A pipe of diameter 1.8m is required to transport oil of sp.gr. 0.8 and viscosity 0.04 poise at the rate of 4 m³/s. Tests were conducted on a 20cm diameter pipe using water at 20°C. Find the velocity and rate of flow in the model. Viscosity of water at 20°C = 0.01 poise.

Fluid Mechanics Problem Solution

Problem Statement

Given Data

Diameter of prototype (D_p) 1.8 m

Specific gravity of oil 0.8

Density of oil (ρ_p) 0.8 × 1000 = 800 kg/m³

Viscosity of oil (μ_p) 0.04 poise = 0.004 Pa·s

Discharge for prototype (Q_p) 4 m³/s

Diameter of model (D_m) 0.2 m

Density of water (ρ_m) 1000 kg/m³

Viscosity of water (μ_m) 0.01 poise = 0.001 Pa·s

Solution Approach

To determine the velocity and flow rate in the model pipe, we’ll use Reynolds’ model law, which ensures dynamic similarity between the prototype and model. We’ll calculate the prototype velocity first, then use the Reynolds number equality to find the model velocity and discharge.

Calculations

Prototype Velocity Calculation

Step 1: Calculate the cross-sectional area of the prototype pipe:

A_p = π/4 × D_p² = π/4 × (1.8)² = 2.545 m²

Step 2: Calculate the velocity in the prototype pipe using the continuity equation:

V_p = Q_p/A_p = 4/2.545 = 1.572 m/s

Model Velocity Using Reynolds’ Model Law

Step 1: By Reynolds’ model law, the Reynolds numbers must be equal for dynamic similarity:

Re_model = Re_prototype

(ρ_m × V_m × D_m)/μ_m = (ρ_p × V_p × D_p)/μ_p

Step 2: Substituting the given values:

(1000 × V_m × 0.2)/0.001 = (800 × 1.572 × 1.8)/0.004

Step 3: Solving for V_m:

(1000 × V_m × 0.2)/0.001 = (800 × 1.572 × 1.8)/0.004

200,000 × V_m = (800 × 1.572 × 1.8)/0.004

200,000 × V_m = (800 × 1.572 × 1.8 × 1000)/4

200,000 × V_m = 565,920

V_m = 565,920/200,000 = 2.83 m/s

Model Discharge Calculation

Step 1: Calculate the cross-sectional area of the model pipe:

A_m = π/4 × D_m² = π/4 × (0.2)² = 0.0314 m²

Step 2: Calculate the discharge in the model pipe:

Q_m = V_m × A_m = 2.83 × 0.0314 = 0.0889 m³/s

Velocity in the model (V_m) = 2.83 m/s

Rate of flow in the model (Q_m) = 0.0889 m³/s = 88.9 liters/s

Detailed Explanation

Model-Prototype Similarity in Fluid Mechanics

In this problem, we applied Reynolds’ model law to ensure dynamic similarity between the prototype (oil pipeline) and the model (water pipe test). The Reynolds number is a dimensionless parameter that characterizes the ratio of inertial forces to viscous forces in a fluid flow system.

Reynolds Number Significance

By maintaining equal Reynolds numbers between the model and prototype, we ensure that the ratio of inertial to viscous forces is the same in both systems. This similarity allows test results from the smaller model to be accurately scaled up to predict the behavior of the larger prototype.

Understanding the Results

The calculated model velocity (2.83 m/s) is higher than the prototype velocity (1.572 m/s) despite the model being smaller. This occurs because:

Water (the model fluid) has a lower viscosity than oil (0.001 Pa·s vs. 0.004 Pa·s)
Water has a higher density than the oil (1000 kg/m³ vs. 800 kg/m³)
The diameter ratio between prototype and model is significant (1.8 m vs. 0.2 m)

Practical Applications

This type of modeling is crucial in engineering design for several reasons:

It allows testing of large-scale designs in smaller, more economical laboratory settings
It helps predict full-scale performance before construction
It enables optimization of designs to minimize energy losses and maximize efficiency
It can identify potential issues or limitations in the proposed design

Flow Regime Analysis

Let’s calculate the Reynolds numbers to verify the flow regime:

Re_p = (ρ_p × V_p × D_p)/μ_p = (800 × 1.572 × 1.8)/0.004 = 565,920

Re_m = (ρ_m × V_m × D_m)/μ_m = (1000 × 2.83 × 0.2)/0.001 = 566,000

Both Reynolds numbers are well above 4000, confirming that both flows are in the turbulent regime. This is important because similarity laws apply differently depending on whether the flow is laminar or turbulent.

Limitations of the Model

While Reynolds similarity ensures dynamic similarity, it’s important to note some limitations:

Surface roughness effects may not scale perfectly
Minor losses at fittings and bends may have different relative impacts
Temperature effects and fluid compressibility are not accounted for
Wall effects may be relatively different due to the scale difference

This problem demonstrates the fundamental principles of dimensional analysis and similarity in fluid mechanics, which are essential tools for engineers designing fluid systems across various scales.

A pipe of diameter 1.8m is required to transport oil of sp.gr. 0.8 and viscosity 0.04 poise at the rate of 4 m³/s. Tests were conducted on a 20cm diameter pipe using water at 20°C. Find the velocity and rate of flow in the model.