Mathcad - 05-16

Mathcad - 05-16 - 24.998kVA = i.e., 25 kVA kVA kW ≡ kVAR...

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I s 104.16A = arg I s ( 29 29.541 - deg = The transformer delivers rated voltage so: V fl V srated = V fl 240V = The primary voltage, referred to the secondary side at full load, is calculated by adding the drop across the equivalent impedance (caused by the load current) to the secondary full load voltage V referredp V fl I s Z eqs + = V referredp 251.011 6.331j + V = V referredp 251.091V = which is the no load voltage V nl V referredp = VR V nl V fl - V fl = VR 4.621% = The rated kVA can be found by multiplying the secondary current by the voltage S rated I L V srated = S rated
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Unformatted text preview: 24.998kVA = i.e., 25 kVA kVA kW ≡ kVAR kW ≡ Problem 5-16: V prated 480 V ⋅ = V srated 240 volt ⋅ = Note: The given equivalent impedance is referred to the low side (secondary) Z eqs 0.062 j 0.105 ⋅ + ( ) Ω ⋅ = F p 0.87 = lagging I L 104.16 A ⋅ = Solution θ acos F p ( 29-= θ 29.541-deg = angle is negative since lagging F p The secondary load current is given in complex form by: I s I L e j θ ⋅ ⋅ = I s 90.619 51.356j-A =...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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