ELEC-E8409 HIGH VOLTAGE ENGINEERING Exercise 1 ELEC-E8409 HIGH VOLTAGE ENGINEERING

Question 1: An insulator is composed of two parallel connected plane capacitors C1 = 100 nF and C2 = 200 nF with loss angles (dissipation angles) = 4.6º and = 5.7º. Electrode distance is 3mm. VRMS is 6 kV at 50 Hz. What is the insulator’s dissipation factor (loss tangent)? Calculate the insulator’s dielectric losses and capacitive reactive power Q.

Ci Gi jB δ G  Dissipation Factor:  Capacitance:  Susceptance: a.) Dissipation Factor: Ci Gi jB Y G δ  Capacitance:  Susceptance:  Conductance:  Dissipation Factor: b.) DIELECTRIC LOSSES Pd and CAPACITIVE REACTIVE POWER Qc:  Admittance Y:  Apparent Power:  Dielectric Losses:  Capacitive Reactive Power:

Question 2: A capacitor’s electrodes are coated with insulation sheets with permittivity ε1 and ε2 and respective electric field strength E1 and E2. The electric field for the air between the electrodes is E0. Both plates have an area of 0.2 dm2 and voltage between the plates is 100V. What are the insulator’s relative permittivity values and electric field strength of each layer when E1 : E0 : E2 = 2 : 8 : 3. U = 100 V A = 0.2 dm2 ɛ1 ɛ0 ɛ2 1 mm E2 E0 E1

Voltage over electrodes : SERIES INSULATION: 1 2 ε1 ε2 D1 D2 The normal component of electric flux density D = ɛE is constant at the material boudnaries (D1n = D2n) Remove ɛ0: and Voltage over electrodes :

Question 3: The conductor of a bushing has a diameter of 30 mm. The conductor is surrounded by 15 mm thick hard board shell with relative permittivity of 4.5. The shell is enclosed in an oil container with 100 mm diameter. Relative permittivity of oil is 2.2. How many percent does the field strength on the surface of the metal conductor decrease due to the hard board shell? How is applied voltage distributed amongst the insulation layers when U = 50 kV. 30 mm 60 mm 100 mm

Decrease in electric field due to hard board layer:  Without hard board (no ɛr1): U2 U1 U r1 r2 r3 ε1 ε2  With hard board (including ɛr1): Decrease in electric field due to hard board layer:

Voltage distribution in the layers, U = 50 kV: U2 U1 U r1 r2 r3 ε1 ε2 Voltage distribution in the layers, U = 50 kV: From previous slide: U1 U Insert Q into U1