The correct option is
D Law of conservation of mass
Considering the steady flow of fluid through a duct (i.e the inlet and outlet flows do not vary with time). The velocity
V and density
ρ of the fluid will be constant over any cross-sectional area
A, due to steady state flow.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/962235/original_2.png)
Since the flow is steady, there is no mass accumulation, in the considered control volume (CV) i.e.the fluid volume considered for study. Which means that the incoming mass at cross-sectional area
A1 must be equal to the outgoing mass at cross-sectional area
A2 during a short time interval
Δt.
Incoming volume flow at cross-sectional area
A1=A1V1Δt
Outgoing volume flow at cross-sectional area
A2=A2V2Δt
Incoming mass at cross-sectional area
A1=ρA1V1Δt
Outgoing mass at cross-sectional area
A2=ρA2V2Δt
Applying conservation mass,
ρA1V1Δt=ρA2V2Δt
∴ρA1V1=ρA2V2 ...(i)
Now considering ideal fluid
ρ=constant, so Eq.
(i) becomes:
A1V1=A2V2 ...(ii)
Here, Eq.
(i) is general continuity equation, while Eq.
(ii) is continuity equation for ideal fluid (considering steady state flow)
⇒Option
(d) is correct.