Ship Propulsion Problems and Solutions

NA 5007 Original Lecture by Dr.Glover

Calculation & Typing Ye Wint Thu 1

Problem (1)

A large single-screw vessel is to be propelled by a direct-drive diesel engine having a

maximum continuous rating of 23400 kW at 87.5 rpm and 6-bladed propeller. Values of

effective power ,PE, derived from model resistance experiments by means of the ITTC skin

friction formulation are as follows.

V (knots) 12 13 14 15 16 16

PE (KW) 6010 7460 9280 11470 14120 17430

Given that the torque identity wake fraction, wQ=0.40, the thrust deduction fraction, t=0.25,

the relative rotative efficiency, R=1.02, the power prediction factor, (1+x)=1.042, the

shafting efficiency, s=0.98 and using the diagram provided, determine the ship speed on trial

in the fully loaded condition with the engine developing 85% of the maximum continuous

power at 100% nominal r.p.m and the diameter and mean face pitch of the corresponding

optimum propeller.

N= rpm, PD = delivered power in kW, Va = advance speed in knots, D= diameter in metres.

Solution

PE(trial) = PE(service) x (1+x) = 1.042 PE(service)

V (knots) 12 13 14 15 16 16

PE (KW) 6010 7460 9280 11470 14120 17430

PE(trial) 6262.42 7773.32 9669.76 11951.74 14713.04 18162.06

0

5000

10000

15000

20000

12 13 14 15 16 16

PE

(K

W)

V (knots)



PD = 85%MCR s = 0.85 x 23400 x 0.98 = 19492.2 kW

PE = DPD = 19492.2D

Va = (1-w) Vs = 0.6Vs

1st Trial 2

nd Trial 3

rd Trial

Assume D 0.68 0.644 0.631

PE = 6984.71D 13254.7 12.552.98 12299.58

Vs from curve 15.7 15.15 15.1

Va = 0.6Vs 9.42 9.09 9.06

51.94 56.79 57.25

o 0.505 0.495 0.493

D=1.275o 0.644 0.631 0.631

Do= 8.62m

DB= 0.95Do = 8.62 x 0.95 = 8.19m

(P/D)B = 0.765

P = 0.765 x 8.19 = 5.53m

Problem (2)

A 30,000 DTW Product Tanker is to be fitted with 4-bladed propeller driven by a diesel

engine having a Maximum Continuous Power of 8385 kW at 117.4 rpm.

Values of effective power, PE, corresponding to the ship in full load condition as scaled from

model resistance experiments are as follows:



V (knots) 12.64 13.79 14.36 14.94 15.51 16.09

PE (KW) 2890 3850 4460 5180 5980 6950

Given that the torque identity wake fraction, wQ=0.40, the thrust deduction fraction, t=0.20,

the relative rotative efficiency, R=1.02, the power prediction factor, (1+x)=0.939, the

shafting efficiency, s=0.98 and using the diagram provided, determine the ship speed on trial

in the fully loaded condition with the engine developing 85% of the maximum continuous

power and the diameter and mean face pitch of the corresponding optimum propeller.

In fact, the trial are to be run with the ship in ballast condition to which the following PE

value correspond:

V (knots) 12.64 13.79 14.94 15.51 16.09 16.66

PE (KW) 2470 3360 4590 5360 6260 7330

Given that the ballast condition, wQ=0.43, t=0.21, R=1.02, (1+x)=0.939 and s=0.98,

determine the ship speed on ballast trial with the engine developing 85% of the maximum

continuous power and with the prppeller dimensions derived above.

Where, PD=delivered power in KW

N = rate of rotation in rpm

Va=advanced speed in knot

D =propeller diameter in meter

Solution

PE(trial) = PE(service) x (1+x) = 0.939 PE(service)

V (knots) 12.64 13.79 14.36 14.94 15.51 16.09

PE (KW) 2890 3850 4460 5180 5980 6950

PE(trial) 2713.71 3615.15 4187.94 4864.02 5615.22 6526.05



PD = 85%MCR s = 0.85 x 8385 x 0.98 = 6984.71 kW

PE = DPD = 6984.71D

Va = (1-w) Vs = 0.6Vs

1st Trial 2

nd Trial 3

rd Trial

Assume D 0.68 0.699 0.711

PE = 6984.71D 4749.6 4882.31 4966.13

Vs from curve 14.8 15.15 15.25

Va = 0.6Vs 8.88 9.09 9.15

48.35 45.6 44.86

6.95 6.75 6.69

o 0.514 0.523 0.5235

D=1.63o 0.699 0.711 0.711

0

1000

2000

3000

4000

5000

6000

7000

12.64 13.79 14.36 14.94 15.51 16.09

PE

(k

W)

V (knot)



Do= 5.82m

DB= 0.95Do = 5.82 x 0.95 = 5.53m

(P/D)B = 0.81

P = 0.81 x 5.53 = 4.48m

PE(ballast) = PE(service) x (1+x) = 0.939 PE(service)

V (knots) 12.64 13.79 14.94 15.51 16.09 16.66

PE (KW) 2470 3360 4590 5360 6260 7330

PE (ballast) 2319.33 3155.04 4310.01 5033.04 5878.14 6882.87

PD = 85%MCR s = 0.85 x 8385 x 0.98 = 6984.71 kW

PE = DPD = 6984.71D

Va = (1-w) Vs = 0.57Vs

0

1000

2000

3000

4000

5000

6000

7000

8000

12.64 13.79 14.94 15.51 16.09 16.66

PE

(k

W)

V (knot)



1st Trial 2

nd Trial

Assume D 0.7 0.712

PE = 6984.71D 4889.297 4973.11

Vs from curve 15.25 15.3

Va = 0.57Vs 8.69 8.721

51.03 50.58

7.14 7.11

245.11 244.23

o 0.505 0.506

D=1.41o 0.712 0.712

Problem (3)

A large single screwed tanker was originally designed to achieve a service speed of 15 knots

with a steam turbine developing 23900 kW at a propeller rate of rotation of 85 rpm. The 6

bladed propeller corresponding to those conditions has a diameter of 8.23m and mean face

pitch ratio of 0.82.

For economic reasons the service speed of the ship has been reduced to 13.5 knots

corresponding to machinery power of 16800 kW. Assuming that the torque identity wake

fraction, wQ=0.40, the shafting efficiency, s=0.98, and using the diagram provided,

determine the propeller rate of rotation in reduced power condition and the corresponding

open water propeller efficiency.

It is anticipated that further economic gains could be achieved by fitting a new propeller, the

rate of rotation is reduced such that in the reduced power condition. Determine new propeller

rate of rotation, the diameter and mean face pitch of the proposed propeller and the increase

in open water efficiency.



Solution

Original Economic

Vs = 15 knots Vs = 13.5 knots

Ps = 23900 kW Ps = 16800 kW

N = 85 rpm s = 0.98

D = 8.23 m w = 0.4

P/D = 0.82 PD = Ps s=16800x0.98=16464 kW

Va = (1-w) Vs=(1-0.4)Vs=8.1 knot

N Bp

P/D

60 47.76 6.91 199.98 1.15

70 55.72 7.46 233.31 0.985

80 63.68 7.98 266.64 0.78

90 71.64 8.46 299.97 0.66

0

0.2

0.4

0.6

0.8

1

1.2

1.4

60 70 80 90

P/D

N



At P/D=0.82, N=78 rpm

From chart, o=0.985

Form chart, P/D=0.625, o=0.552, =252

D = 252/24.2 = 10.41m

Increase in o = 0.985-0.522 = 0.463

Problem (4)

A large gas carrier has the following values of effective power corresponding to ideal trial

conditions

Va (knot) 17 18 19 20

PE (kW) 12560 15370 18390 22110

A ship is fitted with direct drive machinery and a 4 bladed propeller with diameter D=7.71m,

pitch ratio P/D=0.86.

On trial with the engine developing 29400 kW at 105 rpm, the ship achieved of 19.33 knots

and the analyzed value of the torque identity wake fraction, wQ=0.33.

Assuming that in average service conditions the effective 20% higher than that for trial and

the wake fraction is increased by 10%, calculate, using the following data and the diagram

provided, the speed of the ship in service and the corresponding rate of rotation of the



propeller with the engine developing 29400 kW. Thrust deduction fraction = 0.19, relative

rotative efficiency = 1.01, shafting efficiency=0.98

N= rpm, PD = delivered power in kW, Va = advance speed in knots, D= diameter in metres.

Solution

Since effective power is 20% higher than that for trial, PE(service) = 1.2 PE(trial)

Va (knot) 17 18 19 20

PE (kW) (trial) 12560 15370 18390 22110

PE (kW) (service) 15072 18444 22068 26532

wQ = 1.1 x 0.33 = 0.363

PD = 29400 x 0.98 = 28812 kW

PE = DPD = 28812D

Va = (1-0.363) Vs = 0.637Vs

0

5000

10000

15000

20000

25000

30000

17 18 19 20

PE

(k

W)

V (knot)


Calculation & Typing Ye Wint Thu

10

1st Trial

Assume D 0.68

PE 19592.16

Vs from curve 18.3

Va 11.6

Bp 0.423N

2.71N

N Bp

P/D

90 38.07 6.13 195.3 1.15

100 42.3 6.5 217 0.9

110 46.53 6.82 283.7 0.78

120 50.70 7.12 260.4 0.68

At P/D = 0.86, N = 103

0

0.2

0.4

0.6

0.8

1

1.2

90 100 110 120

P/D

N



11

Bp 43.57

223.51

o 0.513

D 0.65

1st Trial

Assume D 0.65

PE 18727.8

Vs from curve 18.8

Va 11.53

Bp 0.435N

2.19N

N Bp

P/D

90 6.26 197.1 1.045

100 6.59 219 0.88

110 6.92 240.9 0.72

120 7.22 262.6 0.63

0

0.2

0.4

0.6

0.8

1

1.2

90 100 110 120

P/D

N



12

At P/D = 0.86, N = 101

Bp 43.635

221.19

o 0.514

D 0.65

Problem (5)

A research vessel is to be propelled by the direct drive machinery and a 4 bladed propeller.

The engine speed corresponding to the maximum continuous power is 188 rpm and the ship

is required to achieve a speed of 15.5 knots on trial with the engine developing 85% of

maximum continuous power.

Using the diagram provided and using the following information, determine the required

maximum continuous power if the machinery and the diameter, the face pitch, surface area

and BAR of the propeller.

Effective power at 15.5 knots = 2950 kW

Power prediction factor, (1+x) = 1.01

Torque identity wake fraction = 0.23

Thrust deduction wake fraction = 0.17

Relative rotative efficiency = 1.04

Shafting efficiency = 0.98

Immersion to shaft centerline = 3.95

p e = 99629 + 10179 H N/m2

qT = (11.66 x Va)2 + (0.828 x ND )

2 = 310875 N/m

2

Ap = AD x (1.067 0.229 P/D) m2

N= rpm, PD = delivered power in kW, Va = advance speed in knots, D= diameter in metres,

H=immersion to shaft centerline (m), P=mean face pitch (m)



13

Solution

PE = DPD , PD= PE/D=2950/D

Va = (1-w) Vs = (1-0.13) x 15.5 = 11.935 knots

1st Trial 2

nd Trial

Assume D 0.68 0.65

PD 4338.24 4358.46

Bp 29.112 29.78

5.4 5.46

o 0.58 0.579

D 0.65 0.65

At Bp=29.78, (P/D)o=0.8, o=200,

DB = 0.95Do = 0.95 x 3.87 = 3.68m

(P/D)B = 0.92

p e = 99629 + 10179 x 3.95 = 139836.05 N/m2

qT = (11.66 x 11.935)2 + (0.828 x 188 x 3.68 )

2 = 347517 N/m

2

(p e) / qT = = 0.402, c = 0.17

T / Ap = qT x c = 347515 x 0.17 = 59077.5

Ap = 7.46 m2

AD = 7.46 / (1.067-0.229 x 0.92) = 8.712



14

BAR = AD / Ao = 8.71 / ( x 3.682/4) = 0.82

Calculated BAR > Design BAR, Risk of cavitation cannot be minimized.

Problem (6)

The propeller of an oil product tanker is to have diameter and fixed pitch in relation to full

load trial performance and its blade surface area is to be adequate to minimize the risk of

cavitation in service in ballast condition. It is estimated that on trial with the engine

developing 8650 rpm at 212.5 rpm, the ship speed will be 16.03 knots. Using the diagram

provided and following data, determine the diameter and mean face pitch of the propeller to

satisfy these conditions.


Shafting efficiency = 0.98

In the service ballast condition with the engine developing the same power, the ship speed is

predicted to be 16.9 knots and the propeller rate of rotation, 119.0 rpm. Using the following

data and the diagram provided determine the required blade surface area and the

corresponding BAR,


Immersion to shaft centerline, H = 4.25m

Propeller efficiency = 0.53

On the cavitation diagram:

p e = 99629 + 10179 H N/m2

qT = (11.66 x Va)2 + (0.828 x ND )

2 = 310875 N/m

2

Ap = AD x (1.067 0.229 P/D) m2

Solution

For trial condition

PD = 8650 x 0.98 = 8477 kW

Va = (1-w) Vs = (1-0.32) x 16.03 = 10.9 knots



15

At Bp=33.02, (P/D)o=0.76, o=213,

DB = 0.95Do = 0.95 x 5.82 = 5.53m

(P/D)B = 0.86

For service condition

p e = 99629 + 10179 x 4.25 = 14288.75 N/m2

Va = (1-w) Vs = (1-0.4) x 15.5 = 10.14 knots

qT = (11.66 x 10.14)2 + (0.828 x 119 x 5.53 )

2 = 310875 N/m

2

(p e) / qT = = 0.46, c = 0.182

T / Ap = qT x c = 310875 x 0.182 = 56579.25

Ap = 15.2m2

AD = 15.2 / (1.067-0.229 x 0.86) = 17.47

BAR = AD / Ao = 17.47 / ( x 5.532/4) = 0.73

Problem (7)

An ore patrol vessel is to be fitted with two diesel engines geared to a single shaft and driving

a 4 bladed controllable pitch propeller. Each engine has a break power of 1501 kW and the

propeller rate of rotation is kept constant at 225 rpm.

Using the diagrams provided, determine the diameter, mean face pitch and blade surface area

of a propeller suitable for one engine cruise condition for which the ship speed Vs=15 knots

and the torque identity wake fraction, wQ=0.24, immersion to shaft centerline, H=3.05 and

shafting efficiency = 0.98.

In two engine condition the ship speed increased to 16.5 knots and the wake fraction remains

unchanged. Using the propeller diameter previously determine, calculate the increased

propeller pitch and the propeller efficiency and plot the operating condition on the cavitation

diagram. What conclusion would you draw from the result of the later calaulation?



16

p e = 99629 + 10179 H N/m2

qT = (11.66 x Va)2 + (0.828 x ND )

2 = 310875 N/m

2

Ap = AD x (1.067 0.229 P/D) m2

N= rpm, PD = delivered power in kW, Va = advance speed in knots, D= diameter in metres,

H=immersion to shaft centerline (m), P=mean face pitch (m)

Solution

For one-engine condition

Vs = 15 knots, wQ = 0.24, H=0.35, s = 0.98

PD = PB s = 1510x 0.98 = 1479.8 kW

Va = (1-w) Vs = (1-0.24) x 15 = 11.4 knots

0.1739Bp1/2

= 0.89

Form B4-85 chart, 1/J = 1.9

= 101.3 / J = 101.3 x 1.9 = 192.47

Fo cha o=192.47

DB = 0.95Do = 0.95 x 2.97 = 2.8215m

1 / J = 182.7 / 101.3 = 1.8

From chart, at 0.1739Bp1/2

= 0.89 and 1 / J = 1.8,

o = 0.558, (P/B)o = 0.935, P = 0.935 x 2.8215 = 2.64m

p e = 99629 + 10179 x 3.05 = 130675 N/m2

qT = (11.66 x 11.4)2 + (0.828 x 225x 2.64 )

2 = 259567.5 N/m

2

(p e) / qT = = 0.5, form Burrills diagram, c = 0.183



17

T / Ap = qT x c = 259567.5 x 0.183 = 47500.85

Ap = 2.96m2

AD = 2.96 / (1.067-0.229 x 0.935) = 3.47

BAR = AD / Ao = 3.47 / ( x 2.82152/4) = 0.56

For two-engine condition

Vs = 16.5 knots, wQ = 0.24, D = 2.8215

PD = PB s = 2 x 1510 x 0.98 = 2959.6 kW

Va = (1-w) Vs = (1-0.24) x 16.5 = 12.54 knots

0.1739Bp1/2

= 0.877

1 / J = 166.091 / 101.3 = 1.64

From chart, at 0.1739Bp1/2

= 0.877 and 1 / J = 1.64,

o = 0.542, (P/B)o = 1.095, P = 1.095 x 2.8215 = 3.0895m

Increase in pitch = 3.0895 2.64 = 0.4495

qT = (11.66 x 12.54)2 + (0.828 x 225x 2.8215 )

2 = 297682.37 N/m

2

(p e) / qT = = 0.44, form Burrills diagram, c = 0.17

T / Ap = qT x c = 297682.37 x 0.11 = 50606

Ap = 4.91m2

AD = 4.91 / (1.067-0.229 x 1.095) = 6.02

BAR = AD / Ao = 6.02 / ( x 2.82152/4) = 0.96

Design BAR = 0.85 < calculated BAR, Risk of cavitation cannot be minimized.

Problem (8)

The propeller for a large bulk carrier was designed to absorb 9500kW, which corresponds to

90% of maximum continuous power, at 100% nominal speed, 122 rpm and a ship trial speed



18

of 14 knots. In making the design the torque identity wake fraction was to be 0.37 and the

diameter and the mean face pitch ratio were determined as 6.07m and 0.70 respectively.

On trial, with the engine developing 9500 kW, the ship achieved a speed of 14.05 knots, but

the propeller run slow at 118.7 rpm.

Use the following open water data to determine the value of wake fraction corresponding to

the trail conditions.

J 0.25 0.3 0.35 0.40

kQ 0.0268 0.0250 0.0231 0.0211

And use the diagram provided to calculate the pitch reduction required propeller speed on

trial to 122 rpm.

Assume shafting efficiency = 0.98.

N= rpm, PD = delivered power in kW, Va = advance speed in knots, D= diameter in metres

Solution

PD = Ps s = 9500 x 0.98 = 9310 kW

PD = 2 n QD

( )

By interpolation, J = 0.3625

J = Va / ND, Va = 0.3625 x 118.7 x 3.07 / 60 = 4.35 m/s

wT = 1 ( Va / Vs) = 1- (8.45 / 15.05) = 0.399

From B4-65 chart, P/D = 0.69

P = 0.69 x 6.07 = 4.19m

Original pitch = 0.7 x 6.07 = 4.249 m



19

Reduction in pitch = 4.249 4.19 = 3.059m

Problem (9)

(a) Assuming that

is a linear of the advance coefficient J, show that, for a given

constant wake, wT

(b) A twin screw container ship has propellers of 6200mm with a pitch of 7500mm.

On trial the power transmitted by one shaft Ps as measured by a torsion-meter is 40550 HP

(metric) at 122 rpm and transmission efficiency is 98%.

With the dimensions used above the values of A and B in the above relations are 0.0578 and

0.0212 respectively.

Determine the apparent slip and hence the ship speed.

If the Taylor wake wT=0.07 what is the effective of propeller?

Solution

(a)

KQ is a linear function of J.

KQ = a J + b

a

( )

PD = 2 n Q

Q = PD / 2 eq (2)

Vs = P n (1-Sa) eq (3)

Va = Vs (1- wT) ... eq (4)

Sub: eq (3) in (4)

Va = P n (1-Sa) (1- wT) . eq (5)

Sub: eq (5),(2) in (1)

a ( ) ( )



20

PD / n3 = a D

4 2 P (1-Sa) (1- wT) + 2 D b

(b)

Ps = 40550 x2

PD = 0.98 x 40550 x2 = 59290.588 kW

Sa = 0.36

Vs = P n (1-Sa) = 7.5 x 112 x (1-0.36) /60 = 8.96 m/s = 17.39 knots

Va = Vs (1- wT) = (1-0.07) x 17.39 = 16.71 knots

Problem (10)

On trial, at a speed of 21 knots the measured delivered power is 11186 kW at 96 shaft rpm.

The propeller diameter is 6.7 m and the sea water density was 1026 kg / m3.

Form open water curve of the propeller the following pairs of values can be obtained:

10 KQ 0.33 0.21

j 0.7 0.9

What is trial analysis wake?

Solution

PD = 2 n QD

10KQ = 0.313

By interpolation, J = 0.73

J = Va / ND, Va = 0.73 x 1.6 x 6.7 = 7.826 m/s = 15.196 knots

wT = 1 ( Va / Vs) = 1- (15.196 / 21) = 0.276

Ship Propulsion Problems and Solutions

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