Need Help?

Subscribe to Thermodynamics

Subscribe
  • Notes
  • Comments & Questions

 * Vander waals :
 $$
\left(P+\frac{9}{V^{2}}\right)(V-b)=R T
$$

where $$
a = \frac{27 R^{2} T_{cr}^{2}}{64 \  P_{c r}}, \quad b=\frac{R T_{cr}}{8 \  P_{c r}}
$$


*  Beattie - Bridgeman :
 $$
P=\frac{R_{v} T}{\overline{J}^{2}}\left(1-\frac{c}{\overline{v} T^{3}}\right)(\overline{v}+B)-\frac{A}{\overline{v}^{2}}
$$

where $$
A=A_{0}\left(1-\frac{9}{\overline{v}}\right) \quad, \quad B=B_{0}\left(1-\frac{b}{\overline{v}}\right)
$$

$$
* A_{0}, a, B_{0}, b \text { and } C \longrightarrow Table \ 3-4
$$

$$
P=1.6 \mathrm{MPa} \longrightarrow 1600 \mathrm{KPa}
$$

$$
V=0.01343 \quad m^{3} / k g
$$

 (a) ideal gas      $$
p v=R T \longrightarrow T=\frac{P{v}}{R}
$$

\(∴ T=\frac{1600 * 0.01343}{0.08149}=263.688 k \)

 (b) Vander Waals :
    $$
\left(P+\frac{9}{V^{2}}\right)(v-b)=RT
$$

$$
a=\frac{27 R^{2} T_{c r}^{2}}{64 \  P_{c r}}=\frac{27*(0.08149)^{2} (374.2)^{2}}{64*4059}=0.0966
$$

$$
b=\frac{R T_{c r}}{8 P_{c r}}=\frac{0.08149 * 374.2}{8 * 4059}=9.39*10^{-4}
$$

\(∴ T=327347 \mathrm{k} \)

 (c) Tables :
    $$
[A-12] @ P=1600 \mathrm{kPa}
$$

$$
v>v g \rightarrow \text { superheated }
$$

\(∴ A-13 \quad @ P=1.6 \mathrm{MPa} \)

$$
\longrightarrow T=70 \mathrm{c}^{\circ}
$$

$$
273+70=343k
$$

No comments yet

Join the conversation

Join Notatee Today!