Assignment – 3

Copper Disc Dissolution

Copper Disc Dissolution

In the Chemical Engineerig Laboratory, last week I performed the experiment of “Copper Disc Dissolution”. We also had to do the calculations at home for this experiment. In the same week, we got to learn ‘Pandas’ and as it is a good tool for computation, I used it to perform the calculations of my experiment.

This week was really really very hectic for me, we were having Funtech which is our intra college festival and being a part of Technological Association, I was a part of organising team. During last 7 days, I also had 2 company interviews for in plant training and there were thousand more things which I don’t even remember now. For the last 6 days, I was literally running all the time, used to go home very late (by 12.00-12.30 in the night) and I used to leave again in the early morning by 6.45 am. But, whatever I was doing, I was enjoying each and every part of it and there was a thrill in that. Last night was the final night and it was one of the best nights of the college, the fest ended well with a bang! We all went to have dinner together and then I took a train to go home. I took a quick nap in the travel 😋 because I knew after going home I had this assignment still pending of which today was the submission and I will have to stay up and do it othwrwise I will lose marks.

And as usual, I did this at 3 AM !

https://github.com/chiragmule1729/3-AM-Assignments.git

In [16]:

import scipy
import math
import pandas as pd
import matplotlib.pyplot as plt

Observation Table

In [48]:

example = pd.DataFrame(
{
    "Speed":[0,354.3,425.5,777.8,999.9,1183.4],
    "TitrationI":[31.2,31.2,37.9,42.1,48.4,50.0],
    "TitrationF":[31.2,37.9,42.1,48.4,50.0,52.1],
    "Disc diameter":[4.532,4.530,4.525,4.510,4.500,4.492],
    "Thickness":[0.63,0.608,0.60,0.57,0.551,0.541],   
},
    index = ["initial",1,2,3,4,5]
)
example

Out[48]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness
initial	0.0	31.2	31.2	4.532	0.630
1	354.3	31.2	37.9	4.530	0.608
2	425.5	37.9	42.1	4.525	0.600
3	777.8	42.1	48.4	4.510	0.570
4	999.9	48.4	50.0	4.500	0.551
5	1183.4	50.0	52.1	4.492	0.541

Calculation

In [55]:

def Bi(row,a,b,c):
    ti=row["TitrationI"]
    bi=(a*(b-ti))/c
    return bi

In [56]:

example["Bi"]=example.apply(Bi,axis=1, args=(0.1,53.5,10))

In [57]:

example

Out[57]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness	Bi	Bf
initial	0.0	31.2	31.2	4.532	0.630	0.223	0.223
1	354.3	31.2	37.9	4.530	0.608	0.223	0.156
2	425.5	37.9	42.1	4.525	0.600	0.156	0.114
3	777.8	42.1	48.4	4.510	0.570	0.114	0.051
4	999.9	48.4	50.0	4.500	0.551	0.051	0.035
5	1183.4	50.0	52.1	4.492	0.541	0.035	0.014

In [58]:

def Bf(row,a,b,c):
    tf=row["TitrationF"]
    bf=(a*(b-tf))/c
    return bf

In [59]:

example["Bf"]=example.apply(Bf,axis=1, args=(0.1,53.5,10))

In [60]:

example

Out[60]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness	Bi	Bf
initial	0.0	31.2	31.2	4.532	0.630	0.223	0.223
1	354.3	31.2	37.9	4.530	0.608	0.223	0.156
2	425.5	37.9	42.1	4.525	0.600	0.156	0.114
3	777.8	42.1	48.4	4.510	0.570	0.114	0.051
4	999.9	48.4	50.0	4.500	0.551	0.051	0.035
5	1183.4	50.0	52.1	4.492	0.541	0.035	0.014

In [61]:

def ksle(row):
    bi=row["Bi"]
    bf=row["Bf"]
    ksle=248.139*math.log(bi/bf)
    return ksle

In [62]:

example["ksle"]=example.apply(ksle,axis=1)

In [63]:

example

Out[63]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness	Bi	Bf	ksle
initial	0.0	31.2	31.2	4.532	0.630	0.223	0.223	0.000000
1	354.3	31.2	37.9	4.530	0.608	0.223	0.156	88.663976
2	425.5	37.9	42.1	4.525	0.600	0.156	0.114	77.830673
3	777.8	42.1	48.4	4.510	0.570	0.114	0.051	199.596266
4	999.9	48.4	50.0	4.500	0.551	0.051	0.035	93.418768
5	1183.4	50.0	52.1	4.492	0.541	0.035	0.014	227.367466

In [68]:

def Re(row,a,b):
    d=row["Disc diameter"]
    n=row["Speed"]
    Re=(d*d*n*a)/(100*60*b)
    return Re

In [69]:

example["Re"]=example.apply(Re,axis=1, args=(1000,0.00119))

In [70]:

example

Out[70]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness	Bi	Bf	ksle	Re
initial	0.0	31.2	31.2	4.532	0.630	0.223	0.223	0.000000	0.000000e+00
1	354.3	31.2	37.9	4.530	0.608	0.223	0.156	88.663976	1.018285e+06
2	425.5	37.9	42.1	4.525	0.600	0.156	0.114	77.830673	1.220221e+06
3	777.8	42.1	48.4	4.510	0.570	0.114	0.051	199.596266	2.215760e+06
4	999.9	48.4	50.0	4.500	0.551	0.051	0.035	93.418768	2.835851e+06
5	1183.4	50.0	52.1	4.492	0.541	0.035	0.014	227.367466	3.344359e+06

In [79]:

def Sh(row):
    Re=row["Re"]
    Sh=(0.03387*0.62*(Re**0.5))/(0.565)
    return Sh

In [80]:

example["Sh"]=example.apply(Sh,axis=1)

In [81]:

example

Out[81]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness	Bi	Bf	ksle	Re	Sh
initial	0.0	31.2	31.2	4.532	0.630	0.223	0.223	0.000000	0.000000e+00	0.000000
1	354.3	31.2	37.9	4.530	0.608	0.223	0.156	88.663976	1.018285e+06	37.505340
2	425.5	37.9	42.1	4.525	0.600	0.156	0.114	77.830673	1.220221e+06	41.056101
3	777.8	42.1	48.4	4.510	0.570	0.114	0.051	199.596266	2.215760e+06	55.324799
4	999.9	48.4	50.0	4.500	0.551	0.051	0.035	93.418768	2.835851e+06	62.589305
5	1183.4	50.0	52.1	4.492	0.541	0.035	0.014	227.367466	3.344359e+06	67.969623

In [88]:

def kslp(row):
    d=row["Disc diameter"]
    Sh=row["Sh"]
    kslp=(0.0907)*Sh*100/d
    return kslp

In [89]:

example["kslp"]=example.apply(kslp,axis=1)

Calculation Table

In [90]:

example

Out[90]:

	Speed	TitrationI	TitrationF	Disc diameter	Thickness	Bi	Bf	ksle	Re	Sh	kslp
initial	0.0	31.2	31.2	4.532	0.630	0.223	0.223	0.000000	0.000000e+00	0.000000	0.000000
1	354.3	31.2	37.9	4.530	0.608	0.223	0.156	88.663976	1.018285e+06	37.505340	75.093474
2	425.5	37.9	42.1	4.525	0.600	0.156	0.114	77.830673	1.220221e+06	41.056101	82.293665
3	777.8	42.1	48.4	4.510	0.570	0.114	0.051	199.596266	2.215760e+06	55.324799	111.262955
4	999.9	48.4	50.0	4.500	0.551	0.051	0.035	93.418768	2.835851e+06	62.589305	126.152222
5	1183.4	50.0	52.1	4.492	0.541	0.035	0.014	227.367466	3.344359e+06	67.969623	137.240535

In [108]:

n=[0,354.3,425.5,777.8,999.9,1183.4]

In [109]:

ksle=[88.663976,77.830673,199.596266,93.418768,227.367466]

In [110]:

kslp=[0,75.093474,82.293665,111.262955,126.152222,137.240535]

Graph

In [112]:

plt.plot(n,kslp,'g-',label='kslp vs rpm')
plt.ylabel('kslp')
plt.xlabel('rpm')
plt.legend(loc='best')
plt.show()

In [116]:

example.to_excel("Copper Disc Dissolution.xlsx", sheet_name="17CHE123")

Results and Conclusion

From the above graph, we can observe –

• Predicted Mass transfer coefficient and experimental coefficient fairly match
• Rate of mass transfer increases with increase in RPM
• The reaction is phase-4 regime