Resistance + Propulsion: Scaling laws
A 6 m model of a 180 m long ship is towed in a model basin at a speed of 1.61 m/s. The towing pull is 20 N.
The wetted surface of the model is 4 m2. Estimate the corresponding speed for the ship in knots and the effective power PE, assuming resistance coefficients to be independent of scale for simplicity.
Solution
The scale is:

Ls 180

 30
Lm
6
The wetted surface at full scale is:
S s  2  S m  302  4  3600 m2
The Froude number is:
Vm
1.61

 0.21
g  Lm
9.81  6
Fn 
The Froude number gives the full-scale speed:
Vs  Fn  g  Ls  0.21  9.81  180  8.824 m/s
1 kn = 0.5144 m/s. Thus the speed of the full-scale ship is Vs 
8.824
 17.15 kn.
0.5144
A simple scaling law assumes that the resistance coefficients remain constant. (More accurate prediction
methods have a slight speed dependence of the frictional resistance coefficient and introduce a correlation
coefficient.) Then:
cT 
1
2
RT , m
20
1
 541 kN
2
  Vm  Sm 2  1000  1.612  3600
The effective power is:
PE  RT , s  Vs  541  8.824  4771 kW
Resistance + Propulsion: Simple scaling laws
a) A ship of 15000 t displacement has speed 20 kn. Determine the corresponding speed for a similar
ship of displacement 14000 t using geometric and Froude similarity.
b) A ship of 140 m length has speed 15 kn. Determine the corresponding speed for a 7 m model.
c) A ship of 5000 t displacement has speed 18 kn, length 120 m. The towing tank model has 6 m length.
At which speed should the model be tested? What is the ratio of the ship-to-model PE at this speed?
Solution
a) Let's denote the ship with 15000 t displacement with index 1, the other with index 2. The lengths L
of the ships correlate to the displacements  following geometric similarity:
L13 1

L32  2

L1

L2
3
1
3
2
The speeds follow from Froude similarity:
V1
gL1

V2
gL2
 
L
 V2  V1  2  V1   2 
L1
 1 
1/ 6
 14000 
 20  

 15000 
1/ 6
 19.8 kn
b) We keep Froude similarity:
Vm
V
 s
gLm
gLs
 Vm  Vs 
Lm
7
 15 
 3.35 kn = 1.73 m/s
Ls
140
c) We keep Froude similarity:
Vm
V
 s
gLm
gLs
 Vm  Vs 
Lm
6
 18 
 4.02 kn = 2.07 m/s
Ls
120
The effective power follows from PE = RV. The resistance is expressed as R  c  12    V 2  A ;
thus it scales with 3, as the area scales with 2 and the speed in Froude similarity scales with 0.5.
Thus the power scales with 3.5:
PE , s  Ls 
 
PE , m  Lm 
3.5
 120 


 6 
3.5
 35777
Resistance + Propulsion: Power estimate following simple scaling laws
A ship (L = 122.00 m, B = 19.80 m, T = 7.33 m,  = 8700 t) has the following power requirements:
V [kn]
16
17
18
19
20
P [kW] 2420 3010 3740 4620 5710
Estimate the power requirements for a similar ship with 16250 t displacement at speed 19.5 knots.
Solution
We convert the speed of the new ship of 19.5 kn to the corresponding speed for the original ship, following
Froude and geometric similarity:
1/ 6
 
V1  V2 
 V2   1 
gL2
 2 
gL1
 8700 
 19.5  

 16250 
1/ 6
 17.57 kn
For this speed, the power for the original ship is linearly interpolated to:
P1  (17.57  17) 
3740  3010
 3010  3426 kW
18  17
This is scaled to the new ship:
 
P  P1   new 
  old 
3.5 / 3
 16250 
 3426  

 8700 
3.5 / 3
 7088 kW
Explanation: Power scales with speed V3 and area A, which gives for the speed with Froude similarity a scale
with length scale 1.5 and for the area 2, thus together 3.5. If we use the displacement instead of the length,
we divide by 3. Hence the exponent of 3.5/3.
Resistance + Propulsion: Ship vs Car - Scaling and resistance prediction
A car has a wind resistance value of CW = 0.3. The car has a frontal area of 2 m2. The air density is 1.23
kg/m3 and the car weighs 1200 kg.
Now compare the efficiency of a ship to that of the car. A ferry has a resistance of 2000 kN at a speed of 25
kn. The ferry has a mass of 16500 t, length Lpp = 190 m, and a wetted surface of 4400 m2.
a)
b)
c)
d)
e)
Compute the necessary power for car speeds of 10, 20, 30, 40, 100 and 200 km/h.
What scale has to be taken to get a ship model of same mass as the car?
What is the model speed if Froude similarity is kept?
What is the model resistance if the correlation allowance is cA = 0?
How fast could the car go with the same power the ship model needs in towing at design Froude
number?
sea = 1.1610-6 m2/s, fresh = 1.1410-6 m2/s, sea = 1026 kg/m3
Solution
a) The force follows from:
F  CW 
 air
2
V 2  A
The reference area is here the frontal area. Thus:
V
10 km/h
20 km/h
30 km/h
40 km/h
100 km/h
200 km/h
F
2.85 N
11.39 N
25.63 N
45.60 N
284.72 N
1138.9 N
2.78 m/s
5.56 m/s
8.3 m/s
11.11 m/s
27.78 m/s
55.60 m/s
P = FV
8W
63 W
212 W
506 W
7910 W
63272 W
b) Geometric similarity gives:
1/ 3
 
   s 
 m 
1/ 3
 16500 


 1.2 
 23.957
Then the model length is:
Lm 
Ls


190
 7.93 m
23.957
c) The ship speed is:
Vs  25  0.5144  12.861 m/s
The Froude number of the ship is:
Vs
12.861

 0.298
gLs
9.81  190
Fn 
Model and full-scale ship have the same Froude number, thus:
Vm  Fn  gLm  0.298  9.81  7.93  2.63 m/s
d) First, we compute the Reynolds numbers of ship and model:
12.861  190
 21  108
6
 sea
1.16  10
V L
2.63  7.93
 m m
 18.3  106
 fresh 1.14  10 6
Rn , s 
Rn, m
Vs  Ls

The frictional resistance coefficients are:
0.075
0.075

 1.398  10 3
2
2
8
log 10 Rn, s  2 log 10 21 10  2
0.075
0.075


 2.708  10 3
2
2
6
log 10 Rn,m  2 log 10 18.3 10  2
cF , s 
cF , m




The total resistance coefficient for the ship is:
cT , s 
RT , s
2  106

 5.357  10 3
2
2
1
1
 V S
 1026  12.86  4400
2 sea s s
2
We assume the same wave resistance coefficient for full scale and model scale. Thus:
cT , m  cF , m  cF , s  cT , s  (2.708  1.398  5.357)  103  6.676  103
This yields a resistance (with S m  S s / 2 ):
RT ,m  cT ,m 
m
2
 Vm2 
Ss

2
 6.676  10 3 
1000
4400
 2.632 
 177 kN
2
23.957 2
e) The necessary power to tow the model:
P  RT , m  Vm  177  2.63  465.5 W
For the car we have:
P  CW 
air
2
 V 3  Am
 V 3
P
465.5
3
 10.8 m/s
CW   air  A
0.3  12  1.23  2
1
2
= 38.9 km/h
Resistance + Propulsion: Sailing yacht test
A sailing yacht has been tested. The full-scale dimensions are Lpp = 9.00 m, S = 24.0 m2,  = 5.150 m3.
The yacht will operate in sea water of  = 1025 kg/m3,  = 1.19106 m2/s.
The model was tested with scale  = 7.5 in fresh water with  = 1000 kg/m3,  = 1.145106 m2/s.
The experiments yield for the model:
Vm [m/s] 0.5
0.6
0.75 0.85
1.0
1.1
1.2
RT,m [N] 0.402 0.564 0.867 1.114 1.584 2.054 2.751
a) Determine the form factor following Hughes-Prohaska.
b) Determine the form factor following ITTC'78. For simplification assume the exponent n for Fn to be
4 and determine just  and k in regression analysis.
Solution
The model data are:
Lm = Ls/ = 9/7.5 = 1.2 m
Sm = Ss/2 = 24/7.52 = 0.427 m2
We consider only the lowest 4 speeds as for the others a considerable wave influence is to be expected.
a) We compute the total resistance coefficient of the model, using Fn  V / gL , Rn  V  L / ,
cF ,0  0.067 /(log 10 Rn  2)2
Vm
0.50 m/s
0.60 m/s
0.75 m/s
0.85 m/s
RT,m
0.402
0.564
0.867
1.114
Fn
Rn105 cF0 103 cT 103
0.146 5.24
4.843
7.532
0.175 6.29
4.643
7.338
0.219 7.86
4.415
7.219
0.248 8.91
4.295
7.222
cT/cF0 Fn4/cF0
1.555 0.091
1.580 0.197
1.635 0.507
1.681 0.859
Then regression analysis (e.g. using a spreadsheet like Excel) yields  = 0.165 and k = 0.545.
(If only the last 3 points are used,  = 0.190, k = 0.540.)
b) We compute now: cF  0.075 /(log 10 Rn  2) 2
Vm
0.50
0.60
0.75
0.85
RT,m
0.402
0.564
0.867
1.114
Fn
Rn105 cF 103 cT 103
0.146 5.24
5.422 7.532
0.175 6.29
5.198 7.338
0.219 7.86
4.943 7.219
0.248 8.91
4.807 7.222
cT/cF
1.389
1.412
1.461
1.502
Fn4/cF
0.082
0.176
0.453
0.768
The expression for n = 4 is:
cT
Fn4
 (1  k )   
cF
cF
Regression analysis yields  = 0.189 and k = 0.376. The form factor differs from Hughes-Prohaska,
as the ITTC'57 line considers already to some extent a form influence.
Resistance & Propulsion: ITTC’78 for underwater vehicle
An underwater vehicle shall be tested to make a resistance prediction. The streamlined body is tested in a
wind tunnel. The following data are given:
Sm = 1.2 m2,
Lm = 1.5 m,
Vm = 30 m/s,
T = 25°C.
The density and viscosity of air at standard atmospheric pressure are (as function of temperature):
  1.293 
273
kg/m3
T  273
 = 105(1.723+0.0047T) kg/(ms)
where T is the temperature in °C.
The model resistance measured in the wind tunnel is 3.9 N. The full-scale vehicle is 9 m long and operated
in the ocean at 15°C,  = 1.19106 m2/s,  = 1026 kg/m3.
a) What similarity law is most important and what full-scale speed in m/s corresponds to keeping this
law?
b) What is the full-scale resistance at this speed (using same total resistance coefficient)?
c) Predict the full-scale resistance following ITTC'78 assuming cAA = cA = 0 for full-scale speed 2 m/s?
d) Can the accuracy of prediction be improved by changing the temperature in the wind tunnel? How?
Solution
a) For an underwater vehicle, wave resistance (and thus Froude similarity) can be neglected. Thus
Reynolds similarity is most important. Same Reynolds number then yields the full-scale speed:
273
273
 1.293 
 1.1845 kg/m3
T  273
25  273
m  105  1.723  0.0047  T   105  1.723  0.0047  25  105  1.8405 kg/(ms)
 m  1.293 
m 
m 105  1.8405

 1.5538  10 5 m2/s
m
1.1845
Rn , m 
Vm  Lm
m

30  1.5
 2.896  106
5
1.5538  10
Keeping Reynolds similarity (Rn,m=Rn,s) then yields:
Vs 
Rn, s  s Rn, m  s 2.896  106  1.19  106


 0.3829 m/s
Ls
Ls
9
b) The total resistance coefficient is:
cT 
1
2
Rm
3.9
 1
 6.097  10 3
2
2
Vm S m 2  1.1845  30  1.2
The wetted surface at full scale is:
2
L 
 9 
S s  S m   s   1.2     43.2 m2
 1.5 
 Lm 
2
Thus the resistance at full scale (assuming same resistance coefficient) is:
RT  cT 

2
 Vs2  S s  6.097  10 3 
1026
 0.38292  43.2  19.8 N
2
c) The wave resistance (coefficient) is zero. Thus we have simply:
cT  (1  k )  cF
0.075
0.075
cF 

 3.767  10 3
2
2
6
log 10 Rn  2 log 10 2.896  10  2
1 k 
cT 6.097  103

 1.6185
cF 3.767  10 3
Thus, we have the resistance at 2 m/s:
29
 15.12  106
6
s
1.19  10
0.075
0.075
cF 

 2.795  10 3
2
2
6
log 10 Rn  2 log 10 15.12  10  2
Rn 
Vs  Ls



3
cT  (1  k )  cF  1.6185  2.795  10  4.52  103

1026 2
RT  cT   Vs2  S s  4.52  10 3 
 2  43.2  401 N
2
2
d) Reducing the temperature increases the Reynolds number. However, to get the same Reynolds number in this case would require a temperature of 163°C...
Resistance & Propulsion: ITTC’57 full-scale prediction
The full-scale ship is 140 m long and has speed 15 kn, the model 4.9 m. The resistance is measured to 19 N
in the model basin. Following the ITTC'57 approach, what is the predicted full-scale resistance?
The wetted surface of the full-scale ship is 3300 m2. The density of sea water 1025 kg/m3, that of fresh water
1000 kg/m3. m = 1.14106 m2/s , for fresh water, s = 1.19106 m2/s for sea water. Use a correlation coefficient of cA =0.0004.
Solution
The speed for the model follows from Froude similarity:
Vm  Vs 
Lm
4.9
 15  0.5144 
 1.44 m/s
Ls
140
The wetted surface follows from geometric similarity:
S m L2m
 2
Ss
Ls
 Sm  Ss 
L2m
4.9 2

3300

 4.043 m2
2
2
Ls
140
The Reynolds numbers are:
Rn,s
Vm  Lm
1.44  4.9
 6.189  10 6
6
m
1.14  10
V L
(15  0.5144)  140
 s s 
 9.078  108
s
1.19  10 6
Rn,m 

This yields friction resistance coefficients following ITTC'57:
c F ,m 
c F ,s 
0.075
log 10 Rn  2
2
0.075
log 10 Rn  22


log
log
0.075
10
9.078  10  2
8
10

 3.267  10 3

 1.549  10 3
6.189  10  2
0.075
6
2
2
The total resistance of the model follows from the model test:
cT , m 
RT ,m
1
2
 mV S m
2
m

1
2
19
 4.533  10 3
2
 1000  1.44  4.053
The wave resistance coefficient (assumed same for model and ship) is then:
cw  cT ,m  cF ,m  4.533  10 3  3.267  10 3  1.266 10 3
Thus the total resistance coefficient of the ship is:
cT ,s  cF ,s  cw  c A  (1.549  1.266  0.4)  10 3  3.215  10 3
The resistance follows:
RT ,s  cT ,s 
s
2
 Vs2  S s  3.215  10 3 
1025
 7.716 2  3300  323.7 kN
2
Resistance & Propulsion: ITTC’78 full-scale prediction including appendages
A ship model with scale  = 23 was tested in the model basin with 104.1 N total resistance for the bare hull
model at towing speed Vm = 2.064 m/s. The wetted surface of the model is Sm = 10.671 m2 and its length Lm
= 7.187 m. The viscosity is  = 1.14106 m2/s in fresh water and  = 1.19106 m2/s in sea water, density  =
1000 kg/m3 in fresh water, 1025 kg/m3 in sea water.
a) What is the prediction for the total calm-water resistance of the bare hull in sea water of the fullscale ship following ITTC'78 with a form factor k = 0.12? Assume standard roughness ks/Loss = 106,
corresponding roughly to 150 m average surface roughness, neglect air resistance.
b) The appendages of the ship (bilge keels, shaft brackets and rudder) add 8% to the total resistance of
the smooth bare hull. What is the total resistance now?
c) Fouling increases the surface roughness by a factor 10 to ks/Loss = 105. What is the total resistance
now?
Solution
The full-scale ship data are:
Ls    Lm  23  7.187  165.30 m
Vs    Vm  23  2.064  9.899 m/s
S s  2  S m  232  10.761  5645 m2
The Reynolds numbers for model and full scale are:
2.064  7.187
 1.301  107
m
1.14  10 6
V  L 9.899  165.3
 s s 
 1.375  109
6
s
1.19  10
Rn , m 
Rn , s
Vm  Lm

The friction coefficients follow:
0.075
0.075

 2.867  10 3
2
2
7
log 10 Rn  2 log 10 1.301  10  2
0.075
0.075


 1.472  10 3
2
2
9
log 10 Rn  2 log 10 1.375  10  2
cF , m 


cF , s


The frictional resistance for model and full scale follow:
RF , s
m
1000
 2.064 2  10.671  65.2 N
2
2

1025
 cF , s  s  Vs2  S s  1.472  10 3 
 9.8992  5645  417.2 kN
2
2
RF , m  c F , m 
 Vm2  S m  2.867  10 3 
a) We compute the correlation coefficient following ITTC’78:




k
c A  10 3  105  3 s  0.64   10 3  105  3 10 6  0.64  0.41  10 3
Loss


The total resistance coefficient of the model and the model is:
cT , m 
1
2
RT , m
104.1
1
 4.580  103
2
2
mVm Sm 2  1000  2.064  10.671
The total resistance coefficient of the full-scale ship follows then according ITTC'78:
cT , s  (1  k )(cF , s  cF , m )  cT , m  cA  (1  0.12)(1.472  2.867)  4.580  0.41  3.427  103
RT ,s  cT ,s 
s
2
 Vs2  S s  3.427  10 3 
1025
 9.899 2  5645  971.4 kN
2
b) We add 8% for the appendages:
RT ,incl appendages  1.08  RT , s  1.08  971.4  1049 kN
c) The fouling changes the correlation coefficient:




k
c A  10 3  105  3 s  0.64   10 3  105  3 10 5  0.64  1.622  10 3
Loss


cT ,s  (1  k )(cF ,s  cF ,m )  cT ,m  c A  (1  0.12)(1.472  2.867)  4.580  1.622  4.639  10 3
RT ,s  cT ,s 
s
2
 Vs2  S s  4.640  10 3 
1025
 9.899 2  5645  1315 kN
2
This represents an increase of 35% in total bare hull resistance.
Resistance + Propulsion: ITTC'57 and ITTC'78
A ship model (scale  = 23) was tested in fresh water with: RT,m = 104.1 N, Vm = 2.064 m/s, Sm = 10.671 m2,
Lm = 7.187 m. m = 1000 kg/m3, s = 1026 kg/m3, m = 1.14  106 m2/s, s = 1.19  106 m2/s
What is the prediction for the total calm-water resistance in sea water of the full-scale ship following
ITTC'57? Assume cA = 0.0002.
Solution
The full-scale ship data are:
Ls    Lm  23  7.187  165.3 m
Vs    Vm  23  2.064  9.899 m/s
Thus:
S s  2  S m  232  10.761  5645 m2
V L
2.064  7.187
Rn , m  m m 
 1.301  107
m
1.14  10 6
V  L 9.899  165.3
Rn , s  s s 
 1.375  109
6
s
1.19  10
0.075
0.075
cF , m 

 2.867  10 3
2
2
7
log 10 Rn,m  2 log 10 1.301 10  2


0.075
0.075

 1.472  10 3
2
2
9
log 10 Rn, s  2 log 10 1.375 10  2

1000
RF , m  cF , m  m  Vm2  S m  2.867  10 3 
 2.0642  10.671  65.1 N
2
2

1026
RF , s  cF , s  s  Vs2  S s  1.472  10 3 
 9.8992  5645  417.7 kN
2
2
cF , s 


Residual resistance: RR  RT  RF . Thus RR , m  RT , m  RF , m  104.1  65.1  39 N.
Residual resistance coefficient:
cR, m 
1
2
RR, m
39
1
 1.718  103
2
2
m  Vm  Sm 2  1000  2.064  10.671
The residual resistance coefficient is assumed to be constant. Then:
RR , s  cR , s 
s
2
1026
 9.8992  5645  487.5 kN
2
 cF , s  cR, m  cA  (1.472  1.718  0.2) 103  3.390  103
 Vs2  S s  1.78  10 3 
Total resistance coefficient: cT , s
Total calm-water resistance at full scale:
RT , s  cT , s 
s
2
 Vs2  S s  3.390  10 3 
1026
 9.8992  5645  962.9 kN
2
Resistance + Propulsion: Power scaling from model to ship following ITTC'57
A model of 6 m length has total resistance 70 N at 1.8 m/s. The wetted surface of the model is 5 m2. Determine the resistance for a geometrically similar ship of length 130 m using the ITTC'57 approach. Density of
model basin water m = 1000 kg/m3, density of sea water s = 1026 kg/m3. m = 1.14 106 m2/s for fresh
water, s = 1.19 106 m2/s for sea water. Use a correlation coefficient cA = 0.00035.
Solution
The speed for the ship follows from Froude similarity:
Vs  Vm 
Ls
130
 1.8 
 8.38 m/s = 16.3 kn
Lm
6
The wetted surface follows from geometric similarity:
S s  Ls

S m  Lm



2
L
 S s  S m   s
 Lm
2

130 
  5  
  2347 m2
 6 

2
The Reynolds numbers are:
1.8  6
 9.47  106
m
1.14  10 6
V L
8.38  130
 s s 
 9.15  108
6
s
1.19  10
Rn , m 
Rn , s
Vm  Lm

This yields friction resistance coefficients following ITTC'57:
0.075
0.075

 3.028  10 3
2
2
6
log 10 Rn,m  2 log 10 9.47 10  2
0.075
0.075


 1.548  10 3
2
2
8
log 10 Rn, s  2 log 10 9.15 10  2
cF , m 
cF , s




The total resistance coefficient of the model follows from the model test:
cT , m 
1
2
RT , m
70
1
 8.642  103
2
2
mVm Sm 2  1000  1.8  5
The wave resistance coefficient (assumed same for model and ship) is then:
cw  cT , m  cF , m  (8.642  3.028)  103  5.614 103
Thus the total resistance coefficient of the ship is:
cT , s  cF , s  cw  cA  (1.548  5.614  0.35)  103  7.511 103
The resistance follows:
RT , s  cT , s 
s
2
 Vs2  S s  7.511  10 3 
1026
 8.382  2347  635.1 kN
2
Resistance + Propulsion: Resistance prediction with base design following ITTC'57
A base ship (Index O) has the following main dimensions: Lpp,O = 128.0 m, BO = 25.6 m, TO = 8.53 m, CB =
0.565. At a speed VO = 17 kn, the ship has a total calm-water resistance of RT,O = 460 kN. The viscosity of
water is  = 1.19  106 m2/s and density  = 1026 kg/m3.
What is the resistance of the ship with Lpp = 150 m if the ship is geometrically and dynamically similar to the
base ship and the approach of ITTC'57 is used (essentially resistance decomposition following Froude's approach)?


The wetted surface may be estimated by an empirical estimate: S  3.4  1 / 3  0.5  Lwl  1 / 3
Lwl may be estimated by Lwl = 1.01Lpp.
The Reynolds number shall be based on Lpp. The correlation coefficient can be neglected.
Solution
The ships are geometrically similar with scaling factor:  
We have:
Lpp,1
Lpp,O

150
 1.1719
128
O  CB ,O  L pp,O  BO  TO  0.565  128  25.6  8.53  15792.4 m3
Lwl ,O  1.01  L pp, O  1.01  128  129.28 m.




SO  3.4  1 / 3  0.5  Lwl 1 / 3  3.4  15792.41 / 3  0.5  129.28  15792.41 / 3 m2
S1    SO  1.1719  3761.9  5166.4 m
2
2
2
The speed of the base ship is VO = 17 kn = 170.5144 = 8.7448 m/s. Dynamic similarity means:
V1    VO  1.1719  8.7448  9.467 m/s
Then we have:
VO  Lpp,O
8.7448  128
 9.406  108
6

1.19  10
V1  Lpp,1 9.467  150
Rn,1 

 1.193  109
6

1.19  10
0.075
0.075
cF , O 

 1.5423  10 3
2
2
8
log 10 Rn,O  2 log 10 9.406 10  2
Rn,O 

0.075
0.075

 1.4976  10 3
2
2
9
log 10 Rn,1  2 log 10 1.193 10  2

1026
RF ,O  cF ,O   VO2  SO  1.5423  10 3 
 8.74482  3761.9  227.6 kN
2
2

1026
RF ,1  cF ,1   V12  S1  1.4976  10 3 
 9.467 2  5166.4  355.7 kN
2
2
The residual resistance is RR  RT  RF . Thus:
RR ,O  RT ,O  RF ,O  460  227.6  232.4 kN
cF ,1 


Then:
RR,1  RR,O  3  232.4  1.17193  374.0 kN
and
RT ,1  RR ,1  RF ,1  374.0  355.7  729.7 kN
Resistance & Propulsion: Ship-propeller interaction based on model test
Consider the propeller and ship described below:

s
f
W
D
J
n
KT
V
R
0
density of sea water
kinematic viscosity of sea water
kinematic viscosity of fresh water
wake fraction
propeller diameter
advance number (open-water)
propeller rpm
thrust coefficient
ship speed
relative rotative efficiency
open-water efficiency
1026
1.13910-6
1.18810-6
0.135
4.5
0.833
3
0.1594
13
0.95
0.684
kg/m3
m/s2
m/s2
m
1/s
m/s
a) The ship resistance at design speed is 580 kN. The ship is at constant speed. What is the thrust deduction fraction t?
b) Give an estimate for the propulsive efficiency D.
c) For a 1:16 model of the ship a wake fraction w = 0.19 is measured in towing tank tests. What should
be the propeller rpm at the model speed corresponding to full-scale design speed?
d) Compare Reynolds numbers at 0.7R for the model and at full scale. The full-scale propeller chord c
at r/R = 0.7 is 2 m.
Solution
a) The propeller thrust follows from:
T  K T    n 2  D 4  0.1594  1026  32  4.5 4  603.5 kN
Thrust and resistance (without propeller) are coupled by:
T (1  t )  RT
 t  1
RT
580
 1
 0.04
T
603.5
The thrust deduction is thus 4%.
b) Inflow velocity: V A  Vs  (1  w)  13  (1  0.135)  11.245 m/s
RT  Vs
580  13

 1.111
T  VA
603.5  11.245
Propulsive efficiency:  D   0  R  H  0.684  0.95  1.111  0.72
Hull efficiency:  H 
c) Similarity laws give:
Vs
 
Vm
Inflow velocity in model scale: VA,m
Vs
13
 3.25 m/s

16
 Vm  (1  wm )  3.25  (1  0.19)  2.63 m/s
 Vm 

The advance number in model scale is equal to the advance number in full scale. This gives:
n
V A, m
J  Dm

V A, m
J  ( Ds /  )

2.63
 11.2 Hz
0.833  (4.5 / 16)
Thus the model propeller turns at 674 rpm.
2
2
d) Circumferential velocity at r/R=0.7: VR  V A  (0.7nD )
At full scale:
VR,s  11.245 2  (0.7  3  4.5) 2  31.75 m/s
2.632  (0.7  11.2  (4.5 / 16)) 2  7.4 m/s
At model scale: VR ,m 
Reynolds number: Rn 
At full scale:
Rn , s 
VR  c

VR , s  c s
At model scale: Rn ,m 
s

VR ,m  c m
m
31.75  2
 56  10 6
6
1.139  10

7.4  (2 / 16)
 0.78  10 6
6
1.188  10
The model scale Reynolds number is close to the point where laminar/turbulent transition is expected. Unless turbulence is stimulated, some contamination due to partially laminar flow may be
expected.
Resistance + Propulsion: Resistance prediction with model tests
Make a resistance prognosis for a newly designed containership. A model of the ship is towed in a model
basin. Temperature of seawater shall be 15°, its salinity 3.5%, temperature of water in basin 20°. The model
basin measures the values given in the table below.
V [kn] 2.4
2.6
2.8
2.9
3.0
3.1
3.2
3.3
3.4
R [N] 23.24 28.38 33.92 37.53 41.47 44.79 47.49 52.13 58.71
The model has Lpp = 5.6 m at model scale  = 25 and wetted surface S = 8.16 m2.
a) Compute density, viscosity and wetted surface for model and full scale, using the formulae:
  106  0.014  s  (0.000645  t  0.0503)  t  1.75

p1
p2  0.7028423  p1
p1 = 5884.8170366 + t(39.803732+t(-0.3191477+t0.0004291133)) + 26.126277s
p2 = 1747.4508988 + t(11.51588  0.046331033t)  s(38.5429655+0.1353985t))
where s is the salinity and t the temperature in °C.
b) Compute Froude and Reynolds numbers for the model test, the speeds and Reynolds numbers for
full scale.
c) Compute the frictional (ITTC'57), residual and total resistance coefficients for the model.
d) Compute the frictional and total resistance coefficients for full scale as well as the resistance at full
scale following ITTC'57, assuming cA = 0.0003.
Solution
a) Using Ls    Lm , S s  2  S m and the formulas with s = 0.035 and t = 20, we get:
Model
Full scale
Lpp [m]
5.6
140
S [m2]
8.16
5100
0.9983
1.026
 [t/m3]
2
6
 [m /s] 1.00210
1.190106
b) Fn 
V L
V
and Rn 
and 1 kn = 0.5144 m/s. Then we have:

gL pp
Vm [kn] Vm [m/s]
2.4
1.235
2.6
1.337
2.8
1.440
2.9
1.492
3.0
1.543
3.1
1.595
3.2
1.646
3.3
1.698
3.4
1.749
Fn
Rn,m106
0.1666 6.900
0.1804 7.475
0.1943 8.050
0.2013 8.337
0.2082 8.625
0.2151 8.912
0.2221 9.200
0.2290 9.487
0.2360 9.775
Vs [m/s] Vs [kn] Rn,s106
6.173
12.0
778.1
6.687
13.0
842.9
7.202
14.0
907.8
7.459
14.5
940.2
7.716
15.0
972.6
7.973
15.5
1005.0
8.230
16.0
1037.4
8.488
16.5
1069.9
8.745
17.0
1102.3
c)
RT , m
1
  Vm2  Sm
2 m
0.075

log 10 Rn, m  22
cT , m 
cF , m
cR  cT , m  cF , m
Vm [kn] RT [N] cT,m103 cF,m103 cR103
2.4
23.24
3.744
3.203 0.540
2.6
28.38
3.895
3.158 0.738
2.8
33.92
4.014
3.116 0.898
2.9
37.53
4.141
3.097 1.043
3.0
41.47
4.275
3.079 1.197
3.1
44.79
4.324
3.061 1.264
3.2
47.49
4.303
3.044 1.259
3.3
52.13
4.442
3.028 1.414
3.4
58.71
4.712
3.012 1.700
d) cT , s  cF , s  cR  c A
cF , s 
0.075
log 10 Rn, s  22
RT , s  cT , s 
s
2
 Vs2  S s
Vs [kn]
12.0
13.0
14.0
14.5
15.0
15.5
16.0
16.5
17.0
cF,s103
1.579
1.564
1.549
1.542
1.536
1.530
1.524
1.518
1.512
cT,s103 RT [kN]
2.420
242.2
2.602
304.4
2.747
372.7
2.886
420.1
3.033
472.0
3.094
514.6
3.083
546.4
3.232
609.2
3.513
702.9
Resistance + Propulsion: Wageningen propeller behind ship
Consider a ship with speed Vs = 15 kn, thrust deduction fraction t = 0.2, wake fraction w = 0.15. The ship is
equipped with a Wageningen B4-40 propeller with diameter D = 6 m. The water density is  = 1025 kg/m3.
The resistance curve between 10 and 16 kn is given in good approximation by (R in kN, V in kn):
R  10  V 2  185  V  1100
a) What is the required thrust?
b) At what points (J, KT, KQ, 0) does the propeller operate assuming P/D = 0.8, 0.9, 1.0, 1.1, and 1.2,
respectively?
c) What propeller P/D would you chose? What are the corresponding open-water efficiency, torque
and delivered power of the engine?
d) What are then delivered power and open-water efficiency at Vs = 12 kn?
Use the file Wageningen_B4-40.pdf to find the solution.
Solution
a) Resistance at Vs = 15 kn: R  10  V 2  185  V  1100  10  152  185  15  1100  575 kN
RT
575

 718.85 kN
1  t 1  0.2
T
b) The thrust coefficient is defined as:
KT 
  n2  D 4
V
J A
The advance number is defined as:
nD
The inflow velocity is: VA  (1  w)  Vs  (1  0.15)  (15  0.5144)  6.559 m/s
Thrust at Vs = 15 kn:
T
We eliminate the unknown rpm from these two equations:
KT
T
718.75


 0.453  KT  0.453  J 2
2
2
2
J
  VA  D 1.025  6.5592  62
We compute a characteristics curve:
J
0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90
KT 0.137 0.163 0.191 0.222 0.255 0.290 0.327 0.367
Now we plot this function in the diagram for the B4-40 propeller. The intersections of this KT = f(J)
curve with the KT curves of the propeller for the various P/D values (solid lines) gives (approximately taken from the diagram) J and KT. The corresponding values of 0 and 10KQ for this J are listed
as well:
P/D
0.8
0.9
1.0
1.1
1.2
J
0.570
0.620
0.665
0.705
0.745
KT
0.147
0.174
0.200
0.225
0.251
10KQ
0.150
0.265
0.330
0.400
0.475
0
0.620
0.630
0.635
0.630
0.625
c) The best efficiency is for P/D = 1.0 with 0 = 0.635. So we select this propeller. The propeller rpm
VA
V
6.559
 n A 
 1.64 s-1 = 98.6 rpm
nD
J  D 0.665  6
Torque follows from: Q  KQ    n2  D5  0.033  1025  1.642  65  707 kNm
follows from:
J
The necessary delivered power follows from: PD  2  n  Q  2  1.64  707  7285 kW
d) Correspondingly we get for 12 knots:
R12  10  122  185  12  1100  320 kN
R
320
T12  12 
 400 kN
1  t 0.8
VA  (1  0.15)  (12  0.5144)  5.247 m/s
KT ,12  0.394  J 2
J
0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90
KT,12 0.119 0.142 0.166 0.193 0.222 0.252 0.285 0.319
P/D
1.0
J
0.690
KT
0.188
10KQ
0.405
0
0.650
VA
5.247

 1.267 s-1 = 76.0 rpm
J  D 0.690  6
Q  KQ    n2  D5  0.0405  1025  1.2672  65  518 kNm
This yields:
n
PD  2  n  Q  2 1.167  518  4124 kW
Resistance + Propulsion: Containership lengthened – Lap-Keller
A container ship shall be lengthened by adding a parallel midship section of 12.50 m length (40' container
and space between stacks). At full engine power (100% MCR = maximum continuous rating), the ship is
capable of V = 15.6 kn. Ship data (original before conversion):
Lpp
Lwl
B
T
= 117.20 m
= 120.00 m
= 20.00 m
= 6.56 m
rb = 1.5 m (bilge radius)
CB = 0.688
lcb = 0.0 m
Wake fraction and thrust deduction are estimated by: w  0.75  CB  0.24 and t  0.5  CB  0.15
Viscosity and density are:  = 1.1910-6 m2/s, sea = 1025 kg/m3
The ship is equipped with a propeller with 0 = 0.55. The relative rotative efficiency is R = 1.
a) What is the power requirement before conversion?
b) What is the power requirement after conversion, if the propeller is assumed to remain unchanged?
Base your prediction on Lap-Keller1, with correlation coefficient c A  0.35  103 .
Solution
a) Considerations for the original hull
The displacement is:   CB  L pp  B  T  0.688  117.20  20.00  6.56  10579 m3
The wetted surface following Keller (1973) is:
S  (3.4  1 / 3  0.5  Lpp )  1 / 3  (3.4  105791 / 3  0.5  117.20) 105791 / 3  2925 m2
The midship section area is:




Am  B  T  rb2    2   20.00  6.56  1.5 2    2   130.20 m2
2

2

C
0.688
A
130.20
 0.9926 and CP  B 
 0.692 .
Thus Cm  m 
B  T 20  6.56
CM 0.9926
CP = 0.692 and lcb = 0 make this ship a group C ship according to Keller's (1973) Figure 1.
The calculation length for Lap-Keller is Ld  1.01  L pp  1.01  117.20  118.37 m.
( Ld  Lwl for Lwl  1.01L pp , but this is not the case here.)
The prismatic coefficient based on this length is CPd = CP/1.01 = 0.685.
The speed is V = 15.6 kn = 8.025 m/s.
8.025  118.37
 7.983  108
6

1.19  10
0.075
0.075

 1.574  10 3
cF following ITTC'57 is: cF 
2
2
8
log 10 Rn  2 log 10 7.983  10  2
The Reynolds number is: Rn 
V  Ld



The residual coefficient is determined following Lap-Keller. For
V
8.025

 0.891 and CP = 0.692 we read from the diagram in Fig.4 of Keller
CPd  L
0.685  118.37
(1973): r = 0.025. We do not have to apply any correction for CP, as CP < 0.8. Then:
1
Lap, A.J.W. (1954). Diagrams for determining the resistance of single screw ships. Int. Shipb. Progr., p.179; Keller, W.H. auf'm (1973). Extended
diagrams for determining the resistance and required power of single-screw ships. Int. Shipb. Progr., p.253
cR 
Am
130.20
 r 
 0.025  1.111  10 3
S
2925
The correction factor K2 for B/T is given by Lap (1954) as:
B

 20.00

K 2  1  0.05    2.4   1  0.05  
 2.4   1.032
T

 6.56

The total resistance is:
RT  c F  c R  c A   K 2 

2
 V 2  S  1.574  1.111  0.35  1.032 
1025
 8.025 2  2925
2
= 302 kN
PE  RT  V  302  8.025  2424 kW
w  0.75  CB  0.24  0.75  0.688  0.24  0.276
t  0.5  CB  0.15  0.5  0.688  0.15  0.194
1  t 1  0.194
H 

 1.113
1  w 1  0.276
PE
2424
PD 

 3960 kW
 H 0  R 1.113  0.55  1
b) Considerations after ship conversion (We denote values for the new version by a prime.)
The new length is L' pp  L pp  L  117.20  12.50  129.70 m.
Thus L'd  1.01  L' pp  1.01  119.70  131.00 m.
'    Am  L  10579  130.2  12.50  12206.5 m3

12206.5
C 'B 

 0.710
L'B  T 131  20  6.56
C'
0.710
C'
C 'P  B 
 0.715 and C 'Pd  P  0.708 .
1.01
CM 0.9926
Thus the ship is a group D ship according to Keller’s (1973) Figure 1.
The wetted surface according to the Lap-Keller formula is:
S '  (3.4  1 / 3  0.5  Lpp )  1 / 3  (3.4  12206.51 / 3  0.5 117.20)  12206.51 / 3  3296 m2
w'  0.75  C'B 0.24  0.75  0.710  0.24  0.293
t '  0.5  C'B 0.15  0.5  0.710  0.15  0.205
1  t ' 1  0.205
 'H 

 1.124
1  w' 1  0.293
The other data remain unchanged. (The relative rotative efficiency is largely influenced by the
aftbody shape and the propeller. Both do not change here, so the assumption of unchanged R is acceptable.) We now use Keller's (1973) Figure 5.
Rn
cF
V / C 'Pd L'd
8.834  108 r
1.554  103 cR
(cF+cA+cR)  K2
0.833
0.0285
RT
3
PE
1.126  10
3.127  103 PD
340 kN
2730 kW
4416 kW
Thus the power requirement increases by 456 kW or 11.5%.
Note: The method is based on regression analysis of ships without bulbous bows. Such ships are not built
anymore. The method of Lap-Keller must thus be seen with appropriate scepticism!