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Verizon [17]
3 years ago
11

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

parts B & C, on a molar basis. These three components will undergo a reaction when fed into a reactor. The reaction is A+2B 4C-> 2x + 3Y The product Y is an important product, and so we want to maximize (to the extent possible) the conversion to Y. What must the conversion of the limiting reagent be in order to have the molar fraction of Y in the outlet stream from the reactor be 0.10?
Engineering
1 answer:
tigry1 [53]3 years ago
5 0

Answer:

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

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

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.

<em>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</em>

<em></em>

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

<em>from conservation of mass law</em>, 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 = <u>0.533</u>

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3 years ago
Estimate the quantity of soil to be excavated from the borrow pit​
Serggg [28]

Answer

1056

Explanation:

for example

A soil is to be excavated from a borrow pit which has a density of 1.75g/cc and water content of 12%. The G is 2.7 the soil is compacted to that water content of 18% and dry density of 1.65g/cc. for 1000 m3 of soil used in fill estimate

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3 years ago
Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
3 years ago
Which Finance jobs can someone pursue with only a high school diploma? Check all that apply.
Zolol [24]

Answer:

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Explanation:

3 0
3 years ago
Read 2 more answers
2. A F-22 Raptor has just climbed through an altitude of 9,874 m at 1,567 kph when a disk
BabaBlast [244]

The pressure difference across the sensor housing will be "95 kPa".

According to the question, the values are:

Altitude,

  • 9874

Speed,

  • 1567 kph

Pressure,

  • 122 kPa

The temperature will be:

→ T = 15.04-[0.00649(9874)]

→     = 15.04-64.082

→     = -49.042^{\circ} C

now,

→ P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}

→      = 27.074

hence,

→ The pressure differential will be:

= 122-27

= 95 \ kPa

Thus the above solution is correct.

Learn more about pressure difference here:

brainly.com/question/15732832

3 0
3 years ago
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