Tutorial :Mechanic - electrician Topic: Electrical measurements the 2nd. Year Performance measurement DC, AC 1f Prepared by: Ing. Jiří Smílek Projekt Anglicky v odborných předmětech, CZ.1.07/1.3.09/ je spolufinancován Evropským sociálním fondem a státním rozpočtem České republiky.
Work is a physical quantity. ◦ Designation – W- el. work (energy) - A –mechanical work ◦ Unit - 1J = 1 W.s (electrical ) joule (symbol J) - other Power output equals work done in one second. ◦ Designation - P ◦ Unit - watt (symbol W). Power output equals work done in one second P=W/t (W;Ws,s) Electrical power output– electrical power current I at a voltage U ◦ P=U.I ◦ Unit - watt (symbol W).
Direct current power output- DConly active P=U*I, Unit - W - watt Using Ohm’s law I=U/R (U=RI, R=U/I) P=RI 2 P=U 2 /R Alternating current power output- AC 1 f - single-phase apparent S=U*I, unit – V.A – volt-ampere activeP=U*I*cos φ, unit - W - watt idleQ=U*I*sin φ, unit - var – volt-ampere reactive
3 f - three-phase apparent S=√3U*I activeP=√3U U*I*cos φ idleQ=√3U U*I*sin φ
Power output P – common name Power appliances P 1 – power output, which the appliance consumes – the motor voltage and current Power output appliances P 2 – power output, which the appliance puts out – the motor speed and torque Power input(100 %) Power output (85 %) Supplied power power put out P 1 = P 2 + ΔP P 2 = P 1 - ΔP
Loss of power- ΔP=P 1 -P 2 - Unit W The engine heat Efficiency η=P 2 /P Unit % Power input(100 %)Power output (85 %) Supplied powerpower put out P 1 = P 2 + ΔP P 2 = P 1 - ΔP Loss of power (15 %) ΔP = P1 – P2
DC – direct current power output Measurements using the method: indirect – voltmeter and ammmeter R V – internal resistance of the voltmeter RA – internal resistance of the ammeter a) I = I A, U = U V – U A = U V - R A. I A b) U=U V, I = I A – I V = I A - U V / R V P = U. I = U V. I A - R A. I A 2 P = U. I = U V. I A - U V 2 / R V R A > R
AC – Alternating current power output Measurements using the method : direct – wattmeter, To check using a voltmeter and ammeter With the current coil voltage Actual power output P s P S = P – R A I 2 – R WA I 2 R WA – coil current resistance meter
AC – Alternating Measurements using the method: direct – wattmeter, To check using a voltmeter and ammeter With the current coil voltage Actual power output P s P S = P – U 2 / R V – U 2 R WU R WU – coil resistance voltage meter
Name the types of power output: a) d) b) e) c) f) Explain what power input is ◦ Explain how you prove a loss of power ◦ Write how to calculate the motor efficiency ◦
How can you overload: a)A current coil wattmeter, b)A voltage coil wattmeter c)Two coils, Even when the hand shows deflection within the scale?
Mužík, J. Management ve vzdělávání dospělých. Praha: EUROLEX BOHEMIA, ISBN Wikipedie Příručka elektrotechnika, Europa 2000 Elektrické měření - E. Viteček, Vl.HOS - SNTL 1978 Elektřina kolem nás – Zdeněk Opava – Albatros 1981 Operační program Vzdělávání pro konkurenceschopnost, ESF 2007 – Available at: konkurenceschopnosthttp:// konkurenceschopnost