Derivation of the Loss Due to Sudden Enlargement

Oct 13, 2025 by talynt.org 14 views

Introduction

In fluid dynamics, the loss of energy due to a sudden enlargement in a pipe is a crucial concept to understand. This phenomenon occurs when a fluid flows through a pipe and suddenly expands to a larger diameter, resulting in a significant loss of energy. In this article, we will discuss the concept of loss due to sudden enlargement in a pipe and derive the mathematical expression for this loss.

Mathematical Derivation

To derive the expression for the loss due to sudden enlargement, we need to consider the energy equation for a fluid flowing through a pipe. The energy equation is given by:

${ \frac{p_1}{\rho} + \frac{V_1^2}{2} + gz_1 = \frac{p_2}{\rho} + \frac{V_2^2}{2} + gz_2 + h }$

where:

$p_1$ and $p_2$ are the pressures at the inlet and outlet of the pipe, respectively
$\rho$ is the density of the fluid
$V_1$ and $V_2$ are the velocities at the inlet and outlet of the pipe, respectively
$g$ is the acceleration due to gravity
$z_1$ and $z_2$ are the elevations at the inlet and outlet of the pipe, respectively
$h$ is the head loss due to friction and other losses

For a sudden enlargement, the pressure and elevation terms are negligible, and the energy equation simplifies to:

${ \frac{V_1^2}{2} = \frac{V_2^2}{2} + h_e }$

where $h_e$ is the head loss due to the sudden enlargement.

Derivation of the Loss Due to Sudden Enlargement

To derive the expression for the loss due to sudden enlargement, we need to consider the momentum equation for a fluid flowing through a pipe. The momentum equation is given by:

${ \frac{\partial}{\partial x} \left( \rho V^2 \right) = - \frac{\partial p}{\partial x} }$

For a sudden enlargement, the pressure gradient term is negligible, and the momentum equation simplifies to:

${ \frac{\partial}{\partial x} \left( \rho V^2 \right) = 0 }$

Integrating the momentum equation from the inlet to the outlet of the pipe, we get:

${ \rho V_1^2 = \rho V_2^2 }$

Substituting this expression into the energy equation, we get:

${ \frac{V_1^2}{2} = \frac{V_2^2}{2} + h_e }$

Simplifying the expression, we get:

${ h_e = \frac{(V_1 - V_2)^2}{2g} }$

where $h_e$ is the head loss due to the sudden enlargement.

Discussion

The expression for the loss due to sudden enlargement is given by:

${ h_e = \frac{(V_1 - V_2)^2}{2g} }$

This expression shows that the loss due to sudden enlargement is proportional to the square of the difference between the velocities before and after the enlargement.

Conclusion

In conclusion, the loss due to sudden enlargement in a pipe is a crucial concept to understand in fluid dynamics. The mathematical expression for this loss is given by:

${ h_e = \frac{(V_1 - V_2)^2}{2g} }$

This expression shows that the loss due to sudden enlargement is proportional to the square of the difference between the velocities before and after the enlargement.

References

[1] White, F. M. (2011). Fluid Mechanics. McGraw-Hill.
[2] Munson, B. R., Young, D. F., & Okiishi, T. H. (2013). Fundamentals of Fluid Mechanics. John Wiley & Sons.

Appendix

The following appendix provides additional information on the derivation of the expression for the loss due to sudden enlargement.

A.1 Derivation of the Momentum Equation

The momentum equation for a fluid flowing through a pipe is given by:

${ \frac{\partial}{\partial x} \left( \rho V^2 \right) = - \frac{\partial p}{\partial x} }$

For a sudden enlargement, the pressure gradient term is negligible, and the momentum equation simplifies to:

${ \frac{\partial}{\partial x} \left( \rho V^2 \right) = 0 }$

Integrating the momentum equation from the inlet to the outlet of the pipe, we get:

${ \rho V_1^2 = \rho V_2^2 }$

A.2 Derivation of the Energy Equation

The energy equation for a fluid flowing through a pipe is given by:

${ \frac{p_1}{\rho} + \frac{V_1^2}{2} + gz_1 = \frac{p_2}{\rho} + \frac{V_2^2}{2} + gz_2 + h }$

For a sudden enlargement, the pressure and elevation terms are negligible, and the energy equation simplifies to:

${ \frac{V_1^2}{2} = \frac{V_2^2}{2} + h_e }$

Substituting the expression for the momentum equation into the energy equation, we get:

Introduction

In our previous article, we discussed the concept of loss due to sudden enlargement in a pipe and derived the mathematical expression for this loss. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the significance of the loss due to sudden enlargement in a pipe?

A: The loss due to sudden enlargement in a pipe is a crucial concept to understand in fluid dynamics. It occurs when a fluid flows through a pipe and suddenly expands to a larger diameter, resulting in a significant loss of energy. This loss can have a significant impact on the performance of a system, such as a pump or a turbine.

Q: What are the factors that affect the loss due to sudden enlargement in a pipe?

A: The loss due to sudden enlargement in a pipe is affected by several factors, including:

The difference in velocity between the inlet and outlet of the pipe
The diameter of the pipe
The density of the fluid
The acceleration due to gravity

Q: How can the loss due to sudden enlargement in a pipe be minimized?

A: The loss due to sudden enlargement in a pipe can be minimized by:

Using a gradual enlargement instead of a sudden one
Increasing the diameter of the pipe gradually
Using a pipe with a smaller diameter
Increasing the velocity of the fluid

Q: What is the relationship between the loss due to sudden enlargement and the Reynolds number?

A: The loss due to sudden enlargement in a pipe is related to the Reynolds number, which is a dimensionless quantity that characterizes the nature of fluid flow. At high Reynolds numbers, the loss due to sudden enlargement is greater, while at low Reynolds numbers, the loss is smaller.

Q: Can the loss due to sudden enlargement in a pipe be calculated using numerical methods?

A: Yes, the loss due to sudden enlargement in a pipe can be calculated using numerical methods, such as the finite element method or the finite difference method. These methods can provide accurate results, but they require a significant amount of computational resources.

Q: What are the applications of the concept of loss due to sudden enlargement in a pipe?

A: The concept of loss due to sudden enlargement in a pipe has several applications in engineering, including:

Designing pipes and fittings for fluid flow systems
Optimizing the performance of pumps and turbines
Reducing energy losses in fluid flow systems
Improving the efficiency of fluid flow systems

Q: What are the limitations of the concept of loss due to sudden enlargement in a pipe?

A: The concept of loss due to sudden enlargement in a pipe has several limitations, including:

It assumes a sudden enlargement, which may not be the case in reality
It does not take into account the effects of turbulence and other complex phenomena
It requires a significant amount of computational resources to calculate the loss

Conclusion

In conclusion, the loss due to sudden enlargement in a pipe is a crucial concept to understand in fluid dynamics. It occurs when a fluid flows through a pipe and suddenly expands to a larger diameter, resulting in a significant loss of energy. By understanding the factors that affect the loss due to sudden enlargement, engineers can design pipes and fittings that minimize energy losses and improve the efficiency of fluid flow systems.

References

[1] White, F. M. (2011). Fluid Mechanics. McGraw-Hill.
[2] Munson, B. R., Young, D. F., & Okiishi, T. H. (2013). Fundamentals of Fluid Mechanics. John Wiley & Sons.

Appendix

The following appendix provides additional information on the concept of loss due to sudden enlargement in a pipe.

A.1 Derivation of the Momentum Equation

The momentum equation for a fluid flowing through a pipe is given by:

${ \frac{\partial}{\partial x} \left( \rho V^2 \right) = - \frac{\partial p}{\partial x} }$

For a sudden enlargement, the pressure gradient term is negligible, and the momentum equation simplifies to:

${ \frac{\partial}{\partial x} \left( \rho V^2 \right) = 0 }$

Integrating the momentum equation from the inlet to the outlet of the pipe, we get:

${ \rho V_1^2 = \rho V_2^2 }$

A.2 Derivation of the Energy Equation

The energy equation for a fluid flowing through a pipe is given by:

${ \frac{p_1}{\rho} + \frac{V_1^2}{2} + gz_1 = \frac{p_2}{\rho} + \frac{V_2^2}{2} + gz_2 + h }$

For a sudden enlargement, the pressure and elevation terms are negligible, and the energy equation simplifies to:

${ \frac{V_1^2}{2} = \frac{V_2^2}{2} + h_e }$

Substituting the expression for the momentum equation into the energy equation, we get:

${ h_e = \frac{(V_1 - V_2)^2}{2g} }$