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wel
3 years ago
9

For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste

stream under steady-state conditions with a first-order reaction: reactor volume = 280 m3, flow rate = 14 m3 · day−1, and reaction rate coefficient = 0.05 day−1.
Engineering
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

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10. True or False: You should select your mechanic before you experience vehicle failure.
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7 0
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Which one of the following is not an economic want?
Nostrana [21]

The available options are:

a. Want for a television set

b. Want for medical attention

c. Want for friendship

d. Want for new clothing

Answer:

Want for friendship

Explanation:

Given that economic want is what humans desire to have or possess such that they pay money to acquire them.

Hence, from the available options "want for friendship " is not economic want because it can't be bought with money, while other options can be bought with money or monetary transaction.

8 0
3 years ago
The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the
gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, v^{2}=\frac {ks^{2}}{m} hence v=s\sqrt {\frac {k}{m}} where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as 9.81 m/s^{2}

Therefore, F_y=0.75\times 9.81=7.3575 N\approx 7.36 N

The sum of forces in normal direction is given by Ma_n=Ks therefore

Ma_n=200*0.1=20 N

Therefore, <u>normal force on the rod is 20 N</u>

5 0
3 years ago
A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l
Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

E=\frac {\sigma}{\epsilon}

Making \sigma the subject then

\sigma=E\epsilon

where \sigma is the stress and \epsilon is the strain

Since strain is given as 0.025% of the length then strain is \frac {0.025}{100}=0.00025

Now substituting E for 10\times 10^{6} psi then

\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi

(b)

Stress, \sigma= \frac {F}{A} making A the subject then

A=\frac {F}{\sigma}

A=\frac {\pi(d_o^{2}-d_i^{2})}{4}

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that 2t=d_o - d_i where t is thickness

Now substituting

\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}

\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4

(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

But the outer diameter is given as 2 in hence

(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}

d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in

As already mentioned, 2t=d_o - d_i hence t=0.5(d_o - d_i)

t=0.5(2-1.785)=0.1075 in

3 0
3 years ago
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