1 10 Priyanta Notes on Powering

Materials prepared by Dwi Priyanta Department of Marine Engineering – ITS

Surabaya

Notes on Marine Engineering – 2 7-1

MODULE 7 BASIC TURBINE – PROPELLER MATCHING General description of this module In order a ship is able to move at its designed speed, it is important to match main engine delivered power with propeller required power. If between them do not match, we may conclude that the power delivered by the main engine will not fully absorbed by the propeller. This module review fundamental theory on ship powering, how to estimate required power by ship, how to match selected main engine of the ship with the selected propeller, and ho to give considerations in matching between combined engines and propeller. Instructional objectives Having completed this module, students (readers) are expected to be able: to estimate the power required by a ship by considering mechanical losses occurred in power transmission

system to match between engine and propeller to give considerations in matching between combined engine and propeller

Estimated time required to complete the material Self study : 120 minutes Tutorial in class : 6x 50 minutes


Surabaya


MODULE 7 BASIC TURBINE – PROPELLER MATCHING

7.1 INTRODUCTION

Marine propulsion is effected by a device that converts the thermal energy obtained

from consumption of fuel into a thrust. The basic configuration of main propulsion system

consists of three major units. They are prime mover, transmission and propulsor. These

major units must be working together so that they will be able to propel a sip at certain

design speed.

The prime mover of a ship can be diesel engine (ranging from slow speed to high

speed diesel engine, either reversing or non-reversing diesel engine), steam turbine, gas

turbine or the combination among them. These engines will convert thermal energy into

mechanical energy. The mechanical energy of these engines are characterized by their

torque and rotational speed (rpm) of their output shafts.

The main function of the transmission system is to transmit power from prime

mover shaft to the propeller. Commonly, the power from prime mover is transmitted by a

rotating shaft to the propeller. Mechanical reduction gear may be needed in the case of

high-speed prime mover.

The main function of propulsor is to convert mechanical energy into the hydraulic

energy of an accelerated fluid or into a thrust horsepower. Fixed pitch propeller (FPP) or

controllable pitch propeller (CPP) is usually found as a propulsor of ship. Each component

must be so selected and coordinated that they will be working properly.

Engine and propeller form a single unit whose characteristics are determine by the

interactions of the individual characteristics of the two Units. In conventional propulsion

plant designs, the engine or turbine produces a torque, Q, at a rated rpm, N, which is

transmitted by the transmission system which includes the shafting, bearings and possibly

reduction gears to the propeller. We should also note that the characteristics of the hull

affect propulsion characteristics, so that engine, propeller, and hull can also be called a

single unit. Therefore, a marine engineer must be familiar with propeller characteristics, and


Surabaya


with hull influences upon them. This module will review basic principles of engine

characteristics, propeller characteristics and interaction between engine and propeller.

7.2 REVIEW ON POWERING

There are some points that must be considered in determining the main engine of

the ship. Besides mechanical aspects where power delivered from main engine to propeller,

the interaction between propeller and hull must also be considered. Fig. 7.1 shows a

schematic diagram of power definitions.

Fig. 7.1

Schematic diagram of power definitions

7.2.1 Effective Horse Power (EHP)

Effective horsepower (EHP) is the power required to tow a hull without a

propeller. Mathematically it is expressed in the following equation.

RVEHP = (7.1)

where,

R = resistance of ship in Newton

V = design speed of ship in m/s

EHP = effective horsepower in Watts

We may write equation (7.1) in English engineering units.

550RVEHP = (7.2)

where,


Surabaya


R = resistance of ship in lb

V = design speed of ship in ft/s

EHP = effective horse power in horsepower

Note that, the denominator of equation (7.2) is a conversion factor. Remember that

1 hp = 33 000 ft-lb/min = 550 ft-lb/sec

Example 7.1 A ship is designed to sail at 20 knots. From the towing tank test data, it is known that the resistance of this ship is 75,000 lbs. Determine the EHP of this ship at its design speed. 20 knots = 20 x 1.689 ft/s = 33.78 ft/s We may be directly calculated the EHP of this ship by using equation (7.2).

HP 4606ft.lb/s/HP 550

ft/s 78.33lb 000,75550

=×

==RVEHP

Towing a scaled ship model in a towing tank, known as a towing test, is the most

reliable method in determining the resistance of the ship. However, this test is time

consuming and so expensive. The test will be done until the final design is accepted. It is

because the design of the ship is subject to the frequent alteration before final design is

accepted. For preliminary design purpose, the resistance of the ship may be approximated

based upon well-known published series of test results derived from systematically varied

hull forms. These methods are:

The Taylor’s Standard Series and Series 60

Holtrop algorithm

Savitsky method

Delft Series

However, the advance developments in computer technology have led in the

developments of “dry towing tank”. The dry towing tank refers to the utilization of computer

technology in determining resistance of the ships. Computer software is the physical form

of a dry towing tank. Using this software, a designer might be able to design a ship starting

from the lines plan of the ship until the estimation of the resistance of designed ship.

Designers can easily adjust their design and calculate the resistance of the designed ship

almost in the same time only in a few minutes. Maxsurf, Hullspeed, Autoship and

Autopower are examples of software that can be called as dry towing tank.


Surabaya


7.2.2 Thrust Horsepower

When a ship is moving ahead, the propeller will accelerates water sternward. The

acceleration will increase the momentum of water. Considering Newton’s second law, the

force equivalent to the increasing accelerated water momentum is called thrust. The product

of thrust and speed of water relative to the propeller – it is called speed of advance, Va – is

called thrust horsepower (THP). Thus, thrust horsepower is power delivered by the propeller

to the water and it is expressed by

aVTTHP ×= (7.3)

Where,

T = Thrust of propeller in Newtons

Va = Speed of advance in m/s

THP = Thrust horsepower in Watts

We may write equation (7.3) in English engineering units.

550

aVTTHP

×= (7.4)

Where,

T = Thrust of propeller in lbs.

Va = Speed of advance in ft/s

THP = Thrust horsepower in horsepower

7.2.3 Propeller Operation Behind The Hull

Thrust deduction

The presence of the propeller operating behind the hull changes the pressure

distribution on the hull and so is the resistance. Therefore, there is a difference between

total resistance of the ship (R) and thrust of the propeller (T). Thrust of propeller will be

greater than resistance of the ship. The quantity T minus R is called thrust deduction and is

normally expressed as a fraction of the thrust..

TtR

TRTt

)1( −=

−=

(7.5)

where t denotes thrust deduction.


Surabaya


Wake fraction

The presence of the hull ahead the propeller changes the average local velocity of

the propeller. If a ship moves at speed V then the accelerated water sternward by its

propeller will have move at the speed less than the speed of the speed. The accelerated

water will move at the speed of Va, known as speed of advance. The term wake speed is

usually used to quantify the difference between V and Va. It is customarily defined as a

fraction of ship’s speed, V.

VwV

VVV

w

a

a

)1( −=

−=

(7.6)

where w denotes wake fraction.

Combining equation (7.1), (7.3), (7.5), and (7.6) we will have a new equation that

relates between EHP and THP. The relation is expressed by

H

EHPt

wRVTHPη

=−

−=

)1()1( (7.7)

Where ηH represents the hull efficiency of the ship.

)1()1(

wt

H −−

=η (7.8)

In engineering English unit, equation (7.7) may be rewritten as

H

EHPt

wRVTHPη

=−×

−=

)1(550)1( (7.9)

Example 7.2 Recall example 7.1. Assume that thrust deduction and wake fraction are 0.19 and 0.29 respectively. Determine the THP of this ship at its design speed. 20 knots = 20 x 1.689 ft/s = 33.78 ft/s We already have the effective horsepower of this ship, that is 4606 HP. The hull efficiency of this ship is

14.1)29.01()19.01(

)1()1(

=−−

=−−

=wt

Hη

We may employ equation (7.9) to calculate the THP of this ship.

HP 404014.1

4606===

H

EHPTHPη


Surabaya


Fig. 7.2

KQ, KT and Open water efficiency of typical propeller

7.2.4 Open Water Test

Propeller characteristics can be described graphically in several ways. Most

appropriate for the discussion to follow are plots of torque coefficient (Kq) and thrust

coefficient (Kt) plotted as functions of the advance coefficient (J). They are defined by

25nDQKQ ρ

= (7.10)

24nDTKT ρ

= (7.11)

nDV

J a= (7.12)

where Q is torque, T is thrust, D is propeller diameter, n is rotational speed, and ρ is water

density. Units are chosen so that each of the coefficients is non-dimensional. Open water

efficiency (ηO), is usually shown on plots of these parameters. It is expressed by

Q

TO K

JKπ

η2

= (7.13)


Surabaya


A plot of typical KT, KQ, and ηo characteristics as functions of J is shown by Fig. 7.2. A set

of these three curves is given for each of five different pitch ratios over the range 0.6 to 1.4

(pitch ratio = pitch/diameter). These curves represent either a family of fixed-pitch

propellers, or a single propeller whose pitch can be varied in service.

Example 7.3 Suppose that the ship discussed in previous example have a fixed pitch propeller with the diameter of 11 ft. Use Fig. 7.2 to estimate the open water efficiency of the propeller and determine the rpm of propeller. From previous example we have R = 75,000 lb V = 33.78 ft/s t = 0.19 w = 0.29 ρ = 1.9903 lb-sec2/ft4 We may calculate thrust of the propeller by employing equation (7.5)

lbs 92,592)19.01(

000,75)1(

=−

=−

=t

RT

Speed of advance of water flowing through the propeller is Va = (1 – w) V = ( 1 – 0.29) (33.78 ft/s) = 23.98 ft/s Although we have designed the speed of ship at 20 knots, but we know that the ship may not sail at full speed all time. To obtain the open water efficiency of the propeller we have to try several operating condition of the propeller including the operation when the ship sails at full speed. Since the operational condition of the propeller is not constant (n is not fixed), then we can not obtain thrust coefficient (Kt) directly and so can the speed of advance coefficient J. We will construct Kt vs J curve for several propeller operating conditions, and plot it in Fig. 7.2. Since n is not fixed, then we will find the values of Kt by using (Kt/J2)xJ2.

0.668798.231199.1

592,92 22222

22

242 =××

==⋅=aa

t

VDT

VDn

nDT

JK

ρρ

The following table shows the value of (Kt/J2)xJ2 for several values of J.

J (Kt/J2) x J2 0,40 0,107 0,50 0,167 0,60 0,241 0,70 0,328 0,80 0,428 0,90 0,542

We will plot the values listed in the table above in Kt vs J curve as shown in Fig. 7.3.


Surabaya


Fig. 7.3

Plotting of (Kt/J2)J2 v.s J in Kt vs. J curve From Fig. 7.3 we see that Kt - J line crossed several curves at several points. As an example, if we choose propeller having pitch ratio 1.0, the line crossed Kt curve at point a. If we draw a line parallel to the vertical axis, the line crossed J-axis at point a’ and giving value of J = 0.58. The line also crossed propeller open water efficiency curve for pitch ratio 1.0 at point a” and giving value of ηo = 0.59. The following table summarizes the propeller open water efficiency for several pitch ratio.

P/D J ηo n (rps) n (rpm) 1,0 0,58 0,59 3,76 225,51724 1,2 0,65 0,58 3,35 201,23077 1,4 0,70 0,55 3,11 186,85714

Note that, we should check whether there are some cavitations problem occurred in propeller operation.

7.2.5 Delivered Horsepower

Delivered horsepower (DHP) is the power that is delivered by the shaft to the

propeller. However, there is a loss between power delivered from the shaft to the propeller.

This loss is due to the propeller efficiency in transmitting power is less than 100%.

Consequently, it could not absorb all delivered energy from the shaft. Therefore, the thrust

horsepower (THP) which is applied to the water by the propeller, is less than the delivered

horsepower (DHP). The relation between THP and DHP is expressed in the following

equation.


Surabaya


p

THPDHP η

= (7.14)

Where ηP represents the propeller efficiency. However, the available data of propeller

efficiency is the open water efficiency of the propeller, ηO. This efficiency does not cover

the presence of the hull in front of the propeller, as explained in section 7.2.4. We have to

introduce relative rotative efficiency (ηR) to take into account the presence of the hull in

front of the propeller. The relation among propeller efficiency, open water efficiency, and

relative rotative efficiency is expressed in the following equation.

ROP ηηη = (7.15)

Equation (7.14) may be written in term of ηo and ηR, that is

QPCEHPEHPTHPDHP

ROHRO

===ηηηηη

(7.16)

Where, QPC represents quasi-propulsive coefficient.

ROHQPC ηηη= (7.17)

Example 7.4 Recall example 7.1 – 7.3. Assume that relative rotative efficiency (ηR) to take into account the presence of the hull in front of the propeller is 0.99. Determine the DHP of this ship at its design speed and the quasi-propulsive coefficient of the ship. Suppose we choose the propeller having pitch ratio 1.0, then the propeller open water efficiency is 0.59. The propeller efficiency is 0.584199.059.0 =×== ROP ηηη The quasi-propulsive coefficient is 0.66587499.059.014.1 =××== ROHQPC ηηη The delivered horsepower (DHP) is

HP 69175841.0

4040===

RO

THPDHPηη

We may calculate DHP by using QPC, that is

HP 6917665874.04606

===QPCEHPDHP


Surabaya


7.2.6 Shaft Horsepower

Shaft horsepower (SHP) is defined as the power delivered by the engine to the

shaft after the gearing and thrust bearing. There are some power losses between thrust

bearing clutch and propeller. The losses due to there are friction between bearings and

shaft; and also between stern tube bearings and shaft. The relation between SHP and DHP

is given by the following equation.

mSB

DHPDHPSHPηηη

== (7.18)

Where,

ηB = line bearing efficiency

ηS = stern tube efficiency

ηm = mechanical efficiency

Typical values of ηB , ηS , and ηm are listed in Table 7.1.

7.2.7 Brake Horsepower

In presence of reduction gear, the power delivered by the prime mover is higher

than that of delivered by the shaft. The power delivered by the prime mover (main engine)

at its connection flange is called brake horsepower (BHP). The relation between BHP and

SHP is given by the following equation.

G

SHPBHPη

= (7.19)

Where,

ηG = gearing efficiency including thrust bearing

Typical values of ηG is listed in Table 7.1. Table 7.1

Typical value of ηB , ηS , ηG and ηm SOURCE ITEM VALUES

Thrust Bearing 0.5 % loss Single Reduction Gear 1.5 % loss Double reduction Gear 2.5 % loss

SNAME T&R Bulletin 3-27 Marine Diesel

Reversing Gears Add 1 % loss ηB x ηS Machinery Aft 2.0 % loss PNA Chapter 7 ηB x ηS Machinery Amidship 3.0 % loss

Marine engineering Chapter 9 Double Reuction Gears 1.5 – 2 % loss


Surabaya


Example 7.5 Recall example 7.1 – 7.4. Suppose that the ship will use low speed engine. To minimize investment, the owner choose one-direction engine, so that the ship will need reversing gear. The engine room will be placed in aft part of the ship. Determine the SHP and BHP of this ship at its design speed and the quasi-propulsive coefficient of the ship. Referring Table 7.1, the mechanical losses of transmission system for engine room laid in the aft part of the ship is 2%, hence the mechanical efficiency of transmission system is 98%. The SHP of the ship is

HP 705898.0

6917===

m

DHPSHPη

The losses due to presence of reduction reversing gear is 1%, therefore the gearing efficiency is 99%. The BHP of the ship is

HP 713099.0

7058===

Gm

SHPBHPη

Note that, the BHP found in this calculation is a tentative one. It is because we do not have chosen the main engine yet. Hence we do not know exactly what the rpm of the engine is. Once we choose the engine, we should iterate our calculation back to obtain best propeller efficiency at given engine rpm. Therefore engine and propeller is perfectly match to propel the ship at design state.

7.3 REVIEW ON PROPELLER CHARACTERISTICS

An engine supplies torque to a propeller at some rotational speed. Understanding

of combined engine-propeller behavior therefore requires that the propeller characteristics

be expressed as a torque - speed relationship, or as a power – speed relationship. A

deviation of these from KT and KQ plots is easily accomplished with the aid of three

assumptions:

1. hull resistance is proportional to thrust

2. resistance (and hence thrust) is proportional to hull speed squared

3. hull speed is proportional to propeller rotational speed.

The last of these requires that J remain constant as speeds change; if so, then KQ is likewise

constant, and torque in consequence must be proportional to n2. The propeller torque

characteristic is therefore a parabola, torque proportional to n2 also. Since power is the

product of torque and rotational speed, the power characteristic is a cubic curve. These

statements constitute the “propeller law”, which may be summarized as follows:

1. Propeller rpm is proportional to ship speed.

2. Propeller torque is proportional to square of propeller rpm.


Surabaya


3. Propeller power is proportional to the cube of propeller rpm.

This law is useful for discussions of engine-propeller behavior, but it is not an

accurate description of all propulsion situations. For example, its assumptions are highly

unrealistic for planning hulls, and are only fair approximation to actuality for displacement

hulls driven a high speed - length ratios.

The propeller law does not cover transient situations, in which propeller speed and

hull speed deviate radically from the assumed proportionality, nor does it cover changes

from one steady-state condition to another. As examples, a change in draft, a change in hull

roughness, or a change in sea state will alter the proportionality between speed squared and

resistance. Figure 7.3 illustrates the effect of resistance proportionality changes. Three

curves of propeller power vs. rpm are shown, for resistances in the ratio 1.0, 1.5, and 2.0 (at

any rpm, the resistances have these ratios) for a typical propeller. Each curve is a cubic, and

so is obeying the propeller law, but their relative positions are found by manipulation of the

KT, KQ and J characteristics, and not by any rule of the propeller law.

This law also does not account for pitch changes, for such a change also alters the

proportionality between speed of hull and speed of propeller. Figure 7.4 illustrates the

effect of pitch changes for a typical propeller Three curves of propeller power vs. rpm are

shown, for pitch ratios of 1.0, 1.2, and 1.4. Each curve is a cubic, and is individually in

accord with the propeller law, but once again, the difference among curves has been found

by additional means.

Note the similarity between Figure 7.4 and Figure 7.5. Each is a plot of three cubic

power-rpm characteristics, the first figure for different resistances, the other for different

pitch ratios. It appears that a change in pitch can offset a change in resistance. For example,

when a power-rpm curve shifts to the left because worsened sea state has increased

resistance, it can be shifted rightward to its original position by a decrease in pitch (if the

pitch can be changed, of course). This does not imply that all will be the same, for

increased resistance will decrease ship speed for a given effective power, and the pitch

change may alter propeller efficiency.


Surabaya


Fig. 7.4

Propeller power vs. speed (rpm) showing effect of increased resistance

Fig. 7.5

Propeller power vs. speed (rpm) showing effect of pitch ratio

7.4 REVIEW ON ENGINE-PROPELLER MATCHING

The matching of engine and propeller is an application of the principle of

conservation of energy. The principle is like this: power produced by the engine must equal

the power absorbed by the load, in this case propeller is the load. In a simple application,

this statement comes close to being trivial and may be intuitively obvious.

Figure 7.6 is an elementary plot of power-speed characteristics for a propeller and

its driving engine. The operating point is the intersection of the two heavy lines, this being

the only place on the plane where power absorbed by the propeller equals that produced by

the engine at a common rpm. (The small amount of power absorbed by bearings and seals

between the two units is neglected here.)

The preceding section has shown that the propeller power-speed curves are

affected by pitch ratio, with each pitch ratio being represented by a unique curve. Figure

7.7 now demonstrates how this affects the engine. This figure is, essentially, Figure 7.6

modified by the addition of a propeller line for a higher and a lower pitch (diameters the

same).


Surabaya


The pitch ratio that produces the perfect ‘match” of engine to propeller may not be

the one that gives the highest possible propeller efficiency. This situation is encountered in

he design of merchant ships with direct connected engines, most often and most seriously

when the ship requires high power at modest speed, typically bulk carriers of 100,000 dead

weight tons and above. A typical case might find best propeller design requiring 80 rpm,

with low speed engines of suitable power rated at perhaps 110 rpm. The usual solution is to

favor the engine, thereby sacrificing some propeller efficiency.

Turbines and medium speed diesels drive the propeller through reduction gears,

and since the ship designer has some freedom in selecting the gear reduction ratio, this

propeller matching problem can usually be solved by selecting a ratio that allows best rpm

for both engine and propeller. The proponents of these engine types cite this freedom as

an advantage over the low speed engine, claiming propulsive efficiency differences of as

much as 5 percent in typical chip designs.

Fig. 7.6

Basic concept of engine-propeller matching

Fig. 7.7

The effect of engine power capability of propeller pitch too high and too low


Surabaya


Example 7.6 Recall example 7.5. Suppose that the owner decided to buy an engine having MCR (maximum continuous rating) 9000 HP at 225 rpm. The engine is so selected that it will cover additional resistance in the future and to prepare the worst weather during the journey of the ship. Determine the new propeller rpm. From initial calculation we already have that the ship design speed is 20 knots. The BHP of engine at the design speed is 7130 HP, propeller rpm is 225 with the pitch ratio of 1.0. Since the MCR of engine is 9000 HP at 225 rpm, then we have to find the engine rpm so that the engine will deliver power at 7130 HP. The propeller is directly connected to the main engine, therefore the engine rpm is equal to the propeller rpm. Propeller law states that propeller power is proportional to the cube of propeller rpm, therefore

3

//

⎟⎠⎞

⎜⎝⎛===

NNo

BHPBHPo

BHPBHPo

DHPDHPo

rGmr

Gm

r ηηηη

where subscripts o represent MCR and maximum rpm of the engine while subscripts r represent rated power of the engine at rated rpm of the engine.

3225

71309000

⎟⎠⎞

⎜⎝⎛=

N

N = 208 rpm

Example 7.7 Figure 7.8 shows typical gas turbine characteristics presented in Speed vs. rpm curve. The engine has three operational conditions, they are maximum intermittent, maximum continuous, and normal operational mode respectively. If this engine will be installed in a ship as prime mover, explain how to obtain operating point of the engine. First, we have to construct propeller characteristic curve that is equal to the engine characteristic. The engine characteristic is plotted in power vs. rpm plane and so is the propeller characteristic. Since the engine characteristic curve is already available, we may construct propeller characteristic curve in the same plane as shown in Fig. 7.8. Note that, we may use propeller law to construct propeller characteristic. But, we must remember that the propeller law does not account for pitch changes. Therefore, we may construct propeller characteristic by combining propeller law and KT, KQ and J characteristics of the propeller. Please also bear in mind, that the engine may rotate at high rpm. Consequently, we may need reduction gear before we transmit the power from main engine to the propeller. Hence, we have to make appropriate conversion of propeller rpm so that it will be equal to the engine rpm. Suppose that we have been able to construct propeller characteristic and plot it together with engine characteristic as shown in Fig. 7.8. The intersection between engine characteristic and propeller characteristic is the operating point of the engine (engine rating). From fig. 7.8, we may have two possible operating point. If we will operate the engine at normal operational mode, then point A is the operating point of the engine. The engine will rotate at about 3800 rpm and it delivers about 35,000 horsepower. But, in some cases we may operate the engine at its maximum continuous rating. If this is the case then point B is the operating point of the engine. The engine will rotate at about 4000 rpm and it delivers about 41,000 horsepower.


Surabaya


Fig. 7.8

Illustration of gas turbine – propeller matching

Note that, several factors may spoil easy perfection and thereby call for some effort

by the designer in attaining the best possible solution under the circumstances. In gas

turbine engines, the problems that might arise from gas turbine – propeller matching are:

the availability of only limited number of sizes of gas turbine engine

the relative cheapness of stock reduction gears as opposed to special design

the limited number of these stock reduction gear ratios

the general fact that the higher reduction gear ratio –whether stock or special – the

higher the cost of the gear.

7.5 DESIGN FOR RESISTANCE CHANGE

The effect of resistance increases on propeller power characteristics has been noted

in an earlier section, and illustrated by Figure 7.4. The consequence to the engine is shown


Surabaya


by Figure 7.9. Presumably the propeller has been selected to allow the maximum engine

output under smooth-water, clean-bottom conditions. But the inevitable increases in

resistance occur in service, shifting the propeller curve leftward, with the consequent loss in

engine power capability indicated by the figure. Comparison with Figure 7.7 shows that the

effect is the same as that of an over-pitched propeller under smooth water conditions. And

this observation suggests a solution : the propeller should be selected slightly under-pitched

that the resistance increase will move the propeller curve into the desired position rather

than out of it. Although the magnitude of the resistance increase cannot be predicted

accurately, and indeed it continuously changes, underpitching is the usual design response

to expected resistance increases.

Fig. 7.9

The effect on engine power capability of a hull resistance increase

7.6 DESIGN FOR COMBINED ENGINE

The use of multiple engines connected to the same propeller shaft is common. It

may be done merely to obtain the total power needed from available engines, or one of a

pair may be a cruising engine, that is, one much smaller than the other. The purpose in the

latter case is to improve part-load efficiency, when the vessel is at cruising speed, the small


Surabaya


engine provides all of propulsion power, operating at its rating. With multiple engines, the

torque and power characteristics are the sum of the characteristics of the individual

engines.

Fig. 7.10

Basic CODOG-propeller matching

CODOG

A combined diesel or gas turbine (CODOG) plant may be a preferred choice for

diverse service requirements such as those associated with coast guard cutters and small

naval combatants. In these cases a relatively small diesel engine (the cruise engine) would be

used during the low-speed, high-endurance cruising mode, to take advantage of the

economy of diesel engine; to take advantage of the economy of a diesel engine; or, for

maximum-power requirements, the higher-power gas turbine (the boost engine) would be

brought into line. A mechanism such as an overriding clutch would be used to ensure

either diesel or the gas turbine, but not both, drive the propeller.

While the fuel economy and endurance of the gas turbine would be less than those

of the diesel, these would be secondary considerations because of the gas turbine’s

infrequent use. CODOG plants make it possible to minimize the operating hours on the


Surabaya


large gas turbine, which is required only for high power requirements, and the presence of

two prime movers provides a degree of redundancy. These features are advantageous in

many practical situations, and CODOG plants are commonly used.

Figure 7.10 shows basic CODOG-propeller matching. As explained before,

CODOG plant has two operational modes, they are cruising and boosting mode. In

cruising mode, the ship will use diesel engine as prime mover. Therefore, the problem in

cruising mode is to make diesel engine and propeller perfectly match. Since the diesel

engine is designed for cruising mode, then the rated RPM for diesel engine is lower than

that of in boosting mode. In boosting mode, the ship will use gas turbine engine as prime

mover. Therefore, the problem in cruising mode is to make gas turbine engine and

propeller perfectly match.

Figure 7.10 also shows that that the diesel and gas turbine engines cannot both be

on the line at the same time. This is a handicap, since it means that the propulsion plant

cannot take advantage of the output for maximum speed. The reason is, if the engine-

propeller matching is such that each engine operates at its rated rpm, then the sum of

either power would move the equilibrium point to a higher rpm. This will overspeed both

engines.

CODAG

To overcome overspeeding problem in CODOG power plant, we may match the

propeller to the sum of the power of engines. This can be achieved by designing propeller

with higher pitch so that rated rpm is not reached until the rated power of both engines is

applied. This is a basic principle of CODAG (combined diesel and gas turbine). Figure 7.11

shows basic CODAG-propeller matching.

Like CODOG plant, a relatively small diesel engine (the cruise engine) would be used

during the low-speed or in cruising mode. But for boosting mode, both gas turbine and diesel

engines are brought into line together. With both engines required to be in operation to develop

full power, the coordination of the two dissimilar types of power plants presents complex

control problems that limit the practical significance of CODAG plants.

However, CODAG plant also entails problem like CODOG plant. As illustrated in

Fig. 7.11, the diesel must be matched to the same propeller but when it is running alone in


Surabaya


the cruising mode, the intersection of its characteristic and the propeller characteristic is at

much lower rpm, and this engine cannot develop its rated power.

Fig. 7.11

Basic CODAG-propeller matching

We may come to the conclusion that if two engines are to be used to drive a

common propeller shaft, a propeller pitch that is optimal when both are driving is poor

match when only one is driving, and vice versa The solution to the dilemma is a propeller

whose pitch can be changed in service, and the controllable pitch propeller (CPP) is

frequently used.

7.7 DESIGN FOR AUXILIARY LOAD

Suppose that if there is an additional load other than propeller in propulsion plant.

The loads may be generators, hydraulic pumps or others. The power required for these


Surabaya


equipments obviously subtracts from that available for propulsion and should be a factor in

the engine-propeller matching process. If the power is quite small compared to propulsion

power, the designer may safely neglect it, but if the auxiliary power is 20% ( a rough guide

only) or more of rated propulsion power, he or she should analyze its effect on the best

match.

Fig. 7.12

Basic engine-propeller with auxiliary load matching

Suppose that a gas turbine engine rated at certain horsepower drives a fixed pitch

propeller. A hydraulic system is used to drive certain deck machinery, and the system pump

is driven by the propulsion engine. This pump may absorb power that is a significant

fraction of propulsion power. The designer is consequently concerned with the rpm and

power that will remain available for propulsion when the pump is running.

The technique advised for determining the operating point of s complex system of

loads and drivers is to lump them into a single driver and a single load. The first step is,

therefore, to find a load characteristic for the piping, and head flow for the pump with rpm

as a parameter. This curve is then added to the propeller curve to represent the total load as

imposed on engine. Equilibrium occurs where this curve crosses the engine curve. Figure

7.12 shows the technique explained above. If there is no auxiliary load, the nominal

propulsive power is laid in point A. But if there is additional auxiliary load exist on

propulsion plant, then the engine will deliver power to the propeller and the auxiliary load

at the rated power shown by point B. The power taken to drive the pump is subtracted


Surabaya


from the propulsive power. Hence, the presence of auxiliary load will shift down the actual

propulsive power is to point C.

Please bear in mind, if the auxiliary load has been taken into account during the

engine-propeller matching process, the intersection between total load and engine power

characteristics lies below the nominal propulsive power. This will lead to the power losses

due to additional load at about 2% - 5%. The loss would be more serious if the engine were

a diesel engine. Its intersection lies about 15% below nominal propulsive power, as shown

in Fig. 7.12.

7.8 SUMMARY

This module has reviewed basic engine-propeller matching. The discussion is

started by reviewing how to estimate power required by a ship. The power estimation of

the ship starts to calculate effective horse power (EHP) up brake horse power (BHP) of

main engine is obtained. Numerical examples regarding to the power estimation of the ship

are also presented.

This module also reviewed basic propeller characteristic and propeller law. Basic

engine – propeller matching is then discussed by considering both engine and propeller

characteristics. Discussions on applying propeller law in engine – propeller matching are

also presented. The influence of resistance change, different propeller pitch, auxiliary load,

and combined engine have also been discussed.


Surabaya


PROBLEMS

1. A ship is designed to sail at 26 knots. From the towing tank test data, it is known that the resistance of this ship is 165,900 lbs. Determine the EHP of this ship at its design speed.

2. Recall problem no.1. Assume that thrust deduction and wake fraction are 0.095 and 0.04 respectively. Determine the THP of this ship at its design speed.

3. Suppose that the ship discussed in previous problem have a fixed pitch propeller with the diameter of 11 ft. Use Fig. 7.2 to estimate the open water efficiency of the propeller and determine the rpm of propeller.

4. Recall problems 1 –3. Assume that relative rotative efficiency (ηR) to take into account the presence of the hull in front of the propeller is 0.99. Determine the DHP of this ship at its design speed and the quasi-propulsive coefficient of the ship.

5. Recall problem 1 – 4. Suppose that the ship will use low speed engine. To minimize investment, the owner choose one-direction engine, so that the ship will need reversing gear. The engine room will be placed in aft part of the ship. Determine the SHP and BHP of this ship at its design speed and the quasi-propulsive coefficient of the ship.

6. Explain briefly, what propeller law is! 7. Identify the problems that might arise from engine – propeller matching! 8. Recall problem 5. Select the appropriate engine for this typical application. Determine

the MCR at rated rpm of the engine. Consider also additional resistance in the future and to prepare the worst weather during the journey of the ship. Determine the new propeller rpm.

9. Explain how resistance change will affect in engine-propeller matching process. 10. Explain how propeller pitch will affect in engine-propeller matching process. 11. Explain how to assess combined engine-propeller matching process.

1 10 Priyanta Notes on Powering

Documents

Transcript of 1 10 Priyanta Notes on Powering