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Alex73 [517]
3 years ago
15

Consider a 18-mm by 18-mm chip that is cooled on its top surface by a 6 m/s flow of 25 ºC air. Any heat transfer from its bottom

surface to the circuit board is ignored. Because of the chip construction, the electrical power dissipated in the chip results in a uniform heat flux over the surface of the chip. If the maximum temperature that any part of the chip can experience is 85ºC, determine the maximum allowable chip power. Note: the air properties can be determined at the ‘film" (or average) temperature.

Engineering
2 answers:
Alina [70]3 years ago
8 0

Answer:

Explanation:

Length (l) = 10

Width (w) = 10

Velocity =

Ambient Temperature = 25 degrees Celsius

Surface Temperature = 85 degrees Celsius.

Note these values as what we will use to solve for the maximum allowable chip power, then go through the attached files for a step by step explanation of how the answer is gotten.

vladimir2022 [97]3 years ago
7 0

Find the below attachments

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A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Dy345-kV, Y transformer to a 345-kV circ
aniked [119]

Answer:

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Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0
Naily [24]

Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

(a) Find the total current crossing the surface z = 0.1 m in the az direction.

(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.

(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

Answer:

a. -39.8μA

b. -15.81mC/m³

c. 29.05m/s

Explanation:

Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

J = 0.

a. Total current is calculated by.

J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

π = 22/7

φdza = 10^-6

z = 0.1

Total current

= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= −39.7543477278310

= -39.8μA

b. Calculating velocity charge density at (ρv)

Density (J) = ρv * V

Where J = Density = -10^6 * z^1.5

V = 2 * 10^6

z = 0.1

Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)

ρv = -0.1^1.5/(2)

ρv = -0.015811388300841

ρv = -0.01581 --------- Approximated

ρv = -15.81mC/m³

c. Calculating Velocity

Velocity = J/V

Where Velocity Charge Density = -2000 C/m3

Where J = -10^6 * z^1.5

z = 0.15

J = -10^6 * 0.15^1.5

J = -58094.75019311125

Velocity = -58094.75019311125/-2000

Velocity = 29.047375096555625m/s

Velocity = 29.05m/s

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A piston-cylinder apparatus has a piston of mass 2kg and diameterof
iragen [17]

Answer:

M =2.33 kg

Explanation:

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atmospheric pressure 101 kPa

water temperature = 50°C

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