Calculate the power required to pump sulphuric acid (dynamic viscosity \(0.04 \mathrm{~Pa} \cdot \mathrm{s}\), relative density \(1.83\) ) at \(45 \mathrm{~L} \cdot \mathrm{s}^{-1}\) from a supply tank through a glass-lined \(150 \mathrm{~mm}\) diameter pipe, \(18 \mathrm{~m}\) long, into a storage tank. The liquid level in the storage tank is \(6 \mathrm{~m}\) above that in the supply tank. For laminar flow \(f=16 / \mathrm{Re}\); for turbulent flow \(f=0.0014\left(1+100 \mathrm{Re}^{-1 / 3}\right)\) if \(\mathrm{Re}<10^{7}\). Take all losses into account.

Kinematic viscosity is a measure of a fluid's resistance to flow. It represents the ratio of the dynamic viscosity to the density of the fluid. In this problem, we calculate it for sulphuric acid using the formula: \[ u = \frac{\text{dynamic viscosity}}{\text{density}} \] First, we find the density of sulphuric acid from its relative density. Since the relative density is 1.83, and considering the density of water as 1000 kg/m^3, we find: \[ \rho = 1.83 \times 1000 = 1830 \text{ kg/m}^3 \] Then, by plugging in this density and the dynamic viscosity (0.04 Pa.s), we get: \[ u = \frac{0.04 \mathrm{~Pa} \cdot \mathrm{s}}{1830 \mathrm{~kg/m}^3} = 2.19 \times 10^{-5} \mathrm{~m}^2/\mathrm{s} \] This small value indicates that the fluid flows relatively easily, which is important for calculating other parameters like Reynolds number and flow regime.

The Reynolds number (Re) helps us determine the flow regime—whether it is laminar or turbulent. It's calculated using the formula: \[ \text{Re} = \frac{UD}{u} \] Here, U is the flow velocity, D is the pipe diameter, and u is the kinematic viscosity. We first need to find the flow velocity U: \[ U = \frac{Q}{A} = \frac{45 \mathrm{~L/s}}{\pi (0.075 \mathrm{~m})^2} = 2.548 \mathrm{~m/s} \] Then, we calculate Re: \[ \text{Re} = \frac{2.548 \mathrm{~m/s} \times 0.150 \mathrm{~m}}{2.19 \times 10^{-5} \mathrm{~m}^2/\mathrm{s}} = 1.75 \times 10^4 \] Because Re is about 1.75 x 10^4, it indicates turbulent flow—flows with Reynolds number greater than 4000 are typically considered turbulent. Knowing this helps us choose the correct formula for the friction factor in the Darcy-Weisbach equation.

The Darcy-Weisbach equation is used to estimate head loss due to friction in a pipe. It is expressed as: \[ h_f = f \times \frac{L}{D} \times \frac{U^2}{2g} \] where: \begin{itemize} \item \( h_f \) is the head loss due to friction, \item \( f \) is the friction factor, \item \( L \) is the length of the pipe, \item \( D \) is the diameter of the pipe, \item \( U \) is the flow velocity, and \item \( g \) is the acceleration due to gravity (9.81 m/s^2). \end{itemize} In this equation, the friction factor \( f \) varies depending on the nature of the flow. For turbulent flows (as we've determined), \( f \) is calculated using: \[ f = 0.0014 \times (1 + 100 \times \text{Re}^{-1/3}) \] By substituting \( \text{Re} \) into this equation, we get: \[ f = 0.0014 \times (1 + 100 \times (1.75 \times 10^4)^{-1/3}) = 0.0056 \] This friction factor is crucial for calculating the head loss in the next steps.

The friction factor (f) quantifies the resistance caused by friction in a pipe. Its value depends on the flow regime, whether it is laminar or turbulent. When dealing with turbulent flow, as in this example, the friction factor is determined using: \[ f = 0.0014 \times (1 + 100 \times \text{Re}^{-1/3}) \] Inserting the Reynolds number (1.75 x 10^4), we find: \[ f = 0.0056 \] Understanding the friction factor allows us to calculate head loss due to friction, which is part of determining the total head and ultimately the power required for pumping. Accurate calculation of the friction factor is essential for efficient pump design and operation.

Head loss represents the energy loss due to friction as fluid flows through a pipe. It can be calculated using the Darcy-Weisbach equation: \[ h_f = f \times \frac{L}{D} \times \frac{U^2}{2g} \] By substituting the friction factor (f = 0.0056), the length of the pipe (L = 18 m), the diameter (D = 0.150 m), the flow velocity (U = 2.548 m/s), and gravitational acceleration (g = 9.81 m/s^2), we get: \[ h_f = 0.0056 \times \frac{18 \text{ m}}{0.150 \text{ m}} \times \frac{(2.548 \text{ m/s})^2}{2 \times 9.81 \text{ m/s}^2} = 0.55 \text{ m} \] This head loss is part of the total head (H), which also includes the vertical distance the fluid must be pumped. Therefore, the total head is: \[ H = 6 \text{ m} + 0.55 \text{ m} = 6.55 \text{ m} \] Finally, with this total head, we can determine the power required to pump the fluid efficiently.