Exercise $2$

Since fuorinated working fluids have been banned in the EU since $2011,$ several altematives are currently being investigated for use in ORC $($Organic Rankine Cycle$)$ systems. One alternative is supercritical $C{O}_2.$ In this process, subcritical $C{O}_2$ is compressed to a supercritical pressure. Heat is added to the supercritical working fluid. In the turbine, the $C{O}_2$ is expanded back to subcritical pressure and the waste heat is removed. Figure $2\mathrm{.}1$ shows the plant process diagram.

Figure $2\mathrm{.}1$: Process Diagram of the ORC Systems

The individual processes of the ORC system can be calculated as follows:

$(1),$ Isentropic Compression

$(2)(3)$: Isobaric Heat Intake

$(3)(4)$: Isentropic Expansion

$(4)(1)$: Isobaric Heat Outtake

The critical State of $C{O}_2$ is given by:

$T_C=304,12K,p_c=7,38$MPa

The low$-$pressure states have a pressure of $60\mathrm{.}6$ bar and are thus in the subcriti range. In contrast, the pressure of the high$-$pressure states is $91\mathrm{.}2$ bar and is th supercritical range. Accordingly, the ideal gas law cannot be used for the calcul

Therefore, perform the calculation using the Van der Waals equation:

$p=\frac{R*T}{V_m-b}-\frac{a}{V_m^2}$

Here $V_m$ is the molar Volume and $R$ the general Gas Constant: $R=8,314\frac{1}{\text{m}\text{o}\text{l}}$

a$)$ Berive relationship between Van der Waals parameter a and b and critical quantey A$.$ thi lite $($molar$,$ eritical volume$).$ Exploit that the critical point is the saddle point the eritical isotherms of the Van der Waals equation.

b$)$ Calculate $a_{1}b$ und $V_{me}$ for $C{O}_2.$

State $(1)$ corresponds to the state of the saturated liquid and state $(4)$ corresponds to the state of the saturated vapor. Thus, the states are on the same isotherm $-$ for $\frac{60\mathrm{.}6}{b}ar,T$ : K$.$ If these values are substituted into the Van der Waals equation and the molar volume determined, $3$ solutions are obtained:

$V_{m+1}=V_m^t=8,9206*{10}^{-5}\frac{m^3}{\text{m}\text{o}\text{t}},($saturated liquid$)$

$V_{mmm}=1,3177*{10}^{-4}\frac{m^3}{\text{m}\text{o}\text{l}}$

$V_{m4}=V_m^{\text{'}\text{'}\text{'}}=2,197*{10}^{-4}\frac{m^3}{\text{m}\text{o}\text{l}},($saturated vapor$)$

c$)$ Using these values and Maxwell's construction, show that $290K$ is a good approximation for the saturation temperature for the pressure of $\frac{60\mathrm{.}6}{b}ar.$

The change of the molar Entropy for a process of a Van der Waals gas is given by:

$S_{mi}-S_{mj}=R*ln\frac{V_{mi}-b}{V_{mj}-b}+C_{Vm}*ln\frac{T_i}{T_j}$

The molar heat capacity is calculated via:

$C_{Vm}=\frac{3}{2}*R$

d$)$ The isentropic compression from the state $(1)$ to the state results in a tempt of $T_2=310K.$ From this, calculate the molar volume in the state.

The change the molar internal energy of a Van der Waals gas is given by:

$U_{mi}-U_{mj}=C_{Vm}*(T_i-T_j)-a*(\frac{1}{V_{mi}}-\frac{1}{V_{mj}})$

e$)$ Finally, isobaric heat addition yields the state $(3)($:$337,95(K),p_3=91,\frac{2}{b}ar\}.$ Calculate the molar heat added.