**5. Conclusions**

Scour induced by single and twin propeller jets were investigated to present the scour structure and followed by the proposal of empirical scour equations based on the maximum scour depth and its position. Empirical scour distribution equation was proposed by using Gaussian normal distribution. The scour structure can be divided into three zones, which are the small scour hole, primary scour hole and deposition dune. 2D scour model are proposed enabling the prediction of the scour profiles for both single-propeller induced scour and twin-propeller induced scour.

For the single propeller, Equations (5)–(7) are proposed to predict the entire scour profile to establish a 2D single propeller scour model.

$$-0.5 < \mathcal{X}/\mathcal{X}\_m < 0: \ \frac{\varepsilon\_m}{\varepsilon\_{\max}} = -0.02 + \left( -0.138 \ast \exp\left( -0.5 \ast \left( \frac{\frac{\mathcal{X}}{\mathcal{X}\_\S} + 0.23}{0.12} \right)^2 \right) \right) \tag{5}$$

$$0.0 < X/X\_{\mathcal{W}} < 1.8: \frac{\varepsilon\_{\mathcal{K}}}{\varepsilon\_{\text{max}}} = -0.038 + \left( -0.9 \ast \exp\left( -0.5 \ast \left( \frac{\frac{X}{X\_S} - 1.09}{0.449} \right)^2 \right) \right) \tag{6}$$

$$1.8 < \text{X/X}\_m < 3: \frac{\varepsilon\_m}{\varepsilon\_{max}} = 0.21 + \left(0.796 \ast \exp\left(-0.5 \ast \left(\frac{\frac{\varepsilon\_s}{\lambda\_s} - 2.5}{0.22}\right)^2\right)\right). \tag{7}$$

For the twin propeller, Equations (9)–(11) are proposed to predict the entire scour profile to establish a 2D twin propeller scour model.

$$-0.5 < X/X\_{m,twin} < 0: \frac{\varepsilon\_{\text{ff}}}{\varepsilon\_{\text{max}}} = -0.23 \* \exp\left(-0.5 \* \left(\frac{\frac{X}{X\_S} + 0.14}{0.09}\right)^2\right) \tag{9}$$

$$0 < X/X\_{m,twin} < 1.6: \frac{\varepsilon\_m}{\varepsilon\_{max}} = 0.04 + \left( -1.03 \ast \exp\left( -0.5 \ast \left( \frac{\frac{x}{X\_3} - 0.87}{0.36} \right)^2 \right) \right) \tag{10}$$

$$1.6 < \text{X/X}\_{m, \text{tuin}} < 2.6: \frac{\varepsilon\_m}{\varepsilon\_{\text{max}}} = -0.19 + \left(0.91 \* \exp\left(-0.5 \* \left(\frac{\frac{x}{X\_\circ} - 2.08}{0.22}\right)^2\right)\right). \tag{11}$$
