Examples

1. Resistance type nozzle - Diffuser (surface angle change < 8⁰ )

Resistors in series 

ΣU = 0,  U_imput + U_F + U_output = 0,    (U vector, ρ = const.)

Is = constant      A₁ u_s1 = A₂ u_s2,   u_s2 = u_s1 A₁ / A₂  (u_s = supply velocity)

The form voltage (because of surface flow change)

ΔU = U_F = (ΔR) I + (ΔΙ) R = (R₁ - R₂)I = R_F I

the form resistance R_F (flow surface A₁ to surface A₂)

R_F = R₁ - R₂ = (u₁ / ρ A₁) - (u₂ / ρ A₂),            I = ρA₁ u₁ →

U_F = R_F I = u₁² - u₁ u₂ A₁ / A₂ = U₁ - U₂ = u₁²- u₂² → 

→ u₂² = u₁ u₂ A₁ / A₂  →  u₁ A₁ =  u₂ A₂

Potential velocity change at nozzle is proportionate to surface

U_F = u₁²{1 - [A₁ / A₂] ²}  is negative at nozzle, positive at Diffuser (in analogy to PTC- thermistor).

R_F = γ R_1,           γ = 1 - [A₁ / A₂]²

The form voltage is maintained in the system and can be used in propulsion systems such as common Spoiler or nozzle drive (Laval). For supersonic transition velocity the calculation includes also thermal voltage (change of charge density ρ).

2. Hydroelectric Station (flow intensity – supply)

Water with flow supply I = Q/t = 10³ Kg/sec initiate a turbine (L_T Power Turbine shaft). The diameter of water pipe surface A is constant. The input pressure on entrance is P₁ = 500 KPa, on exit P₅ = 123 KPa and the elevation difference is r = 10,5 m. Friction losses amount to L _r = 80 KJ / sec. Density of water ρ = 1000 Kg / m³, thermal coefficient of water c = L_r / I ΔΤ = 4,18 KJ / Kg K. Earth's gravitational field density g = 9,81 V / m.

Requisites: power turbine shaft L_T, temperature variation ΔΤ, efficiency factor n, capacity C, internal resistance R, conductor diameter d.

Solution: ΣU = 0 (resistances in series), flow supply I = constant

1. Input Voltage (+). 2. Elevation Voltage (+).  3. Friction voltage drop (-). 4. Turbine voltage drop (-) 5. Output Voltage (-).

U₁ = P₁ / ρ,  U₂ = g r_,   U₃ = L _r / I = c ΔT,   U₄ = L_T / I,   U₅ = P₅ / ρ

ΣU = 0, U₄ = U₁ + U₂ - U₃ - U₅ = 400V,

ΔΤ = U₃ / c = 0,019 K, L_T = U₄ I = 400Kw, η = L_T / I (U₁ + U₂) = 66,3%,

Capacity C = Q/(U₁ + U₂) = t 10 ³ Kg, by 603 V (min. for t = 10 years, otherwise the investment is not amortized!).

Internal resistance R_i = R₃ = U₃ / I = 0,08 {Ω}

Pipe diameter: P₁ = ρ u₁² → dynamic velocity u₁ = 22,36 m/sec and pipeline diameter from A₁ = πd ² / 4 = L₁/ρ u₁³, (L₁ = U₁I) → d = 0,2386 m

This example clarifies the relationship potential - flow Intensity / supply, the role of dynamic velocity (u₅ = 11,09 m/sec) and the role of the pipe surface A as necessary resistance (equivalent to an electrical).

At the exit (if the ambient pressure = 101.325 Pa) dynamic velocity could be reduced to u₅= 10,07 m/s, corresponding to 21,7 Kw of extra turbine power.

The kinetic energy (transition velocity u_s1) is unable to express voltage. The transition velocity (supply) is at the exit also u_s1 = u_s5 = 22,36 m/sec (A₁ u_s1 = A₅ u_s5 ) but: A₁ u₁ ≠ A₅ u₅) where A = constant.

3. Attempt to calculate the mass density limits

Cases:

3.1  The minimum pressure P is equal to the measure of the gravitational constant G (P = G).

3.2  Cosmic Potential U has the maximum value (U _max = c² / 2 ) of a sphere

3.3   Energy has one form.

Since the potential (within the limits of the system) is maintained, the minimum pressure rise to the minimum charge density equal to:

ρ _min = P / U = 2G / c²

Equation (8) shows the maximum radius to:

r _max  = (3 / π) ^1/2 c² / 4G

i.e. the total mass equivalent:

Σm = (3 / π)^1/2 c⁴ / 8G² = 2,222 10⁵⁵ Kg

In this case (no necessary dark energy) the amount of energy was in the shape of area E = PV, not stars.

3.4  By maximum charge density the dynamic velocity provides maximum angular velocity ω = 2π, so that it can not add another one mass because potential is at the maximum U = c² / 2 (at the surface of the sphere, the intensity of gravity is g = c² π2^1/2).

At maximum concentration from the equation u = ω r = 2πr, is given:

sphere radius r = c / π 8^{1/ 2}

maximum charge density ρ _max = 3π / G

but the mass obtained for maximum capacity is equal to:

m = c³ /2πG 8^1/2 = 2, 28 10³⁴

corresponding to a space V_m = c⁵ /4πG² 8^1/2  charge density of ρ = 2G / c²

This precluded the gathering of all energy in the form of a mass, but in 10²¹ spaces V_m, which are due to their potential in alternating voltage (Universe can not be in a single energy form).