Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.

4 0

2 years ago

I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal

tigry1 [53]

Answer:

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

where, Nₐ₀-Nₐ = is the amount in moles of A used up

Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

from conservation of mass law, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if 3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

5 0

2 years ago

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

$D = 7.5 \times 10^{-11} m^2/s$

for 0.225% carbon concentration following formula is used

$\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})$

where, erf stand for error function

$\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248$

$0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})$

$erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248$

$erf(\frac{x}{2\sqrt{DT}}) = 0.751$

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

$\frac{x}{2\sqrt{DT}} = 0.815$

$x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)$

$x = 2.39\times 10^{-3} m$

x = 0.002395 mm

8 0

3 years ago