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ollegr [7]
3 years ago
7

Assume that one component with 256 I/Os is offered in three forms: QFP, full array PBGA, and fl ip chip with gold bumps. The pit

ch of the QFP is 0.4 mm and the lead length is 1 mm. An edge of the QFP to the center of the nearest lead is assumed to the same as the lead pitch. The pitch of the BGA is 1.27mm and an edge to the center of the nearest solder joints is 2.5 mm. The pitch of the flip chip is 0.4 mm and an edge to the center of the nearest solder joints is1mm. Assume the underll extension is 1 mm.
1. Calculate the PCB surface area requirement for each case.
Engineering
1 answer:
Norma-Jean [14]3 years ago
8 0
The gold are assumed and offered in three
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Give five general principles involved in the process of sewage filtration?​
aleksandr82 [10.1K]

Answer:

Some general principles are given below in the explanation segment.

Explanation:

Sewage treatment seems to be a method to extract pollutants from untreated sewage, consisting primarily of domestic sewage including some solid wastes.

<u>The principles are given below:</u>

  • Unless the components throughout the flow stream become greater than the ports or even the gaps throughout the filter layer, those holes would be filled as either a result of economic detection.
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3 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

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irakobra [83]

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(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e
aksik [14]

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

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