(coefficient & ratio)

Each fluid number is simply the ratio of a particular property (pressure, viscosity, elasticity, surface tension, etc.) with a common value defined by the fluid inertia formula; ρ.L².v² (see **How they Work** below). You use this ratio to calculate the 'Number' (Reynolds, Euler, Froude, etc.) or an unknown variable in a known condition and use the result in the same formula to find an unknown variable in the prediction condition (as shown in the **example** below).

As this is a calculation of ratios, you can mix the units you use in the formula, but those used must be consistent with the units you use in the 'known' and 'prediction' calculations.

These formulas are for dimensionless numbers that represent a ratio between the fluid inertia and its properties as follows:

REYNOLDS (viscosity): ( ρ.L².v² / μ.L.v ) → (ρ.L.v / μ)

FROUDE (gravity) ( ρ.L².v² / ρ.L³.g ) → (v / √(L.g))

EULER (pressure) ( ρ.L².v² / ρ.L² ) → (ρ.v² / p)

PRESSURE COEFFICIENT (pressure difference) ( ρ.L².v² / ρ.L² ) → (2.δρ / ρ.v²)

VELOCITY RATIO (centrifugal) ( ρ.L².v² / ρ.L⁴.ω² ) → (v² / L².ω²) > (v² / D.N)^{#}

CAUCHY (elastic) ( ρ.L².v² / E.L² ) → (ρ.v² / E)

MACH (elastic) ( ρ.L².v² / E.L² ) → (v / √(E/ρ))^{##}

STROUHAL (vibration) ( ρ.L².v² / ρ.L⁴.f² ) → (L.f / v)

WEBER (surface tension) ( ρ.L².v² / σL ) → (ρ.L.v² / σ)

# D = diameter and N = revolutions per second

## E is a measure of elasticity {kg/m/s^{2}}. From the Mach formula above, √(E/ρ) has the units m/s, which is a velocity. The Mach Number sets the denominator (bottom bit of fraction: v/√(E/ρ)) to the speed of sound) thereby defining a velocity ratio with the speed of sound

These numbers are used for dynamic fluid scale modelling. That is; the theoretical prediction of the performance of fluids under specifiable conditions, such as the behaviour of air over a full-size aeroplane wing from wind-tunnel testing on a small-scale model, or to establish the excitation frequency of a beam from vortex shedding, or even to predict the behaviour of a large cylindrical (or spherical) object dropped from a low flying aircraft as it bounces over the water! To explain …

The fluid number (Rᴺᵒ, Eᴺᵒ, Fᴺᵒ, etc.) is the same for both the model and the full size unit.

E.g. to calculate the velocity of a model: Rᴺᵒ = ρ.L.v/μ (full-size) → Rᴺᵒ = ρ.L.v/μ (model)

You calculate the fluid number (Rᴺᵒ) for the full-size unit from the information known (ρ, L, v and μ), then transfer this value to the model formula, fill in the bits you know (Rᴺᵒ, ρ, L and μ) and the fluid numbers calculator will calculate the unknown (v). And it works in both directions; i.e. full-size to model and vice-versa

All of the calculation options work similarly. You enter all values except one (the one for which you require an answer!). The fluid numbers calculator will calculate the unknown when only one value is set to zero or left blank. If you enter too much information, the fluid numbers calculator will ask you to "delete the unknown".

The following is an example using Fluid Numbers to solve a real problem:

This example is similar to that given in reference^{(3)} page 3-48;

1) A submarine is to move submerged through seawater @ 32°F and a speed of 10 knots.

a) At what {representative} speed should a 1:20 scale model be towed in 68°F fresh water?

b) If the thrust on the {1/20^{th} scale} model {at velocity a)} is found to be 42,500lbf, what horsepower will be required to propel the submarine?

Use Fluid Numbers' 'Reynolds Number' (Fig 2) calculation option to find a representative model velocity.

The 'Reynold's Number' for the submarine in seawater @ 32°F...

Rᴺᵒ = _____ (make sure that this Input is deleted, i.e. blank)

ρ = 1.995 kslug/ft³ (density of seawater as specified in the reference document example)

L = 20 (20 times the model length)

v = 10 knots (submarine speed from question above)

μ = 3.94E-5 lb.s/ft² (dynamic viscosity of seawater as specified in the reference document example)

... and you get a Reynolds Number (R) of 10126904.

The velocity of an equivalent 1:20 scale model in freshwater @ 68°F ...

Rᴺᵒ = 10126904 (copied and pasted from the above result)

ρ = 1.937 kslug/ft³ (density of fresh water as specified in the reference document example)

L = 1 (1 times the model length)

v = ______ knots (make sure that this Input is deleted, i.e. blank)

μ = 2.092E-5 lb.s/ft² (dynamic viscosity of water at 68°F as specified in the reference document example ⁽³⁾)

... and you get a velocity (v) for the scale model of 109.372654 [knots].

Use Fluid Numbers' 'Euler Number' calculation (substituting pressure for force) to find the representative propulsion force for the submarine.

The 'Euler Number' for the 1:20 scale model in freshwater @ 68°F...

Eᴺᵒ = ______ (make sure that this Input is deleted, i.e. blank)

ρ = 1.937 kslug/ft³ (density of fresh water as specified in the reference document example)

v = 109.372654 knots (final result from calculation '**a)**' above)

p = 42,500lbf (propulsion force for the model from question above)

... and you get a Euler Number (E) of 0.545203.

The propulsion force for the submarine in seawater @ 32°F ...

Eᴺᵒ = 0.545203 (copied and pasted from the above result)

ρ = 1.995kslug/ft³ (density of seawater as specified in the reference document example)

v = 10 knots ⁽³⁾ (submarine speed from question above)

p = ______ (make sure that this Input is deleted, i.e. blank)

... and you get a pressure (propulsion force) (p) of 365.91875 [lbf], which must now be scaled up.

Given that we have substituted force for pressure (force/area) in the above Euler calculation, we must compensate for this when up-scaling by squaring the effect of 'L' (the difference between the model length and the length of the submarine), i.e. 365.91875lbf x 20^{2} = 146367.5 lbf

The submarine needs 146367.5 lbf of propulsion to travel submerged at 10 knots in seawater at 32°F. This propulsion force should now be converted into horse-power:

Note: Conversion of force to power is not covered by this calculator). However, if you're interested...

If: power = work / time *and* work = force *x* distance *then* power = force *x* distance / time

If: velocity = distance / time *then* power = force *x* velocity

force = 146367.5 lbf *and* velocity = 10 knots = 16.89 ft/s ⁽³⁾

**Therefore:** power = 146367.5lbf *x* 16.89ft/s = 2472147.075ft.lbf/s

2472147.075ft.lbf/s = 4,495.8hp ⁽³⁾

- The input data for fresh water and seawater are taken directly from the example set in the reference publication to help you follow the reasoning.
- When calculating with fluid numbers, the units can be mixed as much as you like, so long as they are matched (see
**How To Use Fluid Numbers**above). - Refer to CalQlata's UniQon calculator.

You will find further reading on this subject in reference publication^{(12)}