# 8.16: Inclined Pipes

## 8.16.1 Pure Liquid

The hydraulic gradient can be determined with:

$\ \mathrm{i}_{\mathrm{l}, \theta}=-\frac{\mathrm{d p}}{\mathrm{d x}} \cdot \frac{\mathrm{A} \cdot \mathrm{L}}{\rho_{\mathrm{l}} \cdot \mathrm{A} \cdot \mathrm{L} \cdot \mathrm{g}}=\frac{\tau_{\mathrm{l}} \cdot \mathrm{O} \cdot \mathrm{L}}{\rho_{\mathrm{l}} \cdot \mathrm{A} \cdot \mathrm{L} \cdot \mathrm{g}}+\frac{\rho_{\mathrm{l}} \cdot \mathrm{A} \cdot \mathrm{L} \cdot \mathrm{g} \cdot \sin (\theta)}{\rho_{\mathrm{l}} \cdot \mathrm{A} \cdot \mathrm{L} \cdot \mathrm{g}}=\mathrm{i}_{\mathrm{l}}+\sin (\theta)$

## 8.16.2 Sliding Bed Regime

The hydraulic gradient can be determined with:

$\ \mathrm{i}_{\mathrm{m}, \mathrm{\theta}}=\mathrm{i}_{\mathrm{l}, \mathrm{\theta}}+\mathrm{R}_{\mathrm{s d}} \cdot \mathrm{C}_{\mathrm{v s}} \cdot\left(\mu_{\mathrm{s f}} \cdot \cos (\theta)+\sin (\theta)\right)$

The relative excess hydraulic gradient Erhg,θ is now:

$\ \mathrm{E}_{\mathrm{r h g}, \mathrm{\theta}}=\frac{\mathrm{i}_{\mathrm{m}, \mathrm{\theta}}-\mathrm{i}_{\mathrm{l}, \mathrm{\theta}}}{\mathrm{R}_{\mathrm{s d}} \cdot \mathrm{C}_{\mathrm{v s}}}=\mu_{\mathrm{s f}} \cdot \cos (\theta)+\sin (\theta)$

## 8.16.3 Heterogeneous Regime

In an inclined pipe the effective terminal settling velocity perpendicular to the pipe wall gives a potential energy term of:

$\ \mathrm{S}_{\mathrm{hr}, \theta}=\mathrm{S}_{\mathrm{hr}} \cdot \cos (\theta)=\frac{\mathrm{v}_{\mathrm{t}} \cdot \cos (\theta) \cdot\left(1-\frac{\mathrm{C}_{\mathrm{vs}}}{\mathrm{\kappa}_{\mathrm{C}}}\right)^{\beta}}{\mathrm{v}_{\mathrm{ls}}}$

For the kinetic energy losses, the angle of attack has to be adjusted in an inclined pipe. The angle of attack is defined as the ratio between the terminal settling velocity and the velocity at the thickness of the viscous sub layer, giving:

$\ \mathrm{S}_{\mathrm{r s}, \mathrm{\theta}}=\mathrm{c} \cdot\left(\frac{\delta_{\mathrm{v}}}{\mathrm{d}}\right)^{2 / 3} \cdot\left(\frac{\mathrm{v}_{\mathrm{t}} \cdot \cos (\theta)}{\mathrm{1 1 .6} \cdot \mathrm{u}_{*}-\mathrm{v}_{\mathrm{t}} \cdot \mathrm{s i n}(\theta)}\right)^{4 / 3} \cdot\left(\frac{\mathrm{v}_{\mathrm{t}}}{\sqrt{\mathrm{g} \cdot \mathrm{d}}}\right)^{\mathrm{2}}$

$\ \begin{array}{left} \mathrm{E}_{\mathrm{r h g}, \theta}=\mathrm{S}_{\mathrm{h r}, \theta}+\mathrm{S}_{\mathrm{r s}, \theta}+\sin (\theta)\\ \mathrm{i}_{\mathrm{m}, \theta}=\mathrm{i}_{\mathrm{l}, \theta}+\left(\mathrm{S}_{\mathrm{h r}, \theta}+\mathrm{S}_{\mathrm{r s}, \theta}+\sin (\theta)\right) \cdot \mathrm{R}_{\mathrm{s d}} \cdot \mathrm{C}_{\mathrm{v} \mathrm{s}}\end{array}$

## 8.16.4 Homogeneous Regime

For an inclined pipe only the lifting of the mixture has to be added, giving:

$\ \begin{array}{left}\mathrm{E}_{\mathrm{r h g}, \theta}&=\alpha_{\mathrm{E}} \cdot \mathrm{i}_{\mathrm{l}}+\sin (\theta)\\ \mathrm{i}_{\mathrm{m}, \theta}&=\mathrm{i}_{\mathrm{l}, \theta}+\left(\alpha_{\mathrm{E}} \cdot \mathrm{i}_{\mathrm{l}}+\sin (\theta)\right) \cdot \mathrm{R}_{\mathrm{s d}} \cdot \mathrm{C}_{\mathrm{v s}}\\ &=\mathrm{i}_{\mathrm{l}} \cdot\left(1+\alpha_{\mathrm{E}} \cdot \mathrm{R}_{\mathrm{s d}} \cdot \mathrm{C}_{\mathrm{v s}}\right)+\sin (\theta) \cdot\left(1+\mathrm{R}_{\mathrm{s d}} \cdot \mathrm{C}_{\mathrm{v s}}\right)\end{array}$

## 8.16.5 Sliding Flow Regime

The method for determining the Sliding Flow Regime is not affected by pipe inclination. Of course the equations for a pipe with inclination for the sliding bed regime and the heterogeneous regime have to be applied.

## 8.16.6 The Limit Deposit Velocity

The Limit of Stationary Deposit Velocity is affected by the pipe inclination. In an ascending pipe, the cross sectional averaged line speed has to be higher compared to a horizontal pipe in order to make a bed start sliding. In a descending pipe this line speed is lower. It is even possible that in a descending pipe the bed will always slide because of gravity. The Limit Deposit Velocity as defined as the line speed above which there is no stationary of sliding bed is determined by either the potential energy losses or a limiting sliding bed. In both cases this is affected by the cosine of the inclination angle. Since in both cases the Limit Deposit Velocity depends on the cube root, the Limit Deposit Velocity will decrease according to:

$\ \mathrm{v}_{\mathrm{l s}, \mathrm{l d v}, \mathrm{\theta}}=\mathrm{v}_{\mathrm{l s}, \mathrm{l d v}} \cdot \cos (\theta)^{1 / 3}$

Because of the cube root, this means that for angles up to 45o the reduction is less than 10%.