mass transport mathematics

MATHEMATICS OF MASS TRANSPORT

• Mass transport refers to the movement of substances from one point to another involving movement with the fluid and dispersion or diffusion. Various computer models are used to predict the fate of pollutants and their effects on the ecosystem.
• Mass balance is a very important tool to track pollutants in the environment and for the design of a treatment component.
• The basis for mass balance is "Law of conservation of mass"
• Mass balance is performed over specific control volumes having well defined system boundaries
• Control Volume (CV) is a specific region in space for which mass balance is written. It defines mass flow rates into and out of the system
• Mass enters and exits the control volume
• There may be an increase or decrease in the amount of mass in the control volume due to physical, chemical and or biological reactions
• Overall mass balance between times t and t+Δt = (mass at time t) + (mass entered system between t and t+Δt) - (mass exited between t and t+Δt) + (mass generated or consumed by reaction processes between t and t+Δt)
• Overall mass accumulation (ΔM/Δt) between times (t and t+Δt) is given by:

((mass at time t+Δt) - (mass at time t)) / Δt

=

(mass entered system between t and (t+Δt)) / Δt

-

(mass exited between t and (t+Δt)) / Δt

+

(mass generated or consumed by processes between t and (t+Δt)) / Δt

 As Δt approaches 0 the overall rate of mass accumulation:

(rate of mass accumulation) = (rate mass influx) - (rate mass outflux) + (rate net mass production or consumption)

dM/dt = dMin/dt - dMout/dt + dMrxn/dt

Mass accumulation term = dm/dt

For:

Steady state:       dm/dt = 0

Unsteady state:   dm/dt not equal to zero

Rate of mass in = dm (in)/dt = Qin * Cin

Rate of mass out = dm (out)/dt = Qout * Cout

Balance for a simple constant volume CSTR:

V dc/dt = Qin * Cin - Qout * Cout

Consider a steady state mass balance on a lake only that has two rivers feeding into it and one river flowing out. Assuming that the lake is well-mixed, it behaves like a CSTR

Considering a steady-state mass balance on the lake water only:

Rivers feeding IN:

C1(in) = 0 mg/L, Q1(in) = 5000 L/min

C2(in) = 0 mg/L, Q2(in) = 1000 L/min

One river flowing OUT:

C(out), Q(out) = Q1(in) + Q2(in) = 6000 L/min

At steady-state, there is no net accumulation or depletion of water in the lake

Consider mass balance on  the polluted lake:

C1(in) = 20 mg/L                                                  C1 carries the pollutant, 'X'
Q1(in) = 5000 L/min
C2(in) = 0 mg/L
Q2(in) = 1000 L/min

C(out) = 6000 L/min
Steady state conditions imply that amount of pollutant and water that flows in the lake should be equal to the amount of pollutant and water that flows out.
Carrying out a mass balance:

C(out)*Q(out) = [C1(in) * Q1(in)] + [ C2(in)+ Q2(IN)]

C(out) * 6000 L/min = [(20 mg/L * 5000 L/min) + (0 mg/L * 1000 L/min)]

Therefore:

C(out) = 20 * 5000/6000 = 16.7 mg/L

This implies that the concentration of the pollutant is diluted to 
16.7 mg/L in the river flowing out