻³)²/2 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/2 × 10¹⁵ cm⁻³

p₀ = 2.205 × 10⁻³ cm⁻³

ii. For the hole concentration, p₀ wit N₁ = donor concentration = 7 × 10¹⁵ cm⁻³ and N₂ =  acceptor concentration = 10¹⁶ cm⁻³,

p₀ = 1/2[(N₂ - N₁) +√[(N₂ - N₁)² + 4n₁²] ]      since N₂ > N₁ and N₂ - N₁ = 10¹⁶ cm⁻³ - 7 × 10¹⁵ cm⁻³ = 3 × 10¹⁵ cm⁻³ >> n₁ = 2.1 × 10⁶ cm⁻³

p₀ ≅ N₂ - N₁ = 3 × 10¹⁵ cm⁻³

The electron concentration n₀ is gotten from

n₀p₀ = n₁²

n₀ = n₁²/p₀ = (2.1 × 10⁶ cm⁻³)²/3 × 10¹⁵ cm⁻³ = 4.41 × 10¹² cm⁻⁶/3 × 10¹⁵ cm⁻³

n₀ = 1.47 × 10⁻³ cm⁻³

c. It means that the minority carrier contribute little to the conductivity of the semi-conductor.

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