We now turn our attention to a particularly important and interesting example of model testing in which complete similarity is impossible to achieve, and we shall examine the methods by which the difficulty has been surmounted. The testing of models is the only way at present known of obtaining a reliable indication of the power required to drive a ship at a particular speed, and also other important information.
Any solid body moving through a fluid experiences a drag, that is, a resist- ance to its motion. Part of the drag results directly from the viscosity of the fluid: at the surface of the body the fluid moves at the same velocity as the surface (the condition of no slip at a boundary) and thus shear stresses are set up between layers of fluid there and those farther away. This part of the total drag due to resolved components of surface shear stresses is usually termed
skin friction drag. In addition, viscous effects cause the distribution of pres-
sure round the body to be different from that to be expected in an inviscid fluid, so providing another contribution to the total drag force. This second component of drag – due to the resolved components of normal pressure – is usually known in an aeronautical context (see Section 8.8) as form drag, but in the nautical world is often referred to as eddy-making resistance.
These two types of resistance – the skin friction and eddy-making resist- ance – are experienced by any solid body moving through any fluid. A ship, however, is only partly immersed in a liquid, and its motion through liquid
Ship resistance 183
gives rise to waves on the surface. The formation of these waves requires energy, and, since this energy must be derived from the motion, the ship experiences an increased resistance to its passage through the liquid.
Waves on the surface of a liquid may be of two kinds. Those of the first type are due to surface tension forces, and are known as capillary waves or ripples; they are of little importance except for bodies that are small in size compared with the waves. In the case of ship resistance it is waves of the second type that are important. These result from the action of gravity on the water that tends to accumulate around the sides of the hull. Usually two main sets of waves are produced, one originating at the bow and the other at the stern of the ship. These both diverge from each side of the hull, and there are also smaller waves whose crests are perpendicular to the direction of motion.
In the formation of waves, some water is raised above the mean level, while some falls below it. When particles are raised work must be done against their weight, so gravity, as well as viscosity, plays a part in the resistance to motion of a ship. In a dimensional analysis of the situation the quantities considered must therefore include the gravitational acceleration g. The complete list of relevant quantities is therefore the total resistance force F, the velocity of the ship u, the dynamic viscosity of the liquidµ, its density , some characteristic length l to specify the size of the ship (the overall length, for example) and g. As already indicated, surface tension forces are negligible (except for models so small as to be most suitable as children’s bath-time toys). We assume that the shape is specified, so that quoting a single length is sufficient to indicate all other lengths for a particular design. The geometric similarity of a model should extend to the relative roughness of the surfaces if complete similarity is to be achieved. (Usually the depth of water is sufficiently large compared with the size of the ship not to affect the resistance. Where this is not so, however, the depth of the water must be included among the lengths that have to satisfy geometric similarity. Likewise, the distance from a boundary, such as a canal bank, might have to be considered.) The most suitable form of the result of dimensional analysis is
F= u2l2φ ul µ , u2 lg that is F= u2l2φ{Re, (Fr)2} (5.19) The total resistance therefore depends in some way both on the Reynolds number and on the Froude number. We remember that dimensional analysis tells us nothing about either the form of the functionφ{ }, or how Re and
Fr are related within it. For complete similarity between a prototype and its
model the Reynolds number must be the same for each, that is
puplp µp =
mumlm µm
184 Physical similarity and dimensional analysis
and also the Froude number must be the same, that is,
up (lpgp)1/2 =
um (lmgm)1/2
(5.21) Equation 5.20 gives um/up= (lp/lm)(νm/νp) where ν = µ/, and eqn 5.21
gives um/up = (lm/lp)1/2 since, in practice, gm cannot be significantly dif-
ferent from gp. For testing small models these conditions are incompatible;
together they require(lp/lm)3/2= νm/νpand, since, both model and proto-
type must operate in water, lmcannot be less than lp. (There is no practicable
liquid that would enableνmto be much less thanνp.) Thus similarity of vis-
cous forces (represented by Reynolds number) and similarity of gravity forces (represented by Froude number) cannot be achieved simultaneously between model and prototype.
The way out of the difficulty is basically that suggested by Froude. The
assumption is made that the total resistance is the sum of three distinct
parts: (a) the wave-making resistance; (b) skin friction; (c) the eddy-making resistance. It is then further assumed that part (a) is uninfluenced by viscosity and is therefore independent of the Reynolds number; also that (b) depends only on Reynolds number. Item (c) cannot be readily estimated, but in most cases it is only a small proportion of the total resistance, and varies little with Reynolds number. Therefore it is usual to lump (c) together with (a).
These assumptions amount to expressing the function of Re and Fr in eqn 5.19 as the sum of two separate functions,φ1(Re) + φ2(Fr). Now the
skin friction, (b), may be estimated by assuming that it has the same value as that for a thin flat plate, with the same length and wetted surface area, moving end-on through the water at the same velocity. Froude and others have obtained experimental data for the drag on flat plates, and this inform- ation has been systematized by boundary-layer theory (see Chapter 8). The difference between the result for skin friction so obtained and the total res- istance must then be the resistance due to wave-making and eddies. Since the part of the resistance that depends on the Reynolds number is separately determined, the test on the model is conducted at the corresponding velocity giving equality of the Froude number between model and prototype; thus dynamic similarity for the wave-making resistance is obtained.
Example 5.3 A ship 125 m long (at the water-line) and having a
wetted surface of 3500 m2is to be driven at 10 m· s−1 in sea-water. A model ship of 1/25th scale is to be tested to determine its resistance.
Solution
The velocity at which the model must be tested is that which gives dynamic similarity of the wave-making resistance. That is, the Froude number of prototype and model must be the same.
um (glm)1/2 = up (glp)1/2 ∴ um= up lm lp 1/2 =10 m√· s−1 25 = 2 m · s −1
Ship resistance 185
To be tested, the model must therefore be towed through water at 2 m· s−1. There will then be a pattern of waves geometrically similar to that with the prototype.
Suppose that, at this velocity, the total resistance of the model (in fresh water) is 54.2 N and that its skin friction is given by 12u2mAm× CF, where Amrepresents the wetted surface area of the model and CF
its mean skin-friction coefficient which is given by
CF=
0.075
(log10Re− 2)2
This is a widely used empirical formula for ships with smooth surfaces. (Model ships are usually constructed of, for example, polyurethane so that modifications to the shape may easily be made.)
The Reynolds number Re in the formula is based on the water- line length, which for the model is(125/25)m = 5 m. The kinematic viscosity of fresh water at, say, 12◦C is 1.235 mm2· s−1and so Re for the model is 2 m· s−1× 5 m 1.235× 10−6m2· s−1 = 8.10 × 106 Hence CF= 0.075 (4.9085)2 = 3.113 × 10 −3
and the skin-friction resistance is 1 2× 1000 kg · m −2(2 m · s−1)2 3500 m2 252 3.113× 10−3= 34.87 N However, the total resistance was 54.2 N, so (54.2 − 34.87) N = 19.33 N must be the wave resistance+ eddy-making resistance of the model. This quantity is usually known as the residual resistance.
Now if the residual resistance is a function of Froude number but not of Reynolds number, eqn 5.19 becomes, for the residual resistance
only, Fresid= u2l2φ(Fr). Therefore (Fresid)p (Fresid)m =
pu2plp2 mu2mlm2
(Since Fr is the same for both model and prototype the functionφ(Fr) cancels.) Hence (Fresid)p= (Fresid)m p m u p um 2l p lm 2 = (Fresid)m p m lp lm lp lm 2 = 19.33 1025 1000 253N= 3.096 × 105N
186 Physical similarity and dimensional analysis
For the prototype ship Re is 10 m· s−1 × 125 m/(1.188 × 10−6m2· s−1) = 1.025 × 109 (using a value of kinematic viscosity for sea-water of standard temperature and salinity). For the proto- type, which, even in new condition, has a rougher surface than the model, CFis given by 0.075 (log10Re− 2)2+ 0.0004 = 0.075 (7.0220)2 + 0.0004 = 1.921 × 10 −3
However, the magnitude of the skin friction is not easy to determine accurately for the prototype ship because the condition of the surface is seldom known exactly. The surface of a new hull is quite smooth, but after some time in service it becomes encrusted with barnacles and coated with slime, so the roughness is rather indeterminate. The naval architect has to make allowance for these things as far as possible. For, say, six months in average conditions CFwould be increased by about
45%. Taking CF in the present example as 1.45× 1.921 × 10−3 =
2.785× 10−3 we then find that the ship’s frictional resistance is
1 2pu
2
pApCF=12× 1025 kg · m−3(10 m · s−1)23500 m2× 2.785 × 10−3
= 5.00 × 105N
The total resistance of the prototype is therefore (3.096 + 5.00)105N= 810 kN.
2
Example 5.4 A production torpedo has a maximum speed of
11 m· s−1 as originally designed. By introducing a series of design changes, the following improvements were achieved:
1. the cross-sectional area was reduced by 12%; 2. the overall drag coefficient was reduced by 15%; 3. the propulsion power was increased by 20%.
What was the maximum speed of the redesigned torpedo?
Solution
Use suffix 1 to denote the original design, and suffix 2 to denote the revised design. Then
A2= A1− 0.12A1= 0.88A1 CD2= CD1− 0.15CD1= 0.85CD1
P2= P1+ 0.20P1= 1.20P1
Since
Ship resistance 187 it follows that P2 P1 = D2V2 D1V1 = 1 2V2ACD 2V2 1 2V2ACD 1V1 = V2 V1 3A 2 A1 CD2 CD1 Hence V2 V1 = P 2A1CD1 P1A2CD2 1/3 and V2= 11 m · s−1× 1.2 (0.88)(0.85) 1/3 V2= 12.88 m · s−1 2
We may note that, for a submarine travelling at a sufficient depth below the surface that no surface waves are formed, the total resistance is that due only to skin friction and eddy-making.
Details of methods used for calculating the skin friction vary somewhat, but all methods are in essence based on the assumption that the drag is equal to that for a flat plate of the same length and wetted area as the hull, and moving, parallel to its own length, at the same velocity. This assumption cannot be precisely true: the fact that eddy resistance is present indicates that the flow breaks away from the surface towards the stern of the vessel, and thus there must be some discrepancy between the skin friction of the hull and that of a flat plate. Indeed, the entire method is based on a number of assumptions, all of which involve a certain simplification of the true state of affairs.
In particular, the complete independence of skin friction and wave-making resistance is, we recall, an assumption. The wave-making resistance is in fact affected by the viscous flow round the hull, which in turn is dependent on the Reynolds number. Moreover, the formation of eddies influences the waves generated near the stern of the vessel. We may note in passing that the wave- making resistance of a ship bears no simple relation to the Froude number (and thus to the speed). At some speeds the waves generated from the bow of the ship reinforce those produced near the stern. At other speeds, however, the effect of one series of waves may almost cancel that of the other because
188 Physical similarity and dimensional analysis
the crests of one series are superimposed on the troughs of the other. As a result the graph of wave-making resistance against speed often has a sinuous form similar to that of Fig. 5.1. Nevertheless, the overall error introduced by the assumptions is sufficiently small to make the testing of models extremely valuable in the design of ships.
PROBLEMS
5.1 A pipe of 40 mm bore conveys air at a mean velocity of 21.5 m· s−1. The density of the air is 1.225 kg· m−3 and its dynamic viscosity is 1.8× 10−5 Pa· s. Calculate that volume flow rate of water through the pipe which would correspond to the same value of friction factor f if the dynamic viscosity of water is 1.12× 10−3 Pa· s. Compare the piezometric pressure gradient in the two cases.
5.2 Derive an expression for the volume flow rate of a liquid (of dynamic viscosityµ, density and surface tension γ ) over a V notch of given angleθ. Experiments show that for water flowing over a 60◦V notch a useful practical formula is Q= 0.762 h2.47 for metre-second units. What limitation would you expect in the validity of this formula? Determine the head over a similar notch when a liquid with a kinematic viscosity 8 times that of water flows over it at the rate of 20 L· s−1.
5.3 A disc of diameter D immersed in a fluid of density and vis- cosityµ has a constant rotational speed ω. The power required to drive the disc is P. Show that P = ω3D5φ(ωD2/µ). A
disc 225 mm diameter rotating at 144.5 rad· s−1 (23 rev/s) in water requires a driving torque of 1.1 N· m. Calculate the corresponding speed and the torque required to drive a sim- ilar disc 675 mm diameter rotating in air. (Dynamic viscosities: air 1.86× 10−5 Pa· s; water 1.01 × 10−3 Pa· s. Densities: air 1.20 kg· m−3; water 1000 kg· m−3.)
5.4 The flow through a closed, circular-sectioned pipe may be metered by measuring the speed of rotation of a propeller having its axis along the pipe centre-line. Derive a relation between the volume flow rate and the rotational speed of the propeller, in terms of the diameters of the pipe and the pro- peller and of the density and viscosity of the fluid. A propeller of 75 mm diameter, installed in a 150 mm pipe carrying water at 42.5 L· s−1, was found to rotate at 130 rad· s−1(20.7 rev/s). If a geometrically similar propeller of 375 mm diameter rotates at 64.5 rad· s−1 (10.9 rev/s) in air flow through a pipe of 750 mm diameter, estimate the volume flow rate of the air. The density of the air is 1.28 kg· m−3 and its dynamic vis- cosity 1.93× 10−5Pa· s. The dynamic viscosity of water is 1.145× 10−3 Pa· s.
Problems 189
5.5 A torpedo-shaped object 900 mm diameter is to move in air at 60 m· s−1 and its drag is to be estimated from tests in water on a half-scale model. Determine the necessary speed of the model and the drag of the full-scale object if that of the model is 1140 N. (Fluid properties as in Problem 5.3.)
5.6 What types of force (acting on particles of fluid) would you expect to influence the torque needed to operate the rudder of a deeply submerged mini-submarine? To investigate the oper- ation of such a rudder, tests are conducted on a quarter-scale model in a fresh-water tunnel. If, for the relevant temperat- ures, the density of sea water is 2.5% greater than that of fresh water, and the dynamic viscosity 7% greater, what velocity should be used in the water tunnel to correspond to a velocity of 3.5 m· s−1 for the prototype submarine? If the measured torque on the model rudder is 20.6 N· m, what would be the corresponding torque on the full-size rudder?
5.7 Show that, for flow governed only by gravity, inertia and pres- sure forces, the ratio of volume flow rates in two dynamically similar systems equals the 5/2 power of the length ratio. 5.8 The drag on a stationary hemispherical shell with its open, con-
cave side towards an oncoming airstream is to be investigated by experiments on a half-scale model in water. For a steady air velocity of 30 m· s−1determine the corresponding velocity of the water relative to the model, and the drag on the proto- type shell if that on the model is 152 N. (Fluid properties as in Problem 5.3.)
5.9 The flow rate over a spillway is 120 m3· s−1. What is the max- imum length scale factor for a dynamically similar model if a flow rate of 0.75 m3· s−1 is available in the laboratory? On a part of such a model a force of 2.8 N is measured. What is the corresponding force on the prototype spillway? (Viscosity and surface tension effects are here negligible.)
5.10 An aircraft is to fly at a height of 9 km (where the temper- ature and pressure are−45◦C and 30.2 kPa respectively) at 400 m· s−1. A 1/20th-scale model is tested in a pressurized wind tunnel in which the air is at 15◦C. For complete dynamic similarity what pressure and velocity should be used in the wind-tunnel? (For air at T K,µ ∝ T3/2/(T + 117).)
5.11 In a 1/100th-scale model of a harbour what length of time should correspond to the prototype tidal period of 12.4 hours? 5.12 Cavitation is expected in an overflow siphon where the head is −7 m, the water temperature 10◦C and the rate of flow 7 m3· s−1. The conditions are to be reproduced
on a 1/12th-scale model operating in a vacuum chamber with water at 20◦C. If viscous and surface tension effects may be neglected, calculate the pressure required in the vacuum chamber and the rate of flow in the model (Hint: In the absence of friction the velocity in the siphon is