EFN and Sustainable Growth Redo Problem 23 using sales growth rates of 30 and 35 pe...

Following is the graphical representation showing the relationship between the external funds needed and the sales growth rate:

The above graph shows the change in external fund requirement with the change in the sales growth rate. - At 20 % sales growth rate the need for external financing is $ 13,036.40. - As the sales growth rate changed from 20 % to 30 %, the addition to retained earnings upraised to $ 241,395.60 and hence, the external fund requirement increased to $ 51,040.60 - With 35 % rise in sales growth, the addition to retained earnings ha increased to $ 251,005.80 and hence, the external funds upraised to $ 70,641.62. The graph presents a direct relationship between the sales growth rate and the external funds needed.

d)

Mole fraction of methanol in liquid phase \(\left(x_{1}\right)\) is \(0.8\)

From subpart (a) the bubble pressure \((P)\) at mole fraction of \(0.8\) is \(1.489\) bar. At the bubble point, the state of the system will be liquid.

Fugacity of a component at liquid state is defined as the product of mole fraction of component in liquid phase, activity coefficient of the component, and saturated pressure of the component.

$$ f_{1}=x_{1} \gamma_{1} P_{1}^{5 a t} $$

Substitute \(1.81\) bar for \(P_{1}^{\text {sat }}, 0.992\) for \(\gamma_{1}\), and \(0.8\) for \(x_{1}\) in equation (3) to calculate the fugacity of methanol.

$$ \begin{aligned} f_{1} &=x_{1} \gamma_{1} P \\ &=0.8 \times 0.992 \times 1.81 \mathrm{bar} \\ &=1.436 \mathrm{bar} \end{aligned} $$

Therefore, the required fugacity is \(1.436 \mathrm{bar}\).

e)

Write the equation for chemical potential for methanol as follows:

$$ \mu_{1}=G_{1}+R T \ln \left(\gamma_{1} x_{1}\right) $$

Here, \(\mu_{1}\) is chemical potential of methanol, \(G_{1}\) Gibbs free energy of methanol, \(R\) is gas constant, \(x_{1}\) is mole fraction of methanol, and \(\gamma_{1}\) is activity coefficient.

Mole fraction of methanol in liquid phase \(\left(x_{1}\right)\) is \(0.8\).

Pressure of the system \((P)\) is \(0.4\) bar.

Temperature of the mixture \((T)\) is \(353.15 \mathrm{~K}\).

From subpart (a), \(0.8\) mole fraction and \(353.15 \mathrm{~K}\) activity coefficient of methanol \(\left(\gamma_{1}\right)\) is \(0.992\)

Since the system is pure ethanol liquid, the Gibbs free energy is neglected.

Substitute the corresponding values in the above equation to calculate the chemical potential.

$$ \begin{aligned} \mu_{1} &=G_{1}+R T \ln \left(\gamma_{1} x_{1}\right) \\ &=8.314 \mathrm{~J} / \mathrm{molK} \times 353.15 \mathrm{~K} \times \ln (0.992 \times 0.8) \\ &=-678.8 \mathrm{~J} / \mathrm{mol} \end{aligned} $$

Therefore, the required chemical potential is \(-678.8 \mathrm{~J} / \mathrm{mol}\)

f) The new reference pressure for pure methanol is its saturated pressure and is equal to 1.81 bar The chemical potential is independent of pressure and thus the chemical potential is constant as in subpart (e).

Therefore, the required chemical potential is -678.8 J / mol.