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alexgriva [62]
3 years ago
11

What is the Damkohler number? b) What is the significance of a system with a low Damkohler number?

Chemistry
1 answer:
elena-s [515]3 years ago
8 0

Explanation:

(a)  A number which is dimensionless and provides us an estimate of the degree of conversion which can be achieved in CSTR, that is, continuous stirred tank reactor is known as Damkohler number.

This number is denoted as Da.

Mathematically,     Da = \frac{\text{reaction rate}}{\text{convection rate}}

             Da = \frac{-rA \times V}{Fa_{o}}

Now, for first order system, Da = \frac{-rA \times V}{Fa_{o}}

                        = \frac{k \times CA_{o} \times V}{v \times CA_{o}} = Tk

where,      rA = rate of reaction

                V = volume of reactor

                Fa_{o} = molar flow rate of component A

                 k = rate constant

               CA_{o} = initial concentration of A

                  v = volumetric flow rate of A

                  T = residence time

(b)   Since, from a given Damkohler number we can figure out the possible conversion of CSTR, that is, continuous stirred tank reactor.

So, if we have a low Damkohler number then the system will give us a less conversion formula. As the conversion is as follows.

                       X = \frac{Da}{Da + 1}

Hence, we can conclude that Da \leq 0.1 will give less than 10% conversion as calculated by using above formula.

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deff fn [24]
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4 0
2 years ago
The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure
AlekseyPX

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1 = initial pressure = 198 kPa

P_2 = final pressure = 225 kPa

T_1 = initial temperature = 27^oC=273+27=300K

T_2 = final temperature = ?

Now put all the given values in the above equation, we get:

\frac{198kPa}{300K}=\frac{225kPa}{T_2}

T_2=340.9K\approx 341K

Therefore, the temperature of the air in the tire is, 341 K

5 0
3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
What type of reaction is shown below? <br>a) Addition reaction <br>b) Esterification​
Gemiola [76]

Answer:

a) Addition reaction, is your answer

6 0
3 years ago
A group of giraffes feeds on leaves and grasses during the daytime. Right now, it is dark out and the giraffes are not eating. I
Zanzabum

Answer:

Carbon dioxide is moving out of the living things.

Explanation:

The food materials eaten by living things contain carbon in the form of complex organic matter. When living things feed, they ingest this complex organic material into their bodies.

During the process of digestion, this complex organic material is broken down to give glucose. Glucose is the energy molecule in living things. Excess glucose in the body is stored as glycogen.

During cellular respiration, glucose is broken down to release carbon dioxide. Hence, at night when the giraffe has stopped eating, cellular respiration continues to occur and carbon dioxide is released, that is, carbon dioxide continues to move out of living things at night.

7 0
3 years ago
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