Question

Assuming that there is no transformation of mass to energy or vice versa, it is possible to write a continuity equation for mass. ( \(a\) ) If we use the continuity equation for charge as our model, what quantities correspond to \(\mathbf{J}\) and \(\rho_{v}\) ? (b) Given a cube \(1 \mathrm{~cm}\) on a side, experimental data show that the rates at which mass is leaving each of the six faces are \(10.25,-9.85,1.75,-2.00\), \(-4.05\), and \(4.45 \mathrm{mg} / \mathrm{s}\). If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center.

Short answer

The quantities that correspond to J and 𝜌_v in the continuity equation for mass are mass current density (𝑱_m) and mass density (𝜌_m), respectively. 𝑱_m represents the mass flow per time per area through a given surface, and 𝜌_m represents the mass per unit volume at a given position and time. What is the time rate of change of density at the center of a 1 cm cube given the mass rates at which mass is leaving each face of the cube? The time rate of change of density at the center of the 1 cm cube is approximately -0.0005 kg/m³/s.

Step-by-step solution

Quantity corresponding to \(\mathbf{J}\) and \(\rho_{v}\) in the continuity equation for mass

When using the continuity equation for charge as our model, let's consider the following correspondence relationship: 1. The current density \(\mathbf{J}\) corresponds to the mass current density \(\mathbf{J_m}\), where \(\mathbf{J_m}(x,y,z,t)\) represents the mass flow per time per area through a given surface at position \((x,y,z)\) and time \(t\). 2. The volume charge density \(\rho_{v}\) corresponds to mass density \(\rho_m\), where \(\rho_m(x,y,z,t)\) represents the mass per unit volume at position \((x,y,z)\) and time \(t\). With these correspondences, we can establish the continuity equation for mass that follows: $$\frac{\partial \rho_m}{\partial t} + \nabla \cdot \mathbf{J_m} = 0$$

Applying the continuity equation for mass to the cube

To determine the time rate of change of density at the center of a 1 cm cube, we need to find \(\frac{\partial \rho_m}{\partial t}\) at the center. Given the rates at which mass is leaving each face of the cube \((10.25, -9.85, 1.75, -2.00, -4.05, 4.45)\) mg/s, we'll use the principle of conservation of mass: $$\frac{\partial \rho_m}{\partial t} V = -\sum_{i=1}^6 M_i$$ where: - \(V\) is the volume of the cube, \(1\;\text{cm}^3 = 10^{-3}\;\text{m}^3\) - \(M_i\) is the mass leaving the \(i\)-th face of the cube (given in mg/s) - In order to sum the given values, convert all mass rates to kg/s: \(10^{-6}\; \text{kg/mg}\) Therefore: $$\frac{\partial \rho_m}{\partial t} = -\frac{\sum_{i=1}^6 M_i}{V}$$

Calculating the time rate of change of density

Now we can calculate the time rate of change of density at the center of the cube: $$\frac{\partial \rho_m}{\partial t} = -\frac{(10.25 -9.85 +1.75 -2.00 -4.05 + 4.45) \times 10^{-6}}{10^{-3}} \frac{\text{kg}}{\text{m}^3\text{s}}$$ $$\frac{\partial \rho_m}{\partial t} = -0.0005\; \frac{\text{kg}}{\text{m}^3\text{s}}$$ Hence, the approximate value for the time rate of change of density at the center of the cube is \(-0.0005\;\frac{\text{kg}}{\text{m}^3\text{s}}\).