# Plume Calculations

Scott D. Brown

## Mass Rate Conversion

This calculation outlines the conversion from the concentration rate in ppm/s input to the Blackadar model to a mass rate in gm/s.

### Derivation

The Blackadar is bouyancy driven, and the Briggs buoyancy flux parameter (\(F_b\)) can be computed in various ways. The most practical is from the following equation:

\begin{equation} F_b = g \frac{V_s}{\pi} \frac{T_s - T_a}{T_s} = g v_s r^2 \frac{T_s - T_a}{T_s} \end{equation}

where \(g\) is the gravity constant, \(V_s\) is the stack volume flow rate [m3/s], \(v_s\) is the stack gas exit velocity [m/s], \(r\) is the stack radius [m], \(T_s\) is the stack gas temperature [K] and \(T_a\) is the ambient air temperature [K].

Since \(F_b\), \(r\), \(T_s\) and \(T_a\) are provided was inputs, we can compute the volume flow rate \(V_s\) as:

\begin{equation} V_s = \frac{F_b \pi}{g} \frac{T_s}{T_s - T_a} \end{equation}

The release mass rate is provided as a concentration rate [ppm/s]. If we want to compute the mass flow rate, then we need to do some calculations to compute the relative volume of the gas.

The release concentration rate (\(\dot{C}\)) is ratio of the volume rate of the gas, \(V_g\), and the volume rate of the air, \(V_a\):

\begin{equation} \dot{C} = \frac{V_g [m^3/s]}{V_a [m^3/s] \cdot (1 \times 10^6)} \end{equation}

If we assume the gas is a small fraction of the overall volume, we can assume \(V_a\) equals \(V_s\) and solve for the volume rate of the gas \(V_g\) as:

\begin{equation} V_g = \frac{\dot{C}}{V_s \cdot (1 \times 10^6)} \end{equation}

The Ideal Gas Law (\(PV = nRT\)) allows us to relate the volume of gas to the number of moles (\(n\)) of the gas:

\begin{equation} n = \frac{PV}{RT} \end{equation}

where, \(P\) is the pressure in Pa, \(R\) is the universal gas constant (8.31446261815324 joules/K/mole) and \(T\) is the temperature in K. Since the pressure is not explicitly provided as an input, we will assume standard pressure Pstd = 1 x 105 Pa (100 kPa) or 1 x 105 N/m2. It is assumed that temperature of the released air/gas mixture is the release temperature, \(T = T_s\). Finally, since we are ultimately solving for the mass rate \(\dot{m}\) [gm/s], we will modify this equation to compute the molar rate \(\dot{n}\) [moles/s] by using the gas volume rate (\(V_g\)) rather than just a volume:

\begin{equation} \dot{n}~[moles/s] = \frac{P_{std} V_g}{R T_s} = \frac{P_{std} \dot{C}}{R T_s V_s (1 \times 10^6)} = \frac{P_{std} \dot{C} g (T_s - T_a)}{R T_s^2 F_b \pi (1 \times 10^6)} \end{equation}

Finally, the mass rate \(\dot{m}\) can be computed by folding in the molecular weight (\(m\)) in grams per mole. Note that the molecular weight is not included as an input to the model and must be externally obtained:

\begin{equation} \dot{m}~[gm/s] = \frac{P_{std} \dot{C} m g (T_s - T_a)}{R T_s^2 F_b \pi (1 \times 10^6)} \end{equation}

Conversely, if we want to compute the concentration rate, \(\dot{C}\), from a mass rate, \(\dot{m}\):

\begin{equation} \dot{C}~[ppm/s] = \frac{\dot{m} R T_s^2 F_b \pi (1 \times 10^6)}{P_{std} m g (T_s - T_a)} \end{equation}

### Example #1

Computing the mass release rate, \(\dot{m}\), of a methane (CH4) plume with a release concentration rate of \(\dot{C}\) = 100 ppm/s, a molecular mass \(m\) of 16.04 gm/mole, a Briggs Bouyancy Flux of \(F_b\) = 10.0 m4/s3, a release temperature of \(T_s\) = 350 K and an air temperature of \(T_a\) = 300 K:

\begin{equation} \dot{m}~[gm/s] = \frac{(1 \times 10^5)(100)(16.04)(9.8)(350 - 300)} {(8.3144)(350^2)(10) \pi (1 \times 10^6)} = 2.45 \times 10^{-3} ~[gm/s] = 8.84 ~[gm/hr] \end{equation}

### Example #2

Computing the concentration release rate, \(\dot{C}\), of a methane (CH4) plume with a mass release rate of \(\dot{m}\) = 100 kg/hr (22.777 gm/s), a molecular mass \(m\) of 16.04 gm/mole, a Briggs Bouyancy Flux of \(F_b\) = 10.0 m4/s3, a release temperature of \(T_s\) = 350 K and an air temperature of \(T_a\) = 300 K:

\begin{equation} \dot{C}~[ppm/s] = \frac{(22.77)(8.3144)(350^2)(10) \pi (1 \times 10^6)} {(1 \times 10^5)(16.04)(9.8)(350 - 300)} = 1793.4 ~[ppm/s] \end{equation}