## Continuity equation

In the field of foundry technology, the term continuity equation refers to the balance equation for a volume subject to a stationary pipe flow (e.g. melt flowing into a downsprue).

The flow volume in a pipe of variable diameter (Fig. 1) is constant if the pipe is passed by an incompressible fluid (water, liquid metal). If the diameter decreases, the flow rate must increase, and vice versa. Therefore, the continuity equation for incompressible liquids according to eq. 1 applies:

Eq. 1:$I_V_1 = I_V_2 \rightarrow A_1 \cdot v_1 = A_2 \cdot v_2$

In case of gases, the balance is established for masses instead of volumes. Since the pipe now has capacity for the mass parameter as a result of gas compression, the continuity equation only applies to stationary flows according to eq. 2:

Eq. 2:$I_m_1 = I_m_2 \rightarrow \rho_1 \cdot A_1 \cdot v_1 = \rho_2 \cdot A_2 \cdot v_2$

Derivation of the continuity equation

The continuity equation is based on the principle of the conservation of mass. It states those conditions under which the mass of a liquid or gas flow remains constant in a certain volume. In order to calculate the change in the flow rate of liquid particles passing through a variable cross section, it is therefore assumed that the flow is laminar and that the compressibility κ is zero.

Now, a liquid flowing through a narrowing pipe is considered, as shown in Fig. 2. The following applies to a volume element dV of this liquid according to eq. 3:

Eq. 3:$dV = A_1 \cdot ds_1 = A_2 \cdot ds_2$

At a constant density ρ, the mass dM contained in this volume element then is (eq. 4):

Eq. 4:$dM = \rho \cdot dV$

Then, according to eq. 5, the following mass flows through the surface area A1 in the time dt:

Eq. 5:$\frac{dM}{dt} = \rho \cdot dV \cdot \frac{1}{dt}$

dV1 = A1 · ds1 gives eq. 6:

Eq. 6:$\frac{dM}{dt} = \rho \cdot A_1 \cdot \frac{ds_1}{dt}$

The conservation of mass implies that the same mass per time must also flow through the smaller surface area A2, resulting in eq. 7:

Eq. 7:$\rho \cdot A_1 \cdot \frac{ds_1}{dt} = \rho \cdot A_2 \cdot \frac{ds_2}{dt}$

Under the condition that density is constant, i.e. compressibility κ = 0, and with vi = ds1/dt, this results in the known continuity equation:

Eq. 8: $A_1 \cdot v_1 = A_2 \cdot v_2\ bzw.\ \frac{A_1}{A_2} = \frac{v_2}{v_1}$

Therefore, the cross-sectional area is inversely proportional to the flow rate.

The reason for the increase in flow rate is the pressure difference Δp = p2 - p1.