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borishaifa [10]
3 years ago
6

To enhance the effective surface, and hence the chemical reaction rate, catalytic surfaces often take the form of porous solids.

One such solid may be visualized as consisting of a larger number of cylindrical pores, each of diameter D and length L.
Consider conditions involving a gaseous mixture of A and B for which species A is chemically consumed at the catalytic surface. The reaction is known to be first order k1CA. Under steady state, flow over the porous solid is known to maintain a fixed value of the molar concentration CA0 at the pore mouth.
Beginning from fundamentals, obtain the differential equation that governs the variation of CA with distance x along the pore. Applying appropriate boundary conditions, solve the equation to obtain an expression for CA(x).

Physics
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

See solution with all the conditions considered. A gaseous mixture of A and B for which species A is chemically consumed at the catalytic surface.

The total pore reaction rate is stated below and it can be inferred by applying the bellow analogy.

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Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo
ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

F_{net}=ma

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

a=9.50 m/s^2 (acceleration)

So, the net force is

F_{net}=(94.0)(9.50)=893 N

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

v=u+at

where

v is the  final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

a=9.50 m/s^2 (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

v=0+(9.50)(0.890)=8.5 m/s

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

W=K_f - K_i = \frac{1}{2}mv^2-0

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

K_i = 0 J is the initial kinetic energy

So the work done is

W=\frac{1}{2}(94.0)(8.5)^2=3396 J

The power expended is given by

P=\frac{W}{t}

where

t = 0.890 s is the time elapsed

Substituting,

P=\frac{3396}{0.890}=3816 W

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

F_1=893 N

We know that the average force for the whole race is

F_{avg}=820 N

Which can be rewritten as

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}

And solving for F_2, we find the average force exerted by Bolt on the ground during the second phase:

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N

The net force exerted by Bolt during the second phase can be written as

F_{net}=F_2-D (1)

where D is the air drag.

The net force can also be rewritten as

F_{net}=ma

where

a=\frac{v-u}{t} is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

a=\frac{12.4-8.5}{8.69}=0.45 m/s^2

So we can now find the average drag force from (1):

D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

\Delta E=W=Ds

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m

And so,

\Delta E=Ds=(770.2)(90.6)=69780 J

e)

The power that Bolt must expend just to voercome the drag force is given by

P=\frac{\Delta E}{t}

where

\Delta E is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

\Delta E=69780 J

t = 8.69 s is the time elapsed

Substituting,

P=\frac{69780}{8.69}=8030 W

And we see that it is about twice larger than the power calculated in part c.

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