Emissions From Filling a 5-Gallon Can with Gasoline A new (empty) 5-gallon can (0.019 m³) is filled with gasoline on a day when the temperature of the air and gasoline are 31.5 °C (304.65 K). Gasoline enters the can at a volumetric flow rate of 2.5 GPM. Assume that the properties of gasoline can be

Question

Emissions From Filling a 5-Gallon Can with Gasoline
A new (empty) 5-gallon can (0.019 m³) is filled with gasoline on a day when the temperature of the air and gasoline are 31.5 °C (304.65 K). Gasoline enters the can at a volumetric flow rate of 2.5 GPM. Assume that the properties of gasoline can be approximated by the properties of octane, M = 114, [latex]P_\mathbf{v}[/latex] (at 31.5 °C) = 20 mm Hg (2.657 kPa). Your supervisor asks you to estimate the rate at which gasoline vapor is emitted to the air. Your inclination is to use emission factors (Appendix A.3) and to check the results with Eq. (4-10), but your supervisor distrusts these methods. Since you both graduated from the same university and are proud of its high academic standards, she asks you to predict the emissions from an equation derived from fundamental engineering principles, and then to compare to the prediction from emission factors.

[latex]S_{f,j}=f\frac{VP_{v,j}M_jL_r}{R_uT} [/latex] (4-10)

To do: Use two methods to estimate the rate at which gasoline vapor is emitted to the air: (a) from fundamental engineering principles, and (b) from emission factors.

Accepted Answer

(a) As an upper limit, assume that as soon as filling begins, the new empty tank contains air with gasoline vapor at a partial pressure equal to the gasoline vapor pressure [latex](P_{\mathbf{v},j})[/latex]. Gasoline (denoted as species j) entering the can displaces air and vapor from within the tank. Envision the rising level of gasoline as an upward moving piston. The mass flow rate of vapor leaving the can [latex](\dot{m}_j )[/latex] is equal to the velocity of the rising liquid level (v) times the cross-sectional area of the can (A) times the mass concentration of vapor [latex](c_j)[/latex]:

[latex]\dot{m}_j=\mathbf{v}Ac_j[/latex]

The velocity of the rising liquid (v) is equal to the volumetric flow rate of gasoline entering the can divided by the cross-sectional area (A). Thus

[latex]\dot{m}_j=\frac{Q}{A} Ac_j=Qc_j[/latex]

The mass concentration of gasoline vapor can be expressed as

[latex]{c}_j =\frac{P_jM_j}{R_uT} =\frac{P_{\mathbf{v},j}M_j}{R_uT}[/latex]

since it has been assumed that the vapor partial pressure [latex]P_j[/latex] is equal to the gasoline vapor pressure [latex]P_{\mathbf{v},j}[/latex]. Thus,

[latex]\dot{m}_j=QM_j\frac{P_{\mathbf{v},j}}{R_uT}[/latex]

Multiplying and dividing by volume V and rearranging,

[latex]\dot{m}_j=\left[\frac{QP_{\mathbf{v},j}M_j}{R_uT} \right] \left(\frac{V}{V} \right) =\left[\frac{VP_{\mathbf{v},j}M_j\left(\frac{Q}{V} \right) }{R_uT} \right] =\frac{VP_{\mathbf{v},j}M_jL_r}{R_uT}[/latex] (4-20)

where [latex]L_r[/latex] is the loading rate,

[latex]L_r=\frac{Q}{V}[/latex] (4-21)

which can be thought of as the number of containers loaded per unit time. Here, [latex]L_r[/latex] is

[latex]L_r=\frac{2.5\frac{gal}{min} }{5 gal} =0.50\frac{1}{min}[/latex]

If Eq. (4-20) is multiplied by an empirical (unitless) loading factor f, reflecting the unique physical conditions pertaining to the filling operation, an equation identical to Eq. (4-10) results,

[latex]\dot{m}_j=S_{f,j} =f\frac{VP_{\mathbf{v},j}M_jL_r}{R_uT}[/latex]

For example, it can be assumed that f = 1 when the can is filled from the top (splash loading, as in Figure 4.3a), and the falling liquid mixes the air and liquid inside the tank, which suggests that the partial pressure [latex](P_j)[/latex] is equal to the gasoline vapor pressure, [latex]P_{\mathbf{v},j}[/latex]. Alternatively, f = ½ represents "bottom filling" when gasoline enters the can gently from the bottom and causes the air above the liquid level to be only partially saturated with vapor.

Thus Eq. (4-10), which was introduced as an empirical equation, is in fact borne out by fundamental principles. If gasoline is assumed to be octane, [latex]P_{\mathbf{v},j}[/latex] (at 31.5 °C) = 20.0 mm Hg (2.657 kPa), the emission rate is calculated from the above equation,

[latex]\dot{m_j} =1.0\left[\frac{(2.657 Kpa)(0.019 m^3)(114.\frac{Kg}{Kmol} )}{8.314\frac{KJ}{Kmol\cdot K} (304.65 K)} \right] 0.50\frac{1}{min}\left(\frac{KN}{m^2Kpa} \right) \left(\frac{KJ}{KN\cdot m} \right) \left(\frac{1000 g}{Kg} \right) =1.1\frac{g}{min}[/latex]

(b) Appendix A.3 indicates that the emission factor for normal splash loading of gasoline is equal to 1.4 kg gasoline vapor per m³ of liquid. Thus the vapor emission rate is

[latex]\dot{m}=\left(1.4\frac{Kg}{m^3} \right) (0.019 m^3)\left(0.50\frac{1}{min} \right) \left(\frac{1000 g}{Kg} \right) =13.\frac{g}{min}[/latex]

This emission rate predicted by emission factors is more than an order of magnitude larger than that predicted by Eq. (4-10)!
Discussion: The disparity in the two methods is discomforting, but clearly demonstrates the inherent inaccuracy and uncertainty of emission factors. The principal factor affecting the emission rate is the vapor pressure of the gasoline. If gasoline were more volatile than octane, or if the temperature were higher, such that its vapor pressure was larger than what was assumed, the disparity in predictions would be smaller. Secondly it is not clear that the emission factor for "transfer of hydrocarbons by tank cars and trucks - splash loading, normal service" accurately describes filling a 5-gallon gasoline can. Certainly if the stream of gasoline entering the can possessed a large surface from which evaporation could occur before the liquid even enters the can, the additional evaporation could perhaps account for the disparity. Lastly, nothing in the emission factor accounts for temperature, and since vapor pressure is very sensitive to temperature, the emission factor should not be constant over a wide range of temperatures. Nevertheless, emission factors are useful for obtaining "ballpark estimates."

Question 4.6: Emissions From Filling a 5-Gallon Can with Gasoline A new (e...

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