Water is flowing in fully developed conditions through a 3 -cm-diameter smooth tube with a mass flow rate of \(0.02 \mathrm{~kg} / \mathrm{s}\) at \(15^{\circ} \mathrm{C}\). Determine \((a)\) the maximum velocity of the flow in the tube and \((b)\) the pressure gradient for the flow.

First, let's convert the given quantities into SI units, mainly the diameter of the tube, which is given in centimeters. Diameter: \(3 cm = 0.03 m\)

Now, let's write down all the given data in one place: - Diameter (D): \( \ensuremath{0.03 m}\) - Mass flow rate (m˙): \(\ensuremath{0.02 kg/s}\) - Temperature (T): \(\ensuremath{15^{\circ}C} = \ensuremath{288.15 K}\)

To obtain the water properties (density - \(\rho\) and dynamic viscosity - \(\mu\)) at the given temperature (\(15^{\circ}C\)), we can refer to a standard water properties table or use an online calculator. At \(15^{\circ}C\): - Density (\(\rho\)): \(\ensuremath{999 kg/m^3}\) - Dynamic viscosity (\(\mu\)): \(\ensuremath{1.13 * 10^{-3} \,Pa\cdot s}\)

In a fully developed flow, the velocity profile follows a parabolic shape where the maximum velocity (V_max) is at the center of the tube. To find the maximum velocity, we can use the mass flow rate and the parabolic velocity profile equation: $$V_{max} = \frac{2m˙}{\rho\,\pi\left(\frac{D}{2}\right)^2}$$ Plugging in the given values: $$V_{max} = \frac{2(0.02)}{999\pi \left(\frac{0.03}{2}\right)^2} = \ensuremath{5.62 m/s} $$ The maximum velocity of the flow in the tube is \(\ensuremath{5.62 m/s}\).

To determine the pressure gradient, we will use the Hagen-Poiseuille equation for fully developed flow in a smooth tube: $$ \frac{dP}{dx} = \frac{-32\mu Q}{\pi D^4}$$ We also know that the volume flow rate (Q) can be calculated from the mass flow rate (m˙) and density (\(\rho\)) as follows: $$ Q = \frac{m˙}{\rho}$$ By plugging in the given values, we can find the volume flow rate: $$ Q = \frac{0.02}{999} = \ensuremath{2*10^{-5} m^3/s} $$ Now, we can find the pressure gradient: $$ \frac{dP}{dx} = \frac{-32(1.13 * 10^{-3})(2*10^{-5})}{\pi (0.03)^4} = \ensuremath{-30952 Pa/m}$$ The pressure gradient for the flow is \(\ensuremath{-30952 Pa/m}\). To summarize: \((a)\) The maximum velocity of the flow in the tube is \(\ensuremath{5.62 m/s}\). \((b)\) The pressure gradient for the flow is \(\ensuremath{-30952 Pa/m}\).