PROPELLER DESIGN USING WAGENINGEN B SERIES
Design a propeller for a bulk-carrier with the following detailsLBP(m) = 135.34 m
B(m) = 19.3 m
T(m) = 9.16 m
CB= 0.704
VS(service) (knot)= 15 knots
δV = 1
Trial speed range= 2
Sea margin = 1.2 AE/A0 ? Z 4
SOLUTION
STAGE 1
VS RT PE(trial) PE(service) (knots) kN kW kW 13.7 281.9528 1987 2384.4 14.75 337.9287 2564 3076.8 15.8 413.0411 3357 4028.4 16.86 523.4767 4540 5448 17.91 648.4383 5974 7168.8Maximum permissible propeller diameter = 0.6 T
Maximum continous power at= 0.85
relative-rotative efficiency ηR= 1
shaft transmission efficiency ηS = 0.98
Propeller diameter behind hull Dmax=DB = 5.496 5.5 m
y = 135.16x2 - 3327.5x + 22214 0 1000 2000 3000 4000 5000 6000 7000 8000 12 13 14 15 16 17 18 19 20 Trial Power Service Power Poly. (Trial Power)
w = 0.304
t = 0.214
open water diameter D0=DB/0.95 5.79 m
Keller's formula
RT= 434.3604 kN at Vs=16 knots
T=RT/(1-t)= 552.6214 kN
h=D/2+0.2 (height of shaft centre-line above base)
Atmospheric pressure, Patm= 101300 N/m2 101300 N/m
2
Vapour pressure of water at 15 °C, PV= 1646 N/m2 1646 N/m 2
H= T-h 6.212 m
P0=Patm+ρgH 163763.2 N/m 2
K= 0.2 for single screw
AE/A0 0.482127
Wageningen B-4.55 propeller chosen
VS(trial) = 16 knots PE(trial) = 3574.96 3575 kW Assume ηD = 0.7 PD =PE/ηD 5107.143 5107 kW VA = VS(trial) (1-w) 11.136 knots Bp=1.158(NxPD1/2/VA2.5) δ=3.2808(NxD0/VA)
To find out rpm, select a range of propeller rpm, e.g. N=80~120 rpm, and calculate Bp-δ and read-off propeller efficiency, ηo at corresponding Bp-δ from the diagram:
N(rpm) Bp δ ηο assumed 80 15.99796 136.3527 0.62 90 17.9977 153.3968 0.624 100 19.99744 170.4408 0.626 110 21.99719 187.4849 0.622 120 23.99693 204.529 0.605 y = -1E-08x4 + 4E-06x3 - 0.0005x2 + 0.0282x + 0.006 0.6 0.605 0.61 0.615 0.62 0.625 0.63 80 85 90 95 100 105 110 115 120 125
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Optimum N 100 RPM
maximum η0 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 ηD is converged
Brake power PB=(PE/ηDηS) 5160 kW
Installed maximum continous power =PB/0.85 6070.696844 6071 kW
Delivered power PD=PBηS 5056.89 kW
Therefore Bp= 19.89882
δ = 161.9188
From Bp-δ diagram at [19.89,161.92] read-off Pb/DB 1
Mean face pitch= 5.50 m
Stage 2 Engine selection
calculated optimum rpm 100 Brake power(85% MCR) 5160 kW Installed power(100% MCR) 6071 kW Engine MAN B&W 4S60MC
N rpm Engine Power L1 105 8160 L2 105 5200 L4 79 3920 100 5160 L3 79 6160 70 5160 79 6160 100 6071 105 8160 70 6071 NoptimumPower 100 5160 85% MCR 100 6071 100%MCR 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 70 75 80 85 90 95 100 105 110 N (RPM) Pow er (k Ws )
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 0.3344
ηD (assumed) 0.7
PD=PBηS 5056.89 kW
PE=PDηD 3539.823 kW
From PE(service) vs VS curve at 3539.82 kW obtain Vs(service)
VS(service) = 15.3 knots V PE 15.31 3539.873 15.25 3482.062 VA=VS(1-w) 10.18368 knots Bp = 0.248822 xN δ B= 1.770605 xN For a range of N's N Bp δB 80 19.90575 141.6484 90 22.39396 159.3545 100 24.88218 177.0605 110 27.3704 194.7666 120 29.85862 212.4726
read-off η0 @ intersection of Bp-δ curve with Pb/DB
η0 0.583
ηD 0.688
ηDassumed-ηDcalculated 0.012 if this difference is less than 0.005 there is no need for iteration Let's assume that ηD is converged
PE(service)=PDηDlast 3481.459 kW y = 162.19x2 - 3993x + 26656 0 1000 2000 3000 4000 5000 6000 7000 8000 12 13 14 15 16 17 18 19 20 Trial Power Service Power Poly. (Service Power)
VS(service) = 15.25 knots
From Bp-δ diagram at above intersection point read-off Bp-δ
Bp 24
δ 174
VA 10.1504 knots
N=(δVA/(3.2808D))
N(service) 97.95 rpm
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/m2 101300 N/m
2
Vapour pressure of water at 15 °C, PV= 1646 N/m2 1646 N/m 2
For Trial condition
T = 9.16 m PD = 5056.89 kW N = 100 rpm VA = 11.136 knots P/D = 1 η0 0.626 H= T-h 6.212 m Dynamic pressure qT 224777.6 N/m 2 qT=0.5VR2=0.5[VA2+(0.7πnD)2] P0-Pv 162117.2 N/m 2 Cavitation number σR=(P0-Pv)/qT 0.721234
Referring to Burrill's diagram for upper limit @ σR, the load coefficient, τc is read-off from fig. 4 as:
τc 0.225 By definition T/Ap=τcqT = 50574.96 T=PDη0ηR/VA 552621.4 N ηB=PT/PD=TVA/PD=η0ηR Ap=T/(τcqT) 10.92678 m 2
Developed area from Taylor's relationship
AD=Ap/(1.067-0.229xP/D) 13.03911 m 2
Blade Area Ratio AD ≈ AE
BAR= AE/(πD2/4) 0.55
Selected BAR=0.55 Calculated BAR=0.55