Question

If we have a perfect gas of mass density \(\rho_{m} \mathrm{~kg} / \mathrm{m}^{3}\), and we assign a velocity \(\mathbf{U} \mathrm{m} / \mathrm{s}\) to each differential element, then the mass flow rate is \(\rho_{m} \mathbf{U} \mathrm{kg} /\left(\mathrm{m}^{2}-\mathrm{s}\right)\). Physical reasoning then leads to the continuity equation, \(\nabla \cdot\left(\rho_{m} \mathbf{U}\right)=-\partial \rho_{m} / \partial t .(a)\) Explain in words the physical interpretation of this equation. (b) Show that \(\oint_{s} \rho_{m} \mathbf{U} \cdot d \mathbf{S}=-d M / d t\), where \(M\) is the total mass of the gas within the constant closed surface \(S\), and explain the physical significance of the equation.

Short answer

Answer: The continuity equation, ∇⋅(ρₘU)=-∂ρₘ/∂t, physically represents the conservation of mass for a perfect gas, taking into account changes in density and mass flow rate. The integral expression, ∮ₛρₘU⋅dS=-dM/dt, states that the net mass flow rate through a constant closed surface S is equal to the negative rate of change of the total mass M inside that surface, highlighting the conservation of mass principle for a perfect gas.

Step-by-step solution

Understanding the terms and given equation

Before diving deep into the physical interpretation, let's understand the terms involved: 1. ρₘ: Mass density of the perfect gas (kg/m³) 2. U: Velocity of each differential element (m/s) 3. ρₘU: Mass flow rate (kg/(m²s)) 4. Continuity equation: ∇⋅(ρₘU)=-∂ρₘ/∂t

Physical interpretation of the continuity equation

The continuity equation, ∇⋅(ρₘU)=-∂ρₘ/∂t, expresses the conservation of mass for a perfect gas. It states that the sum of net mass flowing out of a given volume plus the rate of change of density with time should be zero. This is because the mass entering the volume should either stay within it or leave it, thereby keeping the total mass constant. In simpler terms, the equation ensures that mass is conserved in a system when taking into account changes in density and mass flow rate.

Proving the integral expression

Next, we need to prove that ∮ₛρₘU⋅dS=-dM/dt, where M is the total mass of the gas within the constant closed surface S. To do this, we'll use the divergence theorem. The divergence theorem states that: ∮ₛ(F⋅dS) = ∫∫∫_V(∇⋅F)dV In our case, F = ρₘU. Applying the divergence theorem, we get: ∮ₛ(ρₘU⋅dS) = ∫∫∫_V(∇⋅(ρₘU))dV We know from the continuity equation that ∇⋅(ρₘU)=-∂ρₘ/∂t, so: ∮ₛ(ρₘU⋅dS) = -∫∫∫_V(∂ρₘ/∂t)dV Now, the total mass of the gas within the constant closed surface S can be expressed as: M = ∫∫∫_V ρₘdV Differentiating M with respect to time, we get: dM/dt = ∫∫∫_V (∂ρₘ/∂t) dV Therefore, we have: ∮ₛρₘU⋅dS=-dM/dt

Physical significance of the integral expression

The equation ∮ₛρₘU⋅dS=-dM/dt has a clear physical significance. It states that the net mass flow rate through a constant closed surface S is equal to the negative rate of change of the total mass M inside that surface. In other words, if the mass within the volume is increasing, the net mass flow rate entering the surface is larger than the mass flow rate leaving it. Conversely, if the mass within the volume is decreasing, the net mass flow rate leaving the surface is larger than the mass flow rate entering it. This equation essentially highlights the conservation of mass principle for a perfect gas.