Consider the extraction of energy from the tail gases from a nitric acid adsorption tower, such as that described in Chapter 4, Example 4.4. Gas composition, kmol/h: O2 371.5 N2 10,014.7 NO 21.9 NO2 Trace H2O saturated at 250°C If the gases leave the tower at 6 atm, 25°C, and are expanded to, say,

Question

Consider the extraction of energy from the tail gases from a nitric acid adsorption tower, such as that described in Chapter 4, Example 4.4.

Gas composition, kmol/h:

[latex]\begin{array}{lc} O _{2} & 371.5 \N _{2} & 10,014.7 \NO & 21.9 \NO _{2} & ext { Trace } \H _{2} O & ext { saturated at } 250^{\circ} C\end{array}[/latex]

If the gases leave the tower at 6 atm, [latex]25^{\circ} C[/latex], and are expanded to, say, 1.5 atm, calculate the turbine exit gas temperatures without preheat, and if the gases are preheated to [latex]400^{\circ} C[/latex] with the reactor off-gas. Also, estimate the power recovered from the preheated gases.

Accepted Answer

For the purposes of this calculation it will be sufficient to consider the tail gas as all nitrogen, flow 10,410 kmol/h.

[latex]P_{c}=33.5 atm , \quad T_{c}=126.2 K[/latex]

Figure 3.6 can be used to estimate the turbine efficiency.

[latex]\begin{aligned} ext { Exit gas volumetric flow-rate } &=\frac{10,410}{3600} imes 22.4 imes \frac{1}{1.5} \& \simeq 43 m ^{3} / s\end{aligned}[/latex]

from Figure 3.6 [latex]E_{P}=0.75[/latex]

[latex]P_{r} ext { inlet }=\frac{6}{33.5}=0.18[/latex]

[latex]T_{r} ext { inlet }=\frac{298}{126.2}=2.4[/latex]

For these values the simplified equations can be used, equations 3.37a and 3.38a. For [latex]N_{2} \gamma=1.4[/latex]

[latex]m=\frac{1.4-1}{1.4} imes 0.75=0.21[/latex]

[latex]n=\frac{1}{1-m}=\frac{1}{1-0.21}=1.27[/latex]

without preheat [latex]T_{2}=298\left(\frac{1.5}{6.0} ight)^{0.21}=223 K[/latex]

[latex]=-50^{\circ} C[/latex] (acidic water would condense out)

[latex] ext { with preheat } \begin{aligned}T_{2}=673\left(\frac{1.5}{6.0} ight)^{0.21} &=503 K \&=\underline{\underline{230^{\circ} C }}\end{aligned}[/latex]

From equation 3.31, work done by gases as a result of polytropic expansion

[latex]=-1 imes 673 imes 8.314 imes \frac{1.27}{1.27-1}\left\{\left(\frac{1.5}{6.0} ight)^{(1.27-1) / 1.27}-1 ight\}[/latex]

[latex]\begin{aligned} ext { Actual work } &= ext { polytropic work } imes E_{p} \&=6718 imes 0.75=5039 kJ / kmol\end{aligned}[/latex]

[latex]\begin{aligned} ext { Power output } &= ext { work } / kmol imes kmol / s =5039 imes \frac{10,410}{3600} \&=14,571 kJ / s =14.6 MW\end{aligned}[/latex]

Liquid streams

As liquids are essentially incompressible, less energy is stored in a compressed liquid than a gas. However, it is worth considering power recovery from high-pressure liquid streams (>15 bar) as the equipment required is relatively simple and inexpensive. Centrifugal pumps are used as expanders and are often coupled directly to pumps. The design, operation and cost of energy recovery from high-pressure liquid streams is discussed by Jenett (1968), Chada (1984) and Buse (1985).

Question 3.17: Consider the extraction of energy from the tail gases from a...

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