Help with a fluid mechanics assignment on calculating the Molar Volume

In this fluid mechanics assignment help service, the homework helper has compared the molar volume of Cl2 gas at different pressures obtained by the Ideal Gas Law with those obtained by Redlich-Kwong Equation.

The Ideal Gas Law is stated as follows:

Pv=RT

where

  • P is the pressure of the gas,
  • v is the volume of the gas
  • R is the ideal, or universal, gas constant
  • T is the absolute temperature of the gas

The Redlich-Kwong equation is given as:

PvRT=vvbaRT32(v+b){{\frac{Pv}{RT}= \frac{v}{v-b}- \frac{a}{RT^{\frac{3}{2}}(v+b)}}}………………..(1)

      where here a and b are:

Formula

where:

Tc is the temperature at the critical point, and

Pc is the pressure at the critical point.

or equivalently as:

         Formula 2 ………………..(2)

      where

Formula 3

Again it required the assignment helper to show that the give two forms of the Redlich-Kwong equation are indeed equivalent

ParameterTemperature (in K) UniversalGas Constant(in L atm/(K mol)Critical Pressure (in atm) Critical Temperature (in K)
Value 293 0.082057 76.1416.9


Table 1: Values of various parameters used

Pressure(in atm) 11.522.53510152550100

Table 2: Given Pressure values

Assistance with getting a Regular Falsi Method Function on a fluid mechanics homework

Pressure vs Molar Volume


PressureMolar volume (by ideal gas law)Molar volume (by Redlich-Kwong Equation's 1st form)Molar volume (by Redlich-Kwong Equation's 2ndform)
124.05523.76523.51
1.516.03715.74511.602
212.02811.7311.60
2.59.629.329.21
39.017.727.62
54.814.504.44
102.40552.082.04
151.601.241.20
250.9620.05410.124
500.48110.0530.066
1000.240.0530.055

We can infer that the Ideal gas law starts to differ from the Redlich-Kwong Equation at higher pressures. This may be attributed to the greater interaction between the molecules due to higher intermolecular forces in case of higher pressure.

The problem also required the fluid mechanics assignment helper to show that equation (1) and (2), which are two forms of Redlich Kwong equation, are the same.

From Equation (1)

PvRT=vvba(RT32)(v+b){\frac{Pv}{RT}=\frac{v}{v-b}-\frac{a}{(RT^{\frac{3}{2}})(v+b)}}

Simplifying the RHS

PvRT=v(RT32)(v+b)a(vb)(vb)(RT32)(v+b){\frac{Pv}{RT}=\frac{v(RT^{\frac{3}{2}})(v+b)-a(v-b)}{(v-b)(RT^{\frac{3}{2}})(v+b)}}

Cross Multiplying the denominators and Simplifying

Pv=v(RT32)(v+b)a(vb)(vb)(T12)(v+b){Pv=\frac{v(RT^{\frac{3}{2}})(v+b)-a(v-b)}{(v-b)(T^{\frac{1}{2}})(v+b)}}

Pv=v(RT32)(v+b)a(vb)(v2b2)(T12){Pv=\frac{v(RT^{\frac{3}{2}})(v+b)-a(v-b)}{(v^2-b^2)(T^{\frac{1}{2}})}}

pv(v2b2)(T12)=v(RT32)(v+b)a(vb){pv(v^2-b^2)(T^{\frac{1}{2}})=v(RT{\frac{3}{2}})(v+b)-a(v-b)}

PT12v3PT12vb2=v2(RT32)+vb(RT32)av+ab{PT^{\frac{1}{2}}v^3-PT^{\frac{1}{2}}vb^2=v^2(RT^{\frac{3}{2}})+vb(RT^{\frac{3}{2}})-av+ab}

PT12v3PT12vb2=v2(RT32)vb(RT32)+avab=0{PT^{\frac{1}{2}}v^3-PT^{\frac{1}{2}}vb^2=v^2(RT^{\frac{3}{2}})-vb(RT^{\frac{3}{2}})+av-ab=0}

Dividing both sides by PT12 and further simplifying

PT12y3PT12PT12yb2PT12v2(RT32)PT12vb(RT32)PT12+avPT12abPT12=0{\frac {PT^{\frac{1}{2}}y^3}{PT{\frac{1}{2}}}-\frac {PT^{\frac{1}{2}}yb^2}{PT{\frac{1}{2}}}-\frac{v^2(RT^{\frac{3}{2}})}{PT^{\frac{1}{2}}}-\frac{vb(RT^{\frac{3}{2}})}{PT^{\frac{1}{2}}}+\frac{av}{PT^{\frac{1}{2}}}-\frac{ab}{PT^{\frac{1}{2}}}=0}

v3vb2v2(RT)PvbRTP+avPT12abPT12=0{v^3-vb^2-\frac{v^2(RT)}{P}-\frac{vbRT}{P}+\frac{av}{PT^{\frac{1}{2}}}-\frac{ab}{PT^{\frac{1}{2}}}=0}

v3v2(RT)p+v(b2bRTp+avPT12)abPT12=0{v^3-\frac{v^2(RT)}{p}+v(-b^2-\frac{bRT}{p}+\frac{av}{PT^{\frac{1}{2}}})-\frac{ab}{PT^{\frac{1}{2}}}=0}      ………………. (3)

From Equation (2)

PvRT=Z{\frac{Pv}{RT}=Z}

A=aPR2T5/2{A=\frac{aP}{R^2 T^{5/2}}}

B=bPRT{B=\frac{bP}{RT}}

Z3Z2+(ABB2)ZAB=0{Z^3-Z^2+(A-B-B^2)Z-AB=0}

P3y3R3T3P2y2R2T2+PyRT(aPR2T52bPRTb2P2R2T2)abP2R3T32=0{\frac{P^3 y^3}{R^3 T^3}-\frac{P^2 y^2}{R^2 T^2}+\frac{Py}{RT}(\frac{aP}{R^2 T^{\frac{5}{2}}}-\frac{bP}{RT}-\frac{b^2 P^2}{R^2 T^2})-\frac{abP^2}{R^3 T^{\frac{3}{2}}}=0}

R3T3P3P3v3R3T3R3T3P3P2v2R2T2+R3T3P3PvRT(aPR2T52bPRTb2P2R2T2)R3T3P3abP2R3T72=0{\frac{R^3 T^3}{P^3}\frac{P^3 v^3}{R^3 T^3}-\frac{R^3 T^3}{P^3}\frac{P^2 v^2}{R^2 T^2}+\frac{R^3 T^3}{P^3}\frac{Pv}{RT}(\frac{aP}{R^2 T^{\frac{5}{2}}}-\frac{bP}{RT}-\frac{b^2 P^2}{R^2 T^2})-\frac{R^3 T^3}{P^3}\frac{abP^2}{R^3T^{\frac{7}{2}}}=0}

v3RTv2P+R2T2vP2(aPR2T52bPRTb2P2R2T2)abPT12=0{v^3-\frac{RTv^2}{P}+\frac{R^2T^2v}{P^2}(\frac{aP}{R^2T^{\frac{5}{2}}}-\frac{bP}{RT}-\frac{b^2P^2}{R^2T^2})-\frac{ab}{PT^{\frac{1}{2}}}=0}

v3RTv2P+R2T2vP2aPR2T52R2T2vP2bPRTR2T2vP2b2P2R2T2abPT12=0{v^3-\frac{RTv^2}{P}+\frac{R^2T^2v}{P^2}\frac{aP}{R^2T^{\frac{5}{2}}}-\frac{R^2T^2v}{P^2}\frac{bP}{RT}-\frac{R^2T^2v}{P^2}\frac{b^2P^2}{R^2T^2}-\frac{ab}{PT^{\frac{1}{2}}}=0}

v3RTv2P+vPaT12vRTPbvb2abPT12=0{v^3-\frac{RTv^2}{P}+\frac{v}{P}\frac{a}{T^{\frac{1}{2}}}-v\frac{RT}{P}b-vb^2-\frac{ab}{PT^{\frac{1}{2}}}=0}

v3RTv2P+v(aPT12bRTPb2)abPT12=0{v^3-\frac{RTv^2}{P}+v(\frac{a}{PT^{\frac{1}{2}}}-\frac{bRT}{P}-b^2)-\frac{ab}{PT^{\frac{1}{2}}}=0}     ………………. (4)

As we can see, equation (1) upon simplification becomes equations(3), and equation (2) becomes (4), both of which are identical. It is proved that the two forms of Redlich Kwong equation are indeed equivalent

The Reynolds Numberapplication by our Online Fluid Mechanics Tutor

In this question, the our fluid mechanics tutor has plotted a Moody chart by calculating the Darcy Friction Factor for different values of Reynolds Number ranging from 800 to 250,000,000 and also for different values of relative roughness.

The Darcy friction factor can be calculated for given values Reynolds Number and relative roughness by Colebrook equation

(1)                                            formula 4

fD is the Darcy Friction Factor

εD{\frac{\varepsilon}{D}} is the relative roughness value

Re is Reynolds number for the flow

We are using the Regula Falsi method to find the values of fD corresponding to the given Reynolds Number and relative roughness. This is done by finding the roots of the following equation:

1fD=2log10(ϵD3.7+2.51RefD)=0{\frac{1}{\sqrt{f_D}}=-2 \log_{10}(\frac{\frac{\epsilon}{D}}{3.7}+ \frac{2.51}{Re \sqrt f_D})=0}

 Moody Chart

Daigram

Here, we have a pipe in which water has in which water is filled up to depth y in figure 1.

In figure 2, we drop a perpendicular from O to AB to get triangle OAC and OBC

In triangle OAC,

coscos(Θ2)=OCOA{\cos cos(\frac{\Theta}{2})=\frac{OC}{OA}}

We know OD = OC +CD=R from figure 1

And CD = h

Therefore OC=R-h

coscos(Θ2)=RhR{\cos cos(\frac{\Theta}{2})=\frac{R-h}{R}}

Therefore

Θ2=(RhR){\frac{\Theta}{2}=(\frac{R-h}{R})}

Which Implies

Θ=2(RhR){\Theta=2(\frac{R-h}{R})}

We know, the perimeter of an arc = Radius*angle subtended

Therefore Wetted Perimeter(P)is given by

P=R Φ

We knowΦ=2 π-Θ{\Theta} , therefore

P=RπΘ{P=R \pi -\Theta}

We know, wetted area A is given by

A=Area below chord AB

A=Area of sector OAB + are of triangle OAB

A=1/2Angle subtended R2 +1/2base height

A=(2πΘ)R2+12ABOC{A=(2\pi-\Theta)R^2+\frac{1}{2}AB OC}

In triangle OAC sin(θ/2)=AC/R

sin(Θ2)=ACR{sin(\frac{\Theta}{2})=\frac{AC}{R}}

AC=Rsin(Θ2){AC=R sin(\frac{\Theta}{2})}

Similarly

OC=Rcos(Θ2){OC=R cos(\frac{\Theta}{2})}

We know AB=2 AC

A=12(2πΘ)R2+122Rsin(Θ2)Rcos(Θ2){A=\frac{1}{2}(2 \pi-\Theta)R^2+\frac{1}{2}2 R sin(\frac{\Theta}{2})Rcos(\frac{\Theta}{2})}

A=12(2πΘ)R2+12R2(2sin(Θ2)cos(Θ2)){A=\frac{1}{2}(2 \pi-\Theta)R^2+\frac{1}{2}R^2 (2sin(\frac{\Theta}{2})cos(\frac{\Theta}{2}))}

We know

2sin(Θ2)cos(Θ2)=sin(Θ){2sin(\frac{\Theta}{2})cos(\frac{\Theta}{2})=sin(\Theta)}

Therefore.

A=12(2πΘ)R2+12R2sin(Θ){A=\frac{1}{2}(2 \pi-\Theta)R^2+\frac{1}{2}R^2 sin (\Theta)}

Which implies

A=R2(2πΘ+sin(Θ))2{A=\frac{R^2(2 \pi-\Theta+sin(\Theta))}{2}}

For the numerical, the given values are

Length (in m)Depth(in m)Radius of pipe(in m)Volumetric flow rate(m3/sec)Acceleration due to gravity(m/sec2)Density of Water(in kg/m3)Reynolds NumberRelative Roughness
401.81.2129.8199715000000.008


MATLAB Code:

Main Script:

Friction FactorHydraulic Diameter(in m)Average Flow Rate(in m/sec)Head Loss(in m)Pressure Loss(in Pa/m)
0.0352452.89623.29720.269722638