Question

Water is flowing in fully developed conditions through a \(3-\mathrm{cm}\)-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.

Short answer

Question: Determine the maximum velocity of the flow and the pressure gradient for the flow in a horizontal, straight, cylindrical tube with diameter \(3\,\mathrm{cm}\) and length \(1\,\mathrm{m}\), when water at \(15^{\circ} \mathrm{C}\) is transported through it at a mass flow rate of \(0.02\,\mathrm{kg/s}\). Answer: (a) The maximum velocity of the flow is \(0.0565\,\mathrm{m/s}\) and (b) The pressure gradient for the flow is \(-1.08 \times 10^4\,\mathrm{Pa/m}\).

Step-by-step solution

Calculate the cross-sectional area of the tube

To do this, we'll use the formula for the area of a circle, where the diameter is given as \(3\,\mathrm{cm}\). First, we need to convert the diameter to meters: \(3\,\mathrm{cm} = 0.03\,\mathrm{m}\). The area formula is \(A = \pi(d/2)^2\). We'll use this to find the area of the tube. $$ A = \pi\left(\frac{0.03}{2}\right)^2 = 7.069 \times 10^{-4}\,\mathrm{m^2} $$

Calculate the volumetric flow rate

Next, we'll use the mass flow rate and the density of water to calculate the volumetric flow rate, denoted by \(Q\). The density of water at \(15^{\circ} \mathrm{C}\) is approximately \(999\,\mathrm{kg/m^3}\). Using the equation \(Q = \frac{\dot{m}}{\rho}\), where \(\dot{m}\) is the mass flow rate, and \(\rho\) is the density of water, we can find the volumetric flow rate. $$ Q = \frac{0.02\,\mathrm{kg/s}}{999\,\mathrm{kg/m^3}} = 2.002 \times 10^{-5}\,\mathrm{m^3/s} $$

Calculate the maximum velocity

Now that we have the volumetric flow rate and the cross-sectional area of the tube, we can use the equation \(Q = A\bar{v}\), where \(\bar{v}\) is the average velocity of the flow, to find the maximum velocity, denoted by \(v_{max}\). The maximum velocity will be double the average velocity since it's a parabolic velocity distribution in the laminar flow. $$ v_{max} = 2\bar{v} = \frac{2Q}{A} = \frac{2 \times 2.002 \times 10^{-5}\,\mathrm{m^3/s}}{7.069 \times 10^{-4}\,\mathrm{m^2}} = 0.0565\,\mathrm{m/s} $$

Find the dynamic viscosity of water at \(15^{\circ}\mathrm{C}\)

We will need the dynamic viscosity, denoted by \(\mu\), to calculate the pressure gradient. According to literature sources or published tables, the dynamic viscosity of water at \(15^{\circ}\mathrm{C}\) is approximately \(1.137\times10^{-3} \,\mathrm{Pa\cdot s}\).

Calculate the pressure gradient

Using Hagen-Poiseuille's equation for laminar flow through a circular pipe, we can find the pressure gradient. The equation is given by \(\Delta P=-\frac{32 \mu Q L}{\pi D^{4}}\), where \(\Delta P\) is the pressure drop, \(L\) is the pipe length, and \(D\) is the diameter. Rearranging the equation to find the pressure gradient per unit length, denoted by \(\frac{dp}{dz}\), we get \(\frac{dp}{dz}=\frac{-32 \mu Q}{\pi D^{4}}\). $$ \frac{dp}{dz} = \frac{-32 \times 1.137 \times 10^{-3}\,\mathrm{Pa\cdot s} \times 2.002 \times 10^{-5}\,\mathrm{m^3/s}}{\pi \times (0.03)^{4}\,\mathrm{m^4}} = -1.08 \times 10^4\,\mathrm{Pa/m} $$ To sum up, the solutions are: \((a)\) The maximum velocity of the flow is \(0.0565\,\mathrm{m/s}\) and \((b)\) The pressure gradient for the flow is \(-1.08 \times 10^4\,\mathrm{Pa/m}\).