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Unformatted text preview: ECE 333 – Green Electric Energy Spring 2010 Exam #2, Tuesday, April 6, 2010. 11:00AM12:20PM NAME: Solutions Question 1 (10 p) The terminals of a three phase balanced delta connected inductive load are labeled a , b , and c . The voltage across terminals a and b is v ab = √ 2 V cos( ωt ), while the current Fowing into terminal a is i a ( t ) = √ 2 I cos( ωt + θ ) (note the sign of the phase angle) . Let V ab and I a denote the phasors associated with v ab ( t ) and i a ( t ). Let S = P + jQ be the complex power consumed by the load. Among the options below, circle the correct one. Explain your choice and what is wrong with the other options. (1) S = √ 3 V I and S = √ 3 V ab I ∗ a (2) S = √ 3 V I and P = √ 3 V I cos θ (3) S = √ 3 V I and Q = √ 3 V I sin( θ30) (4) S = √ 3 V I and P = √ 3 V I cos( θ + 30) Explanation: Option (4) is the correct one. The line current must lag the phase current 30 ◦ . Given this, and the fact that the load is inductive, it means that theta (as a number) must be negative. Then, since the load is inductive, the phase current must lag the line voltage and therefore, the angle between these two phasors must be smaller in magnitude (absolute value) than theta, it results that the magnitude of this angle must be theta+30. Also, note that this angle is negative as the magnitude of theta must be greater than 30 but we already argued that theta is negative. Question 2 (10 p) As illustrated in the fgure below, the air Fowing in and out o± a wind turbine is contained in a tube, where A u is the tube crosssection upwind the turbine through which air enters, A r is the tube crosssection where the turbine is located, and A d is the tube crosssection downwind the turbine through which air exits. Similarly, is the tube crosssection downwind the turbine through which air exits....
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This note was uploaded on 02/18/2012 for the course ECE 333 taught by Professor Overby during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 Overby
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