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.
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.