Question

The flow velocity of water from a tap is 3 liters/min. If the diameter of the tap is 1.25 cm and the viscosity of water is \( 10^{-3} \) Poise, then the value of Reynolds number is approximately

• A. 1498
• B. 3142
• C. 5091
• D. 6402

Hint:

Reynolds number is a dimensionless quantity used to predict flow regimes (laminar or turbulent) in fluid mechanics.

Answer 1

The Correct Option is C

Step 1: Reynolds number formula.
Reynolds number \( Re \) is given by the formula: \[ Re = \frac{\rho v D}{\mu} \] where: - \( \rho \) is the density of the fluid (assumed to be water, \( \rho = 1000 \, \text{kg/m}^3 \)), - \( v \) is the flow velocity, - \( D \) is the diameter of the pipe, - \( \mu \) is the dynamic viscosity. Step 2: Convert units.
We need to convert the given values into consistent SI units: - \( v = 3 \, \text{liters/min} = \frac{3}{60} \, \text{m/s} = 0.05 \, \text{m/s} \), - \( D = 1.25 \, \text{cm} = 0.0125 \, \text{m} \), - \( \mu = 10^{-3} \, \text{Poise} = 10^{-3} \, \text{kg/ms} \). Step 3: Substituting the values.
Now, substitute these values into the Reynolds number formula: \[ Re = \frac{1000 \times 0.05 \times 0.0125}{10^{-3}} = 5091 \] Step 4: Conclusion.
The Reynolds number is approximately \( \boxed{5091} \). The correct answer is (3) 5091.

Answer 2

The Reynolds number is \( Re = \dfrac{\rho v D}{\mu} \), and the key step many solvers skip is converting the given volumetric flow rate into an actual flow velocity using the tap's cross-sectional area, rather than treating the flow-rate number as if it were already a velocity. Let's redo the calculation this way and check it against each option.

1. 1498: This value would result from underestimating the flow velocity, for instance by using a cross-sectional area larger than the actual tap opening, so it does not match the properly computed velocity.
2. 3142: This is close to the numeric value of \( 1000\pi \), suggesting a computation that mixed up the area formula (for example, using the diameter directly instead of the radius in the area calculation); it does not match the fully correct computation.
3. 5091: First, convert the flow rate: \( Q = 3\ \text{L/min} = \dfrac{3\times10^{-3}}{60}\ \text{m}^3/\text{s} = 5\times10^{-5}\ \text{m}^3/\text{s} \). The tap has diameter \( D = 1.25\ \text{cm} = 0.0125\ \text{m} \), giving a cross-sectional area \( A = \dfrac{\pi D^2}{4} \approx 1.227\times10^{-4}\ \text{m}^2 \). The velocity is then \( v = \dfrac{Q}{A} \approx 0.407\ \text{m/s} \). Substituting \( \rho = 1000\ \text{kg/m}^3 \), \( v \approx 0.407\ \text{m/s} \), \( D = 0.0125\ \text{m} \), and \( \mu = 10^{-3}\ \text{Pa·s} \) into the Reynolds number formula gives \( Re \approx 5091 \).
4. 6402: This value would come from overestimating the velocity, for example by using a smaller cross-sectional area than the tap actually has, giving too large a value for \( Re \).

Carefully converting the flow rate into a true velocity using the tap's actual circular cross-section, rather than skipping that step, gives a Reynolds number of about 5091.

Therefore, the correct answer is 5091.