Let's break down the solution **step by step** as you requested, covering conceptual introduction, formulae, and calculations for each part of the question. --- ## **Step 1: Conceptual Introduction** **Introduction:** An orifice meter is a device used to measure the flow rate of fluids through pipes. It works by introducing a restriction (the orifice) in the pipe, which causes a drop in pressure. By measuring the difference in pressure before and after the orifice and knowing the properties of the fluid and geometry of the system, we can deduce the flow velocity and rate. The main goal is to relate the pressure drop to flow velocity and other parameters using dimensional analysis and fundamental fluid mechanics principles. **Explanation Block:** This introduction highlights the use of orifice meters in fluid mechanics for flow measurement. The pressure drop across the orifice is critical, as it provides a means to compute the fluid velocity. By using both dimensional analysis and Bernoulli’s equation, we can derive key relationships and express them in terms of dimensionless groups, aiding in the generalization and simplification of the problem. --- ## **Step 2: Formulae in the Solution** - **Pressure drop:** \(\Delta P = P_1 - P_2\) - **Pressure coefficient:** \(C_p = \frac{\Delta P}{\rho (V_1^2)}\) - **Bernoulli’s Equation (for incompressible, inviscid flow):** \(P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2\) - **Continuity Equation:** \(A_1V_1 = A_2V_2\), where \(A_1 = \frac{\pi D^2}{4}\), \(A_2 = \frac{\pi d^2}{4}\) **Explanation Block:** The formulas involve fluid properties (density \(\rho\)), geometric parameters (diameters \(D\) and \(d\)), and velocities. The pressure coefficient \(C_p\) is a dimensionless group that links pressure difference and kinetic energy per unit volume. Bernoulli’s equation relates pressures and velocities at two points in the pipe, while the continuity equation ensures mass conservation. --- ## **Step 3: Step-by-step Solution for Part (a) (Dimensional Analysis)** - **Relevant variables:** \(\Delta P, \rho, V_1, D, d\) - **Number of variables (n):** 5 **Fundamental dimensions (m):** 3 (\(M, L, T\)) - **Buckingham Pi Theorem:** Number of dimensionless groups = \(n - m = 2\) - **Choose repeating variables:** \(\rho, V_1, D\) - **Form dimensionless groups:** **For \(\Delta P\):** \[ \Pi_1 = \frac{\Delta P}{\rho V_1^2} \] **For \(d\):** \[ \Pi_2 = \frac{d}{D} \] **Explanation Block:** We identified five variables and three fundamental dimensions, giving two dimensionless groups via Buckingham Pi. By selecting \(\rho, V_1, D\) as repeating variables, we express the pressure drop and orifice diameter as nondimensional ratios, capturing the essential physics without units. The result is that \(C_p = f(d/D)\). --- ## **Step 4: Step-by-step Solution for Part (b) (Bernoulli + Mass Conservation)** 1. **Apply Bernoulli between 1 and 2:** \[ P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2 \] \[ \Delta P = P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2) \] 2. **Continuity equation:** \[ A_1 V_1 = A_2 V_2 \implies V_2 = V_1 \frac{A_1}{A_2} = V_1 \left(\frac{D}{d}\right)^2 \] 3. **Plug into Bernoulli:** \[ \Delta P = \frac{1}{2} \rho \left[ \left(V_1 \frac{D^2}{d^2} \right)^2 - V_1^2 \right] \] \[ = \frac{1}{2}\rho V_1^2 \left( \frac{D^4}{d^4} - 1 \right) \] 4. **Pressure coefficient:** \[ C_p = \frac{\Delta P}{\rho V_1^2} = \frac{1}{2}\left( \frac{D^4}{d^4} - 1 \right) \] **Explanation Block:** Bernoulli’s principle links pressure and kinetic energy changes, while continuity ensures conservation of mass. Substituting velocity ratios from continuity into Bernoulli’s equation, we express the pressure drop in terms of velocities and diameters, leading directly to a functional form for the pressure coefficient as a function of the diameter ratio. --- ## **Step 5: Step-by-step Solution for Part (c) (Compare and Conclude)** - **From (a):** \(C_p = f\left(\frac{d}{D}\right)\) - **From (b):** \(C_p = \frac{1}{2} \left( \frac{D^4}{d^4} - 1 \right)\) - **Functional relationship:** \(C_p = \frac{1}{2} \left( \left( \frac{D}{d} \right)^4 - 1 \right)\) or equivalently, \(C_p = \frac{1}{2} \left( \frac{1}{(d/D)^4} - 1 \right)\) **Explanation Block:** The dimensional analysis predicted a relationship of the form \(C_p = f(d/D)\). The detailed calculation using Bernoulli’s equation provided the explicit function. Thus, the dimensionless pressure coefficient is determined solely by the ratio of orifice to pipe diameter, confirming the result from dimensional analysis. --- ## **Final Answer** \[ \boxed{C_p = \frac{1}{2} \left( \frac{1}{(d/D)^4} - 1 \right)} \] This means the pressure coefficient across an orifice meter in an inviscid flow depends only on the ratio of the orifice diameter to the pipe diameter.