CE 6451- FLUID MECHANICS AND MECHINERY

UNIT I

1. Define specific volume of a fluid and write its unit. [N/D-14]

Volume per unit mass of a fluid is called specific volume.

Unit: m3 / kg.

2. Name the devices that are used to measure the pressure of a fluid. [N/D-14]

The Device used to measuring Fluid Pressure, by the Gauges is,

Piezometer Tube

.Manometer.

3. Calculate the specific weight, density and specific gravity of 1 litre of liquid which

weighs 7 N. [A/M-15]

Solution:

Given V 1litre m

W = 7 N

4. State the concept of pressure measurement used in mechanical gauges[A/M-15]

Many techniques have been developed for the measurement of pressure and vacuum. Instruments

used to measure and display pressure in an integral unit are called pressure gauges or vacuum

gauges. A manometer is a good example as it uses a column of liquid to both measure and

indicate pressure. Likewise the widely used Bourdon gauge is a mechanical device which both

measures and indicates, and is probably the best known type of gauge.

A vacuum gauge is an absolute pressure gauge used to measure the pressures lower than the

ambient atmospheric pressure.

Other methods of pressure measurement involve sensors which can transmit the pressure reading

to a remote indicator or control system.

5. Define Centre of Pressure [M/J-16]

The center of pressure is the point where the total sum of a pressure field acts on a body, causing

a force to act through that point. The total force vector acting at the center of pressure is the value of

the integrated vectorial pressure field. The resultant force and center of pressure location produce

equivalent force and moment on the body as the original pressure field. Pressure fields occur in both

static and dynamic fluid mechanics.

6. Temperature rise, decrease viscosity in liquids but increases it in gases, why?

[M/J-16]

Viscosity is nothing but nature of fluid due to which it offers resistance to gradual deformation. In case of

liquid ,liquid molecules are bonded by inter molecular force of attraction and when we increase the

temperature , this force of attraction gets weaken causing dissociation of bond and molecules becomes

more freely to flow/move, causing decrease in viscosity of liquid ( viscosity of liquid is due cohesion of liquid

molecules i.e. tendency of similar particle to cling one another ),whereas in case of gases ,gaseous

molecules do have weak vander wall force of attraction , as we increase the temperature, energy (in the

form of heat ) transfers to gas molecules, which increase the randomness of molecules and molecules

starts colliding with each other ,retarding the motion of gases ,resulting increase in viscosity of gases.

1. A Liquid has a specific gravity of 0.72. Find its density, specific weight and also the weight per

liter of the liquid. If the above liquid is used for lubrication between a shaft and a sleeve, find the

power lost in liquid for a sleeve length of 100 mm. The diameter of the shaft is 0.5 m and the

thickness of the liquid film is 1 mm. Take the viscosity of fluid as 0.5 N-s/m2and the speed of the

shaft as 200rpm.[N/D-14]

Sol: Volume V = 1 litre = 0.001 m3

Specific gravity S = 0.72

(i) Density ρ = S * 1000 = 0.72 * 1000 = 720 kg/m3

(ii) Specific weight W = ρ*g = 720 * 9.81 = 7063.2 N/m3

(iii) Weight w = W*V = 7063.2 * 0.001 = 7.063 N

(iv) µ = 0.5 Ns/m2

Diameter D = 0.5 m

Speed of shaft N = 200 rpm

Sleeve length L = 100 mm

Thickness of oil film t = 1 mm

u = (πDN) / 60 = (π*0.5*100)/60 = 2.62 m/s

du = 2.62 m/s

dy = 1mm = 0.001 m

τ = µ (du/dy) = 0.5 * (2.62/0.001) = 1310 N/mm2

F = τ * A = 1310 * πDL = 205.77 N

Torque T = F * D/2 = 205.77 * 0.5/2 =51.44 Nm

Power lost = 2πNT/60 = (2*π*100*51.44)/60 = 538.68 W

2. If the velocity distribution over a plate is given by u = 2/3 y – y2 in which u is the velocity in metre

per second at a distance y metre above the plate, determine the shear stress at y = 0 and y = 0.15m.

Take dynamic viscosity of fluid as 8.63 poise. [N/D-14]

3. A 150mm diameter vertical cylinder rotates contacted to the another cylinder of diameter 151

mm. Both the cylinder are 250mm high. If the torque of 12 Nm is required to rotate the inner cylinder

at 100 r.p.m. determine the viscosity of the fluid in the space between the above two

cylinders.[A/M-15]

4. Determine the total pressure on a circular plate of diameter 1.5 m which is placed vertically in

water in such a way that the center of the plate is 3 m below the free surface of water. Also find the

position of centre of pressure.[A/M-15]

5. (a) (i) Differentiate

(1) Real fluids and ideal fluids

(2) Newtonian and non-Newtonian fluids

(4)

(ii) What is the difference between U – tube differential manometer and inverted U- tube differential

manometer?

(4)

(1) Real Fluid:

A fluid, which possesses viscosity, is known as real fluid. All fluids, in actual

practice, are real fluids.

Ideal Fluid:

A fluid, which is incompressible and is having no viscosity, is known as an ideal

fluid. Ideal fluid is only an imaginary fluid as all the fluids, which exist, have some

viscosity.

(2) Newtonian fluids

A Newtonian fluid's viscosity remains constant, no matter the amount of shear applied for a constant

temperature.. These fluids have a linear relationship between viscosity and shear stress.

Non-Newtonian fluids

You can probably guess that non-Newtonian fluids are the opposite of Newtonian fluids. When shear is

applied to non-Newtonian fluids, the viscosity of the fluid changes.

(3) U-Tube Manometer:

It consist a U – shaped bend whose one end is attached to the gauge point ‘A’ and other end is open to the

atmosphere. It can measure both positive and negative (suction) pressures. It contains liquid of specific

gravity greater than that of a liquid of which the pressure is to be measured.

Inverted U-Tube Manometer:

Inverted U-Tube manometer consists of an inverted U – Tube containing a light liquid. This is used to

measure the differences of low pressures between two points where where better accuracy is required. It

generally consists of an air cock at top of manometric fluid type.

6. If the velocity profile of a liquid over a plate is a parabolic with the vertex 20 cm from the plate,

where the velocity is 120 cm/sec. calculate the velocity gradients and shear stress at a distance of

0, 10 and 20 cm from the plate, if the viscosity of the fluid is 8.5 poise.

UNIT II

1. Define circulation and Write its expression. [N/D-14]

If V is the fluid velocity on a small element of a defined curve, and dl is a vector representing

the differential length of that small element, the contribution of that differential length to circulation

is dΓ:

where θ is the angle between the vectors V and dl.

2. Write Euler’s equation. [N/D-14]

Energy is the basis of Euler’s equation. Energy can neither be created nor destroyed. Euler’s

equation is derived by the use of Newton’s second law of motion.

According to Newton’s second law of motion, the time rate of change of momentum of a fluid mass in

any direction is equal to the sum of all the external forces in that direction, i.e.,

F = d(M V)/dt

where F is the force, M is the mass and V is the velocity.

3. Write the continuity equation in three dimensional differential form for compressible

fluids [A/M-15]

Conservation of mass is the basis for continuity equation. Mass can neither be created nor

destroyed. In a given control volume, the net mass rate of inflow into the control volume is equal to

the rate of change of mass.

4. State the impulse momentum principle [A/M-15]

The impulse-momentum theorem states that the change in momentum of an object equals

the impulse applied to it. The impulse-momentum theorem is logically equivalent to Newton's second

law of motion (the force law).

5. Define flow net [M/J-16]

A flow net is a graphical representation of two-dimensional steady-state groundwater flow through

aquifers. Construction of a flow net is often used for solving groundwater flow problems where the

geometry makes analytical solutions impractical.

6. Distinguish between streamline and streak line. [M/J-16]

Streamlines and streak lines are field lines in a fluid flow. They differ only when the flow changes with time,

that is, when the flow is not steady. Considering a velocity vector field in three-dimensional space in the

framework of continuum mechanics, we have that:

Streamlines are a family of curves that are instantaneously tangent to the velocity vector of the flow.

These show the direction in which a massless fluid element will travel at any point in time.

Streak lines are the loci of points of all the fluid particles that have passed continuously through a

particular spatial point in the past. Dye steadily injected into the fluid at a fixed point extends along a

streak line.

1. In a two – two dimensional incompressible flow, the fluid velocity components are given by

u = x – 4y and v= - y – 4x. Show that velocity potential exists and determine its form. Find also

the stream function. [N/D-14]

2. A ventrimeter of inlet diameter 300 mm and throat diameter 150 mm is inserted in vertical pipe

carrying water flowing in the upward direction. A differential mercury manometer connected to the

inlet and throat gives a reading of 200 mm. Find the discharge if the co-efficient of discharge of

meter is 0.98. [N/D-14]

3. Find the density of a metallic body which floats at the interface of mercury of sp. Gr 13.6 and

water such that 40% of its volume is sub-merged in mercury and 60% in water. [A/M-15]

4. An oil of specific gravity 0.8 is flowing through a horizontal venturimeter having a inlet diameter

200 mm and throat diameter 100 mm. The oil – mercury differential manometer shows a reading of

250 mm, calculate the discharge of oil through the venturimeter. Take Cd = 0.98.

[A/M-15]

5. (a) Water flows through a pipe AB 1.2m diameter at 3m/s and then passes through a pipe BC 1.5m

diameter. At C, the pipe branches. Branch CD is 0.8m in diameter and carries one- third of the flow

in AB. The flow velocity in branch CE is 2.5m/s. Find the volume rate of flow in AB, the velocity in

BC, the velocity in CD and diameter of CE. [M/J-16]

6. (b)State Bernoulli’s theorem for steady flow of an incompressible fluid. Derive an expression for

Bernoulli’s equation from first principle and state the assumptions made for such a derivation.

[M/J-16]

UNIT III

1. Sketch the shear stress and velocity distribution for laminar flow across a pipe

section.[N/D-14]

A flow is said to be laminar if Reynolds number is less than 2000 for pipe flow.

Laminar flow is possible only at low velocities and high viscous fluids. In laminar

type of flow, fluid particles move in laminas or layers gliding smoothly over the

adjacent layer.

The formula for velocity distribution is given as

u = - (¼ μ) (∂p/∂x) (R2-r2)

Where R = Radius of the pipe,

r = Radius of the fluid element

2. List the major and minor losses encountered in pipe flow. [N/D-14]

There are two types of energy losses in flow through pipes:

Major energy loss due to frictional resistance offered by the pipe; and

Minor energy losses due to sudden expansion or contraction of the pipe, bend, pipe fittings, etc.

Minor losses are very small in magnitude compared to major losses.

3. Distinguish between hydraulic gradient and energy gradient [A/M-15]

If a piezometer tube is introduced at a section in a pipe flow carrying a liquid, the liquid will rise to a

height equivalent to the pressure head at that section. The sum of pressure head and potential head

is termed as hydraulic or piezometric head. The line joining the points of hydraulic head at various

sections is termed as hydraulic grade line.

H =P/ρg+v2/2g+zP/ρg+v2/2g+z,

where,

H= total head ,

Pressure head = P/ρgP/ρg

Kinetic head = v2/2gv2/2g

Potential head = zz

These stream lines which contains all the mechanical energy (all 3 heads) would be called total

energy lines.

Now subtract the kinetic head from the total energy (H) , then the lines you would get are Hydraulic

Gradient lines (you can also say that these lines contains piezometric head )

4. What are the causes of major and minor losses in pipes [A/M-15]

There will be resistance to the fluid motion due to viscous effects of the fluid as well as due to the

internal surface roughnous of the pipe. This resistance has to be overcome for the flow to be

sustained and hence there will be energy losses taking place in a pipe flows.

5. What is hydraulic gradient line? State its applications. [M/J-16]

If a piezometer tube is introduced at a section in a pipe flow carrying a liquid, the liquid will rise to a

height equivalent to the pressure head at that section. The sum of pressure head and potential head

is termed as hydraulic or piezometric head. The line joining the points of hydraulic head at various

sections is termed as hydraulic grade line.

6. List major and minor losses of flow in pipe. [M/J-16]

There are two types of energy losses in flow through pipes:

Major energy loss due to frictional resistance offered by the pipe; and

Minor energy losses due to sudden expansion or contraction of the pipe, bend, pipe fittings, etc.

Minor losses are very small in magnitude compared to major losses.

1. A liquid of specific weight ρg= 58 lbf/ft3 flows by gravity through a 1-ft tank and a 1-ft

capillary tube at a rate of 0.15 ft3/h. Sections 1 and 2 are at atmospheric pressure. Neglecting

entrance effects, compute the viscosity of the liquid. [N/D-14]

2. At a sudden enlargement of a water main from 240 mm to 480 mm diameter, the

hydraulic gradient rises by 10 mm. Estimate the rate of flow. [N/D-14]

3. A horizontal pipe line 40 m long is connected to a water tank at one end and

discharges freely into the atmosphere at the other end. For the first 25 m of its length from the

tank, the pipe is 150 mm diameter and its diameter is suddenly enlarged to 300 mm. The height

of water level in the tank is 8 m above the centre of the pipe. Considering all losses of head

which occur, determine the rote of flow. Take f =0.01 for both sections of the pipe. [A/M-15]

4. Design the diameter of a steel pipe to carry water having kinematic viscosity v =

10-6 m2/s with a mean velocity of 1m/s. The head loss is to be limited to 5 m per 100 m length

of pipe. Consider the equivalent sand roughness height of pipe k, = 45x10-4cm. Assume that

the Darcy Weisbach friction factor over the whole range of turbulent flow can be expressed as

Where D = Diameter of pipe and Re = Reynolds number. [A/M-15]

5. The rate of flow of water through a horizontal pipe is 0.25 m3/s. The diameter of

the pipe which is 200 mm is suddenly enlarged to 400 mm The pressure intensity in the

smaller pipe is 11.772N/cm3. Determine:

(i) loss of head due sudden enlargement, (ii) pressure intensity in the large pipe,

(ii) power lost due to enlargement. [M/J-16]

6. Determine the difference in the elevations between the water surfaces in the two

tanks which are connected by a horizontal pipe of diameter 300 mm and length 400 m. The

rate of flow of water through the pipe is 300 liters/s. Consider all losses and take the value off

= 008. [M/J-16]

UNIT IV

1. What are the different methods of preventing the separation of boundary layers?[N/D-14]

Control Of Boundary Layer Separation -

The total drag on a body is attributed to form drag and skin friction drag. In some flow

configurations, the contribution of form drag becomes significant.

In order to reduce the form drag, the boundary layer separation should be prevented or delayed so

that better pressure recovery takes place and the form drag is reduced considerably. There are

some popular methods for this purpose which are stated as follows.

i. By giving the profile of the body a streamlined shape.

1. This has an elongated shape in the rear part to reduce the magnitude of the pressure gradient.

2. The optimum contour for a streamlined body is the one for which the wake zone is very narrow

and the form drag is minimum.

ii. The injection of fluid through porous wall can also control the boundary layer separation. This is

generally accomplished by blowing high energy fluid particles tangentially from the location where

separation would have taken place otherwise.

1. The injection of fluid promotes turbulence

2. This increases skin friction. But the form drag is reduced considerably due to suppression of

flow separation

The reduction in form drag is quite significant and increase in skin friction drag can be ignored.

2. Define the terms: Drag and Lift. [N/D-14]

Lift is the force that acts at a right angle to the direction of motion through the air. Lift is created by

differences in air pressure. Thrust is the force that propels a flying machine in the direction of motion.

Engines produce thrust. Drag is the force that acts opposite to the direction of motion.

3. Define – Local Coefficient of Drag [A/M-15]

It is defined as ratio of the shear stress τo to the quantity ρU2/2, which is denoted by CD*

as:

CD* = τo/(ρU2/2)

where τo is shear stress;

ρ is density of the fluid;

and U is main stream velocity.

4. State the boundary conditions for the velocity profiles in a boundary layer. (A/M - 15)

The following are boundary conditions, which must be satisfied by any velocity profile

whether it is laminar or turbulent boundary layer.

(i) At y = 0, u = 0 and du/dy has a finite value;

(ii) At y = δ, u = U; and

(iii) At y = δ, du/dy = 0

5. Define displacement thickness [M/J-16]

Displacement thickness : It is defined as the distance by which the external potential flow is

displaced outwards due to the decrease in velocity in the boundary layer.

6. When a surface will be called as hydro-dynamically smooth? [M/J-16]

When the average depth k of the surface irregularities is less than laminar sub-layer of the

surface δ is called as hydrodynamic ally smooth boundary.

The eddy which formed outside of the laminar sub-layer try to penetrate in the laminar sub-

layer boundary is called as smooth boundary.

When the average depth k of the surface irregularities is greater than laminar sub-layer of

surface δ is called as hydrodynamic ally rough boundary.

The eddy which formed outside of the laminar sub layer penetrates into the laminar sub-layer.

Such boundary is called as rough boundary.

1. A solid cylinder of 10 cm diameter and 40 cm long consists of 2 parts made of different

materials. The first part at the base is 1 cm long, and specific gravity is 6. The other part of the

cylinder is made of the material having specific gravity 0.6. State if it can float vertically in

water. [N/D-14]

2. A ship 70 m long and 10 m broad has a displacement of 19620 W. A weight of 343.35 kN is

moved across the deck through a distance of 6 m The ship is tilted through 6". The moment of

inertia of the ship at water-line about its fore and axis is 75% of M.O.I. of the circumscribing

rectangle. The centre of buoyancy is 2.25 m below water-line. Find the metacentric height and

position of centre of gravity of ship. Specific weight of sea water is 10104 N/ m3 . [N/D-14]

3. Find the displacement thickness and the momentum thickness for the velocity distribution in the

boundary layer given by u/U = 2(y/δ)-(y/δ)2. [A/M-15]

4. The time period of rolling of a ship of weight 29430 kN in sea water is 10 seconds. The centre of

buoyancy of the dip is 1.5 m below the centre of gravity. Find the radius of gravity of the ship if the

moment of inertia of the ship at the water line about fore and aft axis is 1000m4. Take specific

weight of sea water as = 10100 N/m3. [A/M-15]

5. A wooden cylinder of specific gravity 0.6 and circular in cross section is required to float in oil of

specific gravity 0.9. Find the L/D ratio for the cylinder to float with its longitudinal axis vertical in oil,

where L is the height of the cylinder and D is its diameter. [M/J-16]

6. A wooden log of 0.6m diameter and 5m length is floating in river water. Find the depth of the

wodden log in water when the sp. Gravity of the log is 0.7. [M/J-16]

UNIT V

1. Define Froude’s number and write its expression. [N/D-14]

When the forces of gravity can be considered to be the only predominant force

which controls the motion in addition to the force of inertia, the dynamic similarities

of the flow in any two such systems can be established, if the Froude number for

both the system is the same. This is known as Froude Model Law.

Fr(p) = Fr (m)

2. What are the merits of distorted models? [N/D-14]

The results in steeper water surface slopes and magnification of wave heights in

model can be obtained by providing true vertical structure with accuracy.

• The model size can be reduced to lower down the cast.

• Sufficient tractate force can be developed to produce bed movement with a small

model.

3. State Buckingham’s π-Theorem. [N/D-15]

Buckingham π-theorem states that ‘If there are n variables (both independent and dependent)

involved in a physical phenomenon and if these variables involve m fundamental dimensions (M, L,

T) then the variables can be arranged into (n - m) dimensionless terms, which are known as π-terms.

4. What are the advantages of distorted models. [A/M-15]

The results in steeper water surface slopes and magnification of wave heights in

model can be obtained by providing true vertical structure with accuracy.

• The model size can be reduced to lower down the cast.

• Sufficient tractate force can be developed to produce bed movement with a small

model.

5. Define – Reynolds Number [M/J-16]

Reynold’s number is defined as the ratio between the inertial forces of the flowing fluid and the

viscous forces of the fluid. Mathematically, it is given as:

Re = Fi/Fv = ρ V L/ µ

6. What do you know by the term distorted models? [M/J-16]

Distorted model is one which is not geometrically similar to its prototype. Such a model uses different

Scale Ratios for different dimensions. For example, Lr1 = Lp/Lm; Lr2 = Bp/Bm; and Lr3 = Hp/Hm.

1. Obtain an expression in non-dimensional form for the pressure gradient in a horizontal pipe of

circular cross-section. Show how this relates to the familiar expression for frictional head loss.

[N/D-14]

2. Describe briefly the types of forces in moving fluid and the importance of three types of

similarity. [N/D-14]

Forces encountered in flowing fluids include those due to inertia, viscosity, pressure,

gravity, surface tension and compressibility. These forces can be written as follows;

Inertia force: m.a V (dV/dt) V 2 L2

Viscous force: A A du/ dy V L

Pressure force: pA pL2

Gravity force: mg g L3

Surface tension force: L

Compressibility force: Ev A Ev L

3. The tip deflection δ of a cantilever beam is a function of tip load W, beam length l, second

moment of area I and Young’s modulus E. Perform a dimensional analysis of this problem.[A/M-15]

4. Explain briefly the steps involved in Rayleigh’s method of dimensional analysis and also brief the

types of similarities existing between the prototype and its model. [A/M-15]

Modeling and Similitude

A “model” is a representation of a physical system which is used to predict the

behavior of the system in some desired respect. The physical system for which the

predictions are to be made is called “prototype”. Usually, a model is smaller than the

prototype so that laboratory experiments/studies can be conducted. It is less expensive

to construct and operate. However, in certain situations, models are larger than the

prototype e.g. study of the motion of blood cells whose sizes are of the order of

micrometers. “Similitude” is the indication of a known relationship between a model

and prototype. In other words, the model tests must yield data that can be scaled to

obtain the similar parameters for the prototype.

Theory of models: The dimensional analysis of a given problem can be described in

terms of a set of pi terms and these non-dimensional parameters can be expressed in

functional forms;

1 2 , 3 ,..........n (6.2.9)

Since this equation applies to any system, governed by same variables and if the

behavior of a particular prototype is described by Eq. (6.2.9), then a similar

relationship can be written for a model.

1m 2m , 3m ,..........nm (6.2.10)

The form of the function remains the same as long as the same phenomenon is

involved in both the prototype and the model. Therefore, if the model is designed and

operated under following conditions,

2m 2 ; 3m 3............ and

Then it follows that

nm n (6.2.11)

1 1m (6.2.12)

Eq. (6.2.12) is the desired “prediction equation” and indicates that the measured value

of 1m obtained with the model will be equal to the 1 for the

prototype as long as the other pi terms are equal. These are called “model design

conditions / similarity requirements / modeling laws”.

Flow Similarity

In order to achieve similarity between model and prototype behavior, all the

corresponding pi terms must be equated to satisfy the following conditions.

Geometric similarity: A model and prototype are geometric similar if and only if all

body dimensions in all three coordinates have the same linear-scale ratio. In order to

have geometric similarity between the model and prototype, the model and the

prototype should be of the same shape, all the linear dimensions of the model can be

related to corresponding dimensions of the prototype by a constant scale factor.

Usually, one or more of these pi terms will involve ratios of important lengths, which

are purely geometrical in nature.

Kinematic similarity: The motions of two systems are kinematically similar if

homogeneous particles lie at same points at same times. In a specific sense, the

velocities at corresponding points are in the same direction (i.e. same streamline

patterns) and are related in magnitude by a constant scale factor.

Dynamic similarity: When two flows have force distributions such that identical types

of forces are parallel and are related in magnitude by a constant scale factor at all

corresponding points, then the flows are dynamic similar. For a model and prototype,

the dynamic similarity exists, when both of them have same length-scale ratio, timescale

ratio and force-scale (or mass-scale ratio).

In order to have complete similarity between the model and prototype, all the

similarity flow conditions must be maintained. This will automatically follow if all

the important variables are included in the dimensional analysis and if all the

similarity requirements based on the resulting pi terms are satisfied. For example, in

compressible flows, the model and prototype should have same Reynolds number,

Mach number and specific heat ratio etc. If the flow is incompressible (without free

surface), then same Reynolds numbers for model and prototype can satisfy the

complete similarity.

5. A prototype gate valve which will control the flow in a pipe system conveying paraffin is to be studied in a model. List the significant variables on which the pressure drop across the valve would depend. Perform dimensional analysis to obtain the relevant non-dimensional groups. A 1/5 scale model is built to determine the pressure drop across the valve with water as the working fluid. (a) For a particular opening, when the velocity of paraffin in the prototype is 3.0 m s–1 what

should be the velocity of water in the model for dynamic similarity?

(b) What is the ratio of the quantities of flow in prototype and model?

(c) Find the pressure drop in the prototype if it is 60 kPa in the model.

(The density and viscosity of paraffin are 800 kg m–3 and 0.002 kg m–1 s–1 respectively. Take

–6 m2 s–1). [M/J-16]

6. A ship model of scale 1/50 is towed through sea water at a speed of 1 m/s. A force of 2 N is

required to tow the model. Determine the speed of the ship and the propulsive force on the ship. If

the prototype is subjected to wave resistance only. [M/J-16]