Project: Updating the CENG301 Database to Include and/or Change the Values to the Values of Table E.1 in Transport Phenomena (Ed. 2) by Bird, Stewart, and Lightfoot [2002] (Referred to as BS&L for the rest of the project.)

by Lynne Parker

April 26, 2002

For this project I have considered 5% error to be the maximum percentage error allowed for consistency. Consistency means that the values are close enough to be considered interchangeable for use in the problems faced during CENG 402 at Rice University.

Hydrogen

> mw:=2.0158*g/mol; LJones:=[2.915,38]; critT:=33.3*K; critP:=1296.96*kPa; critV:=6.5e-5*m^3/mol; critmu:=3.47e-6*kg/m/s; BS&L (Table E.1)

The following is a comparison between values calculated with Matlab functions using the 301 database values and values listed in Table 9.1-2 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 H2 hydrogen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> ktpcalc(100), ktpcalc(200), ktpcalc(300)
ans =
0.0688
ans =
0.1260
ans =
0.1743
>> cpv
cpv =
17.6300 0.0670 -0.0001 0.0000 -0.0000
>> mw
mw =
2.0160
>> prcalc(100), prcalc(200), prcalc(300)
ans =
0.6899
ans =
0.7189
ans =
0.7329

> restart;

> k100:=0.0688*W/m/K: k200:=0.126*W/m/K: k300:=0.1743*W/m/K: Matlab values (ktpcalc).

> k100book:=0.06799*W/m/K: k200book:=0.1282*W/m/K: k300book:=0.1779*W/m/K: BS&L values.

> %k100:=abs((k100-k100book)/k100book)*100; %k200:=abs((k200-k200book)/k200book)*100; %k300:=abs((k300-k300book)/k300book)*100; % error.

Note that all values contain less than 5% error. Also, the error increases as temperature increases.

> Cpv:=T->17.63+0.067*T-0.0001*T^2: Matlab Cpv value (cpv).

> mol:=2.016*gram: gram:=kg/10^3: Matlab value (mw) and coversion.

> Cpv100:=Cpv(100)*J/mol/K; Cpv200:=Cpv(200)*J/mol/K; Cpv300:=Cpv(300)*J/mol/K;

> Cpv100book:=11192*J/kg/K: Cpv200book:=13667*J/kg/K: Cpv300book:=14316*J/kg/K: BS&L values.

> %Cpv100:=abs((Cpv100-Cpv100book)/Cpv100book)*100; %Cpv200:=abs((Cpv200-Cpv200book)/Cpv200book)*100; %Cpv300:=abs((Cpv300-Cpv300book)/Cpv300book)*100; % error.

All values contain less than 5% value. The percent error decreases as temperature increases.

> Pr100:=0.6899:Pr200:=0.7189:Pr300:=0.7329: Matlab values (prcalc).

> Pr100book:=0.682; Pr200book:=0.724; Pr300book:=0.720; BS&L values.

> %Pr100:=abs((Pr100-Pr100book)/Pr100book)*100; %Pr200:=abs((Pr200-Pr200book)/Pr200book)*100; %Pr300:=abs((Pr300-Pr300book)/Pr300book)*100; % error.

All values are within 5% error. Is there an error trend here? Yes, there is. Since error increased with the k values and decreased with the Cpv values, according to the rise in temperature, the largest errors are seen at the lowest and highest temperatures. (Pr=Cp*mu/k)

The following is a comparison between values calculated with a Matlab function using the 301 database values and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 H2 hydrogen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(90.6), prcalc(273.2), prcalc(673.2)
ans =
0.6859
ans =
0.7302
ans =
0.7394

at T=90.6 K

> restart;

> Pr:=0.6859; Prbook:=0.68; Probs:=0.68; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compared extremely well with both the calculated and observed values in the tables--less than 1% error. Note that the %Pr is negative. This is because the calculated and observed values in the book were equal--so the percent error between them was 0.

at T=273.2 K

> restart;

> Pr:=0.7302; Prbook:=0.73; Probs:=0.70; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The error contained in the value calculated through Matlab using the database conatined less than 5% compared to both the calculated and observed values in the table. However, the %Pr was negative because the value calculated with Matlab was greater than the calculated value in the book, and the observed value was less than both.

at T=673.2

> restart;

> Pr:=0.7394; Prbook:=0.74; Probs:=0.65; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compared well with the value calculated in the book. However, it had almost a 14% error compared to the observed value. It contained less error than the book's value, though. So, it is can still be considered consistent if the Pr value has to be calculated and not observed.

Helium

> mw:= 4.003*g/mol; LJones:=[2.576,10.2]; critT:=5.26*K; critP:=228.99*kPa; critV:=5.78e-5*m^3/mol; critmu:=2.54e-6*kg/m/s; BS&L (Table E.1)

Neon

> mw:= 20.18*g/mol; LJones:=[2.789,35.7]; critT:=35.7*K; critP:=2725.6*kPa; critV:=4.17e-5*m^3/mol; critmu:=1.56e-5*kg/m/s; critk:=0.03314*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 Ne neon

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.6667

> restart;

> Pr:=0.6667; Prbook:=0.667; Probs:=0.66; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated using Matlab and the 301 database was extremely close to both the calculated and observed values in the table. In fact, it was closer to the observed value than the calculated value in the table; %Pr is positive.

Argon

> restart;

> mw:=39.948*g/mol; LJones:=[3.432,122.4]; critT:=150.7*K; critP:=4863.6*kPa; 7.52e-5*m^3/mol; critmu:=2.64e-5*kg/m/s; critk:=0.02971*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 Ar argon

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> LJones
LJones =
3.4180 124.0000 <--Note: These values had to be changed because the
database had not been updated. These values are from Table E.1 in BS&L.

>> LJones=[3.432 122.4]
LJones =
3.4320 122.4000

>> prcalc(273.2)
ans =
0.6665

> restart;

> Pr:=0.6665; Prbook:=0.667; Probs:=0.67; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The errors contained in the value calculated through Matlab using the database were less than 5%--meaning it was consistent with the values in the table. However, the %Pr was negative because the value calculated with Matlab was less than the calculated value in the book, and the observed value was greater than both.

Kryptonite

> mw:=83.8*g/mol; LJones:=[3.675,170.0]; critT:=209.4*K; critP:=5501.9*kPa; critV:= 9.22e-5*m^3/mol; critmu:=3.96e-5; critk:=0.02067*J/m/s/K; BS&L (Table E.1)

Xenon

> mw:=131.3*g/mol; LJones:=[4.009,234.7]; critT:=289.8*K; critP:=5876.9*kPa; critV:=0.0001188*m^3/mol; critmu:=4.9e-5*kg/m/s; critk:=0.01682*J/m/s/K; BS&L (Table E.1)

Air

> mw:=28.964*g/mol; LJones:=[3.617,97.0]; critT:=132.4*K; critP:=3749*kPa; critV:=8.67e-5*m^3/mol; critmu:=1.93e-5*kg/m/s; critk:=0.03799*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using a Matlab function with the 301 database values and values listed in Table 1.1-2 from BS&L.

Table 1.1-2

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 N1.58O.42 air

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> mucalc(0), mucalc(20), mucalc(40), mucalc(60), mucalc(80), mucalc(100)
ans =
1.7176e-05
ans =
1.8126e-05
ans =
1.9052e-05
ans =
1.9947e-05
ans =
2.0826e-05
ans =
2.1674e-05

> mu0:=1.7176e-05*kg/m/s: mu20:=1.8126e-05*kg/m/s: mu40:=1.9052e-05*kg/m/s: mu60:=1.9947e-05*kg/m/s: mu80:=2.0826e-05*kg/m/s: mu100:=2.1674e-05*kg/m/s: Viscosity calculated using mucalc.

> kg:=Pa*s^2*m: Pa:=10^3*mPa: Coversions.

> mu0:=mu0;mu20:= mu20;mu40:=mu40;mu60:=mu60;mu80:=mu80;mu100:=mu100;

> mu0book:=0.01716*mPa*s; mu20book:=0.01813*mPa*s; mu40book:=0.01908*mPa*s; mu60book:=0.01999*mPa*s; mu80book:=0.02087*mPa*s; mu100book:=0.02173*mPa*s; Viscosity from table in BS&L.

> %errorat0:=abs((mu0-mu0book)/mu0book)*100; %errorat20:=abs((mu20-mu20book)/mu20book)*100; %errorat40:=abs((mu40-mu40book)/mu40book)*100; %errorat60:=abs((mu60-mu60book)/mu60book)*100; %errorat80:=abs((mu80-mu80book)/mu80book)*100; %errorat100:=abs((mu100-mu100book)/mu100book)*100; % error.

Note that all values contain less than a 5% error. In fact, all values contain less than .26% error. These values are remarkably close to those in the table.

The following is a comparison between a value calculated using a Matlab function with the 301 database values and a value listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 N1.58O.42 air

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7370

> restart;

> Pr:=0.7370; Prbook:=0.74; Probs:=0.73; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value of Pr calculated using the values from the 301 database was within 5% of the calculated value and the observed value in the book. In fact, it contains less than 0.5% error. Also, the value of the percent error difference (%Pr) is positive. This means, that the value obtained using prcalc was actually closer to the observed value than the calculated value listed in the table.

Nitrogen

> mw:=28.013*g/mol; LJones:=[3.667,99.8]; critT:=126.2*K; critP:=3394.4*kPa; critV:=9.01e-5*m^3/mol; critmu:=0.000018*kg/m/s; critk:=0.03632*J/m/s/K; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

H ere are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 N2 nitrogen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> mucalc(20)
ans =
1.7221e-05

> restart;

> mu:=1.7221e-05*kg/m/s: From mucalc.

> kg:=Pa*s^2*m: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.0175*mPa*s; From Table 1.1-3 from BS&L.

> %error:=abs((mu-mubook)/mubook)*100; % error.

The percent error is less than 5%. This shows that the mu calculated using the Matlab function mucalc is consistent with the table.

The following is an example from BS&L. It illustrates how the critical values in the 301 database are useful when solving problems.

Example 1.3-1

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 N2 nitrogen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> mw
mw =
28.0130
>> critP
critP =
3.3984e+03
>> critT
critT =
126.2700
>> critmu
critmu =
18000000 <--Note: This value had to be changed because the database had not been updated. This value is from Table E.1 in BS&L.

>> critmu=1.8e-5

critmu =

1.8000e-5

> restart;

> pc:=3.3984e3*kPa; Tc:=126.27*K; muc:=1.8e-5*kg/m/s;

> kg:=10*poise*m*s; atm:=101.325*kPa; Conversions.

> p:=854*atm; T:=(50+273.15)*K;

> pr:=p/pc; Tr:=T/Tc;

> mur:=2.39=mu/muc; From Figure 1.3-1.

> mu:=solve(mur, mu);

> mubook:=452*10^(-6)*poise;

> difference:=mu-mubook;

> %error:=abs(difference/mubook)*100; A percent error below 5 is usually considered acceptable.

The value of mu calculated using the values from the 301 database contained less than 5% error when compared with the value in the solution in the book. So, it is reasonable to assume that the value of critmu is consistant with the value calculated using Eq. 1.3-1b.

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 N2 nitrogen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7367

> restart;

> Pr:=0.7367; Prbook:=0.74; Probs:=0.73; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value of Pr calculated using the values from the 301 database contained 5% error when compared to the calculated value and the observed value in the book. In fact, it contains less than 1% error. Also, the value of the percent error difference (%Pr) is positive. This means, that the value obtained using prcalc was actually closer to the observed value than the value listed in the table.

Oxygen

> mw:=31.999*g/mol; LJones:=[3.433,133]; critT:=154.4*K; critP:=5035.9*kPa; critV:=7.44e-5*m^3/mol; critmu:=0.000025*kg/m/s; critk:=0.04406*J/m/s/K; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 O2 oxygen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> mucalc(20)
ans =
2.0280e-05

> restart;

> mu:=2.0280e-05*kg/m/s:

> kg:=Pa*s^2*m: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.0204*mPa*s;

> %error:=abs((mu-mubook)/mubook)*100; % error.

The percent error contained in value calculated by mucalc using the 301 database is less than 1. They are consistent.

The following is a comparison between values calculated using the 301 database values with Matlab functions and values listed in Table 9.1-2 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 O2 oxygen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> ktpcalc(100), ktpcalc(200), ktpcalc(300)
ans =
0.0094
ans =
0.0182
ans =
0.0257
>> cpv
cpv =
Columns 1 through 4
29.88000000000000 -0.01138000000000 0.00004337000000 -0.00000003700000
Column 5
0.00000000001010
>> mw
mw =
31.9990
>> prcalc(100), prcalc(200), prcalc(300)
ans =
0.7371
ans =
0.7366
ans =
0.7392

> restart;

> k100:=0.0094*W/m/K:;k200:=0.0182*W/m/K:k300:=0.0257*W/m/K: Matlab values (ktpcalc).

> k100book:=0.00904*W/m/K:k200book:=0.01822*W/m/K:k300book:=0.02657*W/m/K: BS&L values.

> %k100:=abs((k100-k100book)/k100book)*100; %k200:=abs((k200-k200book)/k200book)*100; %k300:=abs((k300-k300book)/k300book)*100; % error.

These values were all within 5% error--meaning they are consistent with the values in the table.

> cpv:=T->29.88-0.01138*T+0.00004337*T^2-0.000000037*T^3+0.0000000000101*T^4: Matlab Cpv value (cpv).

> mol:=31.999*gram: gram:=kg/10^3: Matlab value (mw) and coversion.

> cpv100:=cpv(100)*J/mol/K; cpv200:=cpv(200)*J/mol/K; cpv300:=cpv(300)*J/mol/K;

> cpv100book:=910*J/kg/K:cpv200book:=911*J/kg/K:cpv300book:=920*J/kg/K: BS&L values.

> %Cpv100:=abs((cpv100-cpv100book)/cpv100book)*100; %Cpv200:=abs((cpv200-cpv200book)/cpv200book)*100; %Cpv300:=abs((cpv300-cpv300book)/cpv300book)*100; % error.

These values contain less than 1% error. However, one has to use 'format long' in matlab to obtain this equation for Cpv. Otherwise, the error contained was over 10%.

> pr100:=0.7371:pr200:=0.7366:pr300:=0.7392: Matlab values (prcalc).

> pr100book:=0.764: pr200book:=0.734: pr300book:=0.716: BS&L values.

> %Pr100:=abs((pr100-pr100book)/pr100book)*100; %Pr200:=abs((pr200-pr200book)/pr200book)*100; %Pr300:=abs((pr300-pr300book)/pr300book)*100; % error.

All of these values that were calculated with Matlab using the database were well within the boundaries of error. They are consistent.

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 O2 oxygen

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7382

> restart;

> Pr:=0.7382; Prbook:=0.74; Probs:=0.74; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compared extremely well with both the calculated and observed values in the tables--less than 1% error. Note that the %Pr is negative. This is because the calculated and observed values in the book were equal--so the percent error between them was 0.

Carbon Monoxide

> mw:=28.01*g/mol; LJones:=[3.59,110]; critT:=132.9*K; critP:=3495.7*kPa; critV:=9.31e-5*m^3/mol; critmu:=0.000019*kg/m/s; critk:=0.03619*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CO carbon monoxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7369

> restart;

> Pr:=0.7369; Prbook:=0.74; Probs:=0.76; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The error contained in the value calculated through Matlab using the database was less than 5%--compared with the calculated and observed values. However, the %Pr was negative because the Matlab value was less than the calculated value in the book, and the observed value was greater than both.

Carbon Dioxide

> mw:=44.01*g/mol; LJones:=[3.996,190]; critT:=304.2*K; critP:=7376.5*kPa; critV:=9.41e-5*m^3/mol; critmu:=3.43e-5*kg/m/s; critk:=0.05104*J/m/s/K; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CO2 carbon dioxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> mucalc(20)
ans =
1.4622e-05

> restart;

> mu:=1.4622e-05*kg/m/s:

> kg:=Pa*s^2*m: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.0146*mPa*s;

> %error:=abs((mu-mubook)/mubook)*100; % error.

Note that the percent error is less than 1. This means that the calculated value for mu using mucalc is consistent with the value in the table.

The following is a comparison between values calculated using the 301 database values with Matlab functions and values listed in Table 9.1-2 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CO2 carbon dioxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> ktpcalc(200), ktpcalc(300)
ans =
0.0098
ans =
0.0162
>> cpv
cpv =
Columns 1 through 4
19.02000000000000 0.07962000000000 -0.00007370000000 0.00000003745000
Column 5
-0.00000000000813
>> mw
mw =
44.0100
>> prcalc(200), prcalc(300)
ans =
0.7565
ans =
0.7817

> restart;

> k200:=0.0098*W/m/K:k300:=0.0162*W/m/K: Matlab values (ktpcalc).

> k200book:=0.0095*W/m/K:k300book:=0.01665*W/m/K: BS&L values.

> %k200:=abs((k200-k200book)/k200book)*100; %k300:=abs((k300-k300book)/k300book)*100; % error.

Both of these values contain less than 5% error. They are consistent with the table.

> cpv:=T->19.02+0.07962*T-0.0000737*T^2+0.00000003745*T^3-0.00000000000813*T^4: Cpv value (cpv).

> mol:=44.01*gram: gram:=kg/10^3: Matlab value (mw) and coversion.

> cpv200:=cpv(200)*J/mol/K; cpv300:=cpv(300)*J/mol/K;

> cpv200book:=734*J/kg/K:cpv300book:=846*J/kg/K: BS&L values.

> %Cpv200:=abs((cpv200-cpv200book)/cpv200book)*100; %Cpv300:=abs((cpv300-cpv300book)/cpv300book)*100; % error.

The contained in these two values is less then 0.1%. Again, this is using 'format long' to run the start301 program for Cpv.

> pr200:=0.7565:pr300:=0.7817: Matlab values (prcalc).

> pr200book:=0.783: pr300book:=0.758: BS&L values.

> %Pr200:=abs((pr200-pr200book)/pr200book)*100; %Pr300:=abs((pr300-pr300book)/pr300book)*100; % error.

Both of these values are within the 5% error limit, and, therefore the values of Pr from Matlab are consistent with those in the table.

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CO2 carbon dioxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7759

> restart;

> Pr:=0.7759; Prbook:=0.78; Probs:=0.78; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The error contained in the value calculated through Matlab using the database was less than 5%--meaning it was consistent with the values in the table. However, the %Pr was negative because the calculated value was equal to the observed value in the table.

Nitric Oxide

> mw:=30.006*g/mol; LJones:=[3.47,119]; critT:=180*K; critP:=6484.8*kPa; critV:=0.000057*m^3/mol; critmu:=2.58e-5*kg/m/s; critk:=0.04945*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with Matlab functions and values listed in Table 9.1-2 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 NO nitric oxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> ktpcalc(200), ktpcalc(300)
ans =
0.0185
ans =
0.0263
>> cpv
cpv =
29.7600 0.0010 0.0000 -0.0000 0.0000
>> mw
mw =
30.0060
>> prcalc(200), prcalc(300)
ans =
0.7438
ans =
0.7459

> restart;

> k200:=0.0185*W/m/K:k300:=0.0263*W/m/K: Matlab values (ktpcalc).

> k200book:=0.01778*W/m/K:k300book:=0.0259*W/m/K: BS&L values.

> %k200:=abs((k200-k200book)/k200book)*100; %k300:=abs((k300-k300book)/k300book)*100; % error.

Both of these values contain less than 5% error. They are consistent with the table.

> cpv:=T->29.76+0.0010*T: Cpv value (cpv).

> mol:=30.006*gram: gram:=kg/10^3: Matlab value (mw) and coversion.

> cpv200:=cpv(200)*J/mol/K; cpv300:=cpv(300)*J/mol/K;

> cpv200book:=1015*J/kg/K;cpv300book:=997*J/kg/K; BS&L values.

> %Cpv200:=abs((cpv200-cpv200book)/cpv200book)*100; %Cpv300:=abs((cpv300-cpv300book)/cpv300book)*100; % error.

These values also contain 5% error. They are also consistent.

> pr200:=0.7438:pr300:=0.7459: Matlab values (prcalc).

> pr200book:=0.781: pr300book:=0.742: BS&L values.

> %Pr200:=abs((pr200-pr200book)/pr200book)*100; %Pr200:=abs((pr300-pr300book)/pr300book)*100; % error.

Both of these values also contain less than 5% error. All of the above values calculated using Matlab and the database are consistent with the table

The following is a comparison between a value calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 NO nitric oxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7453

> restart;

> Pr:=0.7453; Prbook:=0.74; Probs:=0.77; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value of Pr calculated using the values from the 301 database was within 5% of the calculated value and the observed value in the book. Also, the value of the percent error difference (%Pr) is positive. This means, that the value obtained using prcalc was actually closer to the observed value than the value listed in the table.

Nitrous Oxide

> mw:=44.012*g/mol; LJones:=[3.879,220]; critT:=309.7*K; critP:=7265*kPa; critV:=9.63e-5*m^3/mol; critmu:=3.32e-5*kg/m/s; critk:=0.05481*J/m/s/K; BS&L (Table E.1)

Sulfure Dioxide

> mw:=64.065*g/mol; LJones:=[4.026,363]; critT:=430.7*K; critP:=7883.1*kPa; critV:=0.000122*m^3/mol; critmu:=4.11e-5*kg/m/s; critk:=0.041254*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 SO2 sulfur dioxide

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> LJones
LJones =
4.2900 252.0000 <--Note: This value had to be changed because the database had not been updated. This value is from Table E.1 in BS&L.

>> LJones=[4.026 363]
LJones =
4.0260 363.0000
>> prcalc(273.2)
ans =
0.7892

> restart;

> Pr:=0.7892; Prbook:=0.79; Probs:=0.86; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value found using Matlab and the database was within 5% error compared to calculated book value. However, it was above 8% compared to the observed value. However, so is the calculated value of the book. %Pr is negative because both the calculated value from the table and the value calculated using Matlab were below the observed value. However, the Matlab value is a bit lower than the value in the table.

Flourine

> mw:=37.997*g/mol; LJones:=[3.653,112]; BS&L (Table E.1)

Chlorine

> mw:=70.905*g/mol; LJones:=[4.115,357]; critT:=417*K; critP:=7710.8*kPa; critV:=0.000124*m^3/mol; critmu:=0.000042*kg/m/s; critk:=0.040584*J/m/s/K; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 Cl2 chlorine

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7638

> restart;

> Pr:=0.7638; Prbook:=0.76; Probs:=0.76; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The error contained in the value calculated through Matlab using the database was less than 5%--meaning it was consistent with the values in the table. However, the %Pr was negative because the calculated and observed values from the book are equal.

Bromine

> mw:=159.808*g/mol; LJones:=[4.268,520]; critT:=528*K; critP:=10335.2*kPa; critV:=0.000144*m^3/mol; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 Br2 bromine

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C

>> liqmucalc(25)
ans =
0.9340

> restart;

> mu:=0.9340*cp:

> cp:=10^(-3)*Pa*s: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.744*mPa*s;

> %error:=abs((mu-mubook)/mubook)*100; % error.

The value calculated using liqmucalc and the database contains 25% error when compared with the value in the table. This is an unreasonable amount of error. Either the book has the wrong number, the database contains some wrong values, or liqmucalc can not be used for all liquids.

Iodine

> mw:=253.809*g/mol; LJones:=[4.982,520]; critT:=584*K; critP:=10335.2*kPa; critV:=0.000144*m^3/mol; BS&L (Table E.1)

Methane

> mw:=16.04*g/mol; LJones:=[3.78,154]; critT:=191.1*K; critP:=4640.7*kPa; critV:=9.87e-5*m^3/mol; critmu:=1.59e-5*kg/m/s; critk:=0.06611*J/m/s/K; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CH4 methane

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> mucalc(20)
ans =
1.0710e-05

> restart;

> mu:=1.0710e-05*kg/m/s:

> kg:=Pa*s^2*m: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.0109*mPa*s;

> %error:=abs((mu-mubook)/mubook)*100; % error.

The percent error is less than 5%. This means that the value calculated using Matlab and the database is consistent with the value of mu in the table.

The following is a comparison between values calculated using the 301 database values with Matlab functions and values listed in Table 9.1-2 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CH4 methane

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> ktpcalc(100), ktpcalc(200), ktpcalc(300)
ans =
0.0104
ans =
0.0206
ans =
0.0316
>> cpv
cpv =
Columns 1 through 4
38.38000000139699 -0.07366000000000 0.00029090000000 -0.00000026380000
Column 5
0.00000000008006
>> mw
mw =
16.0430
>> prcalc(100), prcalc(200), prcalc(300)
ans =
0.7641
ans =
0.7621
ans =
0.7759

> restart;

> k100:=0.0104*W/m/K:k200:=0.0206*W/m/K:k300:=0.0316*W/m/K: Matlab values (ktpcalc).

> k100book:=0.01063*W/m/K: k200book:=0.02184*W/m/K: k300book:=0.03427*W/m/K: BS&L values.

> %k100:=abs((k100-k100book)/k100book)*100; %k200:=abs((k200-k200book)/k200book)*100; %k300:=abs((k300-k300book)/k300book)*100; % error.

The error contained in values calculated from the database contain an average of 5%. The error grows as the temperature rises.

> cpv:=T->38.38000000139699-0.07366*T+0.0002909*T^2-0.0000002638*T^3+0.00000000008006*T^4: Cpv value (cpv).

> mol:=16.043*gram: gram:=kg/10^3: Matlab value (mw) and coversion.

> cpv100:=cpv(100)*J/mol/K; cpv200:=cpv(200)*J/mol/K; cpv300:=cpv(300)*J/mol/K;

> cpv100book:=2073*J/kg/K: cpv200book:=2087*J/kg/K: cpv300book:=2227*J/kg/K: BS&L values.

> %Cpv100:=abs((cpv100-cpv100book)/cpv100book)*100; %Cpv200:=abs((cpv200-cpv200book)/cpv200book)*100; %Cpv300:=abs((cpv300-cpv300book)/cpv300book)*100; % error.

The percent error is reasonable for all the values. They are consistant with the values from the table. However, the 'format long' command had to be used.

> pr100:=0.7641:pr200:=0.7621:pr300:=0.7759: Matlab values (prcalc).

> pr100book:=0.741: pr200book:=0.721: pr300book:=0.701: BS&L values.

> %Pr100:=abs((pr100-pr100book)/pr100book)*100; %Pr200:=abs((pr200-pr200book)/pr200book)*100; %Pr300:=abs((pr300-pr300book)/pr300book)*100; % error.

The error contained in these values is somewhat reasonable. Though the last one has 10% error; it cannot be considered consistent.

Acetylene

> mw:=26.04*g/mol; LJones:=[4.114,212]; critT:=308.7*K; critP:=6241.6*kPa; critV:=0.000113*m^3/mol; critmu:=2.37e-5*kg/m/s; BS&L (Table E.1)

Ethylene

> mw:=28.05*g/mol; LJones:=[4.228,216]; critT:=282.4*K; critP:=5066.3*kPa; critV:=0.000124*m^3/mol; critmu:=2.15e-5*kg/m/s; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 C2H4 ethylene

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> prcalc(273.2)
ans =
0.7973

> restart;

> Pr:=0.7973; Prbook:=0.80; Probs:=0.80; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compared extremely well with both the calculated and observed values in the tables--less than 1% error. Note that the %Pr is negative. This is because the calculated and observed values in the book were equal--so the percent error was 0.

Ethane

> mw:=30.07*g/mol; LJones:=[4.388,232]; critT:=305.4*K; critP:=4883.9*kPa; critV:=0.000148*m^3/mol; critmu:=0.000021*kg/m/s; critk:=0.084934*J/m/s/K; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

H ere are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 C2H6 ethane

Here are your reactions:
----------------------------------------
No reactions give n

>> Tdeg
Tdeg =
K
>> LJones
LJones =
4.4180 230.0000
>> LJones=[4.388 232] <--Note: This value had to be changed because the
LJones = database had not been updated. This

4.3880 232.0000 value is from Table E.1 in BS&L.
>> prcalc(273.2)
ans =
0.8282

> restart;

> Pr:=0.8282; Prbook:=0.83; Probs:=0.77; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compared well with the value calculated in the book. However, it did contain over 7% error compared to the observed value. It contained less error than the book's value, though; %Pr is positive. So, it is still consistent.

Propyne

> mw:=40.06*g/mol; LJones:=[4.742,261]; critT:=294.8*K; BS&L (Table E.1)

Propylene

> mw:=42.08*g/mol; LJones:=[4.766,275]; critT:=365*K; critP:=4610.3*kPa; critV:=0.000181*m^3/mol; critmu:=2.33e-5*kg/m/s; BS&L (Table E.1)

Propane

> mw:=44.1*g/mol; LJones:=[4.934,273]; critT:=369.8*K; critP:=4245.5*kPa; critV:=0.0002*m^3/mol; critmu:=2.28e-5*kg/m/s; BS&L (Table E.1)

n -butane

> mw:=58.12*g/mol; LJones:=[5.604,304]; critT:=425.2*K; critP:=3799.7*kPa; critV:=2.55e-4*m^3/mol; critmu:=2.39e-5*kg/m/s; BS&L (Table E.1)

i -butane

> mw:=58.12*g/mol; LJones:=[5.393,295]; critT:=408.1*K; critP:=3647.7*kPa; critV:=2.63e-4*m^3/mol; critmu:=2.39e-5*kg/m/s; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 (CH3)2CHCH3 i-butane

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> LJones
LJones =
5.3410 313.0000 <--Note: This value had to be changed because the database had not been updated. This value is from Table E.1 in BS&L.

>> LJones=[5.3930 295]
LJones =
5.3930 295.0000
>> mucalc(23)
ans =
7.6022e-06

> restart;

> mu:=7.6022e-6*kg/m/s:

> kg:=Pa*s^2*m: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.0076*mPa*s;

> %error:=abs((mu-mubook)/mubook)*100; % error

The percent error contained in the value calculated using mucalc and the database was less than 0.1%. The calculated value was consistent with the value in the table.

n -pentane

> mw:=72.15*g/mol; LJones:=[5.85,326]; critT:=468.5*K; critP:=3363.99*kPa; critV:=3.11e-4*m^3/mol; critmu:=2.38e-5*kg/m/s; BS&L (Table E.1)

i -pentane

> mw:=72.15*g/mol; LJones:=[5.812,327]; critT:=460.4*K; critP:=3414.65*kPa; critV:=3.06e-4*m^3/mol; BS&L (Table E.1)

Pentane

> mw:=72.15*g/mol; LJones:=[5.759,312]; critT:=433*K; critP:=3201.9*kPa; critV:=0.000303*m^3/mol; BS&L (Table E.1)

n -hexane

> mw:=86.18*g/mol; LJones:=[6.264,342]; critT:=507.3*K; critP:=3009.3525*kPa; critV:=3.7e-4*m^3/mol; critmu:=2.48e-5*kg/m/s; BS&L (Table E.1)

n -heptane

> mw:=100.2*g/mol; LJones:=[6.663,352]; critT:=540.1*K; critP:=2735.775*kPa; critV:=4.32e-4*m^3/mol; critmu:=2.54e-5*kg/m/s; BS&L (Table E.1)

n -octane

> mw:=114.23*g/mol; LJones:=[7.035,361]; critT:=568.7*K; critP:=2482.4625*kPa; critV:=4.92e-4*m^3/mol; critmu:=2.59e-5*kg/m/s; BS&L (Table E.1)

n -nonane

> mw:=128.26*g/mol; LJones:=[7.463,351]; critT:=594.6*K; critP:=2289.945*kPa; critV:=5.48e-4*m^3/mol; critmu:=2.65e-5*kg/m/s; BS&L (Table E.1)

Cyclohexane

> mw:=84.16*g/mol; LJones:=[6.143,313]; critT:=553*K; critP:=4053*kPa; critV:=0.000308*m^3/mol; critmu:=2.84e-5*kg/m/s; BS&L (Table E.1)

Benzene

> mw:=78.11*g/mol; LJones:=[5.443,387]; critT:=562.6*K; critP:=4924.4*kPa; critV:=0.00026*m^3/mol; critmu:=3.12e-5*kg/m/s; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 1.1-3 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 C6H6 benzene

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
C
>> liqmucalc(20)
ans =
0.6516

> restart;

> mu:=0.6516*cp:

> cp:=10^(-3)*Pa*s: Pa:=10^3*mPa: Conversions.

> mu:=mu;

> mubook:=0.649*mPa*s;

> %error:=abs((mu-mubook)/mubook)*100; % error.

The percent error is very low--less than 1%. This means that the value from liqmucalc consistent with the value in the table. It also means, that liqmucalc does work; the constraints just need to be determined.

Methyl Chloride

> mw:=50.49*g/mol; LJones:=[4.151,355]; critT:=416.3*K; critP:=6677.3*kPa; critV:=0.000143*m^3/mol; critmu:=3.38e-5*kg/m/s; BS&L (Table E.1)

Dichloromethane

> mw:=84.93*g/mol; LJones:=[4.748,398]; critT:=510*K; critP:=6079.5*kPa; BS&L (Table E.1)

Chloroform

> mw:=119.38*g/mol; LJones:=[5.389,340]; critT:=536.6*K; critP:=5471.6*kPa; critV:=0.00024*m^3/mol; critmu:=0.000041*kg/m/s; BS&L (Table E.1)

The following is a comparison between a value calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CHCl3 chloroform

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> LJones
LJones =
5.4300 327.0000 <--These values had to be changed because the database had not been updated. These values are from Table E.1 in BS&L.

>> LJones=[5.389 340]
LJones =
5.3890 340.0000
>> prcalc(273.2)
ans =
0.8598

> restart;

> Pr:=0.8595; Prbook:=0.86; Probs:=0.78; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compares well with the value calculated in the book. However, it contains over 10% error compared to the observed value. It does contain less error than the book's value, though; %Pr is positive. So, it is still consistent.

Carbon Tetrachloride

> mw:=153.82*g/mol; LJones:=[5.947,323]; critT:=556.4*K; critP:=4559.6*kPa; critV:=0.000276*m^3/mol; critmu:=4.13e-5*kg/m/s; BS&L (Table E.1)

The following is a comparison between values calculated using the 301 database values with Matlab functions and values listed in Table 9.1-3 in BS&L.

Table 9.1-3

H ere are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CCl4 carbon tetrachloride

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> LJones
LJones =
5.8810 327.0000 <--Note: This value had to be changed because the database had not been updated. This value is from Table E.1 in BS&L.

>> LJones=[5.947 323]
LJones =
5.9470 323.0000
>> liqmucalc(250), liqmucalc(300), liqmucalc(350)
ans =
2.0557
ans =
0.8273
ans =
0.4657

% listcom

Give the name or formula of the compound
CCl4
/home/ceng301/sdata/carbon_tetrachloride
Liquid: Cp=a0+a1T+a2T^2+a3T^3 J/mol with T in K
a0 a1 a2 a3
12.285 1.0948 -0.31825E-02 0.34252E-05

>> mw
mw =
153.8230

> restart;

> k250book:=0.1092*W/m/K: k300book:=0.09929*W/m/K: k350book:=0.08935*W/m/K: BS&L values.

> mu250:=2.0557*cp: mu300:=0.8273*cp: mu350:=0.4657*cp: Matlab values (liqmucalc).

> cp:=10^(-3)*Pa*s: Conversion.

> mu250:=mu250;mu300:=mu300;mu350:=mu350;

> mu250book:=20.32e-4*Pa*s; mu300book:=8.828e-4*Pa*s; mu350book:=4.813e-4*Pa*s; BS&L values.

> %mu250:=abs((mu250-mu250book)/mu250book)*100; %mu300:=abs((mu300-mu300book)/mu300book)*100; %mu350:=abs((mu350-mu350book)/mu350book)*100; % error.

All of these percent errors are around 5%. However, there is an obvious difference between the two mus at 300 K..

> Cpl:=T->12.285+1.0948*T-0.0031825*T^2+0.34252e-5*T^3: Cpl value (fortran--listcom).

> mol:=153.8230*gram: gram:=kg/10^3: Matlab value (mw) and coversion.

> Cpl250:=Cpl(250)*J/mol/K; Cpl300:=Cpl(300)*J/mol/K; Cpl350:=Cpl(350)*J/mol/K;

> Cpl250book:=0.8617e3*J/kg/K; Cpl300book:=0.8967e3*J/kg/K; Cpl350book:=0.9518e3*J/kg/K; BS&L values.

> %Cpl250:=abs((Cpl250-Cpl250book)/Cpl250book)*100; %Cpl300:=abs((Cpl300-Cpl300book)/Cpl300book)*100; %Cpl350:=abs((Cpl350-Cpl350book)/Cpl350book)*100; % error.

All of these percent errors are around 5%. However, it should be noted that the values calculated using the Cpl in the database rise as the temperature increases, and the values of Cpl decrease as the temperature increases in the table. Could there be an error in one of the signs in the database?

> Pr250:=Cpl250*mu250/k250book: Pr300:=Cpl300*mu300/k300book: Pr350:=Cpl350*mu350/k350book: Finding the Prandtl number using matlab values of Cpl and mu and BS&L values for k because there are no matlab programs to calculate the thermal conductivity of k.

> W:=J/s: kg:=Pa*s^2*m: Conversions.

> Pr250:=Pr250; Pr300:=Pr300; Pr350:=Pr350;

> Pr250book:=16.0; Pr300book:=7.97; Pr350book:=5.13; BS&L values.

> %Pr250:=abs((Pr250-Pr250book)/Pr250book)*100; %Pr300:=abs((Pr300-Pr300book)/Pr300book)*100; %Pr350:=abs((Pr350-Pr350book)/Pr350book)*100; % error.

All of these percent errors are around 5%. The values calculated using the 301 database and the values listed in the book are still consistant, considering the percent in error of the Cpl's and the difference in their accordance with temperature.

The following is a comparison between a value calculated using the 301 database values with a Matlab function and a value listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CCl4 carbon tetrachloride

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> LJones
LJones =
5.8810 327.0000 <--Note: These values had to be changed because the data base had not been updated. These values are from Table E.1 in BS&L.

>> LJones=[5.947 323]
LJones =
5.9470 323.0000
>> prcalc(273.2)
ans =
0.8841

> restart;

> Pr:=0.8841; Prbook:=0.89; Probs:=0.81; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value of Pr calculated using the values from the 301 database was within 5% of the calculated value and 10% of the observed value in the book. Also, the value of the %error difference (%Pr) is positive. This means, that the value obtained using prcalc was actually closer to the observed value than the value listed in the table, so as not to worry that the percent error was larger than 5%. The value is still consistent.

The following is a comparison between a value calculated using the 301 database values with a Matlab function and values listed in Table 9.3-1 from BS&L.

Here are your compounds' formulae and names:
No. Formula Name
----------------------------------------
1 CCl4 carbon tetrachloride

Here are your reactions:
----------------------------------------
No reactions given

>> Tdeg
Tdeg =
K
>> LJones
LJones =
5.8810 327.0000 <--Note: These values had to be changed because the database had not been updated. This value is from Table E.1 in BS&L.

>> LJones=[5.947 323]
LJones =
5.9470 323.0000
>> prcalc(273.2)
ans =
0.8841

> restart;

> Pr:=0.8841; Prbook:=0.89; Probs:=0.81; Matlab (prcalc). BS&L from Eq. 9.3-16. BS&L observed.

> %PrB:=abs((Pr-Prbook)/Prbook)*100; %PrO:=abs((Pr-Probs)/Probs)*100; % error.

> %Pr:=abs((Prbook-Probs)/Probs)*100-%PrO; The difference between the % error of the book's answer from the observed and the % error of Matlab's answer from observed. When %Pr is positive, Matlab's prcalc produces a better answer than the book value.

The value calculated with Matlab using the database compared well with the value calculated in the book. However, it did contain over 9% error compared to the observed value. It contained less error than the book's value, though; %Pr is positive. So, it can still be considered consistent.

Cyanogen

> mw:=52.034*g/mol; LJones:=[4.361,349]; critT:=400*K; critP:=5978.2*kPa; BS&L (Table E.1)

Cyanogen

> mw:=52.034*g/mol; LJones:=[4.361,349]; critT:=400*K; critP:=5978.2*kPa; BS&L (Table E.1)

Carbon Disulfide

> mw:=76.143*g/mol; LJones:=[4.483,467]; critT:=552*K; critP:=7903.4*kPa; critV:=0.00017*m^3/mol; critmu:=4.04e-5*kg/m/s; BS&L (Table E.1)

Dichlorodifluormethane

> mw:=120.91*g/mol; LJones:=[5.116,280]; critT:=384.7*K; critP:=4012.5*kPa; critV:=0.000218*m^3/mol; BS&L (Table E.1)