Presentation on theme: "Exercise 2 S-18.3150 High Voltage Engineering S-18.3146 Suurjännitetekniikka."— Presentation transcript:
Exercise 2 S High Voltage Engineering S Suurjännitetekniikka
Question 1 The figure depicts average breakdown voltage (U 50 ) as function of gas pressure (p) for a coaxial SF 6 insulator. The structure was originally designed for 6 bar. However, it became apparent that it wasn’t economically efficient to design the system for such high pressures. For this reason, pressure was decreased to 4 bar. Design the structure again so that the breakdown voltage and the ratio D/d remains the same. What are the new D and d parameters? d D d=20 mm D=54 mm p [bar] U 50 [kV]
Average breakdown voltage: p [bar] U 50 [kV] (original) (reduced) d D d=20 mm D=54 mm Coaxial Geometry: We want to increase this to equal original U b(6bar) by altering dimensions
Breakdown field strength, p = 4 bar: Denote d´= new inner conductor diameter We want the same breakdown voltage at 4 bar as with original 6 bar D/d ratio remains same (can use original dimensions)
New outer diameter: Fixed ratio: Although, the ratio D/d remains the same, distance between electrodes has increased. 6 bar: D = 54 mm d = 20 mm ∆ = 34 mm 4 bar: D´= 71 mm d´= 26 mm ∆ = 45 mm Originally, breakdown at 4 bar would have occurred at 280 kV. With the new increased distance, voltage has to increase to 370 kV for breakdown to occur Breakdown voltage level is maintained the same with new dimensions and reduced pressure.
Question 2 Calculate the onset of partial discharge for the resin test object when breakdown voltage in the cavity obeys Paschen’s law. Pressure in the cavity is 1013 mbar and its dimensions are 1 mm x 5 mm (h x w). Both cavity and the test object are cylindrical. Calculate the apparent charge and consumed energy of the discharge.
Three-Capacitance Model: Voltage is distributed inversely proportional in relation to capacitance: h a 3 mm 4 r mm 4 r mm hchc CbCb CbCb CcCc CaCa
Paschen’s Law: U b = f(pd) p =1013 mbar, d =1 mm Inception voltage of discharge in cavity u ci = 4.5 kV Inception voltage (applied voltage required to ignite discharge in cavity) u i = 1.5∙ u ci = 6.75 kV ≈ 6.8 kV
~ During discharge, from the point of view of C c representing the capacitance in the void, the rapid change in voltage is distributed between the series connection of C b and C a which are now in parallel with C c Calculating apparent charge: Discharge is much faster than the feeding circuit (external circuit is too slow to react to discharge circuit can be removed) CbCb C´ a CcCc ∆ u c Assuming Extinction voltage = 0 V Apparent charge:
where q = 1.6 nC and u = u i = 6.8 kV Calculating energy: Consumed energy during discharge:
Question 3 The system in question 2 has 50 Hz alternating voltage of 6 kV RMS. Pressure in the cavity is 1013 mbar. Breakdown voltage in the cavity obeys Paschen’s law and the extinction voltage is 0.5 kV. Draw the voltage waveform and calculate the steady state partial discharge frequency.
Voltage over cavity: Applied voltagewhere U = 6 kV Inception voltage: Extinction voltage:
u(t) u c (t) +u ci -u ci +u ce -u ce = 4 · 50 Hz = 200 Hz Number of discharges per period · 50 Hz Discharge frequency =
Question 4 The figure shows a bushing (feed- through insulator). Cylinder 1 is the bushing conductor. Insulation layers 2, 3, and 4 have thin metal sheets between them to improve dielectric strength. Design lengths l 2 and l 3 so that maximum field strength in both sheets is of the same magnitude. What is this maximum field strength value? l 4 l 3 l l 4 l 3 l 2 D D 2 D 3 D 4 D1 = 20 mmD2 = 34 mm D3 = 48 mmD4 = 62 mm L4 = 140 mmU14 = 30 kV
Electric field for a cylinder: r R U r U l Design lengths l 2 and l 3 so that maximum field strength in both sheets is of the same magnitude l 4 and D 1…4 are given: l 4 l 3 l l 4 l 3 l 2 D1D1 D 2 D 3 D 4 1
l 4 l 3 l l 4 l 3 l 2 D1D1 D 2 D 3 D 4 1 Voltage: