Calculation

 

MEC500

EM/ASSIGNMENT(1)2017

Background

The ideal gas law is given by

 

(1)

Where p Where  p is the absolute pressure, V  is   is the volume, n is the number of moles, R moles,  R is  is the universal gas constant and T  is  is the absolute temperature. Although this equation is widely by engineer and scientist, it is accurate over only a limited range of pressure and temperature.

Furthermore, Eqs (1) is more appropriate for some gases that for others. An alternative equation of state for gasses is given by:

   (  )   

(2)

Known as the van der Waals equation, equation, where v=V/n  v=V/n  is the molal volume and a  and b  are empirical constants that depend on the particular gas.

The Task

An engineering design project requires that you accurately estimate the molal volume (n (n)of acetone (a (a= 14.09 and b = 0.00994) for a number of different temperature from 300k, 400k and 500K and  p   p  of 2.5 atm so that appropriate containment vessels can be selected. It is also of interest to examine how well acetone conforms to the ideal gas law by comparing the molal volume as calculated by Eqs (1) and (2).

Instruction

1. By using any of the numerical roots of equations method, solve the task above and determine the roots for val der Waals equations using equations using acetone gas (CO2/PO1/C3) 2. Show justification for your choice of method to solve task above. (CO2/PO1/C3) GIVEN : R = 0.082054 L atm/(mol K)

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EM/ASSIGNMENT(1)2017

MEC500

INTRODUCTION

Roots of equation is one of the numerical methods which widely used in engineering field as in engineering, we rarely come across a simple and straight forward situations which can be solve analytically. The problems are usually very complex and difficult to solve. Therefore, alternative approach provide by numerical method can be used to solve by approximating the true solution using computational implementation.

Roots of equation can be classified into two fundamental approaches : 1. Bracketing Method 

Bisection



False position

2. Open method 



Fixed-point iteration  Newton Rhapson



Secant method



Roots of polynomials

Bracketing methods start with two intial guesses that bracket the true roots while Open Method need no initial guesses to bracket the true root. Open method usually more efficient compared with bracketing method but not always work.

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MEC500

EM/ASSIGNMENT(1)2017

CALCULATION

From the question, the ideal gas law is :

 

(1)

Where  p

:

absolute pressure

V 

:

volume

n

:

the number of moles

 R

:

the universal gas constant

T 

:

the absolute temperature

From equation (1), the molal volume (v) can be determined:

         The given values are :

            Therefore, the molal volume (v) for each temperatures are :

Pressure, p (atm)

Gas constant, R

Temperature (K)

Molal volume,

v

(L/mol)

2.5

0.082054

300

9.846480

2.5

0.082054

400

13.128640

2.5

0.082054

500

16.410800

*These values will be assumed as the initial guess for each temperature.

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EM/ASSIGNMENT(1)2017

MEC500

An alternative equation of state for gases which is known as van der Waals is given by :

   (  )  

(2)

Where :

v=

 

: molal volume.

a and b : empirical constants that depend on the particular gas.

From equation (2), we can simplified the equation into :

  () ()  ()    () 

(3)

By differentiate the simplified equation above, we obtained :

 ()   ()  

(4)

To solve the task given, the method of numerical equation chosen is Newton Rhapson where the first derivative is equivalent of the slope given b y:

 ()     ( ) 

Firstly, we need to find the roots from equation (3) by substituting all the values given. The values given are :

P = 2.5 atm, a = 14.09, b= 0.00994, R = 0.082054 4

 

MEC500

EM/ASSIGNMENT(1)2017

For T=300 K ,

 ()      From this equation, the roots obtained are Since

=0.010118882,  =9.247621045,  =0.598680072.

=9.247621045 is the nearest to the initial guess we obtained from equation (1), therefore

 =9.247621045 is the true root for T=300 K. Initial guess,   = 9.846480 f(  )



f’(  )



εa (%)

εt (%)

9.314463

136.187600

255.983600

-

6.475816

9.314463

9.248599

13.551340

205.747900

0.712149

0.722803

2

9.248599

9.247621

0.195441

199.824000

0.010576

0.010578

3

9.247621

9.247621

0.000043

199.736500

0.000002

0.000002

4

9.247621

9.247621

0.000000

199.736500

0.000000

0.000000

i



0

9.846480

1



+1

By using acetone gas, the root for val der Waals equations at T=300K are 9.247621.

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EM/ASSIGNMENT(1)2017

MEC500

For T=400 K ,

 ()      From this equation, the roots obtained are Since

=12.69497233,  =0.010181469,  =0.433426199.

=12.69497233 is the nearest to the initial guess we obtained from equation (1), therefore

=12.69497233 is the true root for T=400 K. Initial guess,   = 13.128640 f(  )



f’(  )

12.722287

180.559300

444.340500

12.722286

12.695091

10.667360

392.250400

0.214218944

0.21515739

2

12.695091

12.694972

0.046226

388.852700

0.000936417

0.00093644

3

12.694972

12.694972

0.000000

388.837900

0.000000

0.000000

i



0

13.128640

1



+1



εa (%)

3.41605841

By using acetone gas, the root for val der Waals equations at T=400K are 12.694972.

6

εt (%)

 

EM/ASSIGNMENT(1)2017

MEC500

For T=500 K ,

 ()      From this equation, the roots obtained are

=16.07024466, =0.010245653,  =0.340247655.

 =16.070244669 is the nearest to the initial guess therefore  =16.070244669 is the true root for T=500 K. Initial guess,   = 16.410800 Since

f(  )



f’(  )

16.083960

224.395700

686.560275

16.083959

16.070270

8.675450

633.740575

0.085184

0.085331

2

16.070270

16.070247

0.014906

631.563254

0.000147

0.000147

3

16.070247

16.070247

0.000000

631.559551

0.000000

0.000000

i



0

16.410800

1



we obtained from equation (1),

+1



εa (%)

2.119154

By using acetone gas, the root for val der Waals equations at T=500K are 16.070247.

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εt (%)

 

EM/ASSIGNMENT(1)2017

MEC500

DISCUSSION

The van der Waals equation is basically a complex adjustment to the ideal gas as the ideal gas law works only with ideal gases and it is not too accurate in real life situations. However, people choose to work with the ideal gas law rather than the van der Waals equation as in most cases, the difference between these two equation is as low as 1% or less. In this task, both equations are used to determine the roots of molal volume of acetone gas at three different temperatures (300K, 400K and 500K).

Temperature (K)

Ideal Gas law

van der Waals

300

9.846480

9.247621

400

13.128640

12.694972

500

16.410800

16.070247

True roots

=0.010118882 =9.247621045  =0.598680072 =12.69497233  =0.010181469  =0.433426199 =16.07024466  =0.010245653  =0.340247655

The table above shows the values or roots obtained using ideal gas law, van der Waals and also the true roots. During the calculation using the Newton Rhapson method, the values obtained by ideal gas law are used as the initial guess. Comparing the molal volume obtained from both equation with the true roots, the values of root obtained using van der Waals  are closer to the true root than the values obtained using ideal gas law.

From the answers gained, there is slight difference of values obtained from ideal gas law and van der Waals equation. This is due to the properties of the formula that gives the answer a different value. For this task, a Newton Raphson method is used to determine the roots. The reason Newton Raphson is chosen is because this method is easily converged. When the initial guess is nearest to the true value, the roots will easily converge. 8

 

EM/ASSIGNMENT(1)2017

MEC500

CONCLUSSION

 Numerical methods are another approach that provides solution for many engineering problems as some cases cannot be solved using mathematical methods. There are few numerical methods and one of it is roots of equation methods which have been used to solve this task. There are two classes to determine the roots of equation which is bracketing method and open method. The chosen method for this task is Newton Rhapson which is from open method as it is an efficient method. To obtain the answers correctly, Microsoft Excel is used to solve this task. Other than Microsoft Excel, other software such as MATLAB can also be used.

REFERENCES

1. http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/waal.html  2. MEC500 NUMERICAL METHOD CHAPTER 2 “Roots of Equations – Bracketing Method” th 3. Steven C. Chapra,  Numerical Methods for Engineers, 6   edition,McGraw Hills, 2010.

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EM/ASSIGNMENT(1)2017

APPENDIX

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MEC500