Ship Project A - Aalto
Transcript of Ship Project A - Aalto
Ship Project A
2014
ASSIGNMENT ONE
RAN XIAO 467643 & THEIR TOMAS 224051
Resistance estimaππ’on
Methodology brief
As our ship concept is in form of catamaran, it is not proper to simply introduce the method and
diagrams of monohull ship series. For catamarans, the interaction of waves generated by twin hulls
cannot be ignored and thus affects the friction resistance as well as wave pattern resistance.
Different from the methodology used in the previous report where Delft Series (98β) and John
Winterβs empirical resistance diagram are introduced, here we apply the method and diagrams
deduced in <Resistance Experiments on a Systematic Series of High Speed Displacement
Catamaran Forms: Variation of Length-Displacement Ratio and Breadth-Draught Ratio>.
By this method, resistance coefficient is defined as follow:
πΆπ‘πππ‘ππππππ= (1 + β β π) β π β πΆπ + π β πΆπ
Where β is related to the pressure field change and Ο is about wave resistance intereface
factor. For simplification, the formula is transformed into:
πΆπ‘πππ‘ππππππ= (1 + π½ β π) β πΆπ + π β πΆπ
Based on this equation, the practical resistance expression is deduced as:
πΆπ‘πππ‘ππππππ= πΆπΉπ βππ
+ πΆπ πππππβ π½ β π β (πΆπΉπππππ
β πΆπΉπ βππ)
According to our design, parameters are summarized as:
CB CP Cw L/ββ3
S/L B/T
0.495 0.62 0.7 6.34 0.3 1.8
Then we can select the corresponding diagram and mother ship from <Resistance Experiments
on a Systematic Series of High Speed Displacement Catamaran Forms: Variation of Length-
Displacement Ratio and Breadth-Draught Ratio>. Here model are classified as:
We choose 3b* since it is the most similar one. Relevant parameters of 3b* are:
Here are some apparent differences in terms of CP, which may lead to some unexpected estimation
result. As CP stands for how full the underwater part of hull is, a relatively larger value can make
the predicted resistance larger than it should be. But since it is a rough estimation, the estimation
error is acceptable.
Determination of coefficients
Source: <Resistance experiments on a systematic series of high speed displacement catamaran
forms: variation of length-displacement ratio and breadth-draught ratio>
As mentioned previously, we know πΆπ‘πππ‘ππππππ= (1 + π½ β π) β πΆπ + π β πΆπ and in this figure, the
curves of πΆπ‘πππ‘ππππππ and (πΆπ‘πππ‘ππππππ
β πΆπ) have been given, together there are also curves of
πΆπ and (1.65) β πΆπ. Apparently, illustrated in the figure, the curve of (πΆπ‘πππ‘ππππππβ πΆπ)
matches very well with that of (1.65) β πΆπ outside the Froude Number zone of [0.2, 0.65].
Unfortunately the resistance we are studying is in this zone, so we have to estimate the value of
1 + π½ β π one by one according to the graph above, and we yield following graph:
And from the graph below we can also conclude all the πΆπ πππππ= πΆπ π βππ
= πΆπ at corresponding
speed..
Source: <Resistance experiments on a systematic series of high speed displacement catamaran
forms: variation of length-displacement ratio and breadth-draught ratio>
8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16
1+beta*k 1.77 1.88 1.94 2.02 2 2.25 2.38 2.5 2.63 2.88 3.03 2.95 2.76 2.56 2.43 2.16 1.98
1.77 1.881.9412.02 22.25
2.3752.5
2.6252.875
3.0252.9492.763
2.562.432
2.1621.975
0
0.5
1
1.5
2
2.5
3
3.5
1+b
eta
*k
speed (knots)
1+beta*k
1+beta*k
Resistance prediction
Now we have all the values needed for resistance calculation. According to the resistance
expression πΆπ‘πππ‘ππππππ= πΆπΉπ βππ
+ πΆπ πππππβ π½ β π β (πΆπΉπππππ
β πΆπΉπ βππ) as well as formulas
πΆπΉπππππ=
0.075
[log10(πΉπΓ5.56Γ106)β2}2 and ITTCβ57 Correlation line: πΆπΉ =0.075
[log10 π πβ2]2, we have the
following graph:
The coefficient climbs greatly after the speed exceeds 12 knots due to the drastic increase of πΆπ
at the corresponding speed.
Now lastly, we assume our wet surface is 75 m2. We are able to calculate the drag according to
the previous formula and draw the curve in the figure below:
8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16
Ct 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.01 0.01 0.01 0.01
0.0027 0.0023
0.0031 0.0036 0.0039 0.0038 0.0039 0.0042
0.0055
0.0070
0.0085
0.0097
0.0113 0.0115 0.0115 0.0110
0.0106
0.0000
0.0020
0.0040
0.0060
0.0080
0.0100
0.0120
0.0140
Re
sist
amce
co
eff
icie
nt
Speed (knots)
Resistacne Coefficient prediction
Rotation Speed & Propeller Design
Propeller
The ship will be using two propellers, one under each hull at the rear, as seen in figure 1. However,
unlike the ship in the picture, our ship will be using azimuth thrusters. The general idea will still be
the same, though, as the number of propellers and their operation points will be the same. The
number of propellers was chosen because each hull contains a separate engine, and with each
engine driving one propeller we will get some control advantages, and this means that the area
between the hulls can be kept clear.
1785.0682821655.0019242527.373656
3347.5404293994.6657954222.1068984749.460218
5590.806383
7991.980875
11158.21932
14579.67337
17922.69001
22436.49238
24514.3639
26380.9974826984.75826
27601.9221
0
5000
10000
15000
20000
25000
30000
8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16
resi
stan
ce (
N)
speed (knots)
Resistance prediction
Fig. 1, A catamaran (Aluma Marine ManaβO II) with one propeller under each hull [1]
In the initial concept design we calculated that the desired speed could be achieved using two
azimuth thrusters of 100 kW (the power needed was approximately 180 kW, so to ensure sufficient
power 2 x 100 kW could be used). By checking this to available solutions on the market we can
estimate the size of the propeller in this case.
Parameters & assumptions
According to the report and previous calculation, we can find out the following information:
Engine type: Type: Standard azimuth thrusters - type US 55P4
Max input power: 330 KW
Propeller diameter: 1050 mm
Vessel design speed: 15 knots
Resistance at 15 knots: 26381 N
Now as we are going to design the propeller initially, we have to make some assumption for further
calculation and weβll go back to check if our assumptions are reasonable. So here we assume:
t = 0,125
Ap = 0,55
w = 0,45
ππ»
= 1βπ‘
1βπ€= 1,591 and π
π = 0,98 (taken from empirical statistics from vessels whose
engines are located at the rear part), ππΊ
= 0,96 for gear box efficiency
Propeller design
Based on the resources we have at hand, we choose Wageningen B4-55 as our propeller type. As
we already have the thrust deduction and wake factor, the table below can be work out for
application of the B-series diagram and data interpolation.
Speed (knot)/(m/s) D(m) N(round/s) J KT VA(m/s) α΅Ή
15 7,716667 1,05 3 1,347354 1,344416 4,244167 0,742195
15 7,716667 1,05 3,5 1,469841 0,987734 4,244167 0,865894
15 7,716667 1,05 4 1,286111 0,756234 4,244167 0,989594
15 7,716667 1,05 4,5 1,14321 0,597518 4,244167 1,113293
15 7,716667 1,05 5 1,028889 0,48399 4,244167 1,236992
15 7,716667 1,05 5,5 0,828456 0,399992 4,244167 1,360691
15 7,716667 1,05 6 0,759418 0,336104 4,244167 1,48439
15 7,716667 1,05 6,5 0,701001 0,286385 4,244167 1,60809
15 7,716667 1,05 7 0,65093 0,246934 4,244167 1,731789
We can interpolate the α΅Ή values that we have into the diagram Fig. 2 to find out the
corresponding values of π0, pitch ratio and BP with which we can calculate the deliver power and
effective power.
Fig. 2 B4-55 diagram
Source: <Ship Principle>
N(round/s) α΅Ή P/D πΌπ Bp
4 0,989594 1,17 0,734 0,41
4,5 1,113293 1,07 0,729 0,43
5 1,236992 0,99 0,712 0,45
According to the formula given in figure 1: π΅π2 = ππ·
0,5
π·βππ΄1,5 , here comes the deliver power. Using the
efficiency we have assumed and π0 we get, the corresponding effective power and engine power
are able to be listed out.
N(round/s) Bp PD Ps PE
4 0,41 77,63311 82,52 90,65
4,5 0,43 85,3918 90,77 99,04
5 0,45 93,51996 99,40 105,93
Since in previous calculation we have known that the needed power for our design at 15 knots is
about 101,8 KW for each propeller, now we can make a graph to see where the needed power and
effective power meet so that we can determine the optimal pitch ratio and RPM for our propeller.
Apparently, the optimal configuration for our design is RPM = 278 with P/D = 1,05
Assumption Examination
Previously we assume w = 0,45. Now we know our pitch ratio and KT is available, we can check if it
makes sense.
As T = 26381 in total, KT = T/(2*1025*π*N2*D4)=0,56 corresponds to the J0=0,65. Therefore, w = 1-
J0/J = 0,41 which is quite close to our estimated value.
4 4.5 5
engine power 82.52 90.77 99.40
effective power 90.65 99.04 105.93
needed power 101.8 101.8 101.8
0.00
20.00
40.00
60.00
80.00
100.00
120.00
po
we
r (K
W)
propeller speed (round/s)
RPM against Power
engine power effective power needed power
Source: <Ship Principle>
Propeller coefficient
π0 π
π» π
π π
πΊ
0.718 1.591 0.98 0.96
2.2 Machinery
In this part, the choice of machinery will be discussed. We will try to explain why a certain piece of
machinery has been chosen, and what this means for the ship.
The power for the propulsion will be generated using hydrogen cells, and the energy generated will
be stored on on-board batteries as needed. With this system the ship will be able to achieve the
image of a green ship, which will appeal to both commuters and tourists. Thanks to the use of fuel
cells the energy can be generated only when needed, e.g. when the power in the batteries goes
under a certain amount, and if there is sufficient power available only the batteries will drive the
engines. This means that the shipβs machinery consists of four different components have to be
defined: The electric engines, the fuel cells, the hydrogen tanks, and the batteries.
The electric engine needs to be chosen so that it can produce sufficient energy for the propeller it
is driving. When accounting for the efficiency of the propeller the power of the engine should be
at least 150 kW. Here we could use e.g. the Siemens PEM-Motor 1DB2016 β WS54, which has a
power of 160 kW with a weight of < 350 kg [2].
The fuel cells are hard to find information on, but there are definitely fuel cells available at the
power we wish to use. The German company proton motor, for example, manufactures fuel cells
of up to several hundred kW for maritime use [3]. The price is hard to estimate, though, as the
technology improves constantly, and the price varies greatly depending on the product and its
specifications (custom/serial production, power-weight ratio etc.).
For the hydrogen tanks we calculated that we need a capacity of 70 kg. At 350 bar hydrogen has a
density of 70 kg/m3, which means that we need a total of 1000 l of tank space. This means that we
need a total of twenty 50 l tanks. The total weight of these is about 1000 kg.
Finally, the batteries need to store excess energy created by the fuel cells. In the case of the
Zemships 2 concept (which is a ship of a similar size to ours), the ship has 7 x 80 V lead-gel batteries,
giving 360 Ah. Since our ship uses a bit more power (the Zemships concept only has one 48 kW
fuel cell) we would probably need a bit more batteries. A linear approximation puts the amount at
roughly 42 x 80 V batteries. This can still be changed in the future, though, in case we e.g. want to
change to lithium-ion batteries, or get a more precise value on the needed amount of batteries.
Reference
1 Custom aluminum work boats, crew boats, rib boats, deck boats, response vessels marine
fabrication; http://www.alumamarine.com/workboat_files/alumcatamaran/2.jpg; 24.9.2014
2 https://www.industry.usa.siemens.com/drives/us/en/electric-drives/hybrid-
drives/Documents/elfa-components-data-sheets.pdf
3 http://www.eurosfaire.prd.fr/7pc/documents/1374147913_sebastian_dirk___proton_moto
r.pdf
4 Molland, A.F., Wellicome, J.F. and Couser, P.R. (1994) Resistance experiments on a systematic
series of high speed displacement catamaran forms: variation of length-displacement ratio
and breadth-draught ratio. Southampton, UK, University of Southampton, 84pp.
5 Sverre Steen and Knut Minsaas, Ship Resistance, Department of Marine Technology, NTNU,
2013.
6 Zhengbang Sheng, Zhongying Liu, Shanghai Transportation University, <Ship Principal>, 2003.