Tutorial :Mechanic - electrician Topic: Electrical measurements the 2nd. Year Performance measurement DC, AC 1f Prepared by: Ing. Jiří Smílek Projekt.

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2 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.1.07/1.3.09/04.0002 je spolufinancován Evropským sociálním fondem a státním rozpočtem České republiky.

3  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).

4  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

5  3 f - three-phase  apparent S=√3U*I  activeP=√3U U*I*cos φ  idleQ=√3U U*I*sin φ

6  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

7  Loss of power- ΔP=P 1 -P 2 - Unit W  The engine heat  Efficiency η=P 2 /P 1. 100 - 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

8  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

9  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

10  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

11  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 ◦

12  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?

13  Mužík, J. Management ve vzdělávání dospělých. Praha: EUROLEX BOHEMIA, 2000. ISBN 80-7361-269-7.  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 – 2013.  Available at: http://www.msmt.cz/eu/provadeci-dokument-k-op-vzdelavani-pro- konkurenceschopnosthttp://www.msmt.cz/eu/provadeci-dokument-k-op-vzdelavani-pro- konkurenceschopnost