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 ef-fective power PE, assuming resistance coefficients to be independent of scale for simplicity. Solution The scale is:

306

180

m

s

LL

The wetted surface at full scale is: 360043022 ms SS m2

The Froude number is:

21.0681.9

61.1

m

mn Lg

VF

The Froude number gives the full-scale speed: 824.818081.921.0 sns LgFV m/s

1 kn = 0.5144 m/s. Thus the speed of the full-scale ship is 15.175144.0824.8

sV kn.

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:

541360061.11000

202

212

21

,

mm

mTT SV

Rc

kN

The effective power is: 4771824.8541, ssTE VRP 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:

32

31

2

1

2

132

31

LL

LL

The speeds follow from Froude similarity:

8.19150001400020

6/16/1

1

21

1

212

2

2

1

1

VLLVV

gLV

gLV

kn

b) We keep Froude similarity:

35.3140

715 s

msm

s

s

m

m

LLVV

gLV

gLV

kn = 1.73 m/s

c) We keep Froude similarity:

02.4120

618 s

msm

s

s

m

m

LLVV

gLV

gLV

kn = 2.07 m/s

The effective power follows from PE = RV. The resistance is expressed as AVcR 221 ;

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:

357776

120 5.35.3

,

,

m

s

mE

sE

LL

PP

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:

57.171625087005.19

6/16/1

2

12

2

121

VgLgL

VV kn

For this speed, the power for the original ship is linearly interpolated to:

34263010171830103740)1757.17(1

P kW

This is scaled to the new ship:

70888700

1625034263/5.33/5.3

1

old

newPP 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) Compute the necessary power for car speeds of 10, 20, 30, 40, 100 and 200 km/h. b) What scale has to be taken to get a ship model of same mass as the car? c) What is the model speed if Froude similarity is kept? d) What is the model resistance if the correlation allowance is cA = 0? e) How fast could the car go with the same power the ship model needs in towing at design Froude

number? sea = 1.1610-6 m2/s, fresh = 1.1410-6 m2/s, sea = 1026 kg/m3 Solution

a) The force follows from:

AVCF airW 2

2

The reference area is here the frontal area. Thus:

V F P = FV 10 km/h 2.78 m/s 2.85 N 8 W 20 km/h 5.56 m/s 11.39 N 63 W 30 km/h 8.3 m/s 25.63 N 212 W 40 km/h 11.11 m/s 45.60 N 506 W

100 km/h 27.78 m/s 284.72 N 7910 W 200 km/h 55.60 m/s 1138.9 N 63272 W

b) Geometric similarity gives:

957.232.1

16500 3/13/1

m

s

Then the model length is:

93.7957.23

190

s

mLL m

c) The ship speed is: 861.125144.025 sV m/s

The Froude number of the ship is:

298.019081.9

861.12

s

sn gL

VF

Model and full-scale ship have the same Froude number, thus: 63.293.781.9298.0 mnm gLFV m/s

d) First, we compute the Reynolds numbers of ship and model: 8

6, 10211016.1

190861.12

sea

sssn

LVR

66, 103.18

1014.193.763.2

fresh

mmmn

LVR

The frictional resistance coefficients are:

3

2810

2,10

, 10398.121021log

075.02log

075.0

sn

sF Rc

3

2610

2,10

, 10708.22103.18log

075.02log

075.0

mn

mF Rc

The total resistance coefficient for the ship is:

32

21

6

221

,, 10357.5

440086.121026102

sssea

sTsT SV

Rc

We assume the same wave resistance coefficient for full scale and model scale. Thus: 33

,,,, 10676.610)357.5398.1708.2( sTsFmFmT cccc

This yields a resistance (with 2/sm SS ):

177957.23

440063.22

100010676.62 2

232

2,,

s

mm

mTmTSVcR kN

e) The necessary power to tow the model:

5.46563.2177, mmT VRP W For the car we have:

3

21

3

21

3

223.13.05.465

2 ACPVAVCP

airWm

airW

10.8 m/s

= 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, = 1.19106 m2/s. The model was tested with scale = 7.5 in fresh water with = 1000 kg/m3, = 1.145106 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 gLVFn / , /LVRn , 2

100, )2/(log067.0 nF Rc

Vm RT,m Fn Rn105 cF0 103 cT 103 cT/cF0 Fn4/cF0

0.50 m/s 0.402 0.146 5.24 4.843 7.532 1.555 0.091 0.60 m/s 0.564 0.175 6.29 4.643 7.338 1.580 0.197 0.75 m/s 0.867 0.219 7.86 4.415 7.219 1.635 0.507 0.85 m/s 1.114 0.248 8.91 4.295 7.222 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: 2

10 )2/(log075.0 nF Rc

Vm RT,m Fn Rn105 cF 103 cT 103 cT/cF Fn4/cF

0.50 0.402 0.146 5.24 5.422 7.532 1.389 0.082 0.60 0.564 0.175 6.29 5.198 7.338 1.412 0.176 0.75 0.867 0.219 7.86 4.943 7.219 1.461 0.453 0.85 1.114 0.248 8.91 4.807 7.222 1.502 0.768

The expression for n = 4 is:

F

n

F

T

cFk

cc 4

)1(

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:

Lm = 1.5 m, Sm = 1.2 m2, Vm = 30 m/s, T = 25°C. The density and viscosity of air at standard atmospheric pressure are (as function of temperature):

273273293.1

T

kg/m3

= 105(1.723+0.0047T) 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, = 1.19106 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:

1845.127325

273293.1273

273293.1

Tm kg/m3

8405.110250047.0723.1100047.0723.110 555 Tm kg/(ms)

55

105538.11845.1

8405.110

m

mm

m2/s

65, 10896.2

105538.15.130

m

mmmn

LVR

Keeping Reynolds similarity (Rn,m=Rn,s) then yields:

3829.09

1019.110896.2 66,,

s

smn

s

ssns L

RL

RV

m/s

b) The total resistance coefficient is:

32

212

21

10097.62.1301845.1

9.3

mm

mT SV

Rc

The wetted surface at full scale is:

2.435.1

92.122

m

sms L

LSS m2

Thus the resistance at full scale (assuming same resistance coefficient) is:

8.192.433829.02

102610097.62

232 ssTT SVcR

N

c) The wave resistance (coefficient) is zero. Thus we have simply: FT ckc )1(

326

102

10

10767.3210896.2log

075.02log

075.0

n

F Rc

6185.110767.310097.61 3

3

F

T

cck

Thus, we have the resistance at 2 m/s:

66 1012.15

1019.192

s

ssn

LVR

326

102

10

10795.221012.15log

075.02log

075.0

n

F Rc

33 1052.410795.26185.1)1( FT ckc

4012.4322

10261052.42

232 ssTT SVcR

N

d) Reducing the temperature increases the Reynolds number. However, to get the same Reynolds num-

ber 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 = 1.14106 m2/s , for fresh water, s = 1.19106 m2/s for sea water. Use a correlation coeffi-cient of cA =0.0004. Solution The speed for the model follows from Froude similarity:

44.1140

9.45144.015 s

msm L

LVV m/s

The wetted surface follows from geometric similarity:

043.4140

9.43300 2

2

2

2

2

2

s

msm

s

m

s

m

LL

SSLL

SS

m2

The Reynolds numbers are: 6

6, 10189.61014.1

9.444.1

m

mmmn

LVR

86, 10078.9

1019.1140)5144.015(

s

sssn

LVR

This yields friction resistance coefficients following ITTC'57:

326

102

10, 10267.3

210189.6log075.0

2log075.0

nmF R

c

328

102

10, 10549.1

210078.9log075.0

2log075.0

nsF R

c

The total resistance of the model follows from the model test:

32

212

21

,, 10533.4

053.444.1100019

mmm

mTmT SV

Rc

The wave resistance coefficient (assumed same for model and ship) is then: 333

,, 10266.110267.310533.4 mFmTw ccc Thus the total resistance coefficient of the ship is: 33

,, 10215.310)4.0266.1549.1( AwsFsT cccc The resistance follows:

7.3233300716.72

102510215.32

232,,

sss

sTsT SVcR kN

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 = 1.14106 m2/s in fresh water and = 1.19106 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 full-scale ship following ITTC'78 with a form factor k = 0.12? Assume standard roughness ks/Loss = 106, 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 = 105. What is the total resistance now?

Solution The full-scale ship data are: 30.165187.723 ms LL m

899.9064.223 ms VV m/s

5645761.102322 ms SS m2

The Reynolds numbers for model and full scale are: 7

6, 10301.11014.1

187.7064.2

m

mmmn

LVR

96, 10375.1

1019.13.165899.9

s

sssn

LVR

The friction coefficients follow:

327

102

10, 10867.2

210301.1log075.0

2log075.0

nmF R

c

329

102

10, 10472.1

210375.1log075.0

2log075.0

nsF R

c

The frictional resistance for model and full scale follow:

2.65671.10064.22

100010867.22

232,,

mmm

mFmF SVcR N

2.4175645899.92

102510472.12

232,,

sss

sFsF SVcR kN

a) We compute the correlation coefficient following ITTC’78:

33 633

3 1041.064.0101051064.010510

oss

sA L

kc

The total resistance coefficient of the model and the model is:

32

212

21

,, 10580.4

671.10064.210001.104

mmm

mTmT SV

Rc

The total resistance coefficient of the full-scale ship follows then according ITTC'78: 3

,,,, 10427.341.0580.4)867.2472.1)(12.01())(1( AmTmFsFsT cccckc

4.9715645899.92

102510427.32

232,,

sss

sTsT SVcR kN

b) We add 8% for the appendages: 10494.97108.108.1 ,, sTappendagesinclT RR kN

c) The fouling changes the correlation coefficient:

33 533

3 10622.164.0101051064.010510

oss

sA L

kc

3,,,, 10639.4622.1580.4)867.2472.1)(12.01())(1( AmTmFsFsT cccckc

13155645899.92

102510640.42

232,,

sss

sTsT SVcR kN

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 = 1.14 106 m2/s, s = 1.19 106 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: 3.165187.723 ms LL m

899.9064.223 ms VV m/s

5645761.102322 ms SS m2

Thus: 76, 10301.1

1014.1187.7064.2

m

mmmn

LVR

96, 10375.1

1019.13.165899.9

s

sssn

LVR

3

2710

2,10

, 10867.2210301.1log

075.02log

075.0

mn

mF Rc

3

2910

2,10

, 10472.1210375.1log

075.02log

075.0

sn

sF Rc

1.65671.10064.22

100010867.22

232,,

mmm

mFmF SVcR N

7.4175645899.92

102610472.12

232,,

sss

sFsF SVcR kN

Residual resistance: FTR RRR . Thus 391.651.104,,, mFmTmR RRR N.

Residual resistance coefficient: 32

212

21

,, 10718.1

671.10064.2100039

mmm

mRmR SV

Rc

The residual resistance coefficient is assumed to be constant. Then:

5.4875645899.92

10261078.12

232,,

sss

sRsR SVcR kN

Total resistance coefficient: 33,,, 10390.310)2.0718.1472.1( AmRsFsT cccc

Total calm-water resistance at full scale:

9.9625645899.92

102610390.32

232,,

sss

sTsT SVcR kN

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. Deter-mine 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 = 1.14 106 m2/s for fresh water, s = 1.19 106 m2/s for sea water. Use a correlation coefficient cA = 0.00035. Solution The speed for the ship follows from Froude similarity:

38.86

1308.1 m

sms L

LVV m/s = 16.3 kn

The wetted surface follows from geometric similarity:

23476

1305222

m

sms

m

s

m

s

LLSS

LL

SS

m2

The Reynolds numbers are: 6

6, 1047.91014.168.1

m

mmmn

LVR

86, 1015.9

1019.113038.8

s

sssn

LVR

This yields friction resistance coefficients following ITTC'57:

3

2610

2,10

, 10028.321047.9log

075.02log

075.0

mn

mF Rc

3

2810

2,10

, 10548.121015.9log

075.02log

075.0

sn

sF Rc

The total resistance coefficient of the model follows from the model test:

32

212

21

,, 10642.8

58.1100070

mmm

mTmT SV

Rc

The wave resistance coefficient (assumed same for model and ship) is then: 33

,, 10614.510)028.3642.8( mFmTw ccc Thus the total resistance coefficient of the ship is: 33

,, 10511.710)35.0614.5548.1( AwsFsT cccc The resistance follows:

1.635234738.82

102610511.72

232,,

sss

sTsT SVcR kN

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 = 1.19 106 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 ap-proach)? The wetted surface may be estimated by an empirical estimate: 3/13/1 5.04.3 wlLS Lwl may be estimated by Lwl = 1.01Lpp. The Reynolds number shall be based on Lpp. The correlation coefficient can be neglected. Solution

The ships are geometrically similar with scaling factor: 1719.1128150

,

1, Opp

pp

LL

We have: 4.1579253.86.25128565.0,, OOOppOBO TBLC m3

28.12912801.101.1 ,, OppOwl LL m.

3/13/13/13/1 4.1579228.1295.04.157924.35.04.3 wlO LS m2

4.51669.37611719.1 221 OSS m2

The speed of the base ship is VO = 17 kn = 170.5144 = 8.7448 m/s. Dynamic similarity means: 467.97448.81719.11 OVV m/s

Then we have: 8

6,

, 10406.91019.1

1287448.8

OppO

On

LVR

96

1,11, 10193.1

1019.1150467.9

pp

n

LVR

3

2810

2,10

, 105423.1210406.9log

075.02log

075.0

On

OF Rc

3

2910

21,10

1, 104976.1210193.1log

075.02log

075.0

n

F Rc

6.2279.37617448.82

1026105423.12

232,,

OOOFOF SVcR kN

7.3554.5166467.92

1026104976.12

231

211,1, SVcR FF

kN

The residual resistance is FTR RRR . Thus: 4.2326.227460,,, OFOTOR RRR kN

Then: 0.3741719.14.232 33

,1, ORR RR kN and

7.7297.3550.3741,1,1, FRT RRR kN

Resistance & Propulsion: Ship-propeller interaction based on model test Consider the propeller and ship described below:

density of sea water 1026 kg/m3 s kinematic viscosity of sea water 1.13910-6 m/s2 f kinematic viscosity of fresh water 1.18810-6 m/s2 W wake fraction 0.135 D propeller diameter 4.5 m J advance number (open-water) 0.833 n propeller rpm 3 1/s

KT thrust coefficient 0.1594 V ship speed 13 m/s R relative rotative efficiency 0.95 0 open-water efficiency 0.684

a) The ship resistance at design speed is 580 kN. The ship is at constant speed. What is the thrust de-

duction 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: 5.6035.4310261594.0 4242 DnKT T kN

Thrust and resistance (without propeller) are coupled by:

04.05.603

58011)1( TRtRtT T

T

The thrust deduction is thus 4%.

b) Inflow velocity: 245.11)135.01(13)1( wVV sA m/s

Hull efficiency: 111.1245.115.603

13580

A

sTH VT

VR

Propulsive efficiency: 72.0111.195.0684.00 HRD

c) Similarity laws give: 25.316

13

s

mm

s VVVV

m/s

Inflow velocity in model scale: 63.2)19.01(25.3)1(, mmmA wVV m/s The advance number in model scale is equal to the advance number in full scale. This gives:

2.11)16/5.4(833.0

63.2)/(

,,

s

mA

m

mA

DJV

DJV

n Hz

Thus the model propeller turns at 674 rpm.

d) Circumferential velocity at r/R=0.7: 22 )7.0( nDVV AR

At full scale: 75.31)5.437.0(245.11 22, sRV m/s

At model scale: 4.7))16/5.4(2.117.0(63.2 22, mRV m/s

Reynolds number:

cVR Rn

At full scale: 66

,, 1056

10139.1275.31

s

ssRsn

cVR

At model scale: 66

,, 1078.0

10188.1)16/2(4.7

m

mmRmn

cVR

The model scale Reynolds number is close to the point where laminar/turbulent transition is ex-pected. 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: 75.1)0503.0000645.0(014.010 6 tts

12

1

7028423.0 ppp

p1 = 5884.8170366 + t(39.803732+t(-0.3191477+t0.0004291133)) + 26.126277s p2 = 1747.4508988 + t(11.51588 0.046331033t) s(38.5429655+0.1353985t))

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 ms LL , ms SS 2 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 [t/m3] 0.9983 1.026 [m2/s] 1.002106 1.190106

b) pp

n gLVF and

LVRn

and 1 kn = 0.5144 m/s. Then we have:

Vm [kn] Vm [m/s] Fn Rn,m106 Vs [m/s] Vs [kn] Rn,s106

2.4 1.235 0.1666 6.900 6.173 12.0 778.1 2.6 1.337 0.1804 7.475 6.687 13.0 842.9 2.8 1.440 0.1943 8.050 7.202 14.0 907.8 2.9 1.492 0.2013 8.337 7.459 14.5 940.2 3.0 1.543 0.2082 8.625 7.716 15.0 972.6 3.1 1.595 0.2151 8.912 7.973 15.5 1005.0 3.2 1.646 0.2221 9.200 8.230 16.0 1037.4 3.3 1.698 0.2290 9.487 8.488 16.5 1069.9 3.4 1.749 0.2360 9.775 8.745 17.0 1102.3

c) mmm

mTmT SV

Rc

2

21

,,

2,10, 2log

075.0

mn

mF Rc

mFmTR ccc ,,

Vm [kn] RT [N] cT,m103 cF,m103 cR103 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) ARsFsT cccc ,,

2,10, 2log

075.0

sn

sF Rc

sss

sTsT SVcR 2,, 2

Vs [kn] cF,s103 cT,s103 RT [kN] 12.0 1.579 2.420 242.2 13.0 1.564 2.602 304.4 14.0 1.549 2.747 372.7 14.5 1.542 2.886 420.1 15.0 1.536 3.033 472.0 15.5 1.530 3.094 514.6 16.0 1.524 3.083 546.4 16.5 1.518 3.232 609.2 17.0 1.512 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):

110018510 2 VVR

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: 5751100151851510110018510 22 VVR kN

Thrust at Vs = 15 kn: 85.7182.01

5751

t

RT T kN

b) The thrust coefficient is defined as: 42 DnTKT

The advance number is defined as: Dn

VJ A

The inflow velocity is: 559.6)5144.015()15.01()1( sA VwV m/s We eliminate the unknown rpm from these two equations:

222222 453.0453.0

6559.6025.175.718 JK

DVT

JK

TA

T

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 (approximate-ly taken from the diagram) J and KT. The corresponding values of 0 and 10KQ for this J are listed as well:

P/D J KT 10KQ 0 0.8 0.570 0.147 0.150 0.620 0.9 0.620 0.174 0.265 0.630 1.0 0.665 0.200 0.330 0.635 1.1 0.705 0.225 0.400 0.630 1.2 0.745 0.251 0.475 0.625

c) The best efficiency is for P/D = 1.0 with 0 = 0.635. So we select this propeller. The propeller rpm

follows from: 64.16665.0

559.6

DJVn

DnVJ AA s-1 = 98.6 rpm

Torque follows from: 707664.11025033.0 5252 DnKQ Q kNm

The necessary delivered power follows from: 728570764.122 QnPD kW

d) Correspondingly we get for 12 knots: 3201100121851210 212 R kN

4008.0

3201

1212

tRT kN

247.5)5144.012()15.01( AV m/s 2

12, 394.0 JKT 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 J KT 10KQ 0 1.0 0.690 0.188 0.405 0.650

This yields: 267.16690.0

247.5

DJVn A s-1 = 76.0 rpm

5186267.110250405.0 5252 DnKQ Q kNm

4124518167.122 QnPD 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 = 117.20 m rb = 1.5 m (bilge radius) Lwl = 120.00 m CB = 0.688 B = 20.00 m lcb = 0.0 m T = 6.56 m

Wake fraction and thrust deduction are estimated by: 24.075.0 BCw and 15.05.0 BCt Viscosity and density are: = 1.1910-6 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 31035.0 Ac . Solution

a) Considerations for the original hull The displacement is: 1057956.600.2020.117688.0 TBLC ppB m3 The wetted surface following Keller (1973) is:

292510579)20.1175.0105794.3()5.04.3( 3/13/13/13/1 ppLS m2 The midship section area is:

20.13022

5.156.600.2022

22

bm rTBA m2

Thus 9926.056.620

20.130

TBAC m

m and 692.09926.0688.0

M

BP C

CC .

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 37.11820.11701.101.1 ppd LL m.

( wld LL for ppwl LL 01.1 , 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.

The Reynolds number is: 86 10983.7

1019.137.118025.8

d

nLVR

cF following ITTC'57 is:

328

102

10

10574.1210983.7log

075.02log

075.0

n

F Rc

The residual coefficient is determined following Lap-Keller. For

891.037.118685.0

025.8

LCV

Pd

and CP = 0.692 we read from the diagram in Fig.4 of Keller

(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

310111.1025.02925

20.130 rm

R SAc

The correction factor K2 for B/T is given by Lap (1954) as:

032.14.256.600.2005.014.205.012

TBK

The total resistance is:

2925025.82

1025032.135.0111.1574.12

222 SVKcccR ARFT

= 302 kN 2424025.8302 VRP TE kW

276.024.0688.075.024.075.0 BCw 194.015.0688.05.015.05.0 BCt

113.1276.01194.01

11

wt

H

3960155.0113.1

2424

0

RH

ED

PP

kW

b) Considerations after ship conversion (We denote values for the new version by a prime.)

The new length is 70.12950.1220.117' LLL pppp m.

Thus 00.13170.11901.1'01.1' ppd LL m.

5.1220650.122.13010579' LAm m3

710.056.620131

5.12206'

'

TBLC B

715.09926.0710.0''

M

BP C

CC and 708.001.1'' P

PdCC .

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:

32965.12206)20.1175.05.122064.3()5.04.3(' 3/13/13/13/1 ppLS m2

293.024.0710.075.024.0'75.0' BCw 205.015.0710.05.015.0'5.0' BCt

124.1293.01205.01

'1'1'

wt

H

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 ac-ceptable.) We now use Keller's (1973) Figure 5.

Rn 8.834 108 r 0.0285 RT 340 kN cF 1.554 103 cR 1.126 103 PE 2730 kW

dPd LCV ''/ 0.833 (cF+cA+cR) K2 3.127 103 PD 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!