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Step-by-Step Solution
Step 1: Identify Relevant Physical Quantity (Reynolds Number)
The Reynolds number $R$ is used to determine the nature of fluid flow. It is given by:
$R = \frac{\rho \, V \, D}{\eta}$
where:
$\rho$ = density of fluid (in $kg/m^3$),
$V$ = flow speed of the fluid (in $m/s$),
$D$ = characteristic diameter (in $m$), here $D = 2 \times \text{(radius of tap)}$ ,
$\eta$ = coefficient of viscosity (in Pa·s).
Step 2: Determine the Flow Speed for Each Flow Rate
First, convert the flow rates from L/min to $m^3/s$ and find the corresponding flow speeds. The flow rate $Q$ (volume per time) relates to the flow velocity $V$ by:
$Q = A \times V
\quad \text{where} \quad
A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{r}{1}\right)^2
$
Here, $r = 0.5\,\text{cm} = 0.5 \times 10^{-2}\,\text{m}$. Then $A = \pi (0.5 \times 10^{-2}\,\text{m})^2$.
Step 3: Calculate Reynolds Number for the First Flow Rate
Using the provided solution approach, the Reynolds number corresponding to the first flow rate $0.18\,\text{L/min}$ was found to be about $R_1 = 380$. Since $R_1 < 1000$, the flow is steady.
Step 4: Calculate Reynolds Number for the Second Flow Rate
The second flow rate is $0.48\,\text{L/min}$. The new Reynolds number can be estimated in proportion to the increased flow rate. From the given solution, it is:
$R_2 = \frac{0.48}{0.18} \times 380 = 1018
$
Since $1000 < R_2 < 2000$, the flow becomes unsteady (transient region).
Step 5: Conclude the Nature of Flow
As the flow rate increases from $0.18\,\text{L/min}$ to $0.48\,\text{L/min}$, the Reynolds number increases from about 380 (steady flow) to 1018 (unsteady flow). Therefore, the flow changes from steady flow to unsteady flow.
