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3 years ago

What is integrated circuit package

2 answers:

7 0

In electronics manufacturing, integrated circuit packaging is the final stage of semiconductor device fabrication, in which the block of semiconductor material is encapsulated in a supporting case that prevents physical damage and corrosion.

enyata [817]3 years ago

4 0

Answer:

Explanation:

, integrated circuit packaging is the final stage of semiconductor device fabrication, in which the block of semiconductor material is encapsulated in a supporting case that prevents physical damage and corrosion.

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Ideas for an invention for a 9 year old

allochka39001 [22]

> Try to conduct electricity for your whole home using domestic waste.

> A home made small washing machine to wash small clothes like socks

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4 years ago

Read 2 more answers

Engineer Smith, who is licensed in several jurisdictions, recently had his license acted upon by a licensing authority in one of

Nimfa-mama [501]

Answer:

Assuming the infraction in the other jurisdiction is an infraction in Florida, having approximately the same penalty imposed by the FBPE as was imposed in the other jurisdiction

Explanation:

The idea of 8 hours continuation in the options is irrelevant to this case. Rather, Engineer Smith should expect about the same penalty imposition.

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3 years ago

A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct superheated steam in a strip

hram777 [196]

Answer:

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

$\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}$ ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

$X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167$

and

x₁ = 0.2% = 0.002

$X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}$

Now we have to use the balance equation a

$\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}$ .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

$L_s$ = 300(1 - 0.04) = 288 kmol pure oil

G = $G_s$ = 11.42 kmol

$Y_1$ = 0 , solvent free steam

substitute into the equation a

$\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}$

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ <em>Number of trays = 6</em>