4.1. Effect of Trim Angle
The resistance and resistance reduction rate under different trim angles are studied by numerical simulation.
is the trim angle of the SWATH.
is positive when the SWATH is in the state of trim by the head. In different numerical simulation cases, the speed
or the airflow rates
are changed.
is the airflow rate of a single injection gap, and the
q of all injection gaps are the same in a case.
Table 4 shows the information regarding the numerical simulation conditions for different cases.
The total resistance
and the total resistance reduction rate
are shown in
Figure 5.
, where
is the total resistance of
L/s and
is the total resistance of
L/s or 6 L/s.
It can be seen that the change in trim angle can affect the total resistance under different speeds and airflow rates, as shown in
Figure 5a. The effect of the trim angle on the total resistance is more obvious in high-speed cases. In the cases without airflow, the minimum total resistance appears at the trim angle of −1 deg. In the high-speed cases with airflow, the minimum total resistance appears at the trim angle of 0 deg. As the airflow rate increases, the
of trim by the head is gradually smaller than that of the trim by the stern. At the low speed with airflow, the total resistance at trim angles of 0 deg and 1 deg is close. As shown in
Figure 5b, the reduction rate of total resistance with a positive trim angle is greater than that with a negative trim angle. This means the trim by the head is more conducive to total resistance reduction. For example, at the conditions of
m/s and
L/s, the total resistance reduction rate is 38.6% when the angle of trim by the head is 2 deg, while it is 29.04% when the angle of trim by the stern is −2 deg. The larger angle of trim by the head is more conducive to the total resistance reduction in the cases at low speed with airflow. The total resistance reduction rate is the largest at high speed when the angle of trim by the head is 1 deg. In summary, the slight trim by the head is more conducive to the total resistance reduction with airflow.
The resistance components in the cases of different trim angles are shown in
Figure 6. Where
is the resistance component about the shear,
is the resistance component about the pressure.
Figure 6a shows that at the conditions of
L/s, the resistance component
gradually increases with the trim angle
getting larger, which means that the
of the trim by the stern is smaller than that of the trim by the head. In the curve of
m/s and
L/s, when the trim angle
is −2 deg, 0 deg, and 2 deg, the corresponding
are 20.676 N, 21.291 N, and 22.118 N, respectively. In the curve of
m/s and
L/s, when the trim angle
is −2 deg, 0 deg, and 2 deg, the corresponding
are 59.354 N, 62.152 N, and 64.933 N, respectively. After the airflow was injected,
decreases at different trim angles. In the cases of
m/s and
L/s,
gradually decreases with the increase of
, and the minimum
is 15.733 N when the trim angle is 1 deg. After increasing the airflow rate to 6 L/s, the minimum
is 14.15 N with the trim angle of 2 deg. In the cases of
m/s and
L/s, the minimum
is 45.198 N with the trim angle of −1 deg. After increasing the airflow rate to 6 L/s, the minimum
is 34.084 N with the trim angle of 1 deg. With the airflow rate increased, the resistance component
with the trim by the head is gradually smaller than that with the trim by the stern. As shown in
Figure 6b, the
has the same trend under different speeds and airflow rates. With the increase of the trim angle
, the resistance component
decreases gradually, and it increases gradually after reaching the valley point. When the trim angle is 0 deg, the
is the smallest. At the speed of 4 m/s, the smallest
under the airflow rates of 0 L/s, 2 L/s, and 6 L/s are 18.04 N, 16.004 N, and 15.353 N, respectively. At the speed of 7 m/s, the smallest
under the airflow rates of 0 L/s, 2 L/s, and 6 L/s are 50.171 N, 39.459 N, and 36.036 N, respectively. It could be seen that the
of the upright hull attitude is the smallest. When the SWATH trim by the head or the stern appears, the pressure force
increases.
Figure 7 shows the air coverage of
m/s and
L/s with different trim angles. It could be that under the action of buoyancy, the airflow tends to float up, which makes the coverage of air in the upper part of the underwater body. With the increase in the trim angle, the sailing attitude gradually changes from the trim by the stern to the trim by the head, and changes of air coverage at the head of the underwater body are obvious. The air layer distribution with the trim angles of 1 deg and 2 deg is better than that of other trim angles, which results in greater total resistance reduction.
Figure 8 shows the air coverage of
m/s and
L/s with different trim angles. It could be that with the trim angle increasing, the air coverage trend is similar to that of the airflow rate 2 L/s. But, the air layer distribution at each trim angle is significantly greater than that of
L/s.
Figure 9 shows the air coverage of
m/s and
L/s with different trim angles. When
0 deg, the air mainly covers the upper part of the underwater body. With the increase of
, the air covering area increases gradually. The air coverage of
0 deg is the best. The trim angle continues to increase, and the SWATH changes to the trim by the head. When the trim angle is 2 deg, the air layer coverage area is smaller than that of
1 deg. This means that in the conditions of
m/s and
L/s, the trim by the head is not conducive to the air covering, and the larger the angle of the trim by the head, the worse the covering of the air.
Figure 10 shows the air coverage of
m/s and
L/s with different trim angles. It could be that the air coverage of
−2 deg and −1 deg is better than that of
0 deg, 1 deg, and 2 deg. This means that in the conditions of
m/s and
L/s, the trim by the stern is not conducive to covering the air. At the trim angle of 2 deg, the air layer distribution of the strut and the underwater body bow is weaker than that at the trim angle of 2 deg. It could be inferred that as the angle of the trim by the head continues to increase, the air coverage of the SWATH will become worse.
A comprehensive analysis of
Figure 7,
Figure 8,
Figure 9 and
Figure 10 shows that no matter the speed and airflow rate, the trim by the stern is not conducive to the air coverage of the SWATH. At a low speed, the air coverage under the trim by the head is best. The air layer under the upright sailing attitude and the slight trim by the head could cover a larger area at a high speed.
The computing resistance and corresponding reduction rate of different areas are shown in
Figure 11. Where
is the resistance of the underwater body,
is the resistance of the strut,
is the resistance reduction rate of the underwater body, and
is the resistance reduction rate of the strut.
The change of the trim angle could affect the underwater body resistance under different speeds and airflow rates, and the influence of the trim angle is more obvious at a high speed, as shown in
Figure 11a. In the working condition of
m/s, the trim angle has a slight influence on the resistance of the underwater body without airflow. In the working condition of
7 m/s, the influence of the trim angle on the underwater body resistance is obvious, especially in the cases with airflow injection. In the working condition of
m/s and
0 L/s, the smallest underwater body resistance is 75.17 N with
0 deg. When the airflow rate increases to 2 L/s, the underwater body resistances under the trim angles of 0 deg and 1 deg are smaller and close, which are 51.56 N and 51.69 N. When the airflow rate increases to 6 L/s, the smallest underwater body resistance is 41.22 N with
1 deg. It can be seen that a small trim angle by the head can lead to a small underwater body resistance with the airflow injection.
The trend of the resistance curve about the trim angles is the same under the same speed and different airflow rates, as shown in
Figure 11b. The underwater body resistance reduction rate of the trim by the head is larger than that of the trim by the stern and upright attitude with
m/s. For the curve of
2 L/s and
6 L/s, the maximum resistance reduction rates of the underwater body are 25.76% and 29.91%, respectively, with
2 deg. In the working condition of
m/s, the underwater body resistance reduction rate with
1 deg is the largest. For the curve of
2 L/s and
6 L/s, the maximum resistance reduction rates of the underwater body are 32.31% and 46.02%, respectively, when the trim angle is 1 deg.
Figure 11c shows the resistance of the strut with different trim angles. It can be seen that with the increase of
, the
increases gradually. The strut resistance under the trim by the stern is smaller than that under the upright attitude and the trim by the head.
The
of the trim angle of 2 deg is the largest with
m/s and
2 L/s, while it is the smallest with
m/s and
6 L/s, as shown in
Figure 11d. This is because it is difficult for the airflow to cover the front part of the strut unless the airflow rate is large enough. In the working condition of
7 m/s, the strut resistance reduction rate of the trim by the head is larger than that of the trim by the stern and the upright attitude. For the curve of
2 L/s and
6 L/s, the maximum
is 32.31% and 46.02%, respectively, when the trim angle is 1 deg.
The airflow affects the air volume fraction of the fin surface, especially where the fins join the hull, and the air volume fraction of the fins is different at different trim angles. For example,
Figure 12 shows the air volume fraction of the fins with
0 deg and
−2 deg in the working conditions of
7 m/s and
6 L/s.
It can be seen that the air volume fraction at the connection between the fins and the hull is between 0.1 and 0.3. When the trim angle is 0 deg, the area affected by the airflow on the front and rear fins is greater than that when the trim angle is −2 deg. The influence of the trim angle on the resistance reduction of the fins is studied, too. The resistance reduction rate of the front fin and the rear fin is shown in
Figure 13, where
is the resistance reduction rate of the front fin and
is the resistance reduction rate of the rear fin.
It can be seen that the resistance reduction rate of the front fin
is greater than 0 in all numerical cases, as shown in
Figure 13a. The resistance reduction effect is different under different speeds and airflow rates. The
of
4 m/s and
6 L/s is greater than that at other speeds and airflow rates. The
is minimal with
7 m/s and
2 L/s. The increase in speed is not conducive to the resistance reduction of the front fin, while the increase in the airflow rate is conducive to the resistance reduction of the front fin. In the working condition of
4 m/s, the
increases gradually with the increase of the trim angle. This means that the trim by the head is advantageous for the resistance reduction of the front fin. In the working conditions of
7 m/s and
6 L/s, the
with the trim angle of 2 deg is the smallest. The resistance reduction effect of the front fin could be reduced if the trim angle by the head is too large at a high speed and high airflow rate.
It can be seen that when the trim angle is
0 deg, the resistance reduction rate of the rear fin
is less than 0, as shown in
Figure 13b. When the SWATH is under the trim by the head and the upright attitude, the airflow could increase the resistance of the rear fin. When the trim angle is >0 deg, the
is >0 and the
of
6 L/s is larger than that of
2 L/s. The trim by the head is beneficial for the resistance reduction of the aft fin, and the resistance reduction of the rear fin could be improved by increasing the airflow rate under the trim by the head.
4.2. Effect of the Draft
This section studies the resistance and resistance reduction rates under different drafts. The numerical simulation conditions for different cases are shown in
Table 5.
The total resistance
and total resistance reduction rate
are shown in
Figure 14.
It can be seen that the difference between the total resistance of
2 L/s and that of
6 L/s under different drafts with
4 m/s is smaller than the difference of
7 m/s, as shown in
Figure 14a. Regardless of the draft, the smaller airflow rate at a low speed could significantly reduce the resistance, and the larger airflow rate at a high speed could lead to a smaller total resistance. As shown in
Figure 14b, the
gradually increases with the increase of the draft at
4 m/s, while that gradually decreases with the increase of the draft at
7 m/s. In the working condition of
4 m/s, the largest
of
2 L/s and
6 L/s is 19.01% and 24.05% with the draft of 0.24 m. When the speed is 6 m/s, the largest
of the airflow rate of 2 L/s and 6 L/s is 19.01% and 24.05% with the draft of 0.16 m. With the increase in the draft, increasing the airflow rates at a high speed is more conducive to improving the total resistance reduction.
The resistance components in the cases of different drafts are also studied, where
is the resistance component about the shear and
is the resistance component about the pressure. Both
and
increase with the draft, which is similar to the total resistance. But, the sensitivity of
and
to the draft are different, as shown in
Table 6.
,
,
, and
are given as follows:
where
and
are the resistance components about the shear at the draft of 0.16 m and 0.24 m, and
and
are the resistance components about the pressure at the draft of 0.16 m and 0.24 m.
As shown in
Table 6, both
and
are positive at any speed and airflow rate. Both
and
increase with the draft.
is between 33.30% and 48.86%, while
is between 7.51% and 14.52%. It can also be seen that
is much bigger than
, as shown in
Table 5. The draft has a greater influence on
than
. The increase in the draft increases the proportion of the
in the total resistance.
The computing resistance and corresponding reduction rate of different areas are shown in
Figure 15.
is less affected by the draft with
0 L/s, as shown in
Figure 15a. However, the effect of the draft on
can still be observed at the airflow rate of 2 L/s and 6 L/s, especially at a high speed of 7 m/s. For example, when the speed is 7 m/s, an increase in the draft from 0.16 m to 0.24 m will increase the resistance of the underwater body by 1.94 N. This results in the phenomenon that
decreases with the increase of the draft at
7 m/s, as shown in
Figure 15b. The surface pressure of the underwater body gets larger by the increasing draft, which makes the air coverage worse and reduces
. As shown in
Figure 15c, it can be seen that with the increase of the draft,
increases obviously. The increase in the draft causes the wet surface area of the strut to increase. It can be seen that
increases gradually by the increasing draft, as shown in
Figure 15d. This is because the increased wet surface area provides a larger potential coverage area for the air layer.