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Blababa [14]
3 years ago
6

You find a publication from a research laboratory that identifies a new catalyst for ammonia synthesis. The article contains the

following details about the process run in the author's lab:
The chemical reaction is N2 3H2 --> 2NH3
The reaction is exothermic
A fixed bed reactor is used with ruthenium catalyst on graphite support
The reactor is run at 430oC and 250 bar with space velocity of 8000 m3 m-3 h-1
At this temperature and pressure, the reactants and products are in the gas phase
The reactants are fed at flow rates of 400 lbmol/h H2 and 100 lbmol/h N2
The extent of reaction is 90 lbmol/h

You are considering replacing the current iron-oxide catalyst used in your ammonia synthesis plant with this newly discovered ruthenium-based catalyst. You plan to start your evaluation with a model in Aspen Plus and an economic analysis in APEA.

Based only on the information above from the literature, which project component would you select to map this reactor in APEA?
Engineering
1 answer:
Sidana [21]3 years ago
6 0

Answer:

Answer: RStoic

Explanation:

The reactor i will select to model this reactor in Aspenplus is RStoic

with the following reason;

Reason: Rstoic is used in Aspen software when the stoichiometry of reaction is known but kinetics isn't available. It can have one or more feed streams attached to it.

Reasons why other reactors are not preferred:

RYield: RYield performs the calculations based on the yield and we are not provided with the yield.

REquil: REquil is used when we are provided with stoichiometry of the reaction and information about equilibrium constant so that to perform chemical and phase equilibrium reactions & we are not provided with the equilibrium information.

RGibbs: Since reactor and products are in the gas phase, so it is not required to use it as it's used to minimise Gibbs free energy.

RCSTR: Reaction kinetics needed but we are not provided with it.

RPlug: Reaction kinetics needed but we are not provided with it.

RBatch: Reaction kinetics needed but we are not provided with it.

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Please write the following code in Python 3. Also please show all output(s) and share your code.
maksim [4K]

Answer:

sum2 = 0

counter = 0

lst = [65, 78, 21, 33]

while counter < len(lst):

   sum2 = sum2 + lst[counter]

   counter += 1

Explanation:

The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>

While there are numbers in the <em>lst</em>,  loop through <em>lst</em>

Add the numbers in <em>lst</em> to the sum2

Increment <em>counter</em> by 1 after each iteration

6 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

\sigma_c=\frac {K}{Y \sqrt {a\pi}} and making Y the subject

Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

\sigma_c=\frac {K}{Y \sqrt {a\pi}}  and making K the subject

K=\sigma_c Y \sqrt {a\pi}  and substituting 260 MPa for \sigma_c  while a is taken as 0.003m and Y is already known

K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0
3 years ago
A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p
Natali [406]

Answer:

Mechanical Efficiency =  83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = V=8ft^3/s\\

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0
3 years ago
Removing Shingles from a roof is called a
torisob [31]
D remodeling roofing
7 0
2 years ago
The two windings of transformer is: a)- Conductively linked. b)- Not linked at all. c)- Inductively linked d)- Electrically link
kondor19780726 [428]

The two windings of transformer is c)- Inductively linked

Hope this helped!

4 0
3 years ago
Read 2 more answers
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