Yes a flow can be incompressible (rather isochoric) and unsteady.
However, the unsteady term in Conservation of Mass equation is cancelled by advection term regardless of whether the flow is incompressible or compressible
CASE 1: Any fluid material that undergoes an incompressible flow
-No mixing of multiple fluids, etc.
1) Integral Form of Conservation of mass (entire control volume) (see wiki):
- D(mass)/Dt = 0 = D/Dt[integral of density over the control volume]
2) Ignore integral over control volume
3) => Differential Form of Conservation of Mass
(differential control volume)
4) Apply the total derivative to density assuming it's a function of space and time; not pressure
Although if you did assume pressure was a variable then the term generated would be zero due to flow restrictions which are independent of the fluid material property (see case 2) i.e. partial derivative of density wrt to pressure is zero
Also, if any terms are independently zero here then so they are also zero in the differential form of Conservation of Mass
5) Convert the total derivative to a material derivative (the differential control volume follows the differential fluid element)
6) Set Conservation of Mass (Differential Form) equal to equation from 5)
7) => divergence of flow velocity is zero
the unsteady flow will have it's unsteady term "balanced" by the advection term (dot product of the gradient of density with the control volume's velocity) to achieve incompressible flow (isochoric flow).
Which implied that the density has to be non-uniform over that control volume i.e. density is function of space and time
Because flow is incompressible: density should not be a function of pressure
- despite if fluid material is compressible (e.g. air)
- else the term generated from total derivative would need to be zeroed (see 4)
Reiterating: the unsteady and advection terms actually zero out the in conservation of mass you could also think of it as they balance out (sum to zero) according to Wiki (though I disagree) Despite in/compressibility and despite if the flow is un/steady and non/uniform
This says nothing about the D/Dt(density) (although it is zero) so the unsteady and advection terms would still be null; as those terms would cancel with themselves after equating the total derivative (coinciding with the control volume) to the material derivative
- But that would say nothing about the in/compressibility of the flow
- nor the in/compressibility of the fluid material
CASE 2 A Compressible Fluid that undergoes Compressible Flow
No mixing of multiple fluids, etc.
Density would be a function of space, time and pressure.
Same analysis as above except .......
Apply the total derivative to the density function you'll see an additional term (not seen in wiki)
Again unsteady and non-uniform terms will be null
the product of density and divergence of flow velocity will be equal to this generated term due to density's dependence on pressure for compressible flow situation
-do very minor algebra and you have the compressibility (beta) formula
rho = density, P = normal pressures
beta = 1/rho *(drho/dP)
it's really cool cuz it connects very well with mechanics of materials
- Recognize that the divergence of the flow velocity is just the normal strain rates
-If you assume the the flow velocity is not a function of time (steady) and you do some more algebra, calculus, and hook's law you can acquire bulk modulus (K) (inverse of compressibility) from mechanics of materials:
K = -E/(3*(1-2nu) which will be equal to the inverted compressibility side
K = 1/beta = rho*dP/drho