Chapter 8, Problem 8.1P

Interpretation Introduction

Interpretation: The expression for Hagen-Poiseuille equation in terms of flow rate and pipe diameter is to be determined and consequently change in flow rate.

Concept introduction:

The Hagen-Poiseuille equation is used to determine the pressure drop in the Laminar flow region. The equation for the same is given as,

  $\frac{\text{ΔP}}{\text{L}}\text{=}\frac{\text{32μ<mover accent="true"> <mo>V</mo> <mo>¯</mo></mover>}}{{\text{D}}^{\text{2}}}$  ........ (1)

The notations used here are,

  $\Delta \text{P}$ = Pressure drop in the pipe

L = Length of the pipe

D = Pipe diameter

  $\text{μ}$ = Fluid viscosity

  $\text{<mover accent="true"> <mo>V</mo> <mo>¯</mo></mover>}$ = Average velocity of the fluid

The average velocity can also be written as,

  $\text{<mover accent="true"> <mo>V</mo> <mo>¯</mo></mover>}\text{= }\frac{\text{Q}}{\text{A}}$  ........ (2)

Q = Flow rate of the fluid

A = Area of the pipe

Answer to Problem 8.1P

The Hagen-Poiseuille equation in terms of flow rate and diameter is $\frac{\text{ΔP}}{\text{L}}\text{=}\frac{\text{128Qμ}}{{\text{πD}}^{\text{4}}}$ and the percentage change in flow rate due to doubling of pipe diameter is 1500%.

Explanation of Solution

The area of the pipe is given as,

  $\text{A}\text{=}\frac{\text{π}}{\text{4}}{\text{D}}^{\text{2}}$  ........ (3)

Substitute equation (3) in equation (2).

  $\text{<mover accent="true"> <mo>V</mo> <mo>¯</mo></mover>}\text{=}\frac{\text{4Q}}{{\text{πD}}^{\text{2}}}$  ........ (4)

Substitute equation (4) in equation (1).

  $\begin{array}{l}\frac{\text{ΔP}}{\text{L}}\text{=}\frac{\text{32μ}}{{\text{D}}^{\text{2}}}\text{×}\frac{\text{4Q}}{{\text{πD}}^{\text{2}}}\\ \frac{\text{ΔP}}{\text{L}}\text{=}\frac{\text{128Qμ}}{{\text{πD}}^{\text{4}}}\mathrm{\hspace{1em}\hspace{1em}.......}\text{(5)}\end{array}$

Let subscript 1 for initial conditions and subscript 2 for final conditions.

Keeping all other things constant, the equation (5) can be written as,

  $\text{Q}\text{α}{\text{D}}^{\text{4}}$

This dependency for the two conditions mentioned can be written as.

  $\frac{{\text{Q}}_{\text{2}}}{{\text{Q}}_{\text{1}}}\text{=}{\left(\frac{{\text{D}}_{\text{2}}}{{\text{D}}_{\text{1}}}\right)}^{\text{4}}$  ........ (6)

As diameter is doubled, so D₂ = 2D₁. From equation (6).

  $\begin{array}{l}\frac{{\text{Q}}_{\text{2}}}{{\text{Q}}_{\text{1}}}\text{=}{\left(\frac{{\text{2D}}_{\text{1}}}{{\text{D}}_{\text{1}}}\right)}^{\text{4}}\\ \frac{{\text{Q}}_{\text{2}}}{{\text{Q}}_{\text{1}}}\text{=}{\text{2}}^{\text{4}}\text{=}\text{16}\\ {\text{Q}}_{\text{2}}\text{=}{\text{16Q}}_{\text{1}}\end{array}$

The percentage change in flow rate = $\frac{{\text{Q}}_{\text{2}}\text{-}{\text{Q}}_{\text{1}}}{{\text{Q}}_{\text{1}}}\text{×}\text{100}$

Substitute Q₂ = 16Q₁,

The percentage change in flow rate = $\frac{\text{16}{\text{Q}}_{\text{1}}\text{-}{\text{Q}}_{\text{1}}}{{\text{Q}}_{\text{1}}}\text{×}\text{100}\text{=}\text{1500}\text{\%}$

Conclusion

The Hagen-Poiseuille equation in terms of flow rate and diameter is $\frac{\text{ΔP}}{\text{L}}\text{=}\frac{\text{128Qμ}}{{\text{πD}}^{\text{4}}}$ and the percentage change in flow rate due to doubling of pipe diameter is 1500%.