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3 years ago

14

What are the 3 stages of reverse engineering?

1 answer:

jasenka [17]3 years ago

6 0

What are the 3 stages of reverse engineering?

Answer:

1. Implementation Recovery (Determine exactly what you want)

2. Design Recovery (Data Acquisition)

3. Analysis Recovery (Data Processing)

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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>

5 0

4 years ago

Read 2 more answers

A 0.01 m3 piston and cylinder contains an ideal gas at 5.92154 atm, 290 K . A constant pressure process produces 54 kJ of work o

Debora [2.8K]

Answer:

Final Volume of the gas = 0.1 m^3

Explanation:

Check the attached file for the calculations involved in this solution

4 0

3 years ago

The inverted U-tube is used to measure the pressure difference between two points A and B in an inclined pipeline through which

JulijaS [17]

Answer:

i) 0.610 m or 610 mm

ii) 0.4 m or 400 mm

Explanation:

The pressure difference between the pipes is

a) Air

Pa + πha +Ha = Pb + πhb +Hb

Pa - Pb = π(hb-ha) + Hb-Ha

Relative density of air = 1.2754 kg /m3

Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm

b) paraffin of relative density of 0.75

Pa - Pb = π(hb-ha) + Hb-Ha

Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm

8 0

3 years ago

A cartridge electrical heater is shaped as a cylinder of length L=200 mm and outer diameter D=20 mm. Under normal operating cond

Setler79 [48]

Answer:

When water is surrounding T_s = 34.17 degree C

When air surrounding T_S = 1434.7 degree C

from above calculation we can conclude that air is less effective than water as heat transfer agent

Explanation:

Given data:

length = 300 mm

Outer diameter = 30 mm

Dissipated energy = 2 kw = 2000 w

Heat transfer coefficient IN WATER = 5000 W/m^2 K

Heat transfer coefficient in air = 50 W/m^2 K

we know that $q_{convection} = P$

From newton law of coding we have

$q_{convection} = hA(T_s - T_{\infity})$

$T_s$ is surface temp.

T - temperature at surrounding

$P = hA(T_s - T_{\infity})[tex]\frac{P}{\pi hDL} = (T_s - T_{\infity})$

solving for[/tex] T_s [/tex] w have

$T_s = T_{\infty} + \frac{P}{\pi hDL}$

$T_s = 20 + \frac{2000}{\pi 5000\times 0.03\times 0.3}$

$T_s = 34.17 degree C$

When air is surrounding we have

$T_s = T_{\infty} + \frac{P}{\pi hDL}$

$T_s = 20 + \frac{2000}{\pi 2000\times 0.03\times 0.3}$

$T_s = 1434.7 degree C$

from above calculation we can conclude that air is less effective than water as heat transfer agent

5 0

3 years ago

13–27. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-k

Feliz [49]

Answer:

See explanation for step by step procedure to get answer.

Explanation:

Given that:

The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

See the attachments for complete steps to get answer.

4 0

3 years ago