Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

Homework #7

Due 10/22/08

Phosphorous transport in a packed bed reactor

The steady state distribution of phosphorus concentration with depth is assumed to follow

𝐂𝐬

𝐂𝟎≅ 𝐞𝐱𝐩(−

𝐤

𝐯𝐳)

(1)

Results are shown in Figure 3 for the four velocities of 0.118, 0.256, 0.539 and 0.900 cm min-1.

1. Estimate the slope k/v from figure 3 for each velocity.

Answer:

According to Figure 3,

𝑆𝑙𝑜𝑝𝑒 = 𝑡𝑔 𝛼 =Δ𝑦

Δ𝑥

v=0.118(cm/min)

𝑧 = 2𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.075

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.39

𝑡𝑔 𝛼 =Δ𝑦

Δ𝑥=

0.39 − 0.075

10 − 2= 0.039375

v=0.256(cm/min)

𝑧 = 2𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.05

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.23

𝑡𝑔 𝛼 =Δ𝑦

Δ𝑥=

0.23 − 0.05

10 − 2= 0.0225

v=0.539(cm/min)

𝑧 = 2𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.14

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.02

𝑡𝑔 𝛼 =Δ𝑦

Δ𝑥=

0.14 − 0.02

10 − 2= 0.015

v=0.900(cm/min)

𝑧 = 2𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.02

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.1

𝑡𝑔 𝛼 =Δ𝑦

Δ𝑥=

0.1 − 0.02

10 − 2= 0.01

Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

2. From (1) estimate the value of k for each velocity.

Answer: 𝐂𝐬

𝐂𝟎≅ 𝐞𝐱𝐩(−

𝐤

𝐯𝐳)

v=0.118 (cm/min)

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.39

−ln Cs

C0 ≅

k

vz ==> 0.39 ≅

k

0.118 cmmin

10 cm

==> 𝑘 =0.39 × 0.118

cmmin

10 cm = 0.004602 (1/min)

v=0.256 (cm/min)

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.23

−ln Cs

C0 ≅

k

vz ==> 0.23 ≅

k

0.256 cmmin

10 cm

==> 𝑘 =0.23 × 0.256

cmmin

10 cm = 0.005888 (1/min)

v=0.539 (cm/min)

𝑧 = 2𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.14

−ln Cs

C0 ≅

k

vz ==> 0.14 ≅

k

0.539 cmmin

10 cm

==> 𝑘 =0.14 × 0.539

cmmin

10 cm = 0.007546 (1/min)

v=0.900 (cm/min)

𝑧 = 10𝑐𝑚 => − ln 𝐶𝑠

𝐶0 = 0.1

−ln Cs

C0 ≅

k

vz ==> 0.1 ≅

k

0.900 cmmin

10 cm

==> 𝑘 =0.1 × 0.900

cmmin

10 cm = 0.009 (1/min)

Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

3. Plot k vs. v on linear graph paper.

Answer:

V (cm/min) K (1/min)

0.118 0.004602

0.256 0.005888

0.539 0.007546

0.9 0.009

Figure 1: Plot of k vs. v on linear-linear graph.

Better plot resolution Figure 5 attached.

□

Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

4. Assume that k vs. v follows that hyperbolic equation

𝐤 =𝐚𝐯

𝐛 + 𝐯

a , b are constants (2)

Rearranging Eq. (2) to the linear form

𝐯

𝐤=

𝐛

𝐚+

𝟏

𝐚𝐯

(3)

Perform linear regression on v/k vs. v to obtain the slope and intercept.

Plot the linear and data points on linear graph paper.

Answer:

V (cm/min) K (1/min) v/k

0.118 0.004602 25.641

0.256 0.005888 43.478

0.539 0.007546 71.428

0.9 0.009 100

Performing linear regression on v/k vs. v yields the below results: Linear model Poly1: (y = p1*x + p2)

Coefficients (with 95% confidence bounds):

p1 = 93.87 (67.26, 120.5)

p2 = 17.59 (3.135, 32.05) R-square: 0.9914 and r=0.9956

Therefore: v

k=

b

a+

1

av

v

k= 17.59 + 93.87v

For this equation, the value for slope (1/a) is 93.87 and the value for intercept (b/a) is 17.59.

A different approach is to directly fit the hyperbolic equation (k =av

b+v) using the v and k data points. This was

done using MATLAB curve fitting toolbox. Below is the result which is pretty much close to the previous results.

General model:

k =a . v

b + v

Coefficients (with 95% confidence bounds):

a = 0.01032 (0.006976, 0.01367) b = 0.1689 (-0.01212, 0.35)

R-square: 0.9611 and r=0.9803

Using this result, the slope of the Eq.3 (1/a) is 96.89 and the intercept (b/a) is 16.366

Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

5. Estimate parameters a and b for Eq. (2).

Plot the estimation curve on (3).

Answer:

Using linear regression approach, we found slope (1/a) equal to 93.87 and the intercept (b/a) equal to 17.59.

Therefore a and b are calculated as below: 1

a= 93.87 ==> 𝐚 = 𝟎. 𝟎𝟏𝟎𝟔𝟓𝟑𝟎𝟑

b

a= 17.59 or

b

0.0106= 17.59 ==> 𝐛 = 𝟎.𝟏𝟖𝟕𝟑𝟖𝟔𝟖𝟏𝟏

k =0.01065303v

0.187386811 + v

Figure 2: Plot of k =

0.01065303 v

0.187386811 +v on k vs. v data.

Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

A different approach is to directly fit the hyperbolic equation (k =av

b+v) using the v and k data point. This was

done using MATLAB curve fitting toolbox. Below is the result, as it can be seen it is pretty much close to the previous results.

General model:

k =a . v

b + v

Coefficients (with 95% confidence bounds):

a = 0.01032 (0.006976, 0.01367)

b = 0.1689 (-0.01212, 0.35)

R-square: 0.9611 and r=0.9803

Using this result, the slope of the Eq.3 (1/a) is 96.89 and the intercept (b/a) is 16.366

k =0.01032 v

0.1689 + v

Figure 3: plot of k= (0.01032 v)/(0.1689+v) on k vs. v data.

Ramin Shamshiri ABE 6986, HW #7 Due 10/22/08

6. Discuss your results. What is the upper limit on k?

Does k vs. v approach a straight line as 𝒗 𝟎 ?

Write an equation for characteristic depth, z, for Eq. (1). What does this mean?

Answer:

The upper limit on k from the data points is 0.009 and from the curve equation is almost 0.01 (min-1

). Yes, according to Figure 4, it can be seen that as v get larger and larger, the k vs. v converges to a straight line.

Cs

C0≅ exp(−

k

vz)

ln Cs

C0 = −

k

vz

z = −v

kln(

Cs

C0)

Dimension analysis:

[cm] = −

cmmin

1

min

[1]

It means that the parameter depth depends on the velocity and the absorption rate. The dimension of depth is

also shown to agree with what we can see from the equation.

Figure 4