**9.2 Adsorption isotherm**

Since the process of adsorption of natural gas on carbon nanotubes is similar to the adsorption of gas by ANG method, it was considered as a possible type of interaction between nanotubes and natural gas. The adsorption equilibrium depends on the pressure, temperature, and geometry of the adsorbent, which makes adsorption isotherms to be used to obtain the adsorption equilibrium. The Sips equation has the highest consistency with the experimental data to explain the adsorption of natural gas in nanotubes; thus, it can be used to describe the behavior of the adsorption process. The equation of the Sips adsorption isotherm is shown below:

$$\mathbf{q}\_{\mathbf{e}} = \frac{\mathbf{q}\_{\mathbf{m}} (\mathbf{b}\rho)^{\frac{1}{n}}}{\mathbf{1} + (\mathbf{b}\rho)^{\frac{1}{n}}} \tag{1}$$

$$\frac{1}{\mathbf{n}} = \frac{1}{\mathbf{n}\_o} + \alpha \left(\mathbf{1} - \frac{\mathbf{T}}{\mathbf{T}\_o}\right) \tag{2}$$

where b is adsorption/desorption constant (Pa�<sup>1</sup> ), n is adsorbate/adsorbent interaction parameter, p is pressure (Pa), qe is adsorption equilibrium(mmol/gr), qm is maximum adsorption (mmol/gr), T is temperature (K), To is reference temperature (K), and bo,no, and α are constant parameters of adsorption isotherm. Eq. (2) proves that the amount of equilibrium adsorption in the original equation is pressure-dependent, and the dependence of the Sips model on parameters b and n on temperature is proved. In addition, the reference temperature is 283.15 K. The other parameters are different depending on the type and structure of the adsorbent, which are shown in **Table 1**.

$$\mathbf{b} = \mathbf{b}\_o \exp\left[\frac{\mathbf{Q}}{\mathbf{R} \mathbf{T}\_\mathrm{O}} \left(\frac{\mathbf{T}}{\mathbf{T}\_o} - \mathbf{1}\right)\right] \tag{3}$$

where Q is isosteric heat adsorption during the half-life of adsorption (kJ/mol), R is universal gas constant (j/mol K), T is temperature (K), and To is reference temperature (K). Absorption is heat, and when the temperature of the adsorption medium increases, isothermal heat has a negative effect on the process and reduces the adsorption capacity. The Clausius-Clapeyron equation is used to calculate isothermal heat. Each of the adsorption equations can be used to calculate the adsorption heat:

*Natural Gas - New Perspectives and Future Developments*


#### **Table 1.**

*Constant parameters of equations [1].*

$$Q\_{st} = RT^2 \left(\frac{\delta ln\rho}{\delta T}\right)\_q \tag{4}$$

$$Q\_{st} = Q - amRT\_o \ln\left(b\_p\right) \tag{5}$$

where Qst is isosteric heat adsorption. Simultaneously with environmental conditions, equilibrium time has a great impact on the process. The kinetic equation (6) is used to obtain the absorption capacity at a given time. This is because after reaching the equilibrium point, the process of gas adsorption by nanoparticles stops, and this time is called the end time of the process:

$$\frac{t}{q\_t} = \frac{1}{k^2 q\_\epsilon^2} + \frac{t}{q\_\epsilon} \tag{6}$$

where qe is adsorption equilibrium (mmol/gr), qt is adsorption capacity (mmol/gr), t is time (min), and k2 is the experimental parameter that varies upon a change in the pressure, and its relationship is obtained by the interpretation of experimental data [1].

#### **9.3 Mass equation**

Natural gas can be stored on carbon nanotubes in the following two ways:


After writing the mass balance according to the hypotheses, the following equation was obtained:

$$\frac{dm\_{\rm g}}{dt} = \pm m\_{\rm g}^{\cdot} - m\_{\rm s} \frac{da}{dt} \tag{7}$$

$$\mathfrak{a} = \mathbf{M}\_{\mathfrak{g}} \times \mathfrak{q}\_{\mathfrak{t}} \tag{8}$$

$$\frac{da}{dt} = \mathbf{M}\_{\mathbf{g}} \times \frac{dq\_t}{dt} \tag{9}$$

*Storage of Natural Gas by CNTs DOI: http://dx.doi.org/10.5772/intechopen.103814*

$$\frac{da}{dt} = \mathcal{M}\_{\text{g}} \frac{dq\_t}{dt} = \mathcal{M}\_{\text{g}} \left[ \frac{q\_t}{t} - \frac{q\_t^2}{q\_e^t} \right] \tag{10}$$

where ṁ<sup>g</sup> is inlet (outlet) gas flow (gr/s), mg is the amount of gas stored in the vessel as gas phase (gr), ms is loaded mass of nanoparticles in the vessel (gr), a is adsorbed gas relative to adsorbent mass (gr gas/gr adsorbent), and Mg is molecular weight (gr/mol) [1].

#### **9.4 Energy equation**

To investigate the thermal behavior of the vessel, the energy equation in one dimension and the radial direction were used based on Eq. (11):

$$r\left(\varepsilon\rho\_{\rm g}\mathbf{C}\_{\rm g} + \rho\mathbf{C} + \rho\_{\rm a}a\mathbf{C}\_{\rm a}\right)\frac{\delta T}{\delta t} + r\rho\_{\rm g}\theta\mathbf{C}\_{\rm g}\frac{\delta T}{\delta r} = \frac{\delta}{\delta r}\left(r\lambda\frac{\delta T}{\delta r}\right) + \theta\mathbf{Q}\_{\rm st}\rho\frac{\delta a}{\delta t} \tag{11}$$

where C is the specific heat kJ/kg. K, r is radius (m), *v* is velocity of fluid (m/s), α is the relationship between viscosity and permeability, ε is overall porosity, λ is overall thermal conductivity (W/mK), and ρ is density (kg/m<sup>3</sup> ). Using the Darcy equation, the mass velocity of the fluid inside the porous medium can be obtained. Because the Reynolds range was small during the process, the Darcy equation is used:

$$
\theta = -\frac{1}{a}\Delta \tag{12}
$$

$$a = \frac{150\mu}{4} \frac{\left(1 - \varepsilon\_b\right)^2}{\varepsilon\_b^3 R\_p^2} \tag{13}$$

where Δρ is pressure difference (Pa), ε<sup>b</sup> is bed porosity. These laboratoryobtained equations correspond to the experimental results in the main fields for spherical nanoparticles, and their equivalent diameter must be used for nonspherical nanoparticles. Eq. (14) can be used to obtain the equivalent radius of the nanotubes:

$$R\_p = \frac{1}{2} \left( \frac{L}{\ln\left(\frac{2L}{d}\right)} \right) \tag{14}$$

where Rp is the radius of the particle (m), d is external diameter of nanotubes (nm), and L is nanotube length (nm) [1].

#### **9.5 Initial and boundary conditions**

To answer the energy equation, we must first determine the boundary values. For this purpose, the boundary conditions of the system, in the initial network and based on the Eq. (15), was calculated:

$$\frac{\delta T}{\delta r}(r\_i, t) = 0 \tag{15}$$

Two thermal resistors named R1 and R2, which show the conductivity and convection resistance, respectively, are used between the end points (vessel wall) and the fluid. The resistors are closed in series, and the resistance of each of them can be obtained by merging with each other. As **Figure 6** shows, the heat conduction and convection in the wall are balanced, and the thermal equilibrium is applied to the Eq. (16):

$$
\lambda\_{\rm eff} \frac{\delta \mathbf{T}}{\delta \mathbf{r}}(\mathbf{r\_o}, \mathbf{t}) = \mathbf{U}(\mathbf{T\_{r\_o}} - \mathbf{T\_f}) \tag{16}
$$

$$\mathbf{U} = \frac{\mathbf{1}}{\mathbf{R}\_s} \tag{17}$$

$$\mathbf{R\_s = R\_1 + R\_2} \tag{18}$$

$$\mathbf{R\_1} = \frac{\mathbf{r\_o} \ln \left(\frac{\mathbf{r\_w}}{\mathbf{r\_o}}\right)}{\lambda\_\mathbf{w}} \tag{19}$$

$$\mathbf{R\_2} = \frac{1}{\mathbf{h\_o}}\tag{20}$$

where ho is convection heat transfer coefficient of surrounding fluid of vessel (w/m.K). ro is external radius of the vessel (m), Rs is equivalent resistance, rw is inner radius of vessel (m), Tf is bath temperature of vessel (K), U is the overall heat transfer coefficient (W/m<sup>2</sup> .k), λeff is effective heat transfer conduction coefficient (W/m.K), and λ<sup>w</sup> is the thermal conductivity of the vessel's wall (W/m.K) [1].

#### **9.6 Nanotube's properties**

The smallest inner diameter of the nanotube has been reported to be 0.4 nm, and the distance between the nanotubes is equal to 0.34 nm. Hence, to obtain the number of the nanotube wall, Eq. (21) is used. As the nanotube diameter increases, the number of wall increases:

$$N\_{wall} = \frac{1}{d\_d}(d\_{\text{ext}} - d\_{\text{int}}) + \mathbf{1} \tag{21}$$

**Figure 6.** *Thermal boundary conditions in external environment of vessel [1].*

*Storage of Natural Gas by CNTs DOI: http://dx.doi.org/10.5772/intechopen.103814*

where NWall is the number of nanotubes wall, dd is the distance between the walls (nm).

$$\rho = \frac{4000}{1315} \left[ \frac{N\_{\text{wall}}}{d\_{\text{ext}}} - \frac{2d \sum\_{i=0}^{N\_{\text{wall}}^{-1}} i}{d\_{\text{ext}}^2} \right] \tag{22}$$

dext is external diameter of the nanotube (nm), and dint is internal diameter of the nanotube (nm). The weight and density of nanotubes depend on the internal and external diameter of nanotubes and the number of walls.

The properties of methane, natural gas, and carbon nanotubes are given in **Tables 2** and **3**. Nanotubes made of silicon can be called a competitor to carbon nanotubes, because there are many similarities between carbon and silicon. In this experiment, which is performed in a laboratory environment, pure methane gas is used, but on a large and industrial scale, natural gas must be used. By looking at **Table 4**, the natural gas composition used in this experiment can be obtained [1].
