Reservoir >> Suction Pipeþþ Flow Is Steady

3 2 7.33 Water is flowing at a rate of 0.25m /s, and it is assumed that hL=1.5V /2g from the reservoir to the gage, where V is the velocity in the 30-cm pipe. What power must the pump supply?

p = 100 kPa 2 Assumptions Elevation = 10m

Reservoir >> suction pipe D = 30 cm

V1  0 flow is steady

1 Flow is turbulent  Elevation = 6m  1   2  1.0 water 40 cm Given T = 10o C

Flow of water   = const Elevation = 2m

1-D energy equation with a pump present:

p V 2 p V 2 h  1   1  z  2   2  z  h p  1 2g 1  2 2g 2 lt

hp = pump head

3 p1g = 0, p2g = 100 kPa, z1 = 6m, z2 = 10m,  = 9.81 kN/m

100 V 2 V 2 1.25 h   2  (10  6)  1.5 2  14.2  V 2 p 9.81 2g 2g 9.81 2

Q 0.25 Q  V A  V   2 2 2 2 A2  (0.3)     4 

V2  3.537 m / sec

1.25 h  14.2  (3.537) 2  15.79 m p 9.81

˙ pump power:Wp   Q hp  9.81(0.25)(15.79)

- 84 - ˙ Wp  38.72 kW

7.14 Water flow from a pressurized tank as shown. The pressure in the tank above the water surface is 100 kPa gage, and the water surface level is 10m above the outlet. The water exit velocity is 9m/s. The head loss in the systerm varies as

V 2 h  K L L 2g

where KL is the head-loss coefficient. Find the value for KL.

Assumptions Air under pressure Tank >> pipe 1

V1  0 water d Flow is turbulent  2

 1   2  1.0

1-D energy Partly open valve equation: p V 2 p V 2 1   1  z  2   2  z  h  1 2g 1  2 2g 2 lt

3 p1g = 100 kPa, p2g = 0, z1 = 10m,  = 9.81 kN/m

100 92 92  10   K 9.81 2  9.81 L 2  9.81

198.1 81  (1 K ) 9.81 2  9.81 L

2  198.1 1 K   K  4.89  1  3.89 L 81 L

- 85 - 7.19 In the figure for Probs. 7.14 and 7.15, suppose that the reservoir is open to the atmosphere at the top. The valve is used to control the flow rate from the reservoir. The head loss across the valve is given as

V 2 h  10 L 2g

where V is the velocity in the pipe. The cross-sectional area of the pipe is 5 cm2. The head loss due to friction in the pipe is negligible. The elevation of the water level in the reservoir above the pipe outlet is 10m. Find the discharge in the pipe.

Patm Assumptions

Tank >> pipeV1  0 1

water Flow is turbulent   1   2  1.0 d 2

1-D energy equation: Partly open valve p V 2 p V 2 1   1  z  2   2  z  h  1 2g 1  2 2g 2 lt

3 p1g = p2g = 0, z1 = 10m,  = 9.81 kN/m

V 2 V 2 10  2  10 2  10  2  9.81  11V 2 2g 2g 2

10  2  9.81 V   4.223 m / s 2 11

4 3 3 Discharge: Q  V2 A2  4.223 510  2.1110 m / sec

7.36 A small-scale hydraulic power system is shown. The elevation difference between the reservoir water surface and the pond water surface downstream of the reservoir, H, is 10 m. The velocity of the water exhausting into the pond is 5 m/s, and the discharge through the system is 1m3/s. The head loss due to friction in the penstock is negligible. Find the power produced by the turbine in kilowatts.

- 86 - 1 Assumptions

Reservoir >> pipeV  0 flow is H 1 steady turbine Flow is turbulent   1   2  1.0 z Given 2 Flow of water   = const

1-D energy equation with turbine present: p V 2 p V 2 1   1  z  h  2   2  z  h  1 2g 1 T  2 2g 2 lt

hT = turbine head

3 p1g = p2g = 0, z1 = 10m, V2 = 5m / s,  = 9.81 kN/m

V 2 (5) 2 h  z  2  10   8.726 m T 1 2g 2  9.81

˙ Turbine power:WT   Q hT  9.81(1)(8.726)

˙ WT  85.6 kW

7.25 For this system, point B is 10m above the bottom of the upper reservoir. The head loss from A to B is 2V2/2g, and the pipe area is 10-4m2. Assume a constant discharge of 7 x 10-4m3/s. For these conditions, what will be the depth of water in the upper reservoir for which cavitation will begin at point B? Vapor pressure = 1.23 kPa and atmospheric pressure = 100 kPa.

- 87 - B

z Water T = 20o C C

Assumptions

Reservoirs >> pipeVA  VC  0  Flow is steady

Flow is turbulent   1   2  1.0

Flow of water   = const

1-D energy equation: p V 2 p V 2 A   A  z  B   B  z  h  A 2g A  B 2g B lt

3 pAa = 100 kPa, pBa = 1.23 kPa, zB = 10m,  = 9.81 kN/m

1.23  100 V 2 V 2 1.5 z   B  10  2 B  10.07  10  V 2 A 9.81 2g 2g 9.81 B

1.5 z  V 2  0.07 A 9.81 B

- 88 - Q 7 104 Continuity: Q  VB AB  VB   4 AB 10

VB  7 m / s

1.572 z   0.07  7.42 m A 9.81

7.38 Neglecting head losses, detedmine what power the pump must deliver to produce the flow as shown. Here the elevations at points A, B, C, and D are 40 m, 65 m, 35 m, and 30 m, respectively. The nozzle area is 30 cm2.

B Assumptions

Tank >> pipeVA  0  Flow is steady

Flow is turbulent        1.0 A C B A nozzle Given water C

Flow of water   = const D

1-D energy equation with pump present: p V 2 p V 2 A   A  z  h  B   B  z  h  A 2g A p  B 2g B lt

(hp = pump head)

3 pAg = pBg = 0 ; zA = 40m, zB = 65m,  = 9.81 kN/m ;

VB = 0 (maximum height of fluid trajectory)

 hp = zB - zA = (65 - 40) m = 25 m

Bernoulli’s equation along a streamline from C to B: p V 2 p V 2 C  C  gz  B  B  gz  2 C  2 B

- 89 - 2 m VC  2g (zB  zC )  VC  29.81(65  35)  24.26 s

Continuity:

4 3 Q  VC AC  24.26  30  10  0.0728 m / sec

Pump power:

˙ Wp   Q hp

˙ Wp  9.81(0.0728)(25)  17.85 kW

Note: 1 kW = 1.341 hp

Problem:

As shown in the figure, the pump supplies energy to the flow such that the upstream pressure (12-in. pipe) is 10 psi and the downstream pressure (6-in. pipe) is 30 psi when the flow of water is 3.92 cfs. What horsepower is delivered by the pump to the flow?

pump out in

Assumptions  Assume flow is steady   0 & m˙ in  m˙ out  m˙  t Assume flow is uniform at inlet and outlet

˙ Neglect friction  uout = uin  Q  0

zout  zin

 = const

Conservation of mass: m˙ in  m˙ out  Vin Ain  Vout Aout

˙ 3 Vin Ain Vout Aout  V  3.92 ft / sec

- 90 - 3.92 3.92 Vin   4.99 ft / sec; Vout   19.96 ft / sec   (1) 2    (0.5) 2       4   4 

˙ Wss  0 by choice of the c.v. [V is zero at walls and  = 0 at the inlet and outlet]

Energy equation for the c.v.

  p   Q˙  W˙  W˙  W˙  e dV  e   V  dA ss s other      t V A  

outlet   ˙  p Ws  e    V  A inlet  

 2 2  ˙ pB Vout p A Vin Ws   Q       2  2 

62.4 30  10(144) (19.96) 2  (4.99) 2  W˙  (3.92)  s  62.4  32.2  32.2 2 

˙ ft  lbf ft  lbf 12,708.3 Ws  11,289.6  1418.7  12,708.3  hp  23.1 hp sec sec 550

1 hp = 550 ft-lbf/sec

If the discharge of water is Q = 0.06 m3/s, what are the pressures at A and B? Is the machine a pump or a turbine? Neglect losses.

A D z1 = 2 m z2 = 4 m D = 30 cm d = 15 cm water z2 T = 10o C d d B z z1 machine

- 91 - No head loss between B and outlet  Bernoulli can be applied between B and the outlet: p V 2 p V 2 B  B  gz  out  out  gz  2 B  2 out

Q VB  Vout  ; pout  patm ; zout  0 Apipe

pB = pout - gz1 = patm - 1000(9.81)(2) pBg = -19.62 kPa

Apply Bernoulli between A and B: p V 2 p V 2 A  A  gz  B  B  gz  2 A  2 B

Q 0.06 m VA    0.849 s   D2    0.30 2           4   4 

Q 0.06 m VB    3.395 s   d 2    0.15 2           4   4 

2 2  g (3.395)  (0.849)  p  19.62  z  z  z    Ag 1000 1 2 1 2(1000)

1000(9.81) (3.395)2  (0.849)2  p  19.62   41000  53.5 kPa Ag 1000 2(1000)

˙ Neglect losses  Q  0  uout = uin  flow is uniform at inlet and outlet

Assumptions

reservoir >> delivery pipe  Vin  0   flow is steady   0  t  = const

- 92 - Energy equation for the c.v.

˙ Wss  0 by choice of the c.v.

 2 2  ˙ pout Vout pin Vin Ws   Q   gzout    gzin    2  2 

2 ˙  (3.395)  Ws  1000(0.06)0   0  0  0  9.81(2  4)  2 

˙ Ws  3185.8 W  3.2 kW

˙ ˙ Ws  3.2 kW ; Ws  0  Machine is a turbine

Flow through a 90 o reducing elbow

z 2 Assumptions g y Flow is steady Fluid is incompressible x CV u & p are uniform at ‘1’ and ‘2’ flow

1 Energy equation for the c.v.

˙ Wss  0 by choice of the c.v.

˙ Wother  0

- 93 - p p V 2   V 2    Q˙  W˙  m˙ (u  u )  m˙ ( 2  1 )  m˙ g(z  z )  2  V  dA  1  V  dA s 2 1 2 1  2 2  1 1   A2 2 A1 2

 p   p  V 2   V 2   W˙  m˙ 2  gz  1  gz  2  V  dA  1  V  dA  m˙ (u  u )  Q˙ s  2   1   2 2  1 1 2 1       A2 2 A1 2

Velocity is not uniform across ‘1’ and ‘2’; this is the case in all viscous flows. An average velocity can, however, in conjunction with a KINETIC ENERGY FLUX COEFFICIENT () (also called KINETIC ENERGY CORRECTION FACTOR), be used.

V 2   V 2   V 2    V  dA     V  dA   m˙ A 2 A 2 2

 = 1 for uniform flow;  > 1 for non-uniform flow  = 2 for fully developed laminar flow;   1.05 for turbulent flow.

 2   2  ˙ ˙ p2 V2 p1 V1  Q Ws  m˙    2  gz2      1  gz1  m˙ (u2  u1 )     2    2   m˙ 

˙ 2 2 ˙ Ws  p2 V2   p1 V1   Q     2  gz2      1  gz1   (u2  u1 )   m˙   2    2   m˙ 

W˙  p V 2   p V 2  s   1   1  gz    2   2  gz   (u  u )  q m˙   1 2 1   2 2 2  2 1

˙ 2 2 Ws  p2 V2   p1 V1  1     2  z2      1 z1  (u2  u1 )  q m˙ g   2g    2g  g

 p V 2   p V 2   1 1   2 2    1  z1   hp     2  z2   hlT   2g    2g 

hp = pump head ; hlT = head loss term

- 94 - ˙ ˙ Ws Ws ˙ ˙ h   (since W˙  0 for pump)  Ws  hp Q  hpmg p m˙ g m˙ g s pump

Turbine

 p V 2   p V 2   1   1 z   h   2   2  z   h   1 2g 1  T   2 2g 2  lT

hT = turbine head

˙ Ws ˙ ˙ h   Ws  hT Q  hT mg T m˙ g turbine

Head loss in an abrupt expansion

Continuity:

V2 V1 A1  V2 A2  A1  A2 V1

Momentum equation:

( p1  p2 ) A2   QV1 V2 

( p1  p2 ) Q V2 A2  V2 V1   V2 V1   V2 V2 V1   A2 A2

1-D energy equation: p V 2 p V 2 1   1 z  2   2  z  h  1 2g 1  2 2g 2 lt

z1 = z2

- 95 - p  p V 2 V 2 h  1 2  1 2 le  2g

 1   2  1

V2 V2 V1  V2 V1  V2 V1   V1 V2   hle   V1 V2   V2   g 2g g  2 

V V 2 h  2 1 le 2g

Abrupt Contraction

Vena Contracta

Bend in a pipe

- 96 - Examples for the application of Rayleigh’s Theorem

1. Period of a simple pendulum t  f l, g,m t  const l  m  g 

Choose LTM

  T  L M  LT 2 

Dimensional homogeneity: L0 M 0 T  L  M  T 2

1  1 1   0;  2  1     ;     0         2  2 2

l  t  const g the constant must be determined experimentally (const = 2)

Note: Rayleigh’s method can be applied without difficulty when the number of independent variables does not exceed the available number of fundamental units. However, when the number of fundamental units, r, is less than the number of independent variables, p, then (p-r) exponents must be chosen arbitrarily. See example #2

2. Consider pressure losses per unit length in pipes due to friction:

 p  f d,V ,v, l

 p  const d  V  v   l

Choose LTM

    MLT 2 L2 L1 L LT 1  L2T 1  ML3 

ML2T 2 L  2 3 T   M 

- 97 - Dimensional homogeneity:

  1;      2    2       2  3  2    2     2  3  2    1 

  p  v  V 2   const d 1 V 2 v   const  l Vd  d const must be determined experimentally and  must be chosen arbitrarily

 p 3.  f d,V , v, , e e = roughness l

 p  const d V  v   e l

Choose LTM

    ML2T 2  L LT 1  L2T 1   ML3  L  L 2 3  T   M

Dimensional homogeneity:

  1;      2    2   ;   2     2  3    2       1    1   

 p  const d 1  V 2 v   e l

   p  v  V 2  e   const     l  dV  d  d  const must be determined experimentally and  and  must be chosen arbitrarily

4. Rate of flow, Q, of a fluid of viscosity, , through a tube of radius, r, and length, l, under a pressure difference, p.

Q  f p, l, , r

Q  const p l    r 

Choose LTM - 98 -   L3T 1   MLT 2 L2  L  ML1T 1  L

Dimensional homogeneity: M 0 L3T 1  M   L    T 2  1       0     ;  2    1       1; 2   1;          3  1   1   3      3   3    p p  l  Q  const  l   r 3  const  r 3      t 

 = chosen arbirtarily const determined experimentally

Dimensional Analysis Procedure using the Buckingham Pi Theorem:

1. List all variables which influence a given problem

2. Choose a set of fundamental dimensions e.g. MLT or FLT

3. List the dimensions of all the variables in terms of the fundamental dimensions

4. Determine the rank of the dimensional matrix

5. Choose from the independent variables a number (equal to the rank of the dimensional matrix) of repeating variables (also known as repeaters). Note that the dependent variable cannot be chosen as a repeating variable

6. Check on the dimensional independence of the chosen repeating variables

7. Set up dimensional equations by combining the repeating variables with each of the remaining variables, including the dependent one, in turn, to form dimensionless (or -) groups. Dimensional homogeneity must be observed hereby.

Quantity MLT FLT p Pressure ML-1T-2 FL-2  Viscosity (dynamic) ML-1T-1 FL-2T  Viscosity (kinematic) L2T-1 L2T-1  surface tension MT-2 F/L  density ML-3 FT2L-4 c velocity LT-1 LT-1 a acceleration LT-2 LT-2 e roughness (absolute) L L g acceleration (due to gravity) LT-2 LT-2

- 99 - F Force MLT-2 F  shear stress ML-1T-2 FL-2 A Area L2 L2 V Volume L3 L3  specific weight ML-2T-2 F/L3 Q discharge L3T-1 L3T-1 (volumetric flow rate) m˙ mass flow rate MT-1 FL-1T hl head loss L L N rpm T-1 T-1  angular speed T-1 T-1 T Torque ML2T-2 FL H Impulse and Momentum MLT-1 FT E Engergy and Work ML2T-2 FL P Power ML2T-3 FLT-1 E Modulus or elasticity ML-1T-2 F/L2

Problems ball D

1. 7.19 The sketch shows an air jet discharging vertically. Experiments show that a ball placed in the jet is suspended in a h stable position. The equilibrium height of the ball in the jet is d found to depend on D, d, V, , , and W, where W is the V weight of the ball. Dimensional analysis is suggested to correlate experimental data. Find the Pi parameters that characterize this situation.

2. The instrument package for a moon landing is instrument casting encased in a viscoelastic liquid as shown. The package acceleration, a, of the package is expected to depend on ℓ, a dimension of the package, m, the mass of the package, E, the modulus of elasticity of the liquid, , cushion the liquid viscosity, and V, the impact speed. liquid V Dimensional analysis is suggested to help design suitable experiments. Determine the dimensionless Moon surface parameters that result.

- 100 - 3. 7.6 Measurements of the liquid height upstream from an obstruction placed in an open channel flow of a liquid can be used to determine volume flow rate. (Such obstruction, designed and calibrated to measure rate of open-channel flow, are called weirs.) Assume the volume flow rate over a weir, Q, is a function of upstream height, h, gravity, g, and channel width, b. Use dimensional analysis to develop an expression for Q.

4. 7.8 Capillary waves are formed on a liquid free surface as a result of surface tension. They have short wavelengths. The speed of a capillary wave depends on the surface tension, , wavelength, , and liquid density, . Use dimensional analysis to express the wave speed as a function of these variables.

5. 7.13 The vorticity, , at a point in an axisymmetric flow field is thought to depend on the

initial circulation, 0, the radius, r, the time, , and the fluid kinematic viscosity, . Find a set of dimensionless parameters suitable for organizing experimental data.

Solution to #1

h = f (D, d, V, , , W)

h D d V   W

h D d V   W L L L LT-1 ML-3 ML-1T-1 MLT

Dimensional matrix

L T M h 1 0 0 D 1 0 0 d 1 0 0 At least one 3 x 3 V 1 -1 0 determinant is  -3 0 1 nonzero   -1 -1 3 Rank = 3 W 1 -2 1

Choose D, V, , as repeaters

Check on independence of the dimensions of the repeaters

- 101 - L T M D 1 0 1 = V 1 -1 0  0  dimensions of the repeaters are independent  -3 0 1

-groups

h   D1V 1   1 h    1 1 D (obtained by inspection)

d   D2 V  2   2 d    2 2 D

3 3  3  3  D V  

  L0T 0 M 0  L3 LT 1  3 ML3  3 ML1T 1  L3 3 3 3 1T 3 1M  3 1

  3  1  0   3  1;  3  1  0   3  1

 3  3 3 3  1  0   3 1 3 3   3 1 31  1  1

    3 VD

4  4  4  4  D V  W

  L0T 0 M 0  L 4 LT 1  4 ML3  4 MLT 2

L0T 0 M 0  L 4  4 3 4 1 T  4 2 M  4 1

 4   4  3 4  1  0

 4  2  0   4  2,  4  1  0   4  1

 4  3 4   4  1  31  2  1  2

W   4 V 2 D 2

1  f  2 ,  3 ,  4 

- 102 - h  d  W   f  , ,   2 2  D  D VD V D 

2. a = f (ℓ, m, E, , V)

a ℓ m E  V

Choose MLT

a ℓ m E  V LT-2 L M ML-1T-2 ML-1T-1 LT-1

Dimensional Matrix

L T M a 1 -2 0 At least a 3 x 3 l 1 0 0 determinant is m 0 0 1 nonzero E -1 -2 1 Rank = 3  -1 -1 1 V 1 -1 0

Choose ℓ, m, V as repeaters Check on the independence of the dimensions of the repeaters

L T M ℓ 1 0 1 = m 0 0 1  0  dimensions of the repeaters are independent V 1 -1 0

-groups

1 1 1 1  ℓ m V a

 1 0 0 0     2  L0T 0 M 0  L1 M 1 LT 1  LT 2  L T M  L 1 1 1 T 1 M 1

  1 0;   1 2  0   1 2;  1  1 1  0   1  1 1  2  1  1

ℓa   1 V 2

- 103 -  2  2  2  2  l m V E

 2 0 0 0   1  2  1 L0T 0 M 0  L 2 M  2 LT 1  ML1T 2  L T M  L 2 2 T 2 M 2

  2  2  0   2  2;  2 1  0   2  1;

 2   2 1  0   2  1  2  1  2  3

ℓ3 E   2 mV

3 3  3  3  ℓ m V 

 3 0 0 0   1  1  1 L0T 0 M 0  L3 M 3 LT 1  ML1T 1  L T M  L 3 3 T 3 M 3

  3 1  0   3 1;  3  1  0   3  1;  3  3 1  0 

 3 1  3 1 1  2

   ℓ2 3 mV

ℓa  ℓ3 E ℓ2     1  f  2 ,  3    f  ,  mV  mV mV 

4. V = f (, , )

V   

Choose MLT

V    LT-1 MT-2 L ML-3

Dimensional Matrix

L T M V 1 -1 0 At least a 3 x 3 determinant  0 -2 1 is nonzero  Rank = 3  1 0 0  -3 0 1

Choose ,  and  as repeaters

Check on the independence of the dimensions of the repeaters

- 104 - L T M  0 -2 1 =  1 0 0  0  dimensions of the repeaters are independent  -3 0 1

1 1  1 1     V

  L0T 0 M 0  MT 2  1 L1 ML3  1 LT 1  L0T 0 M 0  L1 31 1T 21 1M 1  1

1 1 2  1  0    ;    0      ;   3  1  0    3 1 1 2 1 1 1 1 2 1 1 1 1

 1 1    3   1  1  2 2

    V 1       const  V   const or V  const 1   

5.  = f ( 0, r, , )

 0 r  

Choose MLT

 0 r   T-1 L2T-1 L T L2T-1

Dimensional Matrix

L T M  0 -1 1 all 3 x 3 determinants are  2 -1 0 zero 0 at least one 2 x 2 r 1 0 0 determinant is nonzero  0 1 0  Rank = 2 v 2 -1 -

Choose r and  as repeaters

- 105 - Check on the independence of the dimensions of the repeaters

L T = r 1 0 0  dimensions of the repeaters are independent  0 1

 -groups

1 1 1  r      (by inspection)

 2  2  2  r  0

L0T 0  L 2 T  2 L2T 1  L0T 0  L 2 2T  2 1

  2  2  0   2  2;  2  1  0   2  1

    2 r 2 0

   r 3 3 v  v (by inspection) 3 r 2

      f  ,      f   , v 1 2 3  r 2 0 r 2 

DIMENSIONAL ANALYSIS OF A GENERAL FLOW PROBLEM

1. Variables should include such fluid properties as: density surface tension compressibility viscosity gravitaional effect compressibility is most conveniently expressed in terms of its inverse:

dp dp K   dV d  (Bulk modulus of elasticity)  V  

2. Variables should also include the geometry

- 106 - two linear dimensions are used :  (length of pipe in pipe flow, or chord width in flow around an airfoil) d (diameter of pipe or thickness of airfoil)

3. The velocity is used to characterize the mass flow rate or volumetric flow rate

4. Main performance (i.e., dependent) variable

p-pipe flow; drag (or resistance) or lift in external flows

p = f (V, ℓ, d, , , K, , g)

Number of variables: n = 9

Choose MLT as fundamental units

p V ℓ d   K ML-1T-2 LT-1 L L ML-3 ML-1T-1 ML-1T-2

 g MT-2 LT-2

Dimensional Matrix

L T M p -1 -2 1 V 1 -1 0 At least one 3 x 3 l 1 0 0 determinant is nonzero  d 1 0 0 Rank = 3 = m  -3 0 1  -1 -1 1 K -1 -2 1  0 -2 1 g 1 -2 0

Choose v, d,  as repeaters

Check on the independence of the dimensions of the repeaters L T M V 1 -1 0 = d 1 0 0 0  Dimensions of the repeaters are independent  -3 0 1

- 107 - Number of dimensionless groups: N = (n-m) = 6

 groups

  1 1  1 0 0 0 3 1 1 1 1 1 2 1   d V p  M L T  ML  L LT  ML T

M 0 L0T 0  M 1 1L31 1  1 1T  1 2

  1 1  0   1 1;   1  2  0   1  2;  3 1  1   1  1  0

31   1  2  1  0   1  0

 p pressure forces 1   (EULER NUMBER, Eu  )  V 2 inertia forces

 2  2  2  2   d V ℓ   2   2  0,  2  1 (by inspection)

ℓ   -- implies that shape is a controlling factor 2 d

 3  3  3 0 0 0 3  3  3 1  3 1 1  3   d V   M L T  ML  L LT  ML T

M 0 L0T 0  M 3 1L33 3  3 1T  3 1

  3 1  0   3 1;   3  1  0   3  1;  3 3  3   3  1  0

31   3  1 1  0   3  1

   3 Vd

inertial forces = (reciprocal of the REYNOLDS NUMBER, Re  ) viscous forces

   4  4  4 0 0 0 3 4  4 1 4 1 2  4   d V K  M L T  ML  L LT  ML T

M 0 L0T 0  M  4 1L3 4  4  4 1T  4 2

  4  1  0   4  1;   4  2  0   4  2;  3 4   4   4  1  0

- 108 - 31   4  2  1  0   4  0

K inertial forces  4  = (reciprocal of the MACH NUMBER, Ma  ) V 2 elastic forces

  5 5  5 0 0 0 3 5 5 1 5 2  5   d V   M L T  ML  L LT  MT

M 0 L0T 0  M 5 1L35 5  5 T  5 2

  5 1  0   5  1;   5  2  0   5  2;  3 5   5   5  0

31   5  2  0   5  1

 inertial forces  5  = (reciprocal of the WEBER NUMBER, We  ) d V 2 surfacetension forces

  6 6  6 0 0 0 3 6 6 1 6 2  6   d V g  M L T  ML  L LT  LT

M 0 L0T 0  M  6 L36 6  6 1T  6 2

  6 0;   6  2  0   6  2;  3 6  6   6  1  0

30   6  2  1  0   6  1

d g inertia forces   = reciprocal of the FROUDE NUMBER, Fr  ) 6 V 2 gravity forces

- 109 -