Prop Design Example

prop_design_example.xls

PROPELLER DESIGN USING WAGENINGEN B SERIES Design a propeller for a bulk-carrier with the following details LBP(m) = B(m) = T(m) = CB= VS(service) (knot) = δV = Trial speed range= Sea margin = AE/A0 Z

135.34 19.3 9.16 0.704 15 1 2 1.2 ? 4

SOLUTION

m m m knots

STAGE 1

VS (knots) 13.7 14.75 15.8 16.86 17.91

RT kN 281.9528 337.9287 413.0411 523.4767 648.4383

PE(trial) kW 1987 2564 3357 4540 5974

PE(service) kW 2384.4 3076.8 4028.4 5448 7168.8 y = 135.16x 2 - 3327.5x + 22214

8000 7000 6000 5000

Trial Power

4000

Service Power Poly. (Trial Power)

3000 2000 1000 0 12

13

14

15

16

17

Maximum permissible propeller diameter = Maximum continous power at= relative-rotative efficiency ηR = 1 0.98 shaft transmission efficiency ηS = Propeller diameter behind hull Dmax=DB =

18

19

20

0.6 T 0.85

5.496

Page 1

5.5 m

prop_design_example.xls w= t= open water diameter D 0=DB/0.95

0.304 0.214 5.79 m

AE (1 .3 + 0 .3 Z )T = +K A0 ( P0 − PV ) D 2

Keller's formula

434.3604 kN RT= 552.6214 kN T=RT/(1-t)= h=D/2+0.2 (height of shaft centre-line above base)

at Vs=16 knots

Atmospheric pressure, Patm= 101300 N/m 2 Vapour pressure of water at 15 °C, PV= 1646 N/m H= T-h 6.212 m

2

101300 N/m

2

1646 N/m

2

2

P0=Patm+ρgH 163763.2 N/m K= 0.2 for single screw AE/A0 0.482127 Wageningen B-4.55 propeller chosen VS(trial) = PE(trial) = Assume ηD = PD =PE/ηD VA = VS(trial) (1-w)

16 knots 3574.96 3575 kW 0.7 5107.143 5107 kW 11.136 knots

1/2 2.5 Bp=1.158(NxPD /VA ) δ=3.2808(NxD0/VA)

To find out rpm, select a range of propeller rpm, e.g. N=80~120 rpm, and calculate B p-δ and read-off propeller efficiency, ηo at corresponding Bp-δ from the diagram: Bp

δ

ηο

15.99796 17.9977 19.99744 21.99719 23.99693

136.3527 153.3968 170.4408 187.4849 204.529

0.62 0.624 0.626 0.622 0.605

N(rpm) assumed 80 90 100 110 120 0.63 0.625 0.62 0.615 0.61

y = -1E-08x4 + 4E-06x3 - 0.0005x2 + 0.0282x + 0.006

0.605 0.6 80

85

90

95

100

105

110

115

Page 2

120

125


Optimum N maximum η0

100 RPM 0.626

ηD=ηhηRη0=(1-t/1-w)ηRη0

0.707

∈=ηDcalculated - ηDpreviuos

0.007 if it is > 0.005 go back to "assume ηD" and select new value until it is 0.005

Let's assume that η D is converged 5160 kW Brake power PB=(PE/ηDηS) Installed maximum continous power =P B/0.85 Delivered power PD=PBηS

6070.696844

6071 kW

5056.89 kW

Therefore Bp= δ=

19.89882 161.9188

From Bp-δ diagram at [19.89,161.92] read-off P b/DB Mean face pitch= 5.50 m Stage 2 Engine selection calculated optimum rpm Brake power(85% MCR) Installed power(100% MCR)

1

100 5160 kW 6071 kW

Engine MAN B&W 4S60MC N rpm Engine Power 105 8160 105 5200 79 3920 79 6160 79 6160 105 8160

L1 L2 L4 L3

100 70 100 70

NoptimumPower 100 5160 85% MCR 100 6071 100%MCR 9000 8000 Power (kWs)

7000 6000 5000 4000 3000 2000 1000 0 70

75

80

85

90

95

100

N (RPM)

Page 3

105

110

5160 5160 6071 6071


STAGE 3 Prediction of performance in service Prediction of the ship speed and propeller rate of rotation in service with the engine 85% of MCR w in service= 1.1 w in trial ηD (assumed) PD=PBηS PE=PDηD

0.3344 0.7 5056.89 kW 3539.823 kW

From PE(service) vs VS curve at 3539.82 kW obtain Vs(service) VS(service) = 15.3 knots y = 162.19x 2 - 3993x + 26656

8000

V

7000 6000

PE 15.31 3539.873 15.25 3482.062

5000 Trial Power

4000

Service Power 3000

Poly. (Service Power)

2000 1000 0 12

13

14

15

16

17

VA=VS(1-w) Bp = δ B=

10.18368 knots 0.248822 xN 1.770605 xN

For a range of N's N 80 90 100 110 120

Bp 19.90575 22.39396 24.88218 27.3704 29.85862

18

19

20

δB 141.6484 159.3545 177.0605 194.7666 212.4726

read-off η0 @ intersection of Bp-δ curve with Pb/DB η0 0.583 0.688 ηD 0.012 if this difference is less than 0.005 there is no need for iteration ηDassumed-ηDcalculated Let's assume that η D is converged 3481.459 kW PE(service)=PDηDlast From PE(service) vs VS curve at PE(service) read-off Vs(service) Page 4


VS(service) =

15.25 knots

From Bp-δ diagram at above intersection point read-off Bp-δ 24 Bp δ 174 VA N=(δVA/(3.2808D)) N(service)

10.1504 knots

97.95 rpm

Therefore @ 85% MCR vessel's service speeed, V S =15.25 knots N=97.95 rpm

Page 5


STAGE 4. Determination of the blade surface area & B.A.R. (Cavitation control) h=D/2+0.2 (height of shaft centre-line above base) Atmospheric pressure, Patm= 101300 N/m 2

101300 N/m

2

1646 N/m

2

Vapour pressure of water at 15 °C, PV= 1646 N/m 2 For Trial condition T= PD = N= VA = P/D = η0 H= T-h

9.16 5056.89 100 11.136 1 0.626 6.212

m kW rpm knots

m

Dynamic pressure qT

224777.6 N/m

2

P0-Pv

162117.2 N/m

2

Cavitation number σR=(P0-Pv)/qT

2

qT=0.5VR2=0.5[VA2+(0.7πnD) ]

0.721234

Referring to Burrill's diagram for upper limit @ σR, the load coefficient, τc is read-off from fig. 4 as: 0.225 τc

By definition T/Ap=τcqT = T=PDη0ηR/VA

50574.96 552621.4 N

Ap=T/(τcqT)

ηB=PT/PD=TVA/PD=η0ηR

10.92678 m

2

13.03911 m

2

Developed area from Taylor's relationship AD=Ap/(1.067-0.229xP/D) Blade Area Ratio

AD ≈ AE

2

BAR= AE/(πD /4) Selected BAR=0.55 Calculated BAR=0.55 Calculated BAR=0.55 <= Selected BAR=0.55

0.55

Therefore propeller will have low risk of cavitation

Page 6


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Prop Design Example

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