Preliminary Prediction of Power

Algorithm for the Prediction of Power at the

Preliminary Design Stage

Resistance & Propulsion (1)MAR 2010

Rod Sampson - School of Marine Science and Technology - 4th March 2008


Method is appropriate for large ocean going vessels with modern slow speed direct drive

diesel engines

Method is a basic design tool using chart series diagrams

Design Scope


Ship owner require that a ship should achieve an average speed in service at a certain engine power.

Initial acceptance will be based upon demonstration of a higher speed on trial at the same power

Vtrial = Vservice + V

(V ! 1 knot)

Design Scope


Design Scope

Contract stipulate that the ship should achieve trial speed with the engine developing 85% of its maximum continuous power rating (MCR)


Preliminary Prediction of Power

Estimate resistance and effective power for a range of speeds using appropriate Methodical Series

Data or Statistical Analysis Data (Holtrop & Menen)



methodical series or other

PE =

PE trial =

PE service =

(1 + x)PE

1.2 PE trial

From BTTP - 65 procedure

Based upon 20% sea margin

02000

4000

6000

8000

10000

12000

10.0 12.0 14.0 16.0 18.0 20.0Speed (knot)

Pow

er (

kW)

Pe Trial (kW)PE Service (kW)



20% sea margin

Plot trial and service conditions



Lbp = 135.34mB = 19.3mT = 9.16mCb = 0.704

Assumed vessel particulars from previous example:


Optimum RPM, Propeller and Engine Size

Chose maximum permissible propeller diameter with respect to hull clearances

Determine the optimum engine speeds corresponding to trial conditions, the required maximum continuous power

and the mean face pitch.

Select an appropriate engine



Using the equations provided in the handout determine the wake fraction and thrust deduction factor

DB ! 0.6T = 5.5m = DThe required diameter behind the hull is given as

This gives:

w = 0.304 t = 0.214



As the behind condition allows a smaller propeller diameter calculate the equivalent open water

diameter

DB ! 0.6T

Do =DB0.95

Corresponding open water diameter

= (5.79m)



Bp Select diagram on basis of blade number and blade area ratio

For this exercise the B4.55 diagram is to be used



0

2000

4000

6000

8000

10000

12000

10.0 12.0 14.0 16.0 18.0 20.0Speed (knot)

Pow

er (

kW)

Pe Trial (kW)PE Service (kW) At TRIAL speed read off from the

plot the effective power at trial speed

Remember is not known as it requires knowledge of the

propeller hull interaction.

D

D =(1 t)(1 w) o R

D = k1 o



0

2000

4000

6000

8000

10000

12000

10.0 12.0 14.0 16.0 18.0 20.0Speed (knot)

Pow

er (

kW)


Therefore a value is selected and an iteration is performed

until convergence

D = 0.70

Vs (trial) = 16 knots

Pe (trial) = 3550 kW



0

2000

4000

6000

8000

10000

12000

10.0 12.0 14.0 16.0 18.0 20.0Speed (knot)

Pow

er (

kW)


Assume an initial D = 0.7

From the graph:

PD =PE trial

D

PD =35500.7

PD = 5071 kW



0

2000

4000

6000

8000

10000

12000

10.0 12.0 14.0 16.0 18.0 20.0Speed (knot)

Pow

er (

kW)


Flow through the propeller disc

VA = Vtrial(1 w)

VA = 16(1 0.304)

VA = 11.14 knots



Let:

Bp = k2 N

Bp = 1.158 NP12

D

V 2.5A

Bp = 1.158 N 507112

11.142.5



and

= k3 N

=1J

= 3.2808 NDoVA

= 3.2808 N 5.7911.14



Enter the BP delta values onto the diagram provided

For each RPM calculate the open water efficiency from the diagram

Plot the results



For a range of values of N calculate Bp

N (rpm) Bp

80 15.93 136.4 0.62

90 17.91 153.5 0.624

100 19.91 170.5 0.626

110 21.90 187.6 0.622

120 23.89 204.6 0.605

o



0.6

0.605

0.61

0.615

0.62

0.625

0.63

75 85 95 105 115 125

oo= 0.626

N =100



0.55

0.56

0.57

0.58

0.59

0.6

0.61

0.62

0.63

0.64

0.65

65 70 75 80 85 90 95 100

N

N o


D =(1 t)(1 w) o R


More data is now available to update the initial estimate of D

D =(1 0.214)(1 0.304) 0.626 1.0

D = 0.707



To test convergence the difference from the original and new should be less than 0.005D

Continue until convergence to get N and D

0.707 - 0.7 = 0.007

use the new value of and re-calculate the delivered power and repeat BP d diagram



For this exercise the previous calculation is assumed converged, therefore:

D = 0.707

N = 100


Engine Selection

Once the propulsive efficiency is known the power required for sea trials can be calculated

PB(trial) =PE trialD S

PB(trial) =3550

0.707 0.98

PB(trial) = 5123.7 kW


From this Break Power required to satisfy the trial speed, the contract specified that the engine should

only be at 85% of its total rating

Engine Selection

PB(installed) =PB(trial)

0.85


This value can now be used to calculate the required engine size

Engine Selection

PB(installed) = 6028 kW


Finally modify the propeller diameter using the empirical relationships given previously

Re- calculate (as in example 1 of the numerical example the new diameter for the same delivered power

Bp Plot on the diagram and read the P/D ratio

Engine Selection


Engine Selection

PD = PB(trial) s

Calculate the delivered power at the trial condition

PD = 5123.7 0.98PD = 5021.23


Engine Selection

Calculate the delivered power at the trial condition

Bp = 19.81

= 3.2808 100 5.5011.14

= 161.97

Bp = 1.158 100 5021.2312

11.142.5


Engine Selection

Bp = 19.81 = 161.97

P

D= 1.0

Enter the final values onto the diagram

1.0DB = 5.50mean face pitch:


Engine Selection

Calculated optimum rpm = 100

Installed power = 6029 kW

Trial power = 5123.7 kW

Calculate the power per cylinder and use suitable engine diagrams to select an engine


Engine Selection

R1

R2R3

R4

85% MCR

reducing fuel consumption

trial power

installed power

Engine RPM

Engine Power


Engine Selection

R1

R2R3

R4

Engine RPM

Engine Power

1. Assume a number of Cylinders2. Calculate the required number of installed power

per cylinder3. from range of engines select the appropriate engine

with optimum RPM and power range


Engine Selection

R1

R2R3

R4

Engine RPM

Engine Power

4 Cylinders:5123.7 / 4 (trial) = 1280.92 kW/cyl6028.0 / 4 (total) = 1509 kW/cyl

Suitable engines could be RTA68 and RTA62 at 4 cyl.


Prediction of Service Performance

The final stage is to find the new ship service speed and propeller rate of rotation at constant power of:

PD = 0.85 PB s

i.e with the engines developing 85% of their Brake Power (including transmission losses)

PD = 5123.7 0.98 = 5021.23 kW



Caution:This method is different than the calculation for the trial condition.

For the trial condition the propeller was designed to absorb a particular power

In the service condition the propeller design is fixed :

Diameter (behind) = 5.5m Pitch = 1.0



At this new condition there will be a new wake fraction to allow for hull roughness, wind, waves, etc.

This is defined as:

w = 1.1wtrial

(assume previous values for t & )R

w = 1.1 0.304 = 0.3344



The propulsive efficiency will change for this new condition therefore another iteration must be

performed

This iteration of follows a different method than previous to obtain the optimum efficiency

D



Assume an initial value of D

PE service = PD D

Use the plots of trial and service power to read off the value. Read from the plot the service

speed this occurred PE service

PE service = 5021.23 0.7 = 3514.86 kW



0

1000

2000

3000

4000

5000

6000

7000

8000

13 14 15 16 17 18

Speed [kn]

Pow

er [

kW]

PE (service)

PE (trial)

Vs (service) = 15.25 knots



Re-calculate the advance velocity based on this power and new wake fraction

Va = Vs(1 w)Va = 15.25 (1 0.3344)Va = 10.15 knots



Let:

Bp = k4 N

Bp = 1.158 NP12

D

V 2.5A

Bp = 1.158 N 5021.2312

10.152.5



Let:

= k5 N

= 3.2808 NDVA

= 3.2808 N 5.5010.15



As before use a range of shaft rotations and calculate the new coefficients Bp

Plot these values directly on the diagram Bp

Where this curve intersects the P/D of the propeller designed previously is the required value



For range of values of N calculate Bp

N (rpm) Bp

80 19.99 142.16

90 22.49 159.93

100 24.99 177.77

110 27.49 195.47

o

Basic Design - BP delta diagrams

Rod Sampson - School of Marine Science and Technology - 26th February 2008

0

BP

PD

BP

1.0

o = 0.583



As before use the propulsive efficiency formula below and iterate until the difference between successive

iterations is within 0.005

D =(1 t)(1 w) o R

On convergence the ship speed in service condition has been calculated



D =(1 0.214)(1 0.334) 0.583 1.0

D = 0.688

(D)assumed (D)previous = 0.7 0.688 = 0.0011

(no need for iteration)



PE (service) = 5021.23 0.688

PE (service) = 3454.60 kW

PE (service)Read from the power diagram the for 3454.6 kW




0

1000

2000

3000

4000

5000

6000

7000

8000

13 14 15 16 17 18

Speed [kn]

Pow

er [

kW]

PE (service)

PE (trial)




At the intersection of the new line with the pitch line read off from the diagram the Service

values of BP

BP

BP = 24.2

B = 174



Re-calculate the advance velocity for the final service speed of 15.15 knots

VA = VS(1 w)

VA = 15.15(1 0.3344)VA = 10.08 knots



Finally at the service condition calculated read off the and calculate the RPM in serviceB



= 3.2808 NDVA

N = VA

3.2808D

N =174 10.083.2808 5.50

Nservice = 97.2 rpm



Therefore at 85% MCR:

The vessels service speed is 15.15 knots

The propeller rate of rotation is 97.2 rpm


Determination of Blade Areas

Final stage in the design algorithm is to calculate blade surface area and blade area ratio

This is performed for TRIAL condition



T = 9.16m

N = 100 rpm

VA = 16 (1 0.3044) = 11.14 knots

D = 5.50m

P

D= 1.0 o = 0.626

Using the previous trial conditions:

PD = 5021.23 kW



h =D

2+ 0.2

h = 2.95

shaft immersion at centreline

H = T - h H = 9.16 - 2.95

H = 6.21m

Text



p e = 99629 10179Hp e = 99629 10179 6.21p e = 162840.59 N/m2

The static component

Calculation of the cavitation number

r =p eqt



qt = (11.66VA)2 + (0.828 nD)2

The dynamic component

qt = (11.66 11.14)2 + (0.828 100 5.50)2

qt = 224261.2 N/m2



The resultant cavitation number becomes:

r =162840224261.2

r = 0.726



Entering this value onto the Burrill Diagram:

r = 0.726

c = 0.23



T

AP= c qT

T

AP= 0.23 224261.2 = 51580.08



T

AP= 1941.3

PD oVA

AP

AP = 10.62 m2

AP =1941.3 5021.23 0.626

11.14 51580.08

AP =1941.3 PD o

VA TAP



AD =AP

1.067 0.229PD

AD =10.62

1.067 0.229 1.0

AD = 12.67 m2



BAR =ADpiD2

4

BAR =12.67pi5.52

4

BAR = 0.533



AeAo

Selected was 0.55 Ae ADAssuming

AeAD

= 0.533

Selected area was 0.55, therefore the design will provide a low risk of cavitation


Coursework Submission

To satisfy the requirements of the coursework the following is required:

A typed report covering the 4 stages:

1. Effective power prediction2. Design of propeller and engine3. Prediction of performance in service, 4. Blade surface area and BAR



To satisfy the requirements of the coursework the following is required:

The report should include a detailed hand calculation for one speed (e.g. service speed)

Include tables from Excel where necessary and appropriate graphs

If you only present Excel tables and make a mistake, you cannot collect method marks.



Submission date is:

2nd May 2008



End of Presentation

Documents

Preliminary Prediction of Power