Assignment # B.3: Approximate Calculation of Ship Resistance Hotrop and Mennen’s Method
The total resistance of a ship can be
The wetted area of the hull can be
subdivided into:
approximated by:
S = L( T
2
+ B) C (0.453 + 0.4425 C
M
B
RT=RF(1+k1)+RAPP+RW+RB+RTR+RA
Where,
− 0.2862 C − 0.003467 B / T + 0.3696 C ) M
WP
+ 2.38 A / C
BT
B
RF
Frictional resistance according to
where ABT is the transverse sectional area of the ITTC 1957 friction formula
the bulb at the position where the still-water
= 0.5.ρV2SCF.
surface intersects the stem.
CF = 0.075/(Log10Re-2)2
Re Reynold’s
No.=
The appendage resistance can be
ρ VL /µ
determined from
1+k1
Form factor describing the viscous
R
resistance of the hull form in relation
APP=0.5ρV2SAPP(1+k2)eqCF
Where S
to R
APP the wetted area of the
F
appendages,
1+k
R
2 the appendage
APP
Appendage resistance
resistance factor
RW
Wave-making and wave-breaking
resistance
Approximate 1+k
R
2 values
B
Additional pressure resistance due
to bulbous bow near the water
Rudder behind skeg 1.5~2.0
surface
Rudder behind stern 1.3~1.5
R
Twin-screw balance rudders 2.8
TR
Additional pressure resistance of
immersed transom stern
Shaft brackets 3.0
R
Skeg
1.5~2.0
A
Model-ship correlation resistance
Strut bossings 3.0
The form factor of the hull can be predicted Hull bossings 2.0
by:
Shafts 2.0~4.0
Stabilizer fins 2.8
Dome 2.7
1 + k = c
0
{ .93 + c ( B / L )0.92497
1
13
12
R
Bilge keels 1.4
(0.95 − C )−0.521448 1
( − C + 0.0225 lcb)0.6906}
P
P
The equivalent 1+k2 value for a combination In this formula, CP is the prismatic coefficient of appendages is determined from:
based on the waterline length, L and lcb is
∑ 1(+
the longitudinal centre of buoyancy forward k ) S
(1+k
2
APP
of 0.5 L as a percentage of L. Here, L
2)eq =
R is a
∑ S
parameter reflecting the length of the run APP
The wave resistance is determined from:
according to:
d
−
R
= c c c ∇ρ g
m F + m
λ F
W
1 2
5
{
exp
cos(
2
1
n
2
n
}
L
)
R/L=1-CP+0.06CPlcb(4CP-1)
with
C12=(T/L)0.2228446 if T/L > 0.05
.
3 78613
.
1 07961
− 37565
.
1
c =
c
T B
−
=48.20(T/L-0.02)2.078+0.479948 if 2223105
( / )
90
(
i )
1
7
E
0.02<T/L<0.05
33333
.
0
c =
229577
.
0
( B / L)
if B/L<0.11
=0.479948 if T/L<0.02
7
= B/L if 0.11<B/L<0.25
= 0.5-0.0625 L/B if B/L>0.25
Where T is the average moulded draught.
c = exp(− 89
.
1
c
2
3
C13=1+0.003Cstern
c5 = 1-0.48AT/(BTCM)
Cstern will be -10, 0 and +10 if the afterbody
form is of V-shaped, Normal and U shaped λ = 1.446CP-0.03 L/B if L/B < 12
sections respectively.
= 1.446CP-0.36 if L/B>12
Course Teacher: Dr. Md. Mashud Karim, Associate Professor, Dept. of NAME, BUET
m =
0140407
.
0
L / T − 75254
.
1
∇ / L
C =
(
006
.
0
L +
)
100 − 16
.
0
− 00205
.
0
1
A
−
79323
.
4
B / L − c
4
16
+ 003
.
0
L /
5
.
7 C c ( 04
.
0
− c )
B
2
4
c4= TF/L when TF/L ≤ 0.04
2
3
c
= 07981
.
8
C −
8673
.
13
C +
984388
.
6
C
c
16
P
P
P
4= 0.04 when TF/L>0.04
if C
P<0.8
= 1.73014-0.7067C
Problem: The characteristics of a ship is as P if CP > 0.8
follows:
2
m = c C exp(
1
.
0
2
−
− F )
2
15
P
n
c15 = -1.69385 for L3/▼<512
L.O.W L=205.00 m
= 0 for L3/▼>1727
L.B.P. LPP = 200.00 m
=-1.69385+(L/▼1/3-8.0)/2.36
Breadth moulded B = 32.00 m if 512<L3/▼<1727
Draught moulded on F.P, TF=10.00 m d=-0.9
Draught moulded on A. P. TA=10.00 m i = 1 + 89 exp{−( L / B) .080856 1
( − C ) .030484
Displacement volume moulded, ▼=37500 m3
E
WP
Longitudinal centre of buoyancy 2.02% aft of 1
( − C − 0.0225 lcb)0.6367 ( L / B)0 34574
.
P
R
1/2 LPP
100
(
∇ / 3
L ) .016302}
Transverse bulb area ABT = 20.0 m2
Centre of bulb area above keel line h c = 0.56 1 5.
A
/{ BT (0.31 A
+ T − h )}
B = 4.0 m
3
BT
BT
F
B
Midship section coefficient CM = 0.98
Waterplane area coefficient CWP = 0.75
where hB is the position of the centre of the Transom area AT = 16.0 m2
transverse area ABT above the keel line and Wetted area appendages SAPP = 50.0 m2
TF is the forward draught of the ship.
Stern shape parameter, Cstern = 10.0
Propeller diameter, D = 8.0 m
R =
11
.
0
exp(−3
2
P − ) 3
5
.
1
F A
g
ρ 1
/(
2
+ F )
Number of propeller blades Z = 4
B
B
ni
BT
ni
Clearance of propeller with keel line 0.20 m P = 0.56 A
/( T − 1.5 h )
Ship speed V=25.0 knos B
BT
F
B
Density, ρ = 1025.87
2
F = V / g( T − h −
25
.
0
A
) +
15
.
0
V
Kinematic Viscosity, υ = 1.18831e-006
ni
F
B
BT
2
R
= 5
.
0
V
ρ A c
TR
T
6
Find RF , RAPP,RW, RB, RTR, RA, Rtotal.
c =
1
(
2
.
0
− 2
.
0 F ) if F
6
nT
nT<5
= 0 if F
Reference: J. Holtrop and G.G. J. Mennen, nT≥5
1982: An Approximate Power Prediction F
= V / 2 gA /( B + BC ) nT
T
WP
Method, International Shipbuilding Progress, Vol. 29, No. 335.
R
1
=
V 2
ρ SC
A
A
2
Course Teacher: Dr. Md. Mashud Karim, Associate Professor, Dept. of NAME, BUET