Except for the fact that nitrogen is now being considered to be a real, rather than an ideal, gas, the problem here is the same as in Illustration 3.4-5. In fact, Eqs. h–n in the comments to that illustration apply here. The additional equation needed to solve this problem is obtained in the same manner as in Illustration 4.5-2. Thus, for the cylinder initially filled we have
\underline{S} _{1}^{i}=\underline{S}_{1}^{f} (o)
For an ideal gas of constant heat capacity (which is not the case here) this reduces to
\left(\frac{P_{1}^{f}}{P_{1}^{i}}\right)=\left(\frac{T_{1}^{f}}{T_{1}^{i}}\right)^{C_{ P }^{*} / R}
by Eq. 4.4-3. For real nitrogen gas, Eq. o requires that the initial and final states in cylinder 1 be connected by a line of constant entropy in Fig. 3.3-3.
With eight equations (Eqs. h–o) and eight unknowns, this problem can be solved, though the solution is a trial-and-error process. In general, a reasonable first guess for the pressure in the nonideal gas problem is the ideal gas solution, which was
\underline{S}\left(T_{2}, P_{2}\right)-\underline{S}\left(T_{1}, P_{1}\right)=C_{ P }^{*} \ln \left(\frac{T_{2}}{T_{1}}\right)-R \ln \left(\frac{P_{2}}{P_{1}}\right) (4.4-3)
P_{1}^{f}=P_{2}^{f}=20 bar
Locating the initial conditions (P=40 bar =4 MPa \text { and } T=200 K ) \text { in Fig. } 3.3-3 \text { yields }^{11}\hat{H}_{1}^{i} \approx 337 kJ / kg \text { and } \hat{V}_{1}^{i} \approx 0.0136 m ^{3} / kg , so that
M_{1}^{i}=\frac{V_{1}}{\hat{V}_{1}^{i}}=\frac{1 m ^{3}}{0.136 \frac{ m ^{3}}{ kg }}=73.5 kg
Now using Eq. o \text { (in the form } \hat{S}_{1}^{i}=\hat{S}_{1}^{f} \text { ) and Fig. 3.3-3, we find } T_{1}^{f} \approx 165 K , H_{1}^{f}=310 kJ / kg,\text { and } \hat{V}_{1}^{f} \approx 0.0224 m ^{3} / kg , so that
M_{1}^{f}=\frac{V_{1}}{\hat{V}_{1}^{f}}=\frac{1 m ^{3}}{0.0224 \frac{ m ^{3}}{ kg }}=44.6 kg
and M_{2}^{f}=M_{1}^{i}-M_{1}^{f}=28.9 kg, which implies
\hat{V}_{2}^{f}=\frac{1 m ^{3}}{28.9 kg }=0.0346 \frac{ m ^{3}}{ kg }
Locating P=20 \text { bar ( } 2 MPa \text { ) and } \hat{V}=0.0346 m ^{3} / kg \text { on Fig. } 3.3-3 \text { gives a value for } T_{2}^{f} of about 240 K \text { and } \hat{H}_{2}^{f}=392 kJ / kg.
Finally, we must check whether the energy balance is satisfied for the conditions computed here based on the final pressure we have assumed. To do this we must first compute the internal energies of the initial and final states as follows:
\begin{aligned}\hat{U}_{1}^{i} &=\hat{H}_{1}^{i}-P_{1}^{i} \hat{V}_{1}^{i} \\&=337 \frac{ kJ }{ kg }-40 bar \times 0.0136 \frac{ m ^{3}}{ kg } \times 10^{5} \frac{ Pa }{ bar } \times \frac{1 J }{ m ^{3} Pa } \times \frac{1 kJ }{1000 J }=282.6 \frac{ kJ }{ kg }\end{aligned}
Similarly,
\hat{U}_{1}^{f}=265.2 kJ / kg
and
\hat{U}_{2}^{f}=322.8 kJ / kg
The energy balance (Eq. n of Illustration 3.4-5) on a mass basis is
M_{1}^{i} \hat{U}_{1}^{i}=M_{1}^{f} \hat{U}_{1}^{f}+M_{2}^{f} \hat{U}_{2}^{f}
or
73.5 × 282.6 kJ = 44.6 × 265.2 + 28.9 × 322.8 kJ
2.0771 \times 10^{4} kJ \approx 2.1157 \times 10^{4} kJ
Thus, to the accuracy of our calculations, the energy balance can be considered to be satisfied and the problem solved. Had the energy balance not been satisfied, it would have been necessary to make another guess for the final pressures and repeat the calculation.
It is interesting to note that the solution obtained here is essentially the same as that for the ideal gas case. This is not generally true, but occurs here because the initial and final pressures are sufficiently low, and the temperature sufficiently high, that nitrogen behaves as an ideal gas.
Had we chosen the initial pressure to be higher, say several hundred bars, the ideal gas and real gas solutions would have been significantly different (see Problem 4.22).
