Consider a pipe of flowing water. Suppose the pipe has a constant cross section and we consider a straight section of it (not at any bends/junctions), and the water is flowing steadily at a constant rate, under standard conditions. The area A is the cross-sectional area of the pipe. Suppose the pipe has radius r = 2 cm = 2 × 10 m. The area is then A = π π r 2 . To calculate the mass flux j m (magnitude), we also need the amount of mass of water transferred through the area and the time taken. Suppose a volume V = 1.5 L = 1.5 × 10 m passes through in time t = 2 s. Assuming the density of water is ρ = 1000 kg m, we have: Δ Δ m = ρ ρ Δ Δ V m 2 − − m 1 = ρ ρ ( V 2 − − V 1 ) m = ρ ρ V (since initial volume passing through the area was zero, final is V, so corresponding mass is m), so the mass flux is j m = Δ Δ m A Δ Δ t = ρ ρ V π π r 2 t .