Question:

In a duct, if the flow enters at 1 kg/s and exits at 0.5 kg/s, what additional information is needed to use the continuity equation effectively?

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Continuity Equation Usage. Mass Flow Rate \(\dot{m = \rho A v\). To relate mass flow rate to velocity (v) or area (A), the fluid density (\(\rho\)) must be known.
Updated On: May 7, 2025
  • The viscosity of the fluid
  • The density of the fluid
  • The temperature of the fluid
  • The velocity profile of the flow
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The Correct Option is B

Solution and Explanation

The continuity equation represents the conservation of mass.
The mass flow rate (\(\dot{m}\)) is given by \(\dot{m} = \rho A v\), where \(\rho\) is density, A is cross-sectional area, and v is average velocity.
The problem states the mass flow rate entering (\(\dot{m}_{in} = 1\) kg/s) and exiting (\(\dot{m}_{out} = 0.
5\) kg/s).
Since \(\dot{m}_{in} \neq \dot{m}_{out}\), the flow is unsteady (mass is accumulating or depleting within the duct) or there's a mistake in the premise (as continuity usually implies \(\dot{m}_{in} = \dot{m}_{out}\) for steady flow without mass addition/removal).
However, if the intent is to relate mass flow rate to velocity and area (\(\dot{m} = \rho A v\)) to "use the continuity equation effectively" (perhaps to find velocity or area), we need information about the density (\(\rho\)).
Viscosity relates to friction, temperature relates to density and viscosity but isn't fundamental to mass flow rate itself, and velocity profile describes *how* velocity varies across the area, while the continuity equation usually deals with average velocity.
To connect the given mass flow rates (kg/s) to volumetric flow rates (m\(^3\)/s) or velocities (m/s), the density (\(\rho\), in kg/m\(^3\)) is essential (\( \text{Volume flow rate} = \dot{m}/\rho \)).

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